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Old 2nd September 2008, 11:50 AM   #11
flg is offline flg  United States
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Well, I guess I will try to calculate the actuall C from my sim machine, just for fun. Although, I'm not familar with the method N.P. was speaking of above. measuring 1k, 10k, 100k ??? I would use a square wave signal and calc from the rise time dV/dT and I. Wouldn't that method work.
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Old 2nd September 2008, 08:53 PM   #12
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I suppose it would. I used the three different frequencies as a check to
make sure I was really dealing with capacitance. A 10 KHz figure seems
accurate enough.
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Old 2nd September 2008, 10:02 PM   #13
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Quote:
Originally posted by EUVL
Suppose I use ONE single pair, but increase the bias to 400mA per FET, and at the same time reduce the source degeneration resistor by a factor of 4. Assuming of course that dissipation is not an issue, for the sake of the discussion.

At such "low" bias currents, one can count on that transconductance is roughly proportional to bias (quadratic region), i.e. 4x per FET compared to 100mA. In which case I would expect the voltage required to drive the gate to be comparable to 4 pair @ 100mA. Would I then not get close to 135p, give or take 20% ?
It seems logical that you would see that. The potential flaw would be that the
transconductance might not be truly proportional to the bias current. It's
my experience that it's not, so I expect that you should add that 20% rather
than subtract.
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Old 3rd September 2008, 06:30 PM   #14
EUVL is offline EUVL  Europe
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For the sake of understanding, and ignoring subjective sonic imprresion for a while.

So in the follower circuit like F4, by using a single FET to replace the 4 in parallel, and then take away the heat dissipation by using a cascode of say 4x IRFP240s in parallel at say 5V Vds of the single follower device, I would get a roughly 3 times (or perhaps more because of the cascode) increase in bandwidth ?


Patrick
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Old 3rd September 2008, 06:42 PM   #15
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Probably not quite. First off, the F4 uses 3 devices in parallel, but setting that
aside, the capacitance will go up dramatically with a 5 volt Vds.

Let me emphasize that all of these are barely thumbnail estimates. If you
really want to know, it's best that you hook it up under the conditions of
interest and see what the real figures are.

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Old 3rd September 2008, 07:05 PM   #16
AndrewT is offline AndrewT  Scotland
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does one monitor the gate current by measuring volts drop across the gate resistor?
Will the meter impedance modify the circuit currents when connected?
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regards Andrew T.
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Old 3rd September 2008, 11:13 PM   #17
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Quote:
Originally posted by AndrewT
does one monitor the gate current by measuring volts drop across the gate resistor?
Will the meter impedance modify the circuit currents when connected?
An isolated high quality AC voltmeter will do the job.
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Old 25th May 2012, 10:47 AM   #18
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Quote:
Without a load, these figures are 45 pF and 35 pF respectively.
Quote:
An isolated high quality AC voltmeter will do the job.
I'm wondering how big is the gate resistor?

If you are measuring the 35pF value at 1kHz & 8V RMS , even with a 1k gate resistor:
V=IR
V = I * ( 1 / 2*pi*f*C )
8 = I * ( 1 / 6.28k*35p)
8 = I * ( 1 / 220n)
I = 8 * 220n = 1.76uA

If the gate resistor is 100ohm, we will get 0.16mV of voltage.
A good 5 1/2 digit handheld meter like Fluke 187 has resolution down to 1uV, can still be useful.
A good 6 1/2 digit benchtop meter like Agilent 34410A can do 0.1uV, will be even better.

But my worry is that, with the 8V RMS Common mode signal, and such small voltage to be measured, is the ISOLATION good enough to be trusted?

Click the image to open in full size.

A check into the Agilent user manual and I found something.
The common mode 8V rms signal will pass though Ci and generate voltage on the gate resistance which we are measuring (= Vtest here).
Since Ci is 200pF, our calculation will give us 200pF even if the actual gate capacitance is 0pF.

So obviously PASS is not using the Agilent bench top meter.

Handheld meter might help have better isolation to ground and have lower Ci figure, but I'm not sure how much better.

So I'm a bit unsure how well a isolated voltmeter will work to get the actual gate capacitance.

Last edited by zhoufang; 25th May 2012 at 10:58 AM.
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