I have biased the pre-driver stage on my amp easily using a diode and pot. The pot's contacts are on each end of the diode and I can move the wiper to get the correct amount of bias. The transistor's forward voltage is a bit lower than the diode's. It's safe to assume that they will change by the same factor due to temperature. Since the pre-driver barely pushes more than a few mA, it doesn't heat up, and assumes the same temperature as the diodes (room temp).
I am fairly excited about this amplifier as it will have quite a high damping factor and very little phase shift because it contains no capacitors at all. It will also have no resistors on the output stage, I am determined to prove it can work. Progress on the amp will resume as soon as I get a hold of a 7905 voltage regulator.
I am fairly excited about this amplifier as it will have quite a high damping factor and very little phase shift because it contains no capacitors at all. It will also have no resistors on the output stage, I am determined to prove it can work. Progress on the amp will resume as soon as I get a hold of a 7905 voltage regulator.
Hi Richie00boy,
Your concept does not have proper curent sharing in the output devices. You have only one feedback loop via the driver and both output devices get the same Vgs. So current sharing will depend on matching of the output devices. The 0.22Ω drain resistors will not change that.
To improve current sharing you need local feedback for the output devices individually by adding source resistors.
I don't think it is a problem to connect all the drains together to the emitter of the driver and use a single driver emitter resistor to the load. The current can be high, just use a resistor that can take more power.
Steven
Your concept does not have proper curent sharing in the output devices. You have only one feedback loop via the driver and both output devices get the same Vgs. So current sharing will depend on matching of the output devices. The 0.22Ω drain resistors will not change that.
To improve current sharing you need local feedback for the output devices individually by adding source resistors.
I don't think it is a problem to connect all the drains together to the emitter of the driver and use a single driver emitter resistor to the load. The current can be high, just use a resistor that can take more power.
Steven
Solid Snake,
of course it it possible to make a CFP stable, but it's a hard job and in your simulation you are only changing one parameter. With Vt = 26mV the current in the driver doubles every 18mV. Because Vbe is changing with about -2mV/K you are increasing your output current by two when the temerature rises at about 10K. This means that you need a precise temerature control. A temperature control of the die not of the case. When you are looking to the thermal resistance of the case, the heat sinks and so on it is obvious, that it's not so easy to make a precise temperature control. When self heating arises you are not able to detect the die temperature because of the thermal mass of the heat sink. And the Vbe multiplier must have the same temperature characteristic like the driver trannies. And the beta from the output trannies must not change when temperature changes. These is only some reasons why you need a output resistor from the two halves.
Try to build a CFP, you can make it stable (I did it, thermically and electrically with many effects a simple spice simulation does not show you) but it's much more work than a EF stage. And again it has a bad linearity in the crossover region.
When you're using SPICE you should know what you're doing. such a simple modell will fail, you need a more complex simulation model which also includes the thermal effects. Some oscillation are only present with higher temperatures or the amp run away thermically after some hours or days and blow your speakers. No engineerer is able to make a circuit theoretically stable without a lot of tests. SPICE helps a lot, but its just a tool not more. It's only a simplified model of the reality. It's on you (or better your experience) how to use these models.
of course it it possible to make a CFP stable, but it's a hard job and in your simulation you are only changing one parameter. With Vt = 26mV the current in the driver doubles every 18mV. Because Vbe is changing with about -2mV/K you are increasing your output current by two when the temerature rises at about 10K. This means that you need a precise temerature control. A temperature control of the die not of the case. When you are looking to the thermal resistance of the case, the heat sinks and so on it is obvious, that it's not so easy to make a precise temperature control. When self heating arises you are not able to detect the die temperature because of the thermal mass of the heat sink. And the Vbe multiplier must have the same temperature characteristic like the driver trannies. And the beta from the output trannies must not change when temperature changes. These is only some reasons why you need a output resistor from the two halves.
Try to build a CFP, you can make it stable (I did it, thermically and electrically with many effects a simple spice simulation does not show you) but it's much more work than a EF stage. And again it has a bad linearity in the crossover region.
When you're using SPICE you should know what you're doing. such a simple modell will fail, you need a more complex simulation model which also includes the thermal effects. Some oscillation are only present with higher temperatures or the amp run away thermically after some hours or days and blow your speakers. No engineerer is able to make a circuit theoretically stable without a lot of tests. SPICE helps a lot, but its just a tool not more. It's only a simplified model of the reality. It's on you (or better your experience) how to use these models.
Steven,
Many thanks for your response. I was actually thinking about the circuit again last night and considering the individual device Vbe/Vgs and thermal effects, and had come to the exact same conclusion as you. I'm a little surprised that the great Doug Self doesn't use individual emitter resistors for sharing though.
I agree with you that I could simply use a very high-power resistor for the 0.22 ohm. However, do you see any problem with the scheme for multiple lower-power resistors I proposed? Using this scheme may make board layout easier, that is why I'm thinking about it.
On a different note, I am reading a lot on here about CFP being really difficult to bias correctly and being unstable. The things I am reading are in direct conflict with what I have read on Rod Elliott's site and in Doug Self articles. What do people say about this?
Many thanks for your response. I was actually thinking about the circuit again last night and considering the individual device Vbe/Vgs and thermal effects, and had come to the exact same conclusion as you. I'm a little surprised that the great Doug Self doesn't use individual emitter resistors for sharing though.
I agree with you that I could simply use a very high-power resistor for the 0.22 ohm. However, do you see any problem with the scheme for multiple lower-power resistors I proposed? Using this scheme may make board layout easier, that is why I'm thinking about it.
On a different note, I am reading a lot on here about CFP being really difficult to bias correctly and being unstable. The things I am reading are in direct conflict with what I have read on Rod Elliott's site and in Doug Self articles. What do people say about this?
bocka said:
V3 in the above simulation is in essence to simulate the thermal effect. The CFP essentially has the collector resistor as a local feedback to control Iq (a mechanism you don' thave in an EF stage).
Yes, CFP is harder to get to work, but that does not mean that CFP is thermally inferior. As a matter of fact, the simulation (and reasoning) shows that CFP is thermally superior to an EF stage, because of that local feedback mechanism.
unless you can provide a specific argument as to why it is otherwise.
millwood, have you ever build a CFP or is your knowledge only from a simulation? Many other (professional) designers have tried the CFP, only a very few successfull.
SPICE is providing parameters for simulation of temperature effects, this must not be done with adding two voltage sources V1 and V3.
As I've said before, build the CFP and look what happens. BTW the EF stage also contains nearly 100% local feedback. I agree with you that the output transistor is not the main problem. It's hard to measure the junction temperature of the driver transistor because of the low power device and the high thermal resistance of the die mounting. If you're not coupling the Vbe multiplier in a very effective way to the driver your output stage blows up. I can't see any self heating mechanism of your simulation.
Again, I've never said the CFP will not work, of course it does. But you have many effects a simulation will not show you.
SPICE is providing parameters for simulation of temperature effects, this must not be done with adding two voltage sources V1 and V3.
As I've said before, build the CFP and look what happens. BTW the EF stage also contains nearly 100% local feedback. I agree with you that the output transistor is not the main problem. It's hard to measure the junction temperature of the driver transistor because of the low power device and the high thermal resistance of the die mounting. If you're not coupling the Vbe multiplier in a very effective way to the driver your output stage blows up. I can't see any self heating mechanism of your simulation.
Again, I've never said the CFP will not work, of course it does. But you have many effects a simulation will not show you.
Hi Richie00boy,
You can use multiple resistors in parallel to replace the single 0.22 Ohm resistor if that's easier for your layout, but there is no need to do it as complicated as you propose with additional coupling resistors to the emitter of the driver. All these coupling resistors can be just zero Ohm, putting the 0.22 Ohms in parallel. But you should increase the value of these resistors somewhat, for instance to make the combined resistors around 0.22 Ohm. For good thermal stability make the resistors big enough to get 50-100mV across it due to the idle current.
For lowest distortion it should be less, but then you risk thermal runaway. This design is quite stable with respect to idle current, however, as the driver transistors are more easy to keep at a stable temperature than the output transistors, and in this design the drivers determine the idle current. The MOSFETs are in a local feedback loop and their temperature behaviour is of less importance.
The source resistors of the MOSFETs for current sharing should be higher probably than 0.22 Ohm to get good sharing with less perfect MOSFET matching.
Steven
You can use multiple resistors in parallel to replace the single 0.22 Ohm resistor if that's easier for your layout, but there is no need to do it as complicated as you propose with additional coupling resistors to the emitter of the driver. All these coupling resistors can be just zero Ohm, putting the 0.22 Ohms in parallel. But you should increase the value of these resistors somewhat, for instance to make the combined resistors around 0.22 Ohm. For good thermal stability make the resistors big enough to get 50-100mV across it due to the idle current.
For lowest distortion it should be less, but then you risk thermal runaway. This design is quite stable with respect to idle current, however, as the driver transistors are more easy to keep at a stable temperature than the output transistors, and in this design the drivers determine the idle current. The MOSFETs are in a local feedback loop and their temperature behaviour is of less importance.
The source resistors of the MOSFETs for current sharing should be higher probably than 0.22 Ohm to get good sharing with less perfect MOSFET matching.
Steven
Hi Steven,
Good comments again 🙂
I had not considered tying the MOSFET drains together and using parallel resistors of higher value - thanks for the idea. This would make board layout simpler, but would there be any danger of creating some kind of loop around each resistor?
Good comments again 🙂
I had not considered tying the MOSFET drains together and using parallel resistors of higher value - thanks for the idea. This would make board layout simpler, but would there be any danger of creating some kind of loop around each resistor?
There are only two drivers and two output devices, there is nothing to share the current with. It is just a TIP42 feeding a TIP35 on the - side and a TIP41 feeding a TIP36 on the + side. I've tried going without resistors on the output stage and everything seems to be working well so far. I am concerned about the driver stage. I'm afraid that adding emitter resitsors (common emitter) will add a current feedback element which I don't want because the TIP35/36's aren't exactly linear.
richie00boy said:
No, they are just in parallel. All have the same voltage on one side and the same on the other side, so all have the same current direction. No reason for circulating currents or whatsoever.
Steven
Thanks Steven. I know what you are saying, but I was wondering about possible loops as the resistors in question may be 25 mm or so apart. So this and the consequent track inductance and different device Vgs thresholds got me thinking if it could happen.
BTW do you prefer CFP or EF output?
Many thanks
BTW do you prefer CFP or EF output?
Many thanks
richie00boy said:
I think it is a good thing to strive for mounting the components close together (if dissipation permits) and keeping the wires short.
I must confess to have used only EF outputs so far (for power amplifiers), but I've heard good sounding amplifiers that used collector follower outputs. The CFP has the advantage that in principle the bias current can be more stable.
Steven
I have done some experimenting with my amp over the weekend and I was very pleased with the results. I redesigned the output stage.
The output stage is common emitter, the emitters are on the rails, the collectors are tied together and form the speaker out with no resistors. These obviously get very warm as they dissipate most of the energy.
The driver stage has its emitters on the bases of the output stage and the collectors grounded. This stage dissipates a few watts of power, the transistors get a bit warm.
The pre-driver stage's collectors are coupled to the driver stage's bases. The emitters are on the speaker out. The bases are fed the input signal and through diode coupling and a pot, are always kept a X volts apart at all times. This stage dissipates next to no power at all. It always remains at room temperature.
The output and driver stage can get as hot as they want, with 20v rails, they can cook and the bias current will increase by only %0.5. I didn't get an exact damping factor on this amp but it is VERY high. I can play through an 8 ohm speaker and attach a 1 ohm load and the speaker will sound the same - no drop in output. Of course my transformer wasn't too happy about it but it lived. So after an hour of pounding through a <1 ohm load, it got nice and warm and when I removed the signal, idle currents were the same as when it was cold.
The output stage is common emitter, the emitters are on the rails, the collectors are tied together and form the speaker out with no resistors. These obviously get very warm as they dissipate most of the energy.
The driver stage has its emitters on the bases of the output stage and the collectors grounded. This stage dissipates a few watts of power, the transistors get a bit warm.
The pre-driver stage's collectors are coupled to the driver stage's bases. The emitters are on the speaker out. The bases are fed the input signal and through diode coupling and a pot, are always kept a X volts apart at all times. This stage dissipates next to no power at all. It always remains at room temperature.
The output and driver stage can get as hot as they want, with 20v rails, they can cook and the bias current will increase by only %0.5. I didn't get an exact damping factor on this amp but it is VERY high. I can play through an 8 ohm speaker and attach a 1 ohm load and the speaker will sound the same - no drop in output. Of course my transformer wasn't too happy about it but it lived. So after an hour of pounding through a <1 ohm load, it got nice and warm and when I removed the signal, idle currents were the same as when it was cold.
bocka said:
yes, bocka, I have built both. and the cfps I have built are thermally stable.
How about you?
Go back and look at your reasoning. and then let's discuss why changing the V3 figures wouldn't be simulating what you were talking about.
I have and I think a later reply also indicated how well they had performed thermally. Is your CFP unstable thermally? Maybe we can work together to figure out what you did wrong.
Good. Sounds like we are mamking progress, 🙂
On a 2nd thought, i am not sure how an EF stage has 100% feedback.
why do you need to measure the junction temperature for this thing to work?
there are many ways to Rome and there are many ways to achieve thermal stability. It just so happens that everything else being equal, a CFP stage is more stable.
then please read again.
Hum, I seem to read something different, 🙂
sure. a simulation wouldn't be 100% encompassing (otherwise, it wouldn't be a simulation). But the key is "does it simulate the key effect?". The key effect here is the Vbe drop on the output transistor and how that impacts Iq. For that, the simulation did a good job.
Unless you can provide specifics as to why it didn't.
Hi guys,
need some help working out the resistor wattage for an emitter resistor for parallel BJTs for the output stage.
If my quiscent current is at 3A, and I have 8 output devices with 8 emitter resistors at 0.1ohm each.
Would this calculation be correct?
Current through transistor = 3/8 = 0.375A
Resistor rating = 0.375^2 x 0.1 = 0.0014
Thus the rating for the resistor can be a normal 1/4 watt or 1/2 watt metal film resistor?
If I'm wrong, I suppose a 5 watt wire wound resistor would do but that causes oscillations, am I correct?
Thanks in advance
need some help working out the resistor wattage for an emitter resistor for parallel BJTs for the output stage.
If my quiscent current is at 3A, and I have 8 output devices with 8 emitter resistors at 0.1ohm each.
Would this calculation be correct?
Current through transistor = 3/8 = 0.375A
Resistor rating = 0.375^2 x 0.1 = 0.0014
Thus the rating for the resistor can be a normal 1/4 watt or 1/2 watt metal film resistor?
If I'm wrong, I suppose a 5 watt wire wound resistor would do but that causes oscillations, am I correct?
Thanks in advance
Assuming you have a push-pull stage with a total of 8 complementary output devices you should divide Iq by 4 to get the current through each resistor.
You should consider rms current under full load not just Iq but I guess with 3A that this is a class-a amp?
You should consider rms current under full load not just Iq but I guess with 3A that this is a class-a amp?
If you talk about a push-pull stage with your 8 output devices divided in 4 for the upper half and 4 for the lower half, then Richard is right. The bias current per transistor and its emitter resistor would be 3/4=0.75A. But a class A power amp biased at 3A is able to deliver 6A (i.e. twice I-bias) from one half. For very low frequencies you cannot average the current anymore if you look at the thermal time constant of the resistor, so to be safe, I would take 6A/4=1.5A as the maximum current through each resistor, as long as it stays in class A. But the amplifier might shift into class AB and deliver more current if the load impedance drops for whatever reason. So, I think if you use 1W metal film resistors, you can handle sqrt(P/I)=3.2A per device, which is more than enough even if the amp shifts into class AB. And 1W metal film resistors are still cheap. I like to keep the dissipation in resistors below half the rated power to avoid a high temperature.
If you talk about a single ended output stage with 8 transistors in the output and resistor or choke loading, then the maximum current in each resistor is always lower than what is stated above, so 1W is also enough in that case.
Steven
If you talk about a single ended output stage with 8 transistors in the output and resistor or choke loading, then the maximum current in each resistor is always lower than what is stated above, so 1W is also enough in that case.
Steven
Thanks all for the fast reply. The amp i'm building is ESP's DoZ with ON MJ21196 output transistors. Perhaps i should have mentioned that earlier to clarify things better.
I'll pop by the shops on Monday to get a bunch of 0R1 1watt resistors. Couldn't locate any on RS, hopefully I can find some in the shops. Confirmation 1 watt resistors are sufficient to dissipate the heat?
Does anyone know if using wirewound "white coffin" resistors would cause the amplifier to oscillate due to the inductance induced ? I'm not keen on the white coffins as they take up alot of space on the veroboard I used to mount the emitter resistors.
Regards and many thanks.
I'll pop by the shops on Monday to get a bunch of 0R1 1watt resistors. Couldn't locate any on RS, hopefully I can find some in the shops. Confirmation 1 watt resistors are sufficient to dissipate the heat?
Does anyone know if using wirewound "white coffin" resistors would cause the amplifier to oscillate due to the inductance induced ? I'm not keen on the white coffins as they take up alot of space on the veroboard I used to mount the emitter resistors.
Regards and many thanks.
Opps another thing I forgot to raise, perhaps it would be a better idea to parallel two 0R22 1watt resistors to replace the 0R1 1watt resistors?
This ought to increase the heat dissipation and keep the bias current split evenly among the transistors in the upper and lower rails.
This ought to increase the heat dissipation and keep the bias current split evenly among the transistors in the upper and lower rails.
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