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Old 22nd October 2011, 07:29 PM   #1
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Default 500v/uS slewrate, really?

Doing some simulation with a complementary JFET input stage using 2SK246/2SJ103, instead of the usually recommended 2SK170/2SJ74 JFETs.

Mostly for fun and to see what I could get out of it.

Was a little amazed at the insane slewrate I could achieve and still maintain a nice looking 20 KHz squarewave and good phase and gain margins.

Only downside is THD-20 that is around 0.02% at 60 Watt, 8 Ohm load, the 1 W 8 Ohm load THD-20 is a more decent 0.001XXX%.

THD-1 1 W, 8 Ohm load is 0.0002XXX% rising to 0.0017XX% at 60 W, 8 Ohm load.
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File Type: png Experiment-schematic.png (75.8 KB, 565 views)
File Type: png Experiment-SR-100nS-50V-Rising.png (31.9 KB, 538 views)
File Type: png Experiment-SR-100nS-50V-Falling.png (32.3 KB, 491 views)

Last edited by Neutrality; 22nd October 2011 at 07:47 PM.
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Old 22nd October 2011, 07:32 PM   #2
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20 KHz squarewave, phase and gain margin plot.
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File Type: png Experiment-phase-gain margin.png (48.6 KB, 480 views)
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Old 22nd October 2011, 08:55 PM   #3
jcx is offline jcx  United States
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yes it is literally unbelievable

try setting the sim max step size time in your sim to something appropriate to such an edge - like 1 nS

also look at driver bias, output shoot thru - I don't think that 120 Ohm resistor can drain the output Mosfet gate charge that fast
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Old 23rd October 2011, 02:20 AM   #4
Bonsai is offline Bonsai  Taiwan
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The question here is does your simulation in this case represent the real world? 5pf Cdom seems a bit low to me.
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Old 23rd October 2011, 07:00 AM   #5
Elvee is offline Elvee  Belgium
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I find the 20K waveshape incoherent with that kind of performance: it is much too rounded.
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Old 23rd October 2011, 08:32 AM   #6
RJM1 is offline RJM1  United States
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To measure the slew rate of an amplifier you measure the rise and fall times of a pulse waveform. Rise time is measured from 10% above the minimum voltage to 90% of the maximum voltage and fall time is measured from 10% below the maximum voltage to 90% of the minimum voltage. If you are measuring a square wave output of an amplifier the 10 and 90% points should be of equal voltage and opposite polarity
The rising slew rate that you measured in your first post (pic #2) was between -31.33 and 20.3 Volts and the falling slew rate (pic #3) were between 33.97 and -16.11 Volts.
It’s easy to get insane slew rates when you throw out the slower parts of the signal.
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Old 23rd October 2011, 08:43 AM   #7
Elvee is offline Elvee  Belgium
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Quote:
Originally Posted by Elvee View Post
I find the 20K waveshape incoherent with that kind of performance: it is much too rounded.
This, as an example is a 10KHz squarewave from an amplifier having a modest ~24V/µs slewrate:
Click the image to open in full size.
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Old 23rd October 2011, 08:55 AM   #8
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Click the image to open in full size.
Top: Output, 20V/Div, Bottom: In, 1V/Div F: 20KHz

Square performance of my latest design which has about 100V/uS slew.
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Old 23rd October 2011, 09:12 AM   #9
Bonsai is offline Bonsai  Taiwan
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100uV/s is a good slew rate.

(sorry jcx ;-)
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Old 23rd October 2011, 09:25 AM   #10
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Ah well, it looked too good to be true anyway.

By using the 10%/90% method for determining slewrate I get a number of around 70V/uS.

That was obtained by measuring slewrate between +34V/-34V. Maximum voltage being +38V/-38V.

Which sound a lot more ressonable and plausible.
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