Accurate measurement of jfet gm

[iko]

Is there any simple way to accurately measure the jfet transconductance?

I found a paper named "A note on the measurement of FET parameters" by D.B.S.J Prasada, Rao, B.P. Singh, and Dikshitulu Kalluri, Int. J. Electronics 1976, vol. 41, No. 5, 521-524. The image attached is from this paper. (Send pm if you'd like to get this paper.)

I know that if gm = -2 sqrt(Id * Idss) / Vp but measuring the pinch-off voltage accurately seems problematic. I know of two ways of measuring Vp. One, connect a large resistor between gate and source, and apply voltage to drain, ground to gate, and measure Vp across the resistor. The resistor is recommended to be such that Idss * Rs = 10^4. For instance, for a 2sk170bl with Idss 10mA you would want a 1 megaohms resistor.

The second way of measuring Vp is more direct, but I think it is less accurate. The idea is to apply voltage to drain and ground to gate. Then, with a large input impedance voltmeter measure the voltage from gate to source. The article recommends a 10 megaohms impedance voltmeter.

In any case, these two ways of measuring the pinch-off voltage produce results that are sufficiently different that the gm differs quite a bit; try it, it's simple.

So, back to my original question, is there some established way of measuring the jfet gm? I couldn't find more than what I mentioned above.


The reason I would like to be able to accurately measure gm is so that I could accurately predict/design the gain of a common source jfet stage. Perhaps it's not possible.

[iko]

I don't know if my understanding of the previous circuit is correct, but this, attached, is what I tried. Applied a sine wave at the input, about 200mA p-p. Then, while monitoring V(out), adjusted Rf. I can get V(out) small, but not zero, as it says in the article. When V(out) is smallest, I took out Rf and measured it. The idea is that gm = 1/Rf, and I got sane values, in the 20 something mS ballpark. I tried removing Rs, to see if, as another sanity check the value of Rf goes down, the idea being that gm is higher at Idss. Indeed, this happened. But why doesn't V(out) become zero as the article predicts. What happens is that making Rf smaller, at some point V(out) starts increasing in value. I'm puzzled. Anyone?

[jcx]

for a unspec'd depletion mode mosfet I took ~ 5 I,V points with Vds >> 2*Vp and curve fit to the "square law" gm equation

[AndrewT]

Hi,
is gm the slope of the Vgs vs Id curve?
If so then gm varies all the way from Id=1nA to Id=Idss

What about directly measuring the change in Vgs and the change in Id when near Idss and again at near 50% of Idss?

[Bob Cordell]

Originally Posted by ikoflexer

Is there any simple way to accurately measure the jfet transconductance?

I found a paper named "A note on the measurement of FET parameters" by D.B.S.J Prasada, Rao, B.P. Singh, and Dikshitulu Kalluri, Int. J. Electronics 1976, vol. 41, No. 5, 521-524. The image attached is from this paper. (Send pm if you'd like to get this paper.)

I know that if gm = -2 sqrt(Id * Idss) / Vp but measuring the pinch-off voltage accurately seems problematic. I know of two ways of measuring Vp. One, connect a large resistor between gate and source, and apply voltage to drain, ground to gate, and measure Vp across the resistor. The resistor is recommended to be such that Idss * Rs = 10^4. For instance, for a 2sk170bl with Idss 10mA you would want a 1 megaohms resistor.

The second way of measuring Vp is more direct, but I think it is less accurate. The idea is to apply voltage to drain and ground to gate. Then, with a large input impedance voltmeter measure the voltage from gate to source. The article recommends a 10 megaohms impedance voltmeter.

In any case, these two ways of measuring the pinch-off voltage produce results that are sufficiently different that the gm differs quite a bit; try it, it's simple.

So, back to my original question, is there some established way of measuring the jfet gm? I couldn't find more than what I mentioned above.


The reason I would like to be able to accurately measure gm is so that I could accurately predict/design the gain of a common source jfet stage. Perhaps it's not possible.

Hi ikoflexer,

In my view, the thing that really matters in gm of JFET or MOSFET is not gm as a function of gate voltage, but rather gm as a function of drain current. This largely takes Vp out of the picture. Vp is often all over the place anyway.

AC couple a small signal AC source into the gate at 1 kHz and 10 mV rms. Bias the gate through a 100k resistor from an adjustable negative voltage source for an N-channel JFET. Connect the source to ground. Connect the drain to a positive adjustable voltage supply through a 1k resistor and monitor the AC and DC voltages at the drain. Also have a means of monitoring drain current (voltage drop across the drain load resistor can be used to infer drain current, for example).

Adjust the gate voltage to get the test current at which you are taking the gm datapoint. Adjust the drain voltage supply to get the desired Vds for the datapoint(typically 10V). Measure the AC at the drain. 10 mV rms at the drain will correspond to 1ms.

Under some current conditions, a larger or smaller value of drain load resistor may be necessary to keep the numbers reasonable.

Cheers,
Bob

[iko]

Thanks Bob! I'll try it.

I know some people will wonder how your suggestion works, so I'll write it up for everyone's benefit.

The voltage gain Av, for this case when there is no source resistor, is defined as

Av = Vout / Vin = gm * Rd

(We ignore the minus sign on the rhs, because we just need a number.)

So we can solve for gm = Vout / (Vin * Rd)

Bob suggested Rd = 1000 ohms, and Vin = 0.01V therefore

gm = Vout / 10

So in this example if we measured Vout = 10mV, it would mean gm = 1mS

Of course, this can be done nearer the operating point, when Id is not equal to Idss, if a source resistor Rs is used.

If Rs is used, then

Av = Vout / Vin = gm Rd / (1 + gm Rs)

(I left out the minus sign on gm).

Solving for gm

gm = Vout / (Vin * Rd - Vout Rs)

Rd and Vcc (the DC voltage at the top of the load resistor Rd) should be chosen so that the voltage drop across Rd still leaves enough as Vds to make sense.

Somebody correct me if I'm wrong.

[Bob Cordell]

Originally Posted by ikoflexer

Thanks Bob! I'll try it.

I know some people will wonder how your suggestion works, so I'll write it up for everyone's benefit.

The voltage gain Av, for this case when there is no source resistor, is defined as

Av = Vout / Vin = gm * Rd

(We ignore the minus sign on the rhs, because we just need a number.)

So we can solve for gm = Vout / (Vin * Rd)

Bob suggested Rd = 1000 ohms, and Vin = 0.01V therefore

gm = Vout / 10

So in this example if we measured Vout = 10mV, it would mean gm = 1mS

Of course, this can be done nearer the operating point, when Id is not equal to Idss, if a source resistor Rs is used.

If Rs is used, then

Av = Vout / Vin = gm Rd / (1 + gm Rs)

(I left out the minus sign on gm).

Solving for gm

gm = Vout / (Vin * Rd - Vout Rs)

Rd and Vcc (the DC voltage at the top of the load resistor Rd) should be chosen so that the voltage drop across Rd still leaves enough as Vds to make sense.

Somebody correct me if I'm wrong.

I think you've got it right.

BTW, there is an interesting difference between JFETs and BJTs. The gm for a BJT is proportional to Ic as a result of the exponential relationship of collector current to base-emitter voltage. On the other hand, the JFET is a square-law device, and for this reason its gm is proportional to the sqaure root of drain current.

Cheers,
Bob