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I have OPA355 configured as inverting unity gain. My input signal source is a 10MHz sine wave, with vpeak as 1 V. This opamp is a CMOS based and has very low Ibias current (~pA). With a Gain of 1, the 3dB bandwidth is 450MHz. I expect this to work for a signal of 10MHz. Here is the schematic Schematic

Apparently the output is getting clipped (expect the output to swing between 3.5 and 1.5 Volts). Here is the output and input waveform Waveforms

Could anyone suggest, what's the issue here and how I can fix this?

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  • \$\begingroup\$ Widen your voltage rails. The op amp can't output a voltage that isn't between its V+ and V-. \$\endgroup\$
    – Hearth
    Commented May 14, 2017 at 23:58
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    \$\begingroup\$ Did you allow enough time for C1 & C2 to charge up? \$\endgroup\$ Commented May 15, 2017 at 0:10
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    \$\begingroup\$ What happens if you reduce R1, R2, R3 to 3K? \$\endgroup\$ Commented May 15, 2017 at 0:34
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    \$\begingroup\$ I think you will need to reduce the PMT load resistance and add amplification (with adequate bandwidth). Even 1pF of stray capacitance will reduce your BW to 0.5MHz with 300k source resistance. There are ways to reduce the effective capacitance to less than 1pF but they are troublesome in most cases. \$\endgroup\$ Commented May 15, 2017 at 11:18
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    \$\begingroup\$ Sure. The voltage will be proportionally less. Note that you will need a much better amplifier or several stages of amplification- the OPA355 has a GBW of only 200MHz. 300K/50 ohms is 6000:1. \$\endgroup\$ Commented May 16, 2017 at 3:14

2 Answers 2

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Your 300kOhm feedback resistor, combined with the input capacitance of __ pF mentioned in the datasheet (page 3) creates a pole at __ Hz...

I'm not copypasting the values, that's your job, but I'll give you a hint: there is no way this is going to work at 10MHz (unless the opamp model does not include input capacitance, of course).

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  • \$\begingroup\$ Hi Peufeu, Thanks. Having 300K has introduced a pole at 3.5 MHz. I just verified using AC response. \$\endgroup\$
    – Ash
    Commented May 15, 2017 at 0:57
  • \$\begingroup\$ Doesn't mean you can't make it work add a compensation capacitor across R1 of the same magnitude \$\endgroup\$
    – sstobbe
    Commented May 15, 2017 at 1:07
  • \$\begingroup\$ Compensation will reduce the BW substantially, I need to make it work for 10MHz signals. \$\endgroup\$
    – Ash
    Commented May 15, 2017 at 1:14
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    \$\begingroup\$ Yeah you'll need a lower resistor value, make sure you don't go too low (check the available output current and recommended resistor values in datasheet) \$\endgroup\$
    – bobflux
    Commented May 15, 2017 at 10:34
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This is often a slew rate issue. While the bandwidth may be fine, and small signals will be hunky dory, big signals get messed up.

In this case, this is not the issue, as your peak slope for your signal is \$12.6 \frac{V}{\mu s}\$, and the max slew rate for your op amp is \$300 \frac{V}{\mu s}\$, but it's still worth pointing out.

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  • \$\begingroup\$ lots of slew rate and GBW avail. It's a transient input bias issue \$\endgroup\$ Commented May 15, 2017 at 0:12

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