JFET Calculations: Finding Rs & Rg

[nothing909]

Homework Statement

With reference to the diagram attached.

i) What would be a suitable value for Rg and why
ii) If the drain current is 2mA, determine a suitable Q-point on the mutual characteristics and hence obtain a value for Rs[/B]

The Attempt at a Solution

i) I chose a 2.2M ohm resistor and said that the value of the resistor has to be this large in order to keep the impedance high. Is this correct?

ii) I don't understand how to start this question as I don't have enough variables. I have the Vds and the drain current which is the same for the source current, but how do I go from there? Is there a way to get the Vgs from the mutual characteristics graph?[/B]

 

[lewando]

i) I chose a 2.2M ohm resistor and said that the value of the resistor has to be this large in order to keep the impedance high. Is this correct?

Yes, it keeps the input impedance high. In this particular method of biasing the JFET, the exact value of R_G is generally specified as "large".

Is there a way to get the Vgs from the mutual characteristics graph?

Yes. You are constrained to have I_D of 2 mA on the y-axis, so what is the corresponding V_GS on the x-axis (to be on the V_DS = 3 V curve)?

 

[nothing909]

Yes. You are constrained to have I_D of 2 mA on the y-axis, so what is the corresponding V_GS on the x-axis (to be on the V_DS = 3 V curve)?

So if I take Vgs at say 1.4V? How would that allow me to get the source voltage?

 

[nothing909]

Would it be Vs = Vg - Vgs, so that would be Vs = 0 - 1.4V, making the source voltage 1.4V, making Rs = 1.4V/2mA = 700 ohms.

Then...
Vd = Vdd - Vds - Vs = 12 - 3 - 1.4 = 7.6V, so that make Rd = 3800 ohms

 

[lewando]

Would it be Vs = Vg - Vgs, so that would be Vs = 0 - 1.4V, making the source voltage 1.4V

Some signage issues. V_S does equal V_G - V_GS, but I think V_GS is really -1.4 V (per the graph--note the odd leftward-increasing x-axis).

So you would properly say V_S = 0 - (-1.4 V).

Vd = Vdd - Vds - Vs = 12 - 3 - 1.4 = 7.6V, so that make Rd = 3800 ohms

Some notation issues: V_D is actually V_DS + V_S.

I think you meant to say: [the voltage drop across R_D] = V_DD - V_DS - V_S

 

[nothing909]

Some signage issues. V_S does equal V_G - V_GS, but I think V_GS is really -1.4 V (per the graph--note the odd leftward-increasing x-axis).

So you would properly say V_S = 0 - (-1.4 V).
Some notation issues: V_D is actually V_DS + V_S.

I think you meant to say: [the voltage drop across R_D] = V_DD - V_DS - V_S

I have the equation here saying Vds = Vdd - Vd - Vs, which is wrong is should say Vrd, that's where I got messed up.

Thanks for your help on that

edit: does Vs = Vrs?

I have a couple of other questions to work out:
calculate the mutual conductance at drain current 2mA and then proceed the calculate the amplification factor

for mutual conductance would it be gm = Id / Vgs = 2mA / 1.4 = 1.43 mΩ−1
for amplification factor µ = Rd x gm = 3800 x 1.43x10^-3 = 5.43 dB

 

[lewando]

The equation for mutual conductance, g_m, also called transconductance (a more modern term), is not I_D / V_GS. It is ΔI_D / ΔV_GS. Which is like the slope of the characteristic curve at the operating point, previously established. More precisely, when the circuit is driven by a small AC signal, that results in a value for ΔV_GS. From that you can get ΔI_D (graphically from the curve). Since you don't know anything about the applied input signal, the slope is a good substitute.

For the amplification factor, or gain, your equation looks right, but the result of your stated equation is not in dB. You can represent the result in dB if you want, but that is another step.

 

[nothing909]

The equation for mutual conductance, g_m, also called transconductance (a more modern term), is not I_D / V_GS. It is ΔI_D / ΔV_GS. Which is like the slope of the characteristic curve at the operating point, previously established. More precisely, when the circuit is driven by a small AC signal, that results in a value for ΔV_GS. From that you can get ΔI_D (graphically from the curve). Since you don't know anything about the applied input signal, the slope is a good substitute.

For the amplification factor, or gain, your equation looks right, but the result of your stated equation is not in dB. You can represent the result in dB if you want, but that is another step.

I did Id = Idss (1- (Vgs/Vp))^2, so Id = 4mA (1- (-1.4/-6))^2 = 2.37mA

Then referring to the graph using 2.37mA, Vgs would = -1 so transconductance = 2.37mA/1 = 2.37m mho? Is that correct, I watched a few videos and took a go at it. I'm laying in bed rn at 6am trying to figure this out so I'm not 100% sure, haha.

 

[lewando]

I did Id = Idss (1- (Vgs/Vp))^2, so Id = 4mA (1- (-1.4/-6))^2 = 2.37mA

That equation is for the "mutual characteristic" graph that you attached in your OP. If you compare that with this (a plot of the equation):

You might get different numbers, visually. For the above plot, 2 mA looks like it corresponds to -1.75 V. Whereas the image in your OP indicates, what, -1.4 V?
So pick one (the equation or the image) and stick to it and don't jump between them.

Then referring to the graph using 2.37mA, Vgs would = -1 so transconductance = 2.37mA/1 = 2.37m mho? Is that correct,

Again, transconductance is ΔI_D / ΔV_GS (not I_D / V_GS). It is the slope of the line at your 2 mA operating point.

 

[nothing909]

That equation is for the "mutual characteristic" graph that you attached in your OP. If you compare that with this (a plot of the equation):
View attachment 115048
You might get different numbers, visually. For the above plot, 2 mA looks like it corresponds to -1.75 V. Whereas the image in your OP indicates, what, -1.4 V?
So pick one (the equation or the image) and stick to it and don't jump between them.
Again, transconductance is ΔI_D / ΔV_GS (not I_D / V_GS). It is the slope of the line at your 2 mA operating point.

So how do I find the change in Id and Vgs? I'm confused, if it says in the question to work out the mutual conductance at 2mA, then the change in Id is 0? I don't get it.
I'm looking at this separate question attached trying to work this out and it tells you the change in Id and Vgs, but how do I go about working this out for my question.

Sorry, it's probably simple and I'm just not understanding it.

Thanks

 

[lewando]

You can answer the new attached question because they give you ΔI_D and ΔV_GS. But you don't have the same information for your OP question. So, two ways to proceed:

1) Realize that what you are looking for is the slope of mutual characteristic curve at the bias point (interchangeable with Q-Point). Can you get this slope? If so, you are done.

or

2a) Inject a small signal about your V_GS bias point. Say, 1 V peak-to-peak.
2b) This will give you two new points at V_GS,bias + 0.5 V (blue point) and V_GS,bias - 0.5 V (red point).
2c) From these 2 points on the curve, you can get the corresponding I_D values.
2d) You have ΔV_GS = 1 V, because that is what you have injected.
2e) And you also have ΔI_D from what you did in 2c.

Here is an image stolen from the internet to drive the point home:

It's fuzzy because it's free!

 

[nothing909]

2a) Inject a small signal about your V_GS bias point. Say, 1 V peak-to-peak.
2b) This will give you two new points at V_GS,bias + 0.5 V (blue point) and V_GS,bias - 0.5 V (red point).
2c) From these 2 points on the curve, you can get the corresponding I_D values.
2d) You have ΔV_GS = 1 V, because that is what you have injected.
2e) And you also have ΔI_D from what you did in 2c.

so is this right:

if I vary the Vgs 1V, it would make Vgs go to 2.4 from 1.4, which would then make Id = 1.3mA looking at the graph, so the change in Id would be 0.7mA with the change in Vgs being 1V, so the answer using the equation would be 0.7m mho? yes?

 

[lewando]

So...your result is close, but accuracy (and your method) could be improved. The main source of error is the fact that you are not varying V_GS about the operating point (or bias point, or Q-point). You are deviating from instruction 2b. You are using points at V_GS,bias (green point, sorry was black in post #11--changed for consitency...) and V_GS,bias -1 V (orange point).

The slope at that point (the red point) is going to be slightly flatter than the green point. However, if you rounded your result to 1 mS (use millisiemens now-- in 1881 the mho was replaced by siemens), you would have a good enough answer, considering you have 1 significant digit to work with.

 

[nothing909]

ΔVGS = 0.2 V,

on graph ΔID = 0.2mA when ΔVGS = 0.2 V,

mutual conductance = ΔID / ΔVGS = 0.2mA / 0.2V = 1m mho

is that a good answer?

 

[lewando]

We are now on the same page. Alright!

 

[nothing909]

We are now on the same page. Alright!

thanks for your help.

going back a few posts when I said about the amplification factor µ = Rd x gm = 3800 x 1x10^-3 = 3.8, just to clarify, is that right? and what would it be measured in, I keep seeing dB, but as you said in a previous post that it would take an extra step to allow it to be dB so what is my answer measured in?

 

[lewando]

Yes, I would say so. Thanks for sticking this one out.

 

[nothing909]

Yes, I would say so. Thanks for sticking this one out.

what would be the unit for amplification factor, without going to extra step to get dB

 

[lewando]

I thought you just said that μ = 3.8. Looks right. That is an amplification "factor", something that you would multiply your input voltage (in V) to get an output voltage (in V). The factor is unitless. You could represent it as a "dB" if you want. Want to give that a try?

 

[nothing909]

I thought you just said that μ = 3.8. Looks right. That is an amplification "factor", something that you would multiply your input voltage (in V) to get an output voltage (in V). The factor is unitless. You could represent it as a "dB" if you want. Want to give that a try?

so that would make the Vo = 3.8 x 12 = 45.6V? and you're saying that I can if I want put it as 3.8dB?

 

[lewando]

Don't introduce "Vo" when you have a perfectly good V_D for an output.

3.8 is the amplification "factor". You can't just convert it to dB by appending "dB" to it.

Please take a look at this: http://www.ni.com/white-paper/14930/en/ Specifically, example 3.2