# MOSFET calculation help.

Discussion in 'Homework Help' started by Petrucciowns, Jul 22, 2009.

1. ### Petrucciowns Thread Starter Active Member

Jun 14, 2009
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I need help calculating the voltage gain and the voltage from gate to source on the following circuit:

http://img269.imageshack.us/img269/5195/1123232131.jpg

For AV I tried the equation that TNK gave me for a JFET RD/RS, but I guess this is not true for MOSFETS.

For Vgs I assumed it would be gate voltage- source voltage ,but again I was wrong.

Can anyone help me with this?

Thanks.

2. ### t_n_k AAC Fanatic!

Mar 6, 2009
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I think you are meant to find an estimate for gm in each case and use this to adjust the gain equation ...

Av=-gm*Rd/(1+gm*Rs)

Take the case of Id=500uA.

Vgs = 5.49-1.65=3.84V

Estimate gm = Id/Vgs = 500e-6/3.84 = 130.2e-6

Av = - 130.2e-6*22e3/(1+130.2e-6*3300) = - 2.86/(1+.43)=-2.0

So I think the Av values shown on your solution may be incorrect. It seems the Av was found by just using Av = -gm*Rd which does not allow for the gain degeneration caused by including Rs. The low gm values in your example do not allow the use of the simple approximation of Av=-Rd/Rs.

You'd use a similar approach to find the gain when the other FET gives a drain current of 900uA.

3. ### Petrucciowns Thread Starter Active Member

Jun 14, 2009
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I am going to trust you ,because obviously you have more experience than I do. I just have one question. Why is the gain negative?

4. ### t_n_k AAC Fanatic!

Mar 6, 2009
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Given my patchy recent 'track record' I'm not sure I'd trust myself.

No doubt any errors will soon be unearthed by one of the many skilled forum members.

Why negative gain? This simply means there is a phase inversion of the signal at the drain compared with that at the gate. If you think about what happens physically it will become clear that as the gate voltage is increasing (positive slope) so is the drain current. If the drain current is increasing (positive slope) then the drain voltage is decreasing (negative slope). If the gate voltage signal is a sine wave then the drain voltage at low and mid-band frequencies would be an inverted sine wave or 180 degrees out of phase with the gate voltage.

5. ### The Electrician AAC Fanatic!

Oct 9, 2007
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I notice in the first case, written in pencil to the left of the tabular data, is the notation: Av = 2.86

This is the value of gm*Rd in tnk's calculation, and would be the gain of the circuit if the source resistor were fully bypassed.

Likewise, in the second case, Av = 7.86 is also gm*Rd, with the same significance.

6. ### millwood Guest

gm is typically defined as dId/dVgs, a small signal property, unlike hFE.

7. ### t_n_k AAC Fanatic!

Mar 6, 2009
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Hi Millwood,

Yes I certainly agree.

I have been having some doubts about the circuit gain issue - as per Petrucciowns original diagram. In fact it seems the question never really asked what the gain was - someone has penciled in values for Av as an afterthought.

The data supplied is sufficient only to give a single point estimate of gm which is clearly erroneous if applied to the gain. As your equation above denotes, one would require at least two pairs of static bias values - for Vgs and Id - to even roughly estimate gm . If the value of Av was really anticipated in the solution, gm would have simply been quoted in the original problem.

Petrucciowns - there's insufficient data to find the gains - in the end the problem was really about DC bias. Forget the gain in this case and probably in the previous jFET problem you raised.

8. ### The Electrician AAC Fanatic!

Oct 9, 2007
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I thought the same, but I decided to look up the characteristics of a MOSFET that might be relevant.

The current in this circuit is low, and it's hard to find a really low current MOSFET that isn't a VHF amplifier.

Suppose the MOSFET is a general purpose, less than 1 amp rated device, such as the BS170: www.fairchildsemi.com/ds/BS/BS170.pdf

Have a look at figure 5 on the data sheet. The value of gm is the slope of the transfer characteristic at any given drain current. Your calculation assumed a straight line from the origin to a point on the curve. Since the characteristic curve is concave up at low currents, your value will be smaller that the true value.

You can't even see the curve at 500 μA, but clearly the slope continues to decrease at the really low currents, and taking the slope to be that of a straight line drawn from the origin to a given drain current doesn't appear to be that far off at very low drain currents.

Another MOSFET will have a different curve, but all MOSFETs have a concave up characteristic, and a gm that approaches zero at very low currents.

I'm not so sure that "...one would require at least two pairs of static bias values - for Vgs and Id - to even roughly estimate gm." at the low current level of this problem.

For sure, gm won't be less than the value estimated by taking it to be the slope of a straight line through the origin, and the circuit gain won't be less than the value obtained from this gm.

9. ### mik3 Senior Member

Feb 4, 2008
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Multiply Id with Rs to find Vs

Use the voltage divider rule to find Vg (assuming no current is flowing into the gate)

Then Vgs=Vg-Vs

Vd=26-Id*Rd

Vds=Vd-Vs

10. ### millwood Guest

assuming that the mosfet will remain conducting, and small change in Vgs, delta Vgs, will create a small change in Id, delta Id = delta Vgs / Rs. that change in current will create a small change in Vd, delta Vd = delta Id * Rd = delta Vgs * Rd / Rs.

so the voltage gain, Av = delta Vd / delta Vgs = Rd / Rs.

11. ### The Electrician AAC Fanatic!

Oct 9, 2007
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Vgs is not the same as Vg. It's not true that ΔId = ΔVgs/Rs. What is true is that ΔId = gm*ΔVgs because the definition of gm is gm = ΔId/ΔVgs.

The input to this circuit is applied between gate and ground, not between gate and source; there's a source resistor in there.

So a correct analysis goes like this:

Is = Id and ΔIs = ΔId
ΔVgs = ΔVg - ΔIs*Rs
ΔId = gm*ΔVgs = gm*(ΔVg - ΔIs*Rs) = gm*(ΔVg - ΔId*Rs)
ΔId*(1+gm*Rs) = gm*ΔVg
ΔId = (gm*ΔVg)/(1+gm*Rs)
ΔVd = -ΔId*Rd = -(gm*ΔVg*Rd)/(1+gm*Rs)
Av = ΔVd/ΔVg = -(gm*Rd)/(1+gm*Rs)

Av ≠ ΔVd/ΔVgs; rather Av = ΔVd/ΔVg

12. ### millwood Guest

= -Rd/(1/gm + Rs);

pick a typical mosfet's datasheet, compare 1/gm and Rs and then come back for a discussion if you still don't know why 1/gm is dropped.

13. ### t_n_k AAC Fanatic!

Mar 6, 2009
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Thanks Millwood & Electrician & other wise folk,

It makes sense now.

Did you have any thoughts on the Av estimate in the OP's earlier post on a JFET based circuit? Is Av=-Rd/Rs still a reasonable estimate as well?

14. ### t_n_k AAC Fanatic!

Mar 6, 2009
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The reason I asked is that gm (Yfs?) is likely to be significant. Take as an example the general purpose BC264 JFET which has a Yfs=2.5 mmhos.

The Av estimate for the OP's JFET circuit would be

Av=-2.5e-3*5000/(1+2.5e-3*500)=-12.5/(1+1.25) =-5.5 which is long way from -10.

15. ### The Electrician AAC Fanatic!

Oct 9, 2007
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So now you're taking my correct derivation of ΔVd/ΔVg (whereas your derivation was for ΔVd/ΔVgs), performing a little algebra on it and suggesting that I don't know under what circumstances 1/gm might be negligible compared to Rs?

tnk explained it in post #5 of the thread "JFET voltage gain", where he said:

Let me offer you a discussion on how to correctly derive an expression for delta Vd / delta Vgs.

Your derivation, whose final result is "Av = delta Vd / delta Vgs" is for a "gain" that has no relevance to the given circuit. But even if it did, you didn't get it right.

It's not universally true that delta Id = delta Vgs / Rs. Can you show that it is true in general? Even though it could be numerically correct in the very special circumstance that Rs = 1/gm, it won't generally be true.

And finally, you said that Av = delta Vd / delta Vgs = Rd / Rs, which is seldom true; I already explained the special circumstance where it could be numerically correct.

If you want to define a gain ΔVd/ΔVgs then it can be derived like this:

ΔId = gm*ΔVgs (by definition of gm) from which ΔVgs = ΔId/gm
ΔVd = ΔId*Rd

ΔVd/ΔVgs = (ΔId*Rd)/(ΔId/gm) = gm*Rd

this "gain" is independent of Rs, and is totally different than the expression you derived, which was delta Vd / delta Vgs = Rd / Rs. Your general expression is wrong, and the only way it could accidentally be numerically right is if Rs happened to be numerically equal to 1/gm for some particular MOSFET and source resistor.

Besides, as I pointed out, the gain of this circuit is calculated with the input signal applied between the gate and ground, not between the gate and source. If you were using something like a transformer to apply a floating drive signal between gate and source, then you might care about the ratio ΔVd/ΔVgs.

16. ### The Electrician AAC Fanatic!

Oct 9, 2007
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I went looking for a MOSFET that has such a small current carrying capability that it might be used in a circuit like this, with a 22k Rd.

The BS170, whose data sheet link I posted, has a gm of .32 siemens. This is so high that it would be reasonable to take the circuit gain to be Rd/Rs. But, that's at an Id of 200 mA. The OP's circuit has an Id of 500 μA. Even the BS170 would have a much lower gm at an Id of 500 μA.

But, I thought I would see if I could find a more likely MOSFET for this circuit. The Motorola 3N157 has a maximum Id of 30 mA, and a minimum gm of 1000 uS at an Id of 2 mA. The spec sheet has a graph showing minimum gm vs. Id. At Id = 500 μa, gm could be 580 μS, and at Id = 900 μA, gm could be 700 μS.

These values are not negligible in the gain formula.

17. ### t_n_k AAC Fanatic!

Mar 6, 2009
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Thanks Electrician - much appreciated for your thoughts as ever.

The other earlier post by the OP regarding a JFET amplifier question was here

I think I gave an incorrect estimate for Av in the question as -Rd/Rs which with hindsight requires some further analysis - per my last post in this thread. The OP must shake their head in confusion sometimes - no doubt even the 'experts' sometimes have fuzzy thoughts - more often than not in my case!.

18. ### The Electrician AAC Fanatic!

Oct 9, 2007
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Well, you did say that "So if gm*Rs >> 1 then the approximation is reasonable.
".

Really, it's just a question of just how big is gm.

I'm looking at the specs for a Siliconix JFET, P/N J309, with a minimum gm of 10,000 μS. That's pretty respectable. Then gm*rs = 5, which in the JFET circuit gives a gain of 8.333 taking gm into account, and 10 without taking it into account.

Small JFETs certainly don't have the high gm's that higher current MOSFETs have, so getting a handle on gain without knowledge of gm is dicey.

19. ### millwood Guest

jfets in general and jfet switchers in particular have lower gm so the error in omitting 1/gm is bigger.

(vertical) mosfets in general have high gm (from low single digit s to low double digit s), measured at hundreds of ma and up, thanks to their parrallel construction: a large v-mosfet is essential a combination of hundreds of small v-mosfet.

at lower Id, their gm will suffer as well, but still much lower than Rs.

some numbers, all measured at Id=0.5ma:

mosfets:
IRFP140: gm=100ms, 1/gm=10ohm;
IRF510: gm=75ms, 1/gm=13ohm;
BS170: gm=6.1ms, 1/gm=165ohm.

jfet switcher:
J112: gm=3.9ms, 1/gm=259ohm;

with a RS=3.3kohm, > 20x that of the highest 1/gm, the error from omitting 1/gm is minimal.

kind of like we typically assume Vbe is constant in varying Ic for bjts, people typically do NOT factor in Vgs changes in gain calculation for mosfets. and this example shows you why.

20. ### millwood Guest

here is the result of a quick simulation with irf510 running at around .5ma.

with input changing from 5v - 6v, Id goes from 360u - 660u, and Vd goes from 17.1v to 10.5, for a gain of (17.1v - 10.5v)/(5v-6v)=-6.6x, vs. Rd / Rs = 22k/3.3k=6.67x, pretty close.

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