I have confusion about biasing circuit to FET (Field Effect Transistor) amplifier.

Thread Starter

Dong-gyu Jang

Joined Jun 26, 2015
115
Hello.

Please see the following images.

This is JFET amplifier.
JFET amplifier.jpg

And this is only biasing circuit to the amplifier.

JFET amplifier biasing circuit.jpg

Let's say V_DD is 12 V and R1 = R2 = 1 kΩ. Without V_IN, V_G is clearly 6 V, no doubt at all. good.

However, when V_IN applies, how can I determine V_G? Some signal current from V_In should flows along the path from + GND through C1 and R2 so that current times R2 gives V_G. But...V_G is also determined as 6 V from voltage divider. This is contradiction and if V_G is fixed, whole circuit means nothing to do useful job.

There is something I can't see right now. Could you please fix my vision to this circuit?
 

Attachments

Picbuster

Joined Dec 2, 2013
990
Hello.

Please see the following images.

This is JFET amplifier.
View attachment 99763

And this is only biasing circuit to the amplifier.

View attachment 99765

You make things more difficult than needed(I think).
there are two situations the DC part the gate is at 6V.
The DC source current will produce an offset in RS this will lower the gate source resulting in a balance.
The absolute DC input voltage Gs and gain will produce Vrd = Vout. resulting in an offset Vout.
Now you are able to calculate VRS. You want Ac to swing 100% at output and to be clipped.
VRs=(gain.Rs)/(1+gain.Rs)

The ac part:
the cap parallel to Rs should be a sort cut for input freq used allowing a stable offset at Vout.
However; The dc restoration at Rc//cap could move up when input signal is to high forcing S up and creating a feedback.
This gives you the opportunity to fiddle with this cap and make the mechanism freq depended.



















Let's say V_DD is 12 V and R1 = R2 = 1 kΩ. Without V_IN, V_G is clearly 6 V, no doubt at all. good.

However, when V_IN applies, how can I determine V_G? Some signal current from V_In should flows along the path from + GND through C1 and R2 so that current times R2 gives V_G. But...V_G is also determined as 6 V from voltage divider. This is contradiction and if V_G is fixed, whole circuit means nothing to do useful job.

There is something I can't see right now. Could you please fix my vision to this circuit?
 

crutschow

Joined Mar 14, 2008
24,122
Assuming C1 is large enough to have an impedance much smaller than the parallel value of R1 and R2 at the input frequency, then Vg will be the DC bias (6V) plus and minus the input AC voltage.
 

Thread Starter

Dong-gyu Jang

Joined Jun 26, 2015
115
Assuming C1 is large enough to have an impedance much smaller than the parallel value of R1 and R2 at the input frequency, then Vg will be the DC bias (6V) plus and minus the input AC voltage.
Hello.

I think C1 can be fairly small as DC biasing current from VCC power source will not enter even with small C1, biasing voltage is still 6 V.
 

crutschow

Joined Mar 14, 2008
24,122
Hello.

I think C1 can be fairly small as DC biasing current from VCC power source will not enter even with small C1, biasing voltage is still 6 V.
Certainly the size of the capacitor has no effect on the DC bias.
But the capacitor has to be large to pass the AC voltage without significant attenuation.
The AC signal will be attenuated by -3dB when the capacitance reactance at the input frequency equals the parallel value of R1 and R2 and rolloff at -6dB/octave below that frequency.

Since the purpose of this circuit is to amplify AC signals, you don't want to attenuate the AC input signal.
 

hp1729

Joined Nov 23, 2015
2,304
The gate voltage should be 6V plus some AC ripple that comes through C1. There is not contradiction. This is an application of superposition.

http://www.allaboutcircuits.com/textbook/direct-current/chpt-10/superposition-theorem/
Not homework, right?

VGS of 6 Volts? Or <0.6 V including signal. Should you avoid forward biasing the gate? It will be conducting as VGS gets down to about -0.7 V. The more negative VGS the more the JFET turns off. Depletion mode device, right? As voltage is applied to the gate it turns off.
6 V on the gate, yes, as long as > 6.7 V is dropped across the Source resistor.
Jeeez, JFETs has been a long time ago for me. :)
(edited to correct my math and thinking)

So what kind of signal is being applied?
 
Last edited:

hp1729

Joined Nov 23, 2015
2,304
Not homework, right?

VGS of 6 Volts? Or <0.6 V including signal. Should you avoid forward biasing the gate? It will be conducting as VGS gets down to about -0.7 V. The more negative VGS the more the JFET turns off. Depletion mode device, right? As voltage is applied to the gate it turns off.
6 V on the gate, yes, as long as > 6.7 V is dropped across the Source resistor.
Jeeez, JFETs has been a long time ago for me. :)
(edited to correct my math and thinking)

So what kind of signal is being applied?
JFET exercises
One exercise is just the JFET with a Drain resistor to +5 V. V in is varied and the output voltage is noted.
The other is similar to the circuit in this thread.
Included is the results in an Excel file (pdf format).

No, biasing does not have to be at 1/2 of VDD. Many examples are around with the high side resistor being a 220K and the low side being 22K. I have never done an exercise with this configuration.
 

Attachments

Last edited:

hp1729

Joined Nov 23, 2015
2,304
JFET exercises
One exercise is just the JFET with a Drain resistor to +5 V. V in is varied and the output voltage is noted.
The other is similar to the circuit in this thread.
Included is the results in an Excel file (pdf format).

No, biasing does not have to be at 1/2 of VDD. Many examples are around with the high side resistor being a 220K and the low side being 22K. I have never done an exercise with this configuration.
So what happens if we change the voltage divider to 220K and 22K?

Our input voltage range shifts up.

Does this help the question about biasing point?
 

Attachments

Top