Transimpedance amplifier

Jan 11, 2009 4 Replies

Hi, i've simulated a transimpedance amplifier with spice. My circuit is composed of an opamp and a feedback resistor; in parallel to current source a capacitor is present (i want to use a photodiode's equivalent model). I have a question: is it possible that (with an ac sweep), V/I transfer function is greater (at lower frequencies) than opamp open loop gain? Why? Another question: i wanto to use it as input of a sar adc. I've read this interesting article

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and i want to try it. Is it more correct to use another opamp as buffer and rc circuit as its load (as describend in the link) or use rc load directly to transimpedance amplifier? Thanks


The open loop gain spec you are looking at may be a minimum and the model may use the typical.

You didn't say there was a peak but if you see a peak in the curve just before it falls off the answer may be the phase margin of the system.

Beware of EDN articles they are too often bogus. Does your ADC have sample and hold in it? What sort of load does it present to its input?

You may or may not need another op-amp in the system at all.

How can i verify it?No opamp's macromodel is present, because my opamp is realized with transistor...

Yes, a peak is present....i've read about on graeme's book.I've used a feedback capacitor to remove it

It has a sample and hold with r=3Don switch resistor=3D100ohm and c=3Dsampling capacitor=3D5pF thanks

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Making the peak go partly away but not completely can lead to a better system. A slight bump before the gain starts to fall can move the 3db point up.

The output impedance of the driving circuit needs to be far less than the 100 Ohms. The output will actually look inductive. Since you are modeling the system, you can model the sample and hold action too.

Well the "open loop gain" is measured in Volts out / Volts in and the V/I transfer function is measured in Volt out / Amps in..... so you have an apples and oranges problem I would think.

Mark

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