In a bipolar Ie = Ic + Ib
The class of an amp is nothing to do with voltages. It is about the current flowing in each output device. In class A both devices conduct all the time. In class B they are each only just conducting when the output current is zero. In class C neither conduct when the output current is zero.
Class AB is a term to describe a low current in each device when the output current is zero and during some point of the output current one device will switch off. There is no formal definition as far as I know but I think a typical class AB audio amp has a bias current in the range 50mA to 1A.
The class of an amp is nothing to do with voltages. It is about the current flowing in each output device. In class A both devices conduct all the time. In class B they are each only just conducting when the output current is zero. In class C neither conduct when the output current is zero.
Class AB is a term to describe a low current in each device when the output current is zero and during some point of the output current one device will switch off. There is no formal definition as far as I know but I think a typical class AB audio amp has a bias current in the range 50mA to 1A.
ok, so if I've understood correctly
on a class A amp, like the one I want to build
The mosfet will conduct, if Vin=0
and also if Vin is at the highest value that can output the audio source (negative or positive voltage, everywhere on the signal voltage curve)
is this right?
PS: this doesn't help me to find correct resistances values; for a 2SK216
on a class A amp, like the one I want to build
The mosfet will conduct, if Vin=0
and also if Vin is at the highest value that can output the audio source (negative or positive voltage, everywhere on the signal voltage curve)
is this right?
PS: this doesn't help me to find correct resistances values; for a 2SK216
Mosfets are voltage controlled devices. If you have Vin = 0 then you will have no current flow from source to drain (or vice versa, depending on which theory you subscribe to). The 2SK216 is an N Channel Mosfet which requires positive voltage on the gate. The amount of positive voltage on the gate dictates how much current will flow through the device. The mosfet in the Szekeras has a bias voltage of 10 volts on the gate as it was originally drawn. I haven't done the calculations needed to tell you exactly how much current will flow through the 2SK216 with a gate bias of 10 volts but without Rs to control the current I'm sure that the current through the device would exceed it's max Ids. All you have to do is build the circuit as it's drawn in the Headwize Library and it will work fine. You can use a 2SK216 (which has less input capacitance than the IRF520 or 530) or use any other N channel Power mosfet that meets or exceeds the voltage and current specs of the IRF510 and the amp will work fine and sound good. In this instance the circuit is class A because the entire signal is being ouput through one device. Go on ebay or amazon and get a text on basic electronics and learn about the different classes of amplifiers and what sets them apart. Trying to learn basic electronics on a forum is frustrating and most of the time futile. Good luck with the headphone amp. Drop me an email if you need any help and I'll do my best to answer your questions but I highly reccomend getting the text book.
G
G
Biasing mosfets is a little more complicated than biasing bipolars. Lets say that you bias the gate of a fet with 3 volts and you get 10mA through the device. Next you change the bias to 6 volts and you get 30 mA. At this point you realize that the ratio of Vg and Ids is not linear. The biasing curves for different mosfets are different from one another. I don't have the formulas for calculating the Ids of a mosfet right in front of me now but I will try to explain later when I have more time and I locate my formulas notebook. I'm by no means an expert and perhaps some of the more learned members of this forum will pitch in and help with the explaination. If not then I will get back later today and give it a shot.
G
G
Bricolo,
"The mosfet will conduct, if Vin=0
and also if Vin is at the highest value that can output the audio source (negative or positive voltage, everywhere on the signal voltage curve)"
Correct.
"PS: this doesn't help me to find correct resistances values; for a 2SK216"
Also correct.
I assume you want to calculate the value of R4 (in your original circuit) to set the bias current (Ids when Vin=0) to 250mA. This is actually quite tricky when you use a FET because the Vgs will vary a lot with Ids and the datasheet doesn't give you a formula for Ids vs Vgs. Instead the datasheet gives you a curve.
I would say the easiest thing to do is build it with R4=20ohms and then measure the current. Then adjust the value of R4 until you get the current you want.
To work it out in advance you need to draw what is called a "load-line". You basically take the datasheet graph of Ids vs Vgs and draw a line on top of it to show the current through R4 vs Vgs. This will be a straight line. The current through R4 is simply the (gate voltage - Vgs) divided by R4. The gate voltage is fixed by the gate bias resistors. Now, where the two lines meet will be the bias current that you should expect to see for that value of R4. Of course, the Ids vs Vgs curve is an average curve and each FET will vary a little so it won't be exact.
It is simpler if you use bipolars because the Vbc is always very close to 0.7V when the transistor is conducting.
Good luck.
BAM
"The mosfet will conduct, if Vin=0
and also if Vin is at the highest value that can output the audio source (negative or positive voltage, everywhere on the signal voltage curve)"
Correct.
"PS: this doesn't help me to find correct resistances values; for a 2SK216"
Also correct.
I assume you want to calculate the value of R4 (in your original circuit) to set the bias current (Ids when Vin=0) to 250mA. This is actually quite tricky when you use a FET because the Vgs will vary a lot with Ids and the datasheet doesn't give you a formula for Ids vs Vgs. Instead the datasheet gives you a curve.
I would say the easiest thing to do is build it with R4=20ohms and then measure the current. Then adjust the value of R4 until you get the current you want.
To work it out in advance you need to draw what is called a "load-line". You basically take the datasheet graph of Ids vs Vgs and draw a line on top of it to show the current through R4 vs Vgs. This will be a straight line. The current through R4 is simply the (gate voltage - Vgs) divided by R4. The gate voltage is fixed by the gate bias resistors. Now, where the two lines meet will be the bias current that you should expect to see for that value of R4. Of course, the Ids vs Vgs curve is an average curve and each FET will vary a little so it won't be exact.
It is simpler if you use bipolars because the Vbc is always very close to 0.7V when the transistor is conducting.
Good luck.
BAM
ok, so on the Id/Vgs graph, I draw a horizontal line at Id=250mA
the lines cross at Vgs=2.4V (I used the Id/Vgs curve@75°C)
now, how can I calculate the gate voltage?
I've read another article about mosfets: tell me if what I remember is true, and if it's the basics to know
If Vg<Vt+Vs
->mosfet "closed"
If Vg>Vt+Vs
->mosfet "open"
If Vd>Vd(sat)
Vd(sat)=Vg-Vt
-> saturated mode
If Vd<Vd(sat)
->linear mode
the lines cross at Vgs=2.4V (I used the Id/Vgs curve@75°C)
now, how can I calculate the gate voltage?
I've read another article about mosfets: tell me if what I remember is true, and if it's the basics to know
If Vg<Vt+Vs
->mosfet "closed"
If Vg>Vt+Vs
->mosfet "open"
If Vd>Vd(sat)
Vd(sat)=Vg-Vt
-> saturated mode
If Vd<Vd(sat)
->linear mode
say you have +-15 supply
that is 30 volts
Gate is at half= 15 volts
Source is at the top of Resistor that goes to ground
volt G-S is then 2.4 for 250 mA
Source is at 15-2.4= 12.6 volts, that is at top of resistor
R-source then must be like: 12.6V/0.250A= 50 ohm
(this has to be a 5Watt resistor or better 10watt,
because Watt is: 12.6V x 0.250A)
So if you have +-15 volts supply
and have Gate at 0 volt,
The transistor will put 250mA through Rsource at idle
You get another current and another V gate-source, if you
change resistor.
To see what current it is, messure volt over the Source resistor
and divide it by the Ohm.
that is 30 volts
Gate is at half= 15 volts
Source is at the top of Resistor that goes to ground
volt G-S is then 2.4 for 250 mA
Source is at 15-2.4= 12.6 volts, that is at top of resistor
R-source then must be like: 12.6V/0.250A= 50 ohm
(this has to be a 5Watt resistor or better 10watt,
because Watt is: 12.6V x 0.250A)
So if you have +-15 volts supply
and have Gate at 0 volt,
The transistor will put 250mA through Rsource at idle
You get another current and another V gate-source, if you
change resistor.
To see what current it is, messure volt over the Source resistor
and divide it by the Ohm.
If you put 2 resistors from V+ (+30) to V- (0V)
They wiil divide the voltage.
If both have same value, the voltage between them will be half, of total voltage. Half from 30 is 15.
If one resistor is 2x and the other 1x,
they will divide 30 volt into 2 voltages
that also is like 2x and 1x.
One voltage is over first resistor, the other voltage is
over the other resistor.
But the sum of both voltage is always 30.
For 2x and 1x resistor. The 2 voltages will be 20 and 10 volts.
In this case the voltage Between the resistors
can be +10 volts or +20 volts.
It depends if you put 2x or 1x resistor at bottom, from 0 volts
They wiil divide the voltage.
If both have same value, the voltage between them will be half, of total voltage. Half from 30 is 15.
If one resistor is 2x and the other 1x,
they will divide 30 volt into 2 voltages
that also is like 2x and 1x.
One voltage is over first resistor, the other voltage is
over the other resistor.
But the sum of both voltage is always 30.
For 2x and 1x resistor. The 2 voltages will be 20 and 10 volts.
In this case the voltage Between the resistors
can be +10 volts or +20 volts.
It depends if you put 2x or 1x resistor at bottom, from 0 volts
does the 220Ohm Rg resistor cause no voltage drop between the R2-R3 junction ans G?
An externally hosted image should be here but it was not working when we last tested it.
Ok
Rg causes no voltage drop, since no currant flows through it
So:
Vg=(R3/(R2+R3))*Vcc
Vd=Vcc
We chose a bias current to use (say 200mA)
so, in the datasheet, on the graph of Id vs Vds (this graph is made with many curves, we look where Id=200mA
The curve must cross Id=200mA on the near vertical section (when the mosfet isn't saturated)
Now, we see that many curves, for many values of Vgs, cross the I=200mA line
For safety, we'll chose one that is still near vertical on I=300mA (so the mosfet will still not saturate when we have an input signal)
Let's choose Vgs=3V
We see, that on this curve (Vgs=3V), when Id=200mA, Vds=1.27V (you must zoom a lot on the graph to be precise enough)
To obtain Vgs=3V, with a 12V PS
R3/(R3+R2)=3/12
->we can let R2 to 100K; so R3 must be equal to 300K
If Vcc=12V
Vds=1.27V
V4 (voltage on R4) must be 12-1.27=10.73V
I through R4 is 200mA
->R4=10.73/0.2=54R
Is this correct? (have I understood everything correctly?)
PS: that's something I find strange: Vds<Vgs when the mosfet is in non saturated mode
Rg causes no voltage drop, since no currant flows through it
So:
Vg=(R3/(R2+R3))*Vcc
Vd=Vcc
We chose a bias current to use (say 200mA)
so, in the datasheet, on the graph of Id vs Vds (this graph is made with many curves, we look where Id=200mA
The curve must cross Id=200mA on the near vertical section (when the mosfet isn't saturated)
Now, we see that many curves, for many values of Vgs, cross the I=200mA line
For safety, we'll chose one that is still near vertical on I=300mA (so the mosfet will still not saturate when we have an input signal)
Let's choose Vgs=3V
We see, that on this curve (Vgs=3V), when Id=200mA, Vds=1.27V (you must zoom a lot on the graph to be precise enough)
To obtain Vgs=3V, with a 12V PS
R3/(R3+R2)=3/12
->we can let R2 to 100K; so R3 must be equal to 300K
If Vcc=12V
Vds=1.27V
V4 (voltage on R4) must be 12-1.27=10.73V
I through R4 is 200mA
->R4=10.73/0.2=54R
Is this correct? (have I understood everything correctly?
)PS: that's something I find strange: Vds<Vgs when the mosfet is in non saturated mode
oops
just a question:
is the transistor in saturated mode on the vertical section, or on the horizontal one?
just a question:
An externally hosted image should be here but it was not working when we last tested it.
is the transistor in saturated mode on the vertical section, or on the horizontal one?
None of the above. As a general rule 12V (Vgs) will drive a mosfet into saturation operation. This graph can be used as an aid to determine the bias point. If you look at the graph, think of the Vgs lines as possible bias points. For a good bias point consider the voltage swing of your input, choose one of the Vgs lines and see if your peaks will fit in the range. As you can see from the graph the Vgs=1V and Vgs=.5V have a very different distance between them. If your FET operates in this region you will get distortion. Note: the distance between all of the .5v steps is slightly different, and you will get some distortion because of this. Just try to avoid the really bad regions.
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