this is embarassing
The bjt model schematic presented by Mikeks is a small signal simplified approximation. The input impedance presented to the signal source is a pure linear resistance, r-pi. Under this condition the whole issue is moot. A resistor has a straight line I-V curve. Driving it with a pure voltage source or a pure current source produces identical results. The input resistance is r-pi. Apply a voltage source signal as input. the transfer function is Vin*gm* = Iout, but Iout*Rload = Vout, so Vout/Vin = voltage gain = gm*Rout. The Rd output resistance is usually large compared to Rload, and can usually be neglected. This Rd is the reciprocal of "hoe" given on mfr spec sheets, which is output admittance. Since r-pi is linear, the output current is gm times the voltage across r-pi. But the current through r-pi is identical in waveform to the voltage, being pure linear. This current is Vin/r-pi = Iin, so that Vin = V-pi = Iin*r-pi. Therefore Iout = V-pi*gm = Iin*r-pi*gm. What do you suppose r-pi*gm equals? Would you believe "hfe" (beta)? Hence Iout = hfe*Iin. BJT manufacturers always specify beta (dc and signal). The gm factor varies with current to the first power, i.e. gm = 38.9*Ic.
The reason a simple linear approximation works is due to varying the input a small amount in comparison to the quiescent operating point, typically one tenth. The gm large signal transfer function is exponential, Ic varies as Is*exp((Vbe/Vt)-1). Let Vbe/Vt = x. Ic varies as "e to the x power minus 1" or (exp(x))-1. Recall the Taylor series expansion for exp (x):
exp (x) = 1 + x +(x^2)/2! + (x^3)/3! + ... +(x^n)/n! + ...
So that (exp(x) - 1) = x +(x^2)/2! + (x^3)/3! + ...
If Vd/Vt = x, is much less than one, 0.10 or even less, the higher order terms become very small in comparison to the first term, "x", so that:
(exp(x) - 1) roughly equals x.
So, Ic is roughly proportional to Is*(Vbe/Vt), and a linear relation between Ic and Vbe exists for small deviations about the dc operating point.
We could express this transfer parameter as gm or hfe. Why choose gm? Most signal sources driving the base (CE configuration) are voltage sources, as opposed to current sources. The voltage across r-pi multiplied by gm gives the value of the output current source. Multiplying this current by Rload gives output voltage. This is desirable since most loads are voltage driven.
Remember, with a pure resistance input, V-controlled and I-controlled are one and the same thing. It is the SMALL deviation about the Q-point that makes linearity possible. The emitter degeneration resistor neutralizes the variation in hfe, and fixes the dc value of Ic, which fixes gm. With LARGE signal variations, the nonlinear nature of the bjt takes over and distortion increases. With large signal drive, a change in input base current produces a moderate nonlinear change in output collector current. A large change in Vbe base-emitter voltage produces a colossal nonlinear change in Ic, incurring much greater distortion. The idea behind emitter degeneration and bias networks is to linearize by forcing small deviations about a large operating point. Then the input appears as a pure resistance, and the CCVS/VCCS issue is completely MOOT! Have I explained myself well? Best regards to all.
The bjt model schematic presented by Mikeks is a small signal simplified approximation. The input impedance presented to the signal source is a pure linear resistance, r-pi. Under this condition the whole issue is moot. A resistor has a straight line I-V curve. Driving it with a pure voltage source or a pure current source produces identical results. The input resistance is r-pi. Apply a voltage source signal as input. the transfer function is Vin*gm* = Iout, but Iout*Rload = Vout, so Vout/Vin = voltage gain = gm*Rout. The Rd output resistance is usually large compared to Rload, and can usually be neglected. This Rd is the reciprocal of "hoe" given on mfr spec sheets, which is output admittance. Since r-pi is linear, the output current is gm times the voltage across r-pi. But the current through r-pi is identical in waveform to the voltage, being pure linear. This current is Vin/r-pi = Iin, so that Vin = V-pi = Iin*r-pi. Therefore Iout = V-pi*gm = Iin*r-pi*gm. What do you suppose r-pi*gm equals? Would you believe "hfe" (beta)? Hence Iout = hfe*Iin. BJT manufacturers always specify beta (dc and signal). The gm factor varies with current to the first power, i.e. gm = 38.9*Ic.
The reason a simple linear approximation works is due to varying the input a small amount in comparison to the quiescent operating point, typically one tenth. The gm large signal transfer function is exponential, Ic varies as Is*exp((Vbe/Vt)-1). Let Vbe/Vt = x. Ic varies as "e to the x power minus 1" or (exp(x))-1. Recall the Taylor series expansion for exp (x):
exp (x) = 1 + x +(x^2)/2! + (x^3)/3! + ... +(x^n)/n! + ...
So that (exp(x) - 1) = x +(x^2)/2! + (x^3)/3! + ...
If Vd/Vt = x, is much less than one, 0.10 or even less, the higher order terms become very small in comparison to the first term, "x", so that:
(exp(x) - 1) roughly equals x.
So, Ic is roughly proportional to Is*(Vbe/Vt), and a linear relation between Ic and Vbe exists for small deviations about the dc operating point.
We could express this transfer parameter as gm or hfe. Why choose gm? Most signal sources driving the base (CE configuration) are voltage sources, as opposed to current sources. The voltage across r-pi multiplied by gm gives the value of the output current source. Multiplying this current by Rload gives output voltage. This is desirable since most loads are voltage driven.
Remember, with a pure resistance input, V-controlled and I-controlled are one and the same thing. It is the SMALL deviation about the Q-point that makes linearity possible. The emitter degeneration resistor neutralizes the variation in hfe, and fixes the dc value of Ic, which fixes gm. With LARGE signal variations, the nonlinear nature of the bjt takes over and distortion increases. With large signal drive, a change in input base current produces a moderate nonlinear change in output collector current. A large change in Vbe base-emitter voltage produces a colossal nonlinear change in Ic, incurring much greater distortion. The idea behind emitter degeneration and bias networks is to linearize by forcing small deviations about a large operating point. Then the input appears as a pure resistance, and the CCVS/VCCS issue is completely MOOT! Have I explained myself well? Best regards to all.
cfb/vfb illustrative setup
Greetings worldwide friends,
It's been a while (11 mos.?). Regarding this cfb/vfb issue, I've prepared an illustrative schematic which should lay this issue to rest once and for all. The original question posed by the original poster was "Current Feedback Voltage Feedback - How Do I Tell The Difference?" Refer to the attached file to see how one can easily distinguish vfb from cfb.
The fundamental difference lies in the realization of the **error signal**. Both vfb and cfb op amps present a high impedance at the non-inverting input terminal and a low impedance at the output terminal. At the inverting input terminals they differ. The vfb topology inverting input terminal presents a high impedance vs. a low impedance for the cfb type. Nonetheless, for a non-inverting amp configuration, both vfb and cfb types present a high-Z input and a low-Z output, and can be regarded as voltage controlled voltage sources (VCVS). For an inverting amp configuration, both types present an input impedance equal to the value of the input resistor, and a low-Z output. Again, both types can be treated as a VCVS.
Another commonality is the output state, or condition, which is fed back to the input. Both types employ a feedback of the state of the output **voltage**. The difference lies in the realization of the error signal.
With vfb op amps, the feedback information representing the state of the output voltage is realized, or implemented as a voltage. The vfb input terminal is a very high input Z, and very little current can be inputted. The output is usually divided down through a passive network and connected to the inverting input. In the event that one wishes to feed back the state of the output **current**, this current must be converted into a voltage. The most common method employed is a low valued resistor in series with the output load resistance on the ground side. Once the resistor converts the output current to a voltage, this voltage is fed back into the inverting op amp input. This feedback signal present at the inverting input is a **voltage** indeed, but it carries information regarding the output current. Thus we are feeding back the information or "state" of the output current, but implementing this info in the form of a voltage. This has and still is commonly called **current feedback**. This is good terminology because the signal fed back carries info related to output current, although realized in the form of a voltage. In the first case, **voltage feedback** is the terminology employed since the feedback info carries the state of the output voltage, and of course is realized as a voltage, all or part of the actual output voltage. For decades this nomenclature was employed and well understood.
With cfb op amps, the quantity commonly fed back is the state of the output **voltage**, just as in the vfb case. However, the inverting input terminal presents a low Z input, and feedback signal is realized as a **current**, which is carrying info related to output **voltage**. Of course, we could insert a small valued resistor in series with the output load resistance on the ground side and feed this signal back through a passive network. We now would have a feedback signal carrying the output current info, realized as a current. This is where things can become ambiguous.
The attached schematics illustrate the action of the servo loop for vfb and cfb types. In order to understand how the servo loop is closed via the error signal we will first break the feedback loops with a switch placed in series with the feedback resistors. On page 1, with the vfb, the open switch breaks the feedback loop. The output cannot close the loop in this condition. The error signal is Vd, the voltage difference between the non-inverting and inverting inputs, Vp - Vn = Vd. The non-inverting input Vp is driven with Vin, the input voltage source. With the loop open, Vn, the inverting input terminal, is at ground, or zero volts. The error signal is Vd = Vin - 0 = Vin. Whenever the error signal Vd is non-zero, the output of the op amp is driven in the direction so as to force Vd to zero. But because the feedbak loop is open, the inverting terminal voltage Vn, cannot be driven to the value equal to Vp, and the output goes to the rail and just sits there. If Vp is positive, Vout will equal the positive rail (almost). Of course, no current enters either input since they are high-Z always, even when the amp is saturated and Vn does not equal Vp. Thus in the saturated condition, Vd, the **error signal** of the system remains at Vin indefinitely as the feedback loop is open and non-functional. One can easily verify the above with ammeters/voltmeters.
continued next post
Greetings worldwide friends,
It's been a while (11 mos.?). Regarding this cfb/vfb issue, I've prepared an illustrative schematic which should lay this issue to rest once and for all. The original question posed by the original poster was "Current Feedback Voltage Feedback - How Do I Tell The Difference?" Refer to the attached file to see how one can easily distinguish vfb from cfb.
The fundamental difference lies in the realization of the **error signal**. Both vfb and cfb op amps present a high impedance at the non-inverting input terminal and a low impedance at the output terminal. At the inverting input terminals they differ. The vfb topology inverting input terminal presents a high impedance vs. a low impedance for the cfb type. Nonetheless, for a non-inverting amp configuration, both vfb and cfb types present a high-Z input and a low-Z output, and can be regarded as voltage controlled voltage sources (VCVS). For an inverting amp configuration, both types present an input impedance equal to the value of the input resistor, and a low-Z output. Again, both types can be treated as a VCVS.
Another commonality is the output state, or condition, which is fed back to the input. Both types employ a feedback of the state of the output **voltage**. The difference lies in the realization of the error signal.
With vfb op amps, the feedback information representing the state of the output voltage is realized, or implemented as a voltage. The vfb input terminal is a very high input Z, and very little current can be inputted. The output is usually divided down through a passive network and connected to the inverting input. In the event that one wishes to feed back the state of the output **current**, this current must be converted into a voltage. The most common method employed is a low valued resistor in series with the output load resistance on the ground side. Once the resistor converts the output current to a voltage, this voltage is fed back into the inverting op amp input. This feedback signal present at the inverting input is a **voltage** indeed, but it carries information regarding the output current. Thus we are feeding back the information or "state" of the output current, but implementing this info in the form of a voltage. This has and still is commonly called **current feedback**. This is good terminology because the signal fed back carries info related to output current, although realized in the form of a voltage. In the first case, **voltage feedback** is the terminology employed since the feedback info carries the state of the output voltage, and of course is realized as a voltage, all or part of the actual output voltage. For decades this nomenclature was employed and well understood.
With cfb op amps, the quantity commonly fed back is the state of the output **voltage**, just as in the vfb case. However, the inverting input terminal presents a low Z input, and feedback signal is realized as a **current**, which is carrying info related to output **voltage**. Of course, we could insert a small valued resistor in series with the output load resistance on the ground side and feed this signal back through a passive network. We now would have a feedback signal carrying the output current info, realized as a current. This is where things can become ambiguous.
The attached schematics illustrate the action of the servo loop for vfb and cfb types. In order to understand how the servo loop is closed via the error signal we will first break the feedback loops with a switch placed in series with the feedback resistors. On page 1, with the vfb, the open switch breaks the feedback loop. The output cannot close the loop in this condition. The error signal is Vd, the voltage difference between the non-inverting and inverting inputs, Vp - Vn = Vd. The non-inverting input Vp is driven with Vin, the input voltage source. With the loop open, Vn, the inverting input terminal, is at ground, or zero volts. The error signal is Vd = Vin - 0 = Vin. Whenever the error signal Vd is non-zero, the output of the op amp is driven in the direction so as to force Vd to zero. But because the feedbak loop is open, the inverting terminal voltage Vn, cannot be driven to the value equal to Vp, and the output goes to the rail and just sits there. If Vp is positive, Vout will equal the positive rail (almost). Of course, no current enters either input since they are high-Z always, even when the amp is saturated and Vn does not equal Vp. Thus in the saturated condition, Vd, the **error signal** of the system remains at Vin indefinitely as the feedback loop is open and non-functional. One can easily verify the above with ammeters/voltmeters.
continued next post
Now look at the cfb op amp, still on page 1, under identical conditions. The output of the cfb type is still pinned at the positive rail, just like the vfb type, and the non-inverting input terminal voltage, Vp, equals Vin, just as in the vfb case. But - very important - the inverting input terminal voltage **Vn is equal to Vp = Vin**. Thus the input voltage differnce, Vp - Vn = Vd equals **zero**. Although the feedback loop is open and non-functional, Vd remains zero, unlike the vfb case. The reason is that a cfb op amp does not use an emitter coupled differential pair as does a vfb type, but rather a complementary mixed Darlington configured as an **emitter follower**. The non-inverting input to the cfb op amp is the input of the EF, while the inverting cfb op amp input is really the EF **output**. The EF has a forward voltage gain of +1 and since Vin is at the non-inverting input Vp, the EF drives the inverting input terminal voltage Vn to exactly equal Vp which equals Vin. The reason that the output is pinned against the positive rail is as follows. The output of the EF is sourcing current into R3 the input resistor (for a positive Vin). This current is mirrored to the next stage charging the compensation capacitor to a positive voltage which is buffered and outputted. As long as there is current in the inverting input terminal, In, this cap will be charged by a current which is a mirrorred version of In. Of course once the cap is charged to the positive rail, the output is pinned at the rail and stays there as in the vfb. Confirm what I've said with instruments and it is as plain as day.
On page 2, the switches are closed and normal operation resumes. Once the loop is closed on the vfb op amp, the output feeds back a **voltage** to the inverting input via R5/R6, and Vn climbs until it matches Vp. The output Vo, is then held at this value. The error signal Vd is now zero and steady state is reached.
With the cfb, the output can now drive a **current** through R7 reducing the source current providid by In the inverting input (EF output). When the current in R7 climbs to a value equal to R8, then In = 0, since the two resistor currents are equal. When this happens the output will settle at its steady state value.
Thus when a vfb loop is opened, Vd the error signal becomes and remains non-zero. The loop does not function, and cannot minimize Vd. Ip = In = 0 always regardless of loop condition. When a cfb loop is opened, Vd **remains zero**. The loop does NOT drive Vd to zero in a cfb topology. Vd is always zero because of the emitter follower in the first stage. In, the current into or out of the inverting input terminal, becomes and remains non-zero since the loop cannot function.
Once the feedback loops are restored the Vd in a vfb amp is driven to zero, and the In in the cfb amp is driven to zero as well. Again, set up the schematic I've attached and you can easily verify all I've stated.
In a vfb type, the error signal is Vd the difference voltage across the input terminals. In a cfb type, the error signal is In the current being sourced or sinked by the inverting input terminal.
One more thing. I'm not aware of any cfb topology where the resistors "load" a common emitter (CE) stage. One of the reasons a cfb amp has such good speed is the use of emitter followers, which are much faster than common emitters. A CE cannot even approach the speed of an EF due to Miller effect. Refer to the attached National Semiconductor App note "AN Current Feedback Amplifiers", a very good one I recommend reading. Best regards to all.
On page 2, the switches are closed and normal operation resumes. Once the loop is closed on the vfb op amp, the output feeds back a **voltage** to the inverting input via R5/R6, and Vn climbs until it matches Vp. The output Vo, is then held at this value. The error signal Vd is now zero and steady state is reached.
With the cfb, the output can now drive a **current** through R7 reducing the source current providid by In the inverting input (EF output). When the current in R7 climbs to a value equal to R8, then In = 0, since the two resistor currents are equal. When this happens the output will settle at its steady state value.
Thus when a vfb loop is opened, Vd the error signal becomes and remains non-zero. The loop does not function, and cannot minimize Vd. Ip = In = 0 always regardless of loop condition. When a cfb loop is opened, Vd **remains zero**. The loop does NOT drive Vd to zero in a cfb topology. Vd is always zero because of the emitter follower in the first stage. In, the current into or out of the inverting input terminal, becomes and remains non-zero since the loop cannot function.
Once the feedback loops are restored the Vd in a vfb amp is driven to zero, and the In in the cfb amp is driven to zero as well. Again, set up the schematic I've attached and you can easily verify all I've stated.
In a vfb type, the error signal is Vd the difference voltage across the input terminals. In a cfb type, the error signal is In the current being sourced or sinked by the inverting input terminal.
One more thing. I'm not aware of any cfb topology where the resistors "load" a common emitter (CE) stage. One of the reasons a cfb amp has such good speed is the use of emitter followers, which are much faster than common emitters. A CE cannot even approach the speed of an EF due to Miller effect. Refer to the attached National Semiconductor App note "AN Current Feedback Amplifiers", a very good one I recommend reading. Best regards to all.
Sorry claude...
Your figs are in serious erorr.... The current in a 'voltage feedback' amp's feedback resistors are not zero.....
They are of course typically less than those in a 'current' fedback amp., but this does not provide a true distinction betwen the two arrangements, as the feedback resistors can be made as small as required without altering the fact that a nominal voltage output is sampled, and applied in both cases...
Your figs are in serious erorr.... The current in a 'voltage feedback' amp's feedback resistors are not zero.....
They are of course typically less than those in a 'current' fedback amp., but this does not provide a true distinction betwen the two arrangements, as the feedback resistors can be made as small as required without altering the fact that a nominal voltage output is sampled, and applied in both cases...
Re: Sorry claude...
Sorry, but you are the one that's mistaken. On page 1, the switches are open, so the feedback resistors currents are indeed zero. With open switches there can't be any current since the vfb amp input terminal currents are always zero. Page 1 is perfectly correct. If you think otherwise, then please compute the value.
On page 2, with the switches closed, the vfb feedback resistor currents are NOT claimed as zero, but equal to output current I_O. I didn't explicitly give the value for the sake of brevity, but I hoped that anyone familiar with Ohm's law can see that the current in R6 equals Vn/R6 = Vin/R6. This is also the value of current in R7 as well as the vfb op amp output current I_O. If Vin is 1.0 volt, and R6 is 1.0 kohm, then I_R6, I_R5, and I_O are 1.0 milliamp. I figured that was all too obvious so I didn't state it explicitly. I did not, however, state or imply a zero value of feedback resistor currents. How did you see that? I stated that I_R5 = I_R6 = I_O. Nowhere on page 2 (switches closed) were these currents equated to zero.
Examine my schematic carefully and you will find that all circuit principles are upheld.
mikeks said:
Sorry, but you are the one that's mistaken. On page 1, the switches are open, so the feedback resistors currents are indeed zero. With open switches there can't be any current since the vfb amp input terminal currents are always zero. Page 1 is perfectly correct. If you think otherwise, then please compute the value.
On page 2, with the switches closed, the vfb feedback resistor currents are NOT claimed as zero, but equal to output current I_O. I didn't explicitly give the value for the sake of brevity, but I hoped that anyone familiar with Ohm's law can see that the current in R6 equals Vn/R6 = Vin/R6. This is also the value of current in R7 as well as the vfb op amp output current I_O. If Vin is 1.0 volt, and R6 is 1.0 kohm, then I_R6, I_R5, and I_O are 1.0 milliamp. I figured that was all too obvious so I didn't state it explicitly. I did not, however, state or imply a zero value of feedback resistor currents. How did you see that? I stated that I_R5 = I_R6 = I_O. Nowhere on page 2 (switches closed) were these currents equated to zero.
Examine my schematic carefully and you will find that all circuit principles are upheld.
Actually i was refering to figs with switches closed...
'point is, 'current feedback' as described in most app. notes is really a misnomer...
just see pg 4 here:
http://www.passdiy.com/pdf/PLH_amplifier.pdf
'point is, 'current feedback' as described in most app. notes is really a misnomer...
just see pg 4 here:
http://www.passdiy.com/pdf/PLH_amplifier.pdf
mikeks said:
The circuit on page 4 is NOT cfb at all, nothing like a cfb op amp. As I've stated for the umpteenth time, a cfb amp uses a mixed complimentary Darlington emitter follower input stage. A vfb amp uses an emitter coupled pair (ecp) or differential pair if you will. The feedback in a cfb amp is applied to the junction of two emitters, the output of an emitter follower. The fb in vfb amps is applied to the base of an ecp, where the other base connects the input voltage source.
The page 4 circuit input stage is an ecp, just like a vfb op amp. The feedback is applied to the base of Q7, via R1-R2-C1. This is vfb all the way, not cfb at all. Who would even argue otherwise? The base of Q7 presents a high impedance and the output voltage is divided down to a smaller voltage and fed into the Q7 base of the ecp input stage. The fed back "state" or quantity is the output voltage, and the error signal is implemented as a voltage. Any app note will confirm this (Natl Semi, TI, Intersil, LT, ADI, etc.).
My page 2 resistor currents are stated as non-zero, so what's your beef?
For some reason, current-mode concepts give a handful of people extreme heartburn. These, whom I call "VMT", for "voltage mode thinkers", cannot and will not accept the fact that a control signal can be either a current, or a voltage. An error signal can be a current or a voltage, but not to a VMT. The "state" of a system can and does consist of both current and voltage, but a VMT insists on voltage only. Every example I've produced is perfectly consistent with all known laws of circuits, and no one has proved otherwise. The counter-examples given thus far are not evenly remotely relevant I suggest to those who still don't believe the whole semiconductor community whom I am in total agreement with, to download some cfb/vfb app notes. I tried attaching but I hit the size limit. Anyone who wishes may email me and I'll send a zip file of pertinent notes for study. Please learn the topologies before making such bold pronouncements.
Those who still don't see the difference betwen cfb/vfb amps have much homework to do. It is easy to see. I have products in the field which utilize both, but mostly vfb. In 27 years of electonic circuit design, I've never met any technical person who couldn't understand what has just been stated above. No offense to anyone, but one who doesn't get it at this point isn't going to anytime soon. Good day to all.
correction
On page of 4 of the file in question, I read fig. 5 instead of fig. 4. I apologize, but this changes nothing. This only gives us a chance to examine two more examples.
Whereas fig. 5 is indeed vfb as stated in my previous post, fig. 4 has the feedback connected from the output to the **emitter** of Q1, the pnp input bjt. This topology differs vastly from the cfb op amp topology. There are no current mirrors, and the emitter follower is a single device, only sinking current. The current source above it, I1, however, can source current. The EF has a low output impedance, and the feedback network R1-R2-C1 cannot develop a divided down voltage as done in vfb amps. The EF realized by Q1 will output a voltage with low output impedance forcing the emitter node of Q1 to follow the input voltage at the base of Q1. The R2-C1 branch of the fb network is driven by Q1's emitter. Thus R2-R1-C1 cannot develop a "feedback voltage" as the voltage at the R2-R1 junction, also the Q1 emitter node is forced to follow the input voltage at Q1's base.
The servo action will take place as follows. To ease matters, let's operate at a high enough frequency so that C1 and C2 are short circuited. The output voltage will be determined by Vin * (1 + R1/R2). When the output equals Vin * (1 + R1/R2) the loop is closed. If R1 = 1.0 kohm = R2. and Vin = 1.0 V peak, then Vout = 2.0 V peak. From here on in, quantities are peak value. The R1 current is 1.0 mA, and R2 is the same. The emitter of Q1 sources no current to R2. Q1's emitter sinks current from current source I1, which can be 1.0 mA for example. Suppose the input drops to 0.5 V. The steady state value of output will be 1.0 V, but the loop needs time to close. The EF will respond immediately being quite fast. The emitter of Q1 slews to 0.5 V rapidly. With 0.5 V across R2, its current is reduced to 0.5 mA. The R1 current is still 1.0 mA, but R2 current is 0.5 mA. The difference, 0.5 mA, must be sinked by Q1's emitter, in addition to I1, which is 1.0 mA. So Q1 now sinks 1.5 mA instead of 1.0 mA. This increase in current drives Q2's base harder. Being an inverting stage, Q2's collector slews DOWN, reducing Vout. When Vout reaches the steady state value of 1.0 V, the current in R1 is (1.0V - 0.5V)/(1.0 kohm) = 0.5 mA. But the current in R2 is 0.5 mA as well. The DIFFERENCE in currents in R1-R2 is now ZERO just when the error is NULLED. Q2 stops slewing downward and remains at this new steady state value.
Had the input increased instead of decreased, the extra current needed for R2 would have been sourced by the current source I1 and Q1's emitter current would decrease. Q2's base drive would decrease, forcing Q2 emitter to slew UPWARD. Let's say Vin jumps from 1.0 to 1.5 V. Q1 emitter jumps likewise to 1.5 V. R2 current increases to 1.5 mA while R1's current is at (2.0V - 1.5V)/(1.0 kohm) = 0.5 mA. Since R1's current is 0.5 mA and R2's current is 1.5 mA, the difference is 1.0 mA. This 1.0 mA is provided by the I1 current source. The emitter of Q1 sinks zero current. Therefore Q2's base drive slews DOWNWARD. Likewise its collector slews upward increasing Vout. When 3.0 V is reached, R1 current is (3.0V - 1.5V)/(1.0 kohm) = 1.5 mA, which exactly equals R2 's current of 1.5 mA. The difference is now zero and the output settles.
It is all too obvious that in fig. 4, the output slews whenever the currents in the feedback resistor pair are unbalanced. The output settles when the current difference is nulled by the feedback loop. The feedback pair cannot develop a voltage because the junction point is driven by a low-Z voltage source, the output of an EF (Q1) which forces the junction to follow the input voltage. The differential current enters or leaves the junction formed by I1 and Q1, and constitutes the **error signal** of the servo loop. However, the "feedback quantity" or "state variable" is the output **voltage**, not output current. Again, we must be careful because it can be misleading. The error signal carries output **voltage** information, but is implemented in **current** form.
In fig. 5, the error signal is obviously a voltage, being the difference between the inputs (bases) of the emitter coupled differential pair. The state variable being fed back is also a voltage. This is easy and obvious.
Once again, an error signal can be a current, or it can be a voltage.
Aren't you all glad to know this? Best regards to all.
On page of 4 of the file in question, I read fig. 5 instead of fig. 4. I apologize, but this changes nothing. This only gives us a chance to examine two more examples.
Whereas fig. 5 is indeed vfb as stated in my previous post, fig. 4 has the feedback connected from the output to the **emitter** of Q1, the pnp input bjt. This topology differs vastly from the cfb op amp topology. There are no current mirrors, and the emitter follower is a single device, only sinking current. The current source above it, I1, however, can source current. The EF has a low output impedance, and the feedback network R1-R2-C1 cannot develop a divided down voltage as done in vfb amps. The EF realized by Q1 will output a voltage with low output impedance forcing the emitter node of Q1 to follow the input voltage at the base of Q1. The R2-C1 branch of the fb network is driven by Q1's emitter. Thus R2-R1-C1 cannot develop a "feedback voltage" as the voltage at the R2-R1 junction, also the Q1 emitter node is forced to follow the input voltage at Q1's base.
The servo action will take place as follows. To ease matters, let's operate at a high enough frequency so that C1 and C2 are short circuited. The output voltage will be determined by Vin * (1 + R1/R2). When the output equals Vin * (1 + R1/R2) the loop is closed. If R1 = 1.0 kohm = R2. and Vin = 1.0 V peak, then Vout = 2.0 V peak. From here on in, quantities are peak value. The R1 current is 1.0 mA, and R2 is the same. The emitter of Q1 sources no current to R2. Q1's emitter sinks current from current source I1, which can be 1.0 mA for example. Suppose the input drops to 0.5 V. The steady state value of output will be 1.0 V, but the loop needs time to close. The EF will respond immediately being quite fast. The emitter of Q1 slews to 0.5 V rapidly. With 0.5 V across R2, its current is reduced to 0.5 mA. The R1 current is still 1.0 mA, but R2 current is 0.5 mA. The difference, 0.5 mA, must be sinked by Q1's emitter, in addition to I1, which is 1.0 mA. So Q1 now sinks 1.5 mA instead of 1.0 mA. This increase in current drives Q2's base harder. Being an inverting stage, Q2's collector slews DOWN, reducing Vout. When Vout reaches the steady state value of 1.0 V, the current in R1 is (1.0V - 0.5V)/(1.0 kohm) = 0.5 mA. But the current in R2 is 0.5 mA as well. The DIFFERENCE in currents in R1-R2 is now ZERO just when the error is NULLED. Q2 stops slewing downward and remains at this new steady state value.
Had the input increased instead of decreased, the extra current needed for R2 would have been sourced by the current source I1 and Q1's emitter current would decrease. Q2's base drive would decrease, forcing Q2 emitter to slew UPWARD. Let's say Vin jumps from 1.0 to 1.5 V. Q1 emitter jumps likewise to 1.5 V. R2 current increases to 1.5 mA while R1's current is at (2.0V - 1.5V)/(1.0 kohm) = 0.5 mA. Since R1's current is 0.5 mA and R2's current is 1.5 mA, the difference is 1.0 mA. This 1.0 mA is provided by the I1 current source. The emitter of Q1 sinks zero current. Therefore Q2's base drive slews DOWNWARD. Likewise its collector slews upward increasing Vout. When 3.0 V is reached, R1 current is (3.0V - 1.5V)/(1.0 kohm) = 1.5 mA, which exactly equals R2 's current of 1.5 mA. The difference is now zero and the output settles.
It is all too obvious that in fig. 4, the output slews whenever the currents in the feedback resistor pair are unbalanced. The output settles when the current difference is nulled by the feedback loop. The feedback pair cannot develop a voltage because the junction point is driven by a low-Z voltage source, the output of an EF (Q1) which forces the junction to follow the input voltage. The differential current enters or leaves the junction formed by I1 and Q1, and constitutes the **error signal** of the servo loop. However, the "feedback quantity" or "state variable" is the output **voltage**, not output current. Again, we must be careful because it can be misleading. The error signal carries output **voltage** information, but is implemented in **current** form.
In fig. 5, the error signal is obviously a voltage, being the difference between the inputs (bases) of the emitter coupled differential pair. The state variable being fed back is also a voltage. This is easy and obvious.
Once again, an error signal can be a current, or it can be a voltage.
Aren't you all glad to know this? Best regards to all.
Claude Abraham said:
Claude for the umpteenth time you are wrong...see figure 2 and 15 here:
http://www.national.com/an/AN/AN-597.pdf
http://www.diyaudio.com/forums/showthread.php?postid=438964#post438964
http://www.diyaudio.com/forums/showthread.php?postid=438891#post438891
good grief
Mikeks,
Fig. 15 is ???. I've never seen an op amp like that at all, vfb or cfb. Both inputs are feeding a dual gate JFET. I don't see any relevance. Dr. Cherry, as I stated in 2004, said that a cfb reduces to vfb **if the input stage and current mirrors are removed and replaced with an ecp**. Of course he is right. It is only logical then to conclude that if we DON'T replace the input stage and current mirrors, the said topology does NOT reduce to a vfb type. That's not much of an argument against cfb terminology.
With op amps of either type, the sampled output can be either a voltage or a current. Whether the opamp has a high-Z or a low-Z inverting input, it can accept either the output current or voltage as the sampled quantity fed back. However, whether this output quantity, I or V, is implemented as a current or a voltage depends on the input Z of the inverting terminal. That is why op amp makers refer to a low-Z input type as cfb. The quantity fed back is usually the output voltage. But, it could just as well be a current. Ditto for a vfb. But a vfb op amp with its high-Z inverting input can accept either the output voltage or current as feedback info, BUT ONLY in the form of a voltage. Thus it is the **error signal** that determines whether an op amp is cfb or vfb. With vfb op amps, the error signal must be a voltage, Vd, the difference across inputs, and the sampled output can be either I or V. With a cfb op amp, again, the sampled output can be either I or V, but the error signal is a current, In, the value entering or leaving the negative input, the difference of the feedback resistor currents.
I realize that a vfb need not employ an ecp. It could be a single device. If it presents a high impedance to the feedback network, fb applied to the base of a bjt, or gate of a JFET for example, it is still vfb. The input terminal is high-Z, and draws nearly no current from the feedback network. The voltage at the base/gate of the inverting input terminal is determined by the output voltage driving the fb network, and the ratio of resistor values. A feedback voltage is developed, and the error signal is realized as a voltage.
If the inverting input is low-Z, such as the output of an ef, single or Darlington, things change. If the resistive feedback network is connected directly to a low-Z voltage source, such as the output of an ef, the mid-point is actively driven by the ef. The output voltage does not divide down and drive the input. The fb resistor (upper) between input and output is connected directly across two voltage sources, each low-Z, of differing value. The resistor from input to ground (lower) has the ef output voltage impressed upon it. It's voltage is dictated by the input source, buffered by the ef. The inverting input terminal's voltage is determined by the input voltage source and the ef buffer, NOT by the output voltage state and NOT by the fb network. The fb resistor connected across two low-Z voltage sources will carry a current determined by Ohm's law. The grounded resistor will carry a current likewise in accordance with Ohm. When these currents are unequal, the inverting input terminal, actually an output, will source or sink current to make up the difference. This change in current slews the output up or down until the two currents balance and the difference current is nulled. The output settles here. Any EE professor proficient with cfb/vfb op amps will affirm this as will any Natl Semi, TI, Intersil, etc. op amp specialist. The op amp producers know more than anyone about this.
I am at a loss to make it any clearer. Will someone experienced with cfb amps please chime in. A 2nd set of eyes and ears can often be beneficial.
Mikeks, you keep introducing extraneous info not at all related to the topic. "CFB - VFB, how do I see the difference?" is the issue here.
I feel I've said enough. Good day to all.
Mikeks,
Fig. 15 is ???. I've never seen an op amp like that at all, vfb or cfb. Both inputs are feeding a dual gate JFET. I don't see any relevance. Dr. Cherry, as I stated in 2004, said that a cfb reduces to vfb **if the input stage and current mirrors are removed and replaced with an ecp**. Of course he is right. It is only logical then to conclude that if we DON'T replace the input stage and current mirrors, the said topology does NOT reduce to a vfb type. That's not much of an argument against cfb terminology.
With op amps of either type, the sampled output can be either a voltage or a current. Whether the opamp has a high-Z or a low-Z inverting input, it can accept either the output current or voltage as the sampled quantity fed back. However, whether this output quantity, I or V, is implemented as a current or a voltage depends on the input Z of the inverting terminal. That is why op amp makers refer to a low-Z input type as cfb. The quantity fed back is usually the output voltage. But, it could just as well be a current. Ditto for a vfb. But a vfb op amp with its high-Z inverting input can accept either the output voltage or current as feedback info, BUT ONLY in the form of a voltage. Thus it is the **error signal** that determines whether an op amp is cfb or vfb. With vfb op amps, the error signal must be a voltage, Vd, the difference across inputs, and the sampled output can be either I or V. With a cfb op amp, again, the sampled output can be either I or V, but the error signal is a current, In, the value entering or leaving the negative input, the difference of the feedback resistor currents.
I realize that a vfb need not employ an ecp. It could be a single device. If it presents a high impedance to the feedback network, fb applied to the base of a bjt, or gate of a JFET for example, it is still vfb. The input terminal is high-Z, and draws nearly no current from the feedback network. The voltage at the base/gate of the inverting input terminal is determined by the output voltage driving the fb network, and the ratio of resistor values. A feedback voltage is developed, and the error signal is realized as a voltage.
If the inverting input is low-Z, such as the output of an ef, single or Darlington, things change. If the resistive feedback network is connected directly to a low-Z voltage source, such as the output of an ef, the mid-point is actively driven by the ef. The output voltage does not divide down and drive the input. The fb resistor (upper) between input and output is connected directly across two voltage sources, each low-Z, of differing value. The resistor from input to ground (lower) has the ef output voltage impressed upon it. It's voltage is dictated by the input source, buffered by the ef. The inverting input terminal's voltage is determined by the input voltage source and the ef buffer, NOT by the output voltage state and NOT by the fb network. The fb resistor connected across two low-Z voltage sources will carry a current determined by Ohm's law. The grounded resistor will carry a current likewise in accordance with Ohm. When these currents are unequal, the inverting input terminal, actually an output, will source or sink current to make up the difference. This change in current slews the output up or down until the two currents balance and the difference current is nulled. The output settles here. Any EE professor proficient with cfb/vfb op amps will affirm this as will any Natl Semi, TI, Intersil, etc. op amp specialist. The op amp producers know more than anyone about this.
I am at a loss to make it any clearer. Will someone experienced with cfb amps please chime in. A 2nd set of eyes and ears can often be beneficial.
Mikeks, you keep introducing extraneous info not at all related to the topic. "CFB - VFB, how do I see the difference?" is the issue here.
I feel I've said enough. Good day to all.
This thread brings back memories of the Wireless World letters pages 30 years ago.
Is a transistor a voltage or current amplifier was a hot topic back then as was the Quad 405 current dumper feedforward/back debate.
Nothing changes.
By the way, notice that the LM3886 chip amp has emitter follower inputs to buffer the inputs from the Miller capacitance of the LTP.
Is a transistor a voltage or current amplifier was a hot topic back then as was the Quad 405 current dumper feedforward/back debate.
Nothing changes.
By the way, notice that the LM3886 chip amp has emitter follower inputs to buffer the inputs from the Miller capacitance of the LTP.
Re: good grief
http://www.diyaudio.com/forums/showthread.php?postid=423522#post423522
Claude Abraham said:
http://www.diyaudio.com/forums/showthread.php?postid=423522#post423522
Too me the so called cfb is actually vfb. You are still sampling the output voltage. True current feedback results in a transconductance amplifier, not much use for a power amp.
The fuss in this thread is about the different ways of implementing the error amplifier.
I like the GEM amplifie approach of running a LTP with a capacitor base to emitter on the feedack transistor so that you get the dc performance of the LTP and the speed of emitter feedback. The only risk is that you must make sure that the bypassed transistor does not turn into a rf oscillator.
The fuss in this thread is about the different ways of implementing the error amplifier.
I like the GEM amplifie approach of running a LTP with a capacitor base to emitter on the feedack transistor so that you get the dc performance of the LTP and the speed of emitter feedback. The only risk is that you must make sure that the bypassed transistor does not turn into a rf oscillator.
davidsrsb said:
Maybe I'm misunderstanding you, but are you saying that there is no right answer, that it all depends on your point of view, semantics. etc.? If you construct the test circuits per my diagram. you will see that the servo action differs from vfb to cfb. This is not debatable. With a cfb, the input difference voltage, Vd, remains near zero even when the loop is opened, and the inverting input terminal current changes from near zero to larger. The vfb when opened behaves differently since Vd is near zero with the loop closed and increases with the loop open. But, in a vfb amp, the inverting input current is always zero, whether the loop is closed or open. That's quite a difference. There is no debate here.
This is not a legitimate debate at all. Every semiconductor producer can't be wrong. Regarding transistor as an I or V amplifier, the reason for the conflict is that some people cannot accept any device behavior defined in terms of current. They have a mental block against current as a state or control variable, but not with voltage. These people **never** dispute the terminology **voltage controlled** or **voltage feedback**. So with traditional vfb op amps and FET devices whose behaviors are defined in terms of voltage there is no conflict whatsoever. As soon as a device is described in terms of current, the issue becomes **controversial**.
The cfb device is no more controversial than vfb, and bjt parts are no more controversial than their FET counterparts. Some devices, due to their non-linear I-V and/or temperature characteristics, are much better controlled by a current, and others by a voltage.
Have you ever known of a long debate involving vfb op amps, JFETs, MOSFETs, or any **voltage** defined device? It's not the cfb amp or bjt that is controversial, it's the idea that an error signal or control signal can be a current as well as a voltage. Some people accept voltage only. That is the crux of the issue.
Regarding the current vs. voltage amplifier issue it has always been that the bjt and the FET both amplify voltage and current simultaneously. The beauty of active devices is that they provide not only voltage gain and current gain, bot both in unison, or real power gain. FET and bjt devices are charge controlled and need work or energy to move charge through the device. This requires both I and V. This is not debatable. The bjt has an exponential I vs. V curve. This alone makes it very difficult to control by inputting a voltage. What isn't as widely known is that Ies, the base-emitter junction reverse saturation current increases exponentially with temperature. If a constant voltage source were applied directly to the b-e junction of a bjt, or any forward biased p-n junction (LED for example), current would exist, power would be dissipated and the temperature must increase. The increase in temperature will increase Ies (or Is for an LED), resulting in a further increase in current. At a particular value of Vbe (not very large) this becomes a runaway condition. If, however, we drive the b-e junction with a high impedance current source (or a low impedance voltage source plus a sufficiently valued series resistor), current exists in the p-n junction, and the temperature must, of course, increase. The increase in temp increases Is or Ies the saturation current. This does NOT produce an increase in forward current because the forward current is determined by the current source. Instead the forward voltage drop DECREASES. This is why a device such as an LED is always "current driven" and never "voltage driven". Of course, bjt and LED devices require both I and V to operate, but operation is stable and controllable if we control the current and let the voltage be determined by the diode equation (or Ebers-Moll for a bjt). Hence these are referred to as "current mode" devices. Although a current mode device, a bjt amplifies both I and V. It is both a current and a voltage amplifier. Likewise for a FET. Although it must be "voltage driven", it provides current and voltage amplification and needs both I and V to operate.
There is nothing controversial at all. As long as we understand that "cfb, vfb, current controlled and voltage controlled" have no relation to "causality" and merely describe the optimum way to control a device due to its non-linear characteristics, there should be no conflict.
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