Transistor circuit: Determine source resistance

In summary: So you apply AC sources to the input and find the output, divide the output voltage by the input voltage and that is your gain.In summary, to calculate Rs in order to achieve a quiescent current of 4 mA, you can assume that u_in = 0 and solve for Rs using equation 1 and the given values for k and Vt. For the gain A, you can set up an AC analysis by applying an AC source to the input and finding the output, then dividing the output voltage by the input voltage to determine the gain. The capacitor in the circuit can be treated as a short circuit for the AC analysis.
  • #1
beyondlight
65
0

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


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

The Attempt at a Solution



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

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.
 
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  • #2
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.
 
  • #3
phyzguy said:
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?
 
  • #4
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.
 
  • #5
phyzguy said:
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.
 
  • #6
beyondlight said:
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.
 
  • #7
beyondlight said:
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 VGS is appropriate.
 
  • #8
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.
 

Related to Transistor circuit: Determine source resistance

1. What is a transistor circuit?

A transistor circuit is an electronic circuit that uses a transistor as its main active component. Transistors are semiconductor devices that can amplify or switch electronic signals, making them an essential component in many electronic devices.

2. How does a transistor circuit work?

A transistor circuit works by controlling the flow of current between two terminals, known as the collector and emitter, using a third terminal called the base. By varying the voltage applied to the base, the current between the collector and emitter can be amplified or switched on and off.

3. What is source resistance in a transistor circuit?

Source resistance in a transistor circuit refers to the resistance between the power source and the transistor. It is an important parameter that affects the performance of the circuit, as it determines the amount of current that can flow through the transistor.

4. How do you determine the source resistance in a transistor circuit?

The source resistance in a transistor circuit can be determined by measuring the voltage across the transistor and the current flowing through it, using Ohm's law (R=V/I). It can also be calculated by using the transistor's specifications and the circuit configuration.

5. Why is it important to determine the source resistance in a transistor circuit?

Determining the source resistance in a transistor circuit is important because it affects the performance and stability of the circuit. Too high of a source resistance can result in decreased current flow and potential overheating of the transistor, while too low of a source resistance can cause the transistor to draw too much current and potentially damage it.

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