Transistor circuit: Determine source resistance

[beyondlight]

Homework Statement

Calculate Rs so that the draincurrent ID = 4 mA. Vad does the gain A become? uo/ui? For the transistor:

k = 3 mA/V^2
Vt = -2 V

also:

R1 = 50 k ohm
Rd = 2.2 k ohm
C = infinity for signalfrequencies

image of circuit:

http://sv.tinypic.com/view.php?pic=9pyq9z&s=8#.VSbHQfl_vxM

Homework Equations

i_{ds} = \frac{1}{2}k(v_{gs} - V_{t})^{2} (1)

The Attempt at a Solution

I_{D} = 4 mA (2)
v_{d} = E - R_{D}\cdot I_{D} (3)
v_{s} = R_{S}\cdot I_{D} (4)
v_{ds} = ?

From equation 1 we can get v_gs, that is the voltage between gate and source. But I still don't know how to get RS since I don't know u_in or v_ds.

 

[phyzguy]

I think you are supposed to calculate Rs so that the quiescent current Id = 4mA. So you can assume that u_in = 0. Then you know Vg so you can write an equation for Ids (which you are given) in terms of Rs and solve for Rs.

 

[beyondlight]

I think you are supposed to calculate Rs so that the quiescent current Id = 4mA. So you can assume that u_in = 0. Then you know Vg so you can write an equation for Ids (which you are given) in terms of Rs and solve for Rs.

I don't understand why u_in = 0 just because it is the quiescent current?

 

[phyzguy]

Well, I don't know for certain what the problem had in mind, but it looks like a small signal amplifier. So probably u_in is a small AC signal with an average value of zero, which best amplified and the amplified signal appears at u_out. The quiescent current is the current flowing when no input signal is applied, which means that u_in = 0. To be sure, maybe you should ask whoever assigned the problem whether you can assume that u_in = 0 when calculating Rs.

 

[beyondlight]

Well, I don't know for certain what the problem had in mind, but it looks like a small signal amplifier. So probably u_in is a small AC signal with an average value of zero, which best amplified and the amplified signal appears at u_out. The quiescent current is the current flowing when no input signal is applied, which means that u_in = 0. To be sure, maybe you should ask whoever assigned the problem whether you can assume that u_in = 0 when calculating Rs.

From earlier examples I've seen in lectures that u_in is nonzero in quiesent operation. For example: U_gsq = 5 V, U_dsq = 5.5 V and I_dsq = 4.5 mA.

I got the right answer so you are most probably right. But the quiescent current is a DC current so the quiescent input voltage must also to be a DC voltage. How do I know u_in = 0 produces a quiescent current of 4 mA? Since I am calculating on the quiescent current I am only taking the DC voltage part of u_in otherwise I would had to do my calculations on the small-signal model.

 

[phyzguy]

But the quiescent current is a DC current so the quiescent input voltage must also to be a DC voltage.

The quiescent current is defined as the current that flows even when the input voltage and input current are zero. It is a current intentionally set up in the circuit so that the amplifier is biased at a point at which it will function properly.

 

[NascentOxygen]

From earlier examples I've seen in lectures that u_in is nonzero in quiesent operation. For example: U_gsq = 5 V, U_dsq = 5.5 V and I_dsq = 4.5 mA.

Gate-to-source voltage in your circuit will be non=zero too. It makes no difference to the transistor whether source is zero volts and the gate has a negative potential, or whether the gate is zero and the source is at some positive potential (as here). The important figure is that V_GS is appropriate.

 

[STEMucator]

For the DC analysis, you should be assuming $u_{in} = 0$. You can superimpose the AC analysis separately afterwards.

I'm assuming there is a common ground, so this would lead to $V_G = 0$ because $I_G = 0$.

The capacitor is an open circuit for DC, so all that is left is to find $V_{GS}$ using the saturation current $I_D$.

Then simply using the facts $V_{GS} = V_G - V_S$ and $I_D = I_S$, you can solve for $R_S = \frac{V_S}{I_S}$.

As for the gain, simply look at the circuit with the AC source non-zero and all DC sources grounded. The capacitor will behave like a short circuit so its probably a good idea to re-draw the circuit to be careful.