Electrical Engineering - FET Common Source Configuration with Load

[GreenPrint]

Homework Statement

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

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

Homework Equations

The Attempt at a Solution

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

 

[mfb]

What do you want to calculate, and what are all those variables?

V_G will depend on V_I, and I think you will have to treat this as unknown input. Just calculate everything in terms of an unknown V_I (if that is too hard in general, it might be interesting to consider a sine wave).

 

[GreenPrint]

Here's the actual picture from my book.

http://imageshack.com/a/img600/8114/gz12.png

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

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

Below is my small signal equivalent circuit

http://imageshack.com/a/img21/4747/y59u.png

What I get for the $A_{v_{NL}}$ is

http://imageshack.com/a/img268/4091/f6ze.png

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 $v_{p}$ in my equation for $g_{m}$ and the negative sign in my equation for $A_{v_{NL}}$ which I'm adding now.

http://imageshack.com/a/img713/7876/gi2i.png

I'm not sure how he got $A_{v_{NL}}$

I see how he gets $g_{mo}$

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

I also know that

$g_{m} = g_{mo}(1 - \frac{v_{GS}}{v_{P}})$

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

Oh apparently he's finding $A_{v_{NL}} = \frac{v_{o}}{v_{s}}$

 

[GreenPrint]

Oh apparently he's finding $A_{v_{NL}} = \frac{v_{o}}{v_{s}}$

which leaves me with this instead.

http://imageshack.com/a/img811/3217/naxp.png

but still not able to estimate because we don't know $v_{s}$

hm

 

[rude man]

A_VNL sounds like "no-load voltage gain". That would mean not including R_L in your gain computation.

 

[GreenPrint]

I agree, but I am still left with the problem of not knowing how what $v_{s}$. Taking what you say into consideration I get

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

I do I go about getting around this?

This is how I got that formula

http://img89.imageshack.us/img89/8845/ls8c.png

Thanks for the help.

 

[rude man]

JFET circuit

Haven't we discussed this before? You get V_S by equating the expression for FET current to the current flowing thru R_S. 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, $g_{m} = \frac{2I_{DSS}}{|v_{P}|}(1 - \frac{v_{GS}}{v_{p}})$, V_GS and V_P are both dc quantities and so should be capitalized. Use v_gs for ac signals. I found this problem in several of your equations.

What BTW is g_m0? How's it different from g_m? There is only one value of g_m 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 ...

 

[GreenPrint]

Well $v_{s}$ in this problem though is the input AC voltage. I know that $i_{D} = i_{S}$ and that I can find the voltage at the source of the transistor this way using the equation for $i_{D}$ for a FET, also symbolized by $v_{S}$ I do understand this process now. The only problem is that $v_{s}$ in my equation for $V_{GS}$ 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 $v_{S}$ 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 $V_{GS}$ and $V_{P}$ since they aren't AC parameters.

To me $g_{mo}$ is kind of silly. My book also defines
$g_{m} = g_{mo}(1 - \frac{V_{GS}}{V_{P}})$
where
$g_{mo} = \frac{2I_{DSS}}{|V_{P}|}$
which would be the value of $g_{m}$ when $V_{GS}$ is zero

 

[rude man]

In this problem I take $v_{S}$ 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 v_gs in your equivalent circuit is just v_g which is just a voltage divider away from v_in. You should not have labeled the input V_s or v_s. That's very confusing. Label it v_in.

So now surely you can solve for v_o. What did you compute for V_s and g_m?

 

[GreenPrint]

Ya I established that.

so can I do this then.

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

Does this look better for $v_{in}$? If so from here it's easy.

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

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

*edit* I added in some explanations.

 

[rude man]

?

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

You are supposed to solve for v_o, given v_in. A_vNL = v_o/v_in with R_L removed.

I don't understand your equations.

I didn't see numbers for g_m and V_s.

 

[GreenPrint]

?

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

You are supposed to solve for v_o, given v_in. A_vNL = v_o/v_in with R_L removed.

I don't understand your equations.

I didn't see numbers for g_m and V_s.

Wait really I can make up $v_{in}$ 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 $v_{s}$ you mean the the AC voltage? It's just zero because it's connected to ground in the small signal equivalent circuit.

 

[rude man]

Wait really I can make up $v_{in}$ 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 $v_{s}$ 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.

 

[GreenPrint]

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 $V_{G}$ I just use the DC schematic diagram and get

$V_{G} = \frac{R_{G}v_{in}}{R_{sig} + R_{G}}$

So I don't know $v_{in}$ but I should be able to solve without hm.

 

[rude man]

V_in = 0.
You need V_GS to solve for g_m. You know V_G = 0.

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

Then use your formula for g_m.

 

[GreenPrint]

This is starting to make sense, but why do you take Vin = 0?

 

[GreenPrint]

Why can you let $v_{in}$ be anything you want?

 

[GreenPrint]

$A_{V_{NL}} = \frac{v_{O}}{v_{in}}$
$v_{O} = -g_{m}V_{GS}R_{D}$
$A_{V_{NL}} = -\frac{g_{m}V_{GS}R_{D}}{v_{in}}$
$V_{GS} = V_{G} - V_{S}$
$V_{G} = \frac{v_{in}R_{G}}{R_{G} + R_{sig}}$
$V_{GS} = \frac{v_{in}R_{G}}{R_{G} + R_{sig}} - V_{S}$
$V_{S} = I_{S}R_{S}$
$I_{S} = I_{D} = I_{DSS}(1 - \frac{V_{GS}}{V_{P}})^{2}$
$V_{S} = I_{DSS}(1 - \frac{V_{GS}}{V_{P}})^{2}R_{S}$
$V_{GS} = \frac{v_{in}R_{G}}{R_{G} + R_{sig}} - I_{DSS}(1 - \frac{V_{GS}}{V_{P}})^{2}R_{S}$
$V_{GS} = \frac{v_{in}R_{G}}{R_{G} + R_{sig}} - I_{DSS}(1 - \frac{2V_{GS}}{V_{P}} + \frac{V_{GS}^{2}}{V_{P}^{2}})R_{S}$
$V_{GS} = \frac{v_{in}R_{G}}{R_{G} + R_{sig}} - I_{DSS}R_{S} + \frac{2I_{DSS}V_{GS}R_{S}}{V_{P}} - \frac{V_{GS}^{2}R_{S}I_{DSS}}{V_{P}^{2}}$
$0 = \frac{v_{in}R_{G}}{R_{G} + R_{sig}} - I_{DSS}R_{S} + \frac{2I_{DSS}V_{GS}R_{S}}{V_{P}} - \frac{V_{GS}^{2}R_{S}I_{DSS}}{V_{P}^{2}} - V_{GS}$
$0 = -\frac{R_{S}I_{DSS}}{V_{P}^{2}}V_{GS}^{2} + (\frac{2I_{DSS}R_{S}}{V_{P}} - 1)V_{GS} + \frac{v_{in}R_{G}}{R_{G} + R_{sig}} - I_{DSS}R_{S}$
$V_{GS} = \frac{-(\frac{2I_{DSS}R_{S}}{V_{P}} - 1) \underline{+} \sqrt{(\frac{2I_{DSS}R_{S}}{V_{P}} - 1)^{2} + 4(\frac{R_{S}I_{DSS}}{V_{P}^{2}})(\frac{v_{in}R_{G}}{R_{G} + R_{sig}} - I_{DSS}R_{S})}}{2(-\frac{R_{S}I_{DSS}}{V_{P}^{2}})}$
$V_{GS} = \frac{\frac{2I_{DSS}R_{S}}{V_{P}} - 1 \overline{+} \sqrt{(\frac{2I_{DSS}R_{S}}{V_{P}} - 1)^{2} + 4(\frac{R_{S}I_{DSS}}{V_{P}^{2}})(\frac{v_{in}R_{G}}{R_{G} + R_{sig}} - I_{DSS}R_{S})}}{\frac{2R_{S}I_{DSS}}{V_{P}^{2}}}$
$A_{V_{NL}} = -\frac{g_{m}(\frac{\frac{2I_{DSS}R_{S}}{V_{P}} - 1 \overline{+} \sqrt{(\frac{2I_{DSS}R_{S}}{V_{P}} - 1)^{2} + 4(\frac{R_{S}I_{DSS}}{V_{P}^{2}})(\frac{v_{in}R_{G}}{R_{G} + R_{sig}} - I_{DSS}R_{S})}}{\frac{2R_{S}I_{DSS}}{V_{P}^{2}}})R_{D}}{v_{in}}$
$A_{V_{NL}} = -\frac{g_{m}(\frac{2I_{DSS}R_{S}}{V_{P}} - 1 \overline{+} \sqrt{(\frac{2I_{DSS}R_{S}}{V_{P}} - 1)^{2} + 4(\frac{R_{S}I_{DSS}}{V_{P}^{2}})(\frac{v_{in}R_{G}}{R_{G} + R_{sig}} - I_{DSS}R_{S})}R_{D}V_{P}^{2}}{v_{in}2R_{S}I_{DSS}}$
$g_{m} = \frac{2I_{DSS}}{|V_{P}|}(1 - \frac{V_{GS}}{V_{P}})$
$A_{V_{NL}} = -\frac{2I_{DSS}(1 - \frac{V_{GS}}{V_{P}})(\frac{2I_{DSS}R_{S}}{V_{P}} - 1 \overline{+} \sqrt{(\frac{2I_{DSS}R_{S}}{V_{P}} - 1)^{2} + 4(\frac{R_{S}I_{DSS}}{V_{P}^{2}})(\frac{v_{in}R_{G}}{R_{G} + R_{sig}} - I_{DSS}R_{S})}R_{D}V_{P}^{2}}{v_{in}2R_{S}I_{DSS}|V_{P}|}$
$A_{V_{NL}} = -\frac{(1 - \frac{V_{GS}}{V_{P}})(\frac{2I_{DSS}R_{S}}{V_{P}} - 1 \overline{+} \sqrt{(\frac{2I_{DSS}R_{S}}{V_{P}} - 1)^{2} + 4(\frac{R_{S}I_{DSS}}{V_{P}^{2}})(\frac{v_{in}R_{G}}{R_{G} + R_{sig}} - I_{DSS}R_{S})}R_{D}}{v_{in}R_{S}}$
etc... this problem sucks. I'm not sure how to solve this problem when I don't know $v_{in}$. The previous problem, this information was given.

 

[rude man]

This is starting to make sense, but why do you take Vin = 0?

Because that is the dc operating point of your circuit.

 

[rude man]

Why can you let $v_{in}$ 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.