Electrical Engineering - FET Common Source Configuration with Load

In summary: Oops, I mean Vin and Vo. Notice that the input impedance you calculated is infinite.As for the gain, you have to treat gm as a variable, and express the gain as a function of gm. You'll see that gm cancels out.I agree, but I am still left with the problem of not knowing how what v_{s}. Taking what you say into consideration I getA_{v_{NL}} = -\frac{2I_{DSS}R_{D}}{|v_{P}|(R_{G} + R_{sig})}(1 - \frac{R_{G}v_{s}}{v_{p}(R_{G} + R_{sig})})I do
  • #1
GreenPrint
1,196
0

Homework Statement



jfetamp.gif


I'm given this configuration with a AC source at [itex]v_{i}[/itex] with a [itex]R_{sig}[/itex] connect to the source before the capacitor.

[itex]R_{sig} = 0.6 KΩ[/itex]
[itex]R_{G} = 1 MΩ[/itex]
[itex]R_{D} = 2.7 KΩ[/itex]
[itex]R_{L} = 4.7 KΩ[/itex]
[itex]I_{DSS} = 10 mA[/itex]
[itex]v_{p} = -6 V[/itex]

Homework Equations





The Attempt at a Solution



I was able to find [itex]z_{i}[/itex] and [itex]z_{o}[/itex] very easily. I'm trying to find [itex]A_{v}[/itex] and was able to find the formula [itex]A_{v} = g_{m}(R_{D}||R_{L})[/itex]. The only problem is that I don't know how to find [itex]g_{m}[/itex]. I know that [itex]g_{m} = \frac{2I_{DSS}}{|v_{P}|}(1 - \frac{v_{GS}}{v_{p}})[/itex]. I'm not so sure how to find [itex]v_{GS}[/itex]. I know that [itex]v_{GS} = v_{G} - v_{S}[/itex] and that [itex]v_{S} = 0[/itex] so this comes down to finding [itex]v_{G}[/itex]. I seem to be having some problems doing this. I now that [itex]v_{G} = \frac{R_{G}v_{I}}{R_{sig} + R_{G}}[/itex] but this doesn't seem to really help since I don't know the AC input [itex]v_{I}[/itex]. Thanks for any help.
 
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  • #2
What do you want to calculate, and what are all those variables?

VG will depend on VI, and I think you will have to treat this as unknown input. Just calculate everything in terms of an unknown VI (if that is too hard in general, it might be interesting to consider a sine wave).
 
  • #3
Here's the actual picture from my book.

http://imageshack.com/a/img600/8114/gz12.png [Broken]

For part (a) I'm asked to determine [itex]A_{v_{NL}}[/itex], [itex]Z_{i}[/itex], and [itex]Z_{o}[/itex].

Finding [itex]Z_{i}[/itex], and [itex]Z_{o}[/itex] isn't a problem I just have a problem trying to find [itex]A_{v_{NL}}[/itex]

Below is my small signal equivalent circuit

http://imageshack.com/a/img21/4747/y59u.png [Broken]

What I get for the [itex]A_{v_{NL}}[/itex] is

http://imageshack.com/a/img268/4091/f6ze.png [Broken]

Which I don't see how it's wrong. My professor some how gets a value. He doesn't really show how he got his values but

*edit* I realize that I forgot [itex]v_{p}[/itex] in my equation for [itex]g_{m}[/itex] and the negative sign in my equation for [itex]A_{v_{NL}}[/itex] which I'm adding now.

http://imageshack.com/a/img713/7876/gi2i.png [Broken]

I'm not sure how he got [itex]A_{v_{NL}}[/itex]

I see how he gets [itex]g_{mo}[/itex]

[itex]g_{mo} = \frac{2I_{DSS}}{|v_{p}|} = \frac{2(10 mA)}{6 V} ≈ 3.333 mS[/itex]

I also know that

[itex]g_{m} = g_{mo}(1 - \frac{v_{GS}}{v_{P}})[/itex]

But I don't see how he got a value for [itex]g_{m}[/itex] because we don't know what [itex]v_{GS}[/itex] is other than [itex]v_{GS} = \frac{R_{G}v_{S}}{R_{sig} + R_{G}}[/itex] which doesn't really help because we don't know [itex]v_{s}[/itex] which leads me to believe that there must be some other way to find [itex]v_{GS}[/itex] that I'm not seeing.

Oh apparently he's finding [itex]A_{v_{NL}} = \frac{v_{o}}{v_{s}}[/itex]
 
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  • #4
Oh apparently he's finding [itex]A_{v_{NL}} = \frac{v_{o}}{v_{s}}[/itex]

which leaves me with this instead.

http://imageshack.com/a/img811/3217/naxp.png [Broken]

but still not able to estimate because we don't know [itex]v_{s}[/itex]

hm
 
Last edited by a moderator:
  • #5
AVNL sounds like "no-load voltage gain". That would mean not including RL in your gain computation.
 
  • #6
I agree, but I am still left with the problem of not knowing how what [itex]v_{s}[/itex]. Taking what you say into consideration I get

[itex]A_{v_{NL}} = -\frac{2I_{DSS}R_{D}}{|v_{P}|(R_{G} + R_{sig})}(1 - \frac{R_{G}v_{s}}{v_{p}(R_{G} + R_{sig})})[/itex]

I do I go about getting around this?

This is how I got that formula

http://img89.imageshack.us/img89/8845/ls8c.png [Broken]

Thanks for the help.
 
Last edited by a moderator:
  • #7
JFET circuit

Haven't we discussed this before? You get VS by equating the expression for FET current to the current flowing thru RS. I thought you said you understood this.

I also note that you are still using upper case vs. lower case in a confusing manner. For example, [itex]g_{m} = \frac{2I_{DSS}}{|v_{P}|}(1 - \frac{v_{GS}}{v_{p}})[/itex], VGS and VP are both dc quantities and so should be capitalized. Use vgs for ac signals. I found this problem in several of your equations.

What BTW is gm0? How's it different from gm? There is only one value of gm unless you change the circuit.

I think I'll let others take a shot at this since I feel you & I have already gone over this ground pretty thoroughly in another similar problem. Or am I mistaking you for someone else - that's happened too ...
 
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  • #8
Well [itex]v_{s}[/itex] in this problem though is the input AC voltage. I know that [itex]i_{D} = i_{S}[/itex] and that I can find the voltage at the source of the transistor this way using the equation for [itex]i_{D}[/itex] for a FET, also symbolized by [itex]v_{S}[/itex] I do understand this process now. The only problem is that [itex]v_{s}[/itex] in my equation for [itex]V_{GS}[/itex] isn't the voltage at the source of the transistor but the input AC voltage to the circuit, which isn't given.

In this problem I take [itex]v_{S}[/itex] the voltage at the source of the transistor to be zero since in the small signal equivalent it's connected to ground.

I'll be sure to use capital values for [itex]V_{GS}[/itex] and [itex]V_{P}[/itex] since they aren't AC parameters.

To me [itex]g_{mo}[/itex] is kind of silly. My book also defines
[itex]g_{m} = g_{mo}(1 - \frac{V_{GS}}{V_{P}})[/itex]
where
[itex]g_{mo} = \frac{2I_{DSS}}{|V_{P}|}[/itex]
which would be the value of [itex]g_{m}[/itex] when [itex]V_{GS}[/itex] is zero
 
  • #9
GreenPrint said:
In this problem I take [itex]v_{S}[/itex] the voltage at the source of the transistor to be zero since in the small signal equivalent it's connected to ground.

That is correct. So that makes the problem extremely simple. So vgs in your equivalent circuit is just vg which is just a voltage divider away from vin. You should not have labeled the input Vs or vs. That's very confusing. Label it vin.

So now surely you can solve for vo. What did you compute for Vs and gm?
 
  • #10
Ya I established that.

so can I do this then.

For FET
[itex]i_{D} = i_{S}[/itex] [1]
Here I use
[itex]I_{D} = I_{DSS}(1 - \frac{V_{GS}}{V_{P}})^{2}[/itex]
and
[itex]i_{S} = g_{m}V_{GS}[/itex] and plug these into [1]
[itex]I_{DSS}(1 - \frac{V_{GS}}{V_{P}})^{2} = g_{m}V_{GS}[/itex] [2]
I use this formula for [itex]g_{m}[/itex] and plug into [2]
[itex]g_{m} = \frac{2I_{DSS}}{|V_{P}|}(1 - \frac{v_{GS}}{V_{P}})[/itex]
[itex]I_{DSS}(1 - \frac{V_{GS}}{V_{P}})^{2} = \frac{2I_{DSS}}{|V_{P}|}(1 - \frac{V_{GS}}{V_{P}})V_{GS}[/itex][3]
Here I try and find [itex]V_{GS}[/itex]
[itex]V_{GS} = V_{G} - V_{S}[/itex] [4]
I know that
[itex]V_{S} = 0 V[/itex]
so this simplifies [4] to this
[itex]V_{GS} = V_{G}[/itex] [4]
I apply a voltage divider to find [itex]v_{G}[/itex] and get this formula
[itex]V_{G} = \frac{v_{in}R_{G}}{R_{sig} + R_{G}}[/itex]
I plug this into [4]
[itex]V_{GS} = \frac{v_{in}R_{G}}{R_{sig} + R_{G}}[/itex] [4]
I plug this into [3]
[itex]I_{DSS}(1 - \frac{v_{in}R_{G}}{V_{P}(R_{sig} + R_{G})})^{2} = \frac{2I_{DSS}}{|V_{P}|}(1 - \frac{v_{in}R_{G}}{V_{P}(R_{sig} + R_{G})})\frac{v_{in}R_{G}}{R_{sig} + R_{G}}[/itex] [3]
simplify
[itex]1 - \frac{v_{in}R_{G}}{V_{P}(R_{sig} + R_{G})} = \frac{2}{|V_{P}|}\frac{v_{in}R_{G}}{R_{sig} + R_{G}}[/itex]
move second term on LHS to RHS
[itex]1 = \frac{2}{|V_{P}|}\frac{v_{in}R_{G}}{R_{sig} + R_{G}} + \frac{v_{in}R_{G}}{V_{P}(R_{sig} + R_{G})}[/itex]
Factor
[itex]1 = \frac{v_{in}R_{G}}{R_{sig} + R_{G}}(\frac{2}{|V_{P}|} + \frac{1}{V_{P}})[/itex]
solve for [itex]v_{in}[/itex]
[itex]\frac{R_{sig} + R_{G}}{R_{G}}\frac{1}{\frac{2}{|V_{P}|} + \frac{1}{V_{P}}} = v_{in}[/itex]
simplify
[itex](\frac{R_{sig}}{R_{G}} + 1)\frac{1}{\frac{2}{|V_{P}|} + \frac{1}{V_{P}}} = v_{in}[/itex]
plug in values
[itex](\frac{0.6 KΩ}{1 MΩ} + 1)\frac{1}{\frac{2}{|-6 V|} - \frac{1}{6 V}} = v_{in}[/itex]
[itex](6x10^{-4} + 1)\frac{1}{\frac{2}{6 V} - \frac{1}{6 V}} = v_{in}[/itex]
[itex](6x10^{-4} + 1)\frac{1}{\frac{1}{6 V}} = v_{in}[/itex]
[itex](6x10^{-4} + 1)6 V = v_{in}[/itex]
[itex]6.0036 V = v_{in}[/itex]

Does this look better for [itex]v_{in}[/itex]? If so from here it's easy.

Now that I have [itex]v_{in}[/itex] I solve for [itex]V_{GS} = V_{G}[/itex]
[itex]v_{G} = \frac{v_{in}R_{G}}{R_{sig} + R_{G}} = \frac{(6.0036 V)(1 MΩ)}{0.6 KΩ + 1 MΩ} = 6 V[/itex]
I plug this into the equation for [itex]g_{m}[/itex]
[itex]g_{m} = \frac{2I_{DSS}}{|V_{P}|}(1 - \frac{V_{GS}}{V_{P}})^{2} = \frac{2(10 MΩ)}{6 V}(1 + \frac{6 V}{6 V})^{2} = \frac{1}{3000}(1 + 1)^{2} = \frac{1}{3000}2^{2} = \frac{4}{3000} ≈ 1.333 mS[/itex]

This isn't what my professor got for the supposed solution.

*edit* I added in some explanations.
 
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  • #11
?

vin is a given. It can be anything you like - within limits.

You are supposed to solve for vo, given vin. AvNL = vo/vin with RL removed.

I don't understand your equations.

I didn't see numbers for gm and Vs.
 
  • #12
rude man said:
?

vin is a given. It can be anything you like - within limits.

You are supposed to solve for vo, given vin. AvNL = vo/vin with RL removed.

I don't understand your equations.

I didn't see numbers for gm and Vs.

Wait really I can make up [itex]v_{in}[/itex] to be anything I want? So I can just go with one volt? That seems weird, I was able to solve for it some how.

By [itex]v_{s}[/itex] you mean the the AC voltage? It's just zero because it's connected to ground in the small signal equivalent circuit.
 
  • #13
GreenPrint said:
Wait really I can make up [itex]v_{in}[/itex] to be anything I want? So I can just go with one volt? That seems weird, I was able to solve for it some how.

By [itex]v_{s}[/itex] you mean the the AC voltage? It's just zero because it's connected to ground in the small signal equivalent circuit.

You need to understand what is meant by "gain". Gain = v_out/v_in. That ratio is independent of v_in.

I did not say v_s. I said V_s.
 
  • #14
Gain is the output voltage divided by the AC input? I solved for the AC input voltage and got 6.0036 volts. Oh I need the DC voltage V_S and use that value for V_GS

alright.

But wait,

In order to find [itex]V_{G}[/itex] I just use the DC schematic diagram and get

[itex]V_{G} = \frac{R_{G}v_{in}}{R_{sig} + R_{G}}[/itex]

So I don't know [itex]v_{in}[/itex] but I should be able to solve without hm.
 
  • #15
Vin = 0.
You need VGS to solve for gm. You know VG = 0.

Equate the expression for the dc FET current to the dc drain current to get VS.

Then use your formula for gm.
 
  • #16
This is starting to make sense, but why do you take Vin = 0?
 
  • #17
Why can you let [itex]v_{in}[/itex] be anything you want?
 
  • #18
[itex]A_{V_{NL}} = \frac{v_{O}}{v_{in}}[/itex]
[itex]v_{O} = -g_{m}V_{GS}R_{D}[/itex]
[itex]A_{V_{NL}} = -\frac{g_{m}V_{GS}R_{D}}{v_{in}}[/itex]
[itex]V_{GS} = V_{G} - V_{S}[/itex]
[itex]V_{G} = \frac{v_{in}R_{G}}{R_{G} + R_{sig}}[/itex]
[itex]V_{GS} = \frac{v_{in}R_{G}}{R_{G} + R_{sig}} - V_{S}[/itex]
[itex]V_{S} = I_{S}R_{S}[/itex]
[itex]I_{S} = I_{D} = I_{DSS}(1 - \frac{V_{GS}}{V_{P}})^{2}[/itex]
[itex]V_{S} = I_{DSS}(1 - \frac{V_{GS}}{V_{P}})^{2}R_{S}[/itex]
[itex]V_{GS} = \frac{v_{in}R_{G}}{R_{G} + R_{sig}} - I_{DSS}(1 - \frac{V_{GS}}{V_{P}})^{2}R_{S}[/itex]
[itex]V_{GS} = \frac{v_{in}R_{G}}{R_{G} + R_{sig}} - I_{DSS}(1 - \frac{2V_{GS}}{V_{P}} + \frac{V_{GS}^{2}}{V_{P}^{2}})R_{S}[/itex]
[itex]V_{GS} = \frac{v_{in}R_{G}}{R_{G} + R_{sig}} - I_{DSS}R_{S} + \frac{2I_{DSS}V_{GS}R_{S}}{V_{P}} - \frac{V_{GS}^{2}R_{S}I_{DSS}}{V_{P}^{2}}[/itex]
[itex]0 = \frac{v_{in}R_{G}}{R_{G} + R_{sig}} - I_{DSS}R_{S} + \frac{2I_{DSS}V_{GS}R_{S}}{V_{P}} - \frac{V_{GS}^{2}R_{S}I_{DSS}}{V_{P}^{2}} - V_{GS}[/itex]
[itex]0 = -\frac{R_{S}I_{DSS}}{V_{P}^{2}}V_{GS}^{2} + (\frac{2I_{DSS}R_{S}}{V_{P}} - 1)V_{GS} + \frac{v_{in}R_{G}}{R_{G} + R_{sig}} - I_{DSS}R_{S}[/itex]
[itex]V_{GS} = \frac{-(\frac{2I_{DSS}R_{S}}{V_{P}} - 1) \underline{+} \sqrt{(\frac{2I_{DSS}R_{S}}{V_{P}} - 1)^{2} + 4(\frac{R_{S}I_{DSS}}{V_{P}^{2}})(\frac{v_{in}R_{G}}{R_{G} + R_{sig}} - I_{DSS}R_{S})}}{2(-\frac{R_{S}I_{DSS}}{V_{P}^{2}})}[/itex]
[itex]V_{GS} = \frac{\frac{2I_{DSS}R_{S}}{V_{P}} - 1 \overline{+} \sqrt{(\frac{2I_{DSS}R_{S}}{V_{P}} - 1)^{2} + 4(\frac{R_{S}I_{DSS}}{V_{P}^{2}})(\frac{v_{in}R_{G}}{R_{G} + R_{sig}} - I_{DSS}R_{S})}}{\frac{2R_{S}I_{DSS}}{V_{P}^{2}}}[/itex]
[itex]A_{V_{NL}} = -\frac{g_{m}(\frac{\frac{2I_{DSS}R_{S}}{V_{P}} - 1 \overline{+} \sqrt{(\frac{2I_{DSS}R_{S}}{V_{P}} - 1)^{2} + 4(\frac{R_{S}I_{DSS}}{V_{P}^{2}})(\frac{v_{in}R_{G}}{R_{G} + R_{sig}} - I_{DSS}R_{S})}}{\frac{2R_{S}I_{DSS}}{V_{P}^{2}}})R_{D}}{v_{in}}[/itex]
[itex]A_{V_{NL}} = -\frac{g_{m}(\frac{2I_{DSS}R_{S}}{V_{P}} - 1 \overline{+} \sqrt{(\frac{2I_{DSS}R_{S}}{V_{P}} - 1)^{2} + 4(\frac{R_{S}I_{DSS}}{V_{P}^{2}})(\frac{v_{in}R_{G}}{R_{G} + R_{sig}} - I_{DSS}R_{S})}R_{D}V_{P}^{2}}{v_{in}2R_{S}I_{DSS}}[/itex]
[itex]g_{m} = \frac{2I_{DSS}}{|V_{P}|}(1 - \frac{V_{GS}}{V_{P}})[/itex]
[itex]A_{V_{NL}} = -\frac{2I_{DSS}(1 - \frac{V_{GS}}{V_{P}})(\frac{2I_{DSS}R_{S}}{V_{P}} - 1 \overline{+} \sqrt{(\frac{2I_{DSS}R_{S}}{V_{P}} - 1)^{2} + 4(\frac{R_{S}I_{DSS}}{V_{P}^{2}})(\frac{v_{in}R_{G}}{R_{G} + R_{sig}} - I_{DSS}R_{S})}R_{D}V_{P}^{2}}{v_{in}2R_{S}I_{DSS}|V_{P}|}[/itex]
[itex]A_{V_{NL}} = -\frac{(1 - \frac{V_{GS}}{V_{P}})(\frac{2I_{DSS}R_{S}}{V_{P}} - 1 \overline{+} \sqrt{(\frac{2I_{DSS}R_{S}}{V_{P}} - 1)^{2} + 4(\frac{R_{S}I_{DSS}}{V_{P}^{2}})(\frac{v_{in}R_{G}}{R_{G} + R_{sig}} - I_{DSS}R_{S})}R_{D}}{v_{in}R_{S}}[/itex]
etc... this problem sucks. I'm not sure how to solve this problem when I don't know [itex]v_{in}[/itex]. The previous problem, this information was given.
 
  • #19
GreenPrint said:
This is starting to make sense, but why do you take Vin = 0?

Because that is the dc operating point of your circuit.
 
  • #20
GreenPrint said:
Why can you let [itex]v_{in}[/itex] be anything you want?

You have an amplifier. It has a fixed gain vout/vin. And you can make vin anything you want. Otherwise it's not an amplifier.

I am signing off this thread, sorry.
 

1. What is a FET common source configuration with load?

A FET (Field Effect Transistor) common source configuration with load is a type of electrical engineering circuit that uses a FET as the active element and a load resistor to control the output voltage. It is a common type of amplifier circuit used in electronic devices.

2. How does a FET common source configuration with load work?

In this circuit, the FET acts as a variable resistor that is controlled by the input voltage. When the input voltage increases, the FET resistance decreases, causing the output voltage to increase. The load resistor helps to stabilize and control the output voltage.

3. What are the advantages of a FET common source configuration with load?

One advantage is that it has a high input impedance, which means it requires less input current and can be easily connected to other circuits. It also has a low output impedance, making it suitable for driving other circuits. Additionally, it has a wide bandwidth and low distortion.

4. What are the applications of a FET common source configuration with load?

This circuit is commonly used in audio amplifiers, voltage regulators, and signal processing circuits. It is also used in communication systems, such as radio transmitters and receivers, to amplify and filter signals.

5. What are some considerations when designing a FET common source configuration with load?

Some important factors to consider include selecting the appropriate FET for the desired gain and frequency response, choosing the right load resistor to achieve the desired output voltage, and ensuring proper biasing of the FET to prevent distortion. It is also important to consider temperature effects and stability in the circuit design.

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