Hi...
I have started a new problem to try to solve.
The circuit is attached and we are supposed to calculate Q-point for both transistors.
So the first thing I did was to use Thevenin theorem and calculate Vth and Rth for the Voltage divider at the JFET gate.
I found that Vth = 2V and Rth = 90kΩ
Then I've used the device Id equation and the gate net equation to find Id and Vgs.
Assuming that the JFET is working at it's active zone
Id = Idss(1-(Vgs/Vpoff))^2
and
Vth + Rth*Ig + Vgs + Rs*Id = 0
From the FET's theoretcial characteristics we are know Ig to be 0A, so:
Rth*Ig = 0V
So I got 2 solutions from the quadratic Id equation:
Vgs = -9.26V and Id = 7.26mA
and
Vgs = -3.24V and Id = 1.24mA
Now, I'm supposed to find out Vds to confirm or not the initial assumption. It's here that I have a question.
Because there is no voltage drop at the MOSFET (transistor at the right side), can I simply write the JFET output net equation as:
20 = Vds + Rs*Id ????
I have started a new problem to try to solve.
The circuit is attached and we are supposed to calculate Q-point for both transistors.
So the first thing I did was to use Thevenin theorem and calculate Vth and Rth for the Voltage divider at the JFET gate.
I found that Vth = 2V and Rth = 90kΩ
Then I've used the device Id equation and the gate net equation to find Id and Vgs.
Assuming that the JFET is working at it's active zone
Id = Idss(1-(Vgs/Vpoff))^2
and
Vth + Rth*Ig + Vgs + Rs*Id = 0
From the FET's theoretcial characteristics we are know Ig to be 0A, so:
Rth*Ig = 0V
So I got 2 solutions from the quadratic Id equation:
Vgs = -9.26V and Id = 7.26mA
and
Vgs = -3.24V and Id = 1.24mA
Now, I'm supposed to find out Vds to confirm or not the initial assumption. It's here that I have a question.
Because there is no voltage drop at the MOSFET (transistor at the right side), can I simply write the JFET output net equation as:
20 = Vds + Rs*Id ????
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