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I'm trying to measure the bandwidth of a TIA with a gain of 1e9 on the first stage and 10x on the second stage. The first stage OpAmp is the LTC6268 with a 1G Ohm feedback resistor and the second is a LTC2050 with a 10k/100k feedback. I'm getting correct values for measured gain and my noise floor is fairly good, but the next step in testing the performance of this TIA is its bandwidth.

I know that typically, you would hook up a sinewave and sweep it to find the -3dB cutoff, but I've been told this is not entirely correct, since it doesn't compensate for the capacitance.

The set up is a function generator with a 100mVrms sine wave, sweeping from 0Hz to 100kHz, feeding into a 1G Ohm resistor to give a 100pArms current. This is connected directly to the input of the TIA.

I don't understand enough of the physics and elemental electronics to know why this setup is incomplete. Any pointers and literature would be most welcome!

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Because of the limitations Kevin mentions, I would suggest a multi-stage attenuator with lower value components : say 100k:10 ohms stepping 100mVrms down to 10uVrms, then 100K to your input.

These low values should dwarf stray capacitances at your frequencies of interest.

Bandwidth is higher because the first stage has a source impedance of 10 ohms, and the whole network's source Z is 100,010 ohms, into your input capacitance, instead of 1G into the same capacitance. (If that's still not good enough, try a 3-stage network along the same lines).

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    \$\begingroup\$ Good idea. Be careful about the effect of DC offsets. \$\endgroup\$ – Kevin White Sep 8 '15 at 13:57
  • \$\begingroup\$ Right, this is another method suggested, I get heaps more noise than a single 1G, but would this give me a good idea of the -3dB point? I'm not entirely sure why this works and the single 1G doesn't. \$\endgroup\$ – talsit Sep 8 '15 at 23:35
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    \$\begingroup\$ Simulate both approaches with an appropriate input capacitance (1pf if you don't have a better estimate). I believe you'll find the -3dB point of this input network is much higher than your amplifier while the 1G resistor may well be lower. If this network is "noisier" that may just indicate more bandwidth. \$\endgroup\$ – Brian Drummond Sep 9 '15 at 9:44
  • \$\begingroup\$ So I did what was recommended and got this: TIA Bandwidth I believe this is correct, so... THANKS GUYS! \$\endgroup\$ – talsit Sep 10 '15 at 6:00
  • \$\begingroup\$ I think I have some of the calculations wrong ... checking them again... \$\endgroup\$ – talsit Sep 10 '15 at 6:15
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That method looks appropriate although I can think of a couple things to remember to ensure the result is accurate:

1) Make sure the normal input to the TIA (photodiode or whatever) remains connected. This is to ensure that the capacitance at the input node is the same as when in normal use. The capacitance at this point will affect the bandwidth.

2) Be careful of the parallel capacitance of the signal injection resistor. A surface mount or through hole resistor will probably have 0.1pF to 0.2pF of parallel capacitance. That will mean more current is being injected than you are expecting and affect the results.

3) You may want use a lock-in amplifier to measure the output voltage to make sure you are not measuring the signal plus any noise - you can use the averaging function on a scope as a poor-man's LIA to get more accurate results.

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  • \$\begingroup\$ Hey, thanks for the comments. I can't connect the normal input, just yet - this is a biotech application, and the wetware part hasn't yet been ironed out, though we have some initial estimates of the values we would be dealing with. My part is mainly iterating amplifier designs and trying out different things and try to compare iterations. \$\endgroup\$ – talsit Sep 8 '15 at 23:24
  • \$\begingroup\$ Also, regarding the signal injection resistor, it's quite hard to measure its capacitance, given its value (1G Ohm), but I think this is what our "greybeards" are talking about, that since it has capacitance, I will be measuring its capacitance as well as the system. Is there something that can be done to get around that? \$\endgroup\$ – talsit Sep 8 '15 at 23:26
  • \$\begingroup\$ Brian Drummond's approach should reduce the effect of the capacitance across the injection resistor. You may want to experiment with the actual divider ratio to get a good compromise for noise. It will exaggerate any DC offset you have at the input so be careful about that to make sure you don't saturate the output stage. Are you DC or AC coupled to the second stage? \$\endgroup\$ – Kevin White Sep 8 '15 at 23:32

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