Peculiar would me the name that comes to mind - 3 pieces
1. The inverting DC signal gain is very high, -196k/499 =-392V/V. The 11MHz GBP then gives a 132kHz signal bandwidth while the feedback C also puts in a 17kHz pole that will dominate
2. The didoes and JFET are overdrive shunt. As the amplifier overdrives, the input V- voltage increases from ground until it turns on that path. The diode polarity tell me they only expect positive signals.
3. The output is big DC attenuator, 2.2k/(102k) = .01256.so the net DC gain is .01256*392 = 8.45V/V. The shunt 1nf starts to take the output attenuator towards zero starting at 1.59kHz.
I am sure the response shape looks like a bandpass of some sort. If you want a really sharp BandPass should look at the DAMP BP discussed in these e2e occasionally.
Michael Steffes said:Peculiar would me the name that comes to mind - 3 pieces
1. The inverting DC signal gain is very high, -196k/499 =-392V/V. The 11MHz GBP then gives a 132kHz signal bandwidth while the feedback C also puts in a 17kHz pole that will dominate
2. The didoes and JFET are overdrive shunt. As the amplifier overdrives, the input V- voltage increases from ground until it turns on that path. The diode polarity tell me they only expect positive signals.
3. The output is big DC attenuator, 2.2k/(102k) = .01256.so the net DC gain is .01256*392 = 8.45V/V. The shunt 1nf starts to take the output attenuator towards zero starting at 1.59kHz.
I am sure the response shape looks like a bandpass of some sort. If you want a really sharp BandPass should look at the DAMP BP discussed in these e2e occasionally.
Dear Michael Steffes,
Your answer is very detailed.
But I still don't understand item number 2.
How can such polarized diodes suppress the negative signal?
Thanks,
Hai Lai
