# Transimpedance Amp - first pole location

Discussion in 'General Electronics Chat' started by jbriaris, Apr 26, 2013.

1. ### jbriaris Thread Starter New Member

Apr 11, 2013
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Hi All

I'm modeling an example photodiode (transimpedance) amplifier in LTspice using the ADA4505 from Analog Devices. The theory suggests that the location of the first pole (actually a double pole) will be located at frequency $f_{\rm fp}$ where

$f_{\rm fp} = \sqrt{\frac{f_{\rm c}}{2\pi R_{\rm f} C_{\rm i}}}$

where $f_{\rm c}$ is the crossover frequency of the opamp, $R_{\rm f}$ is the feedback resistance of the transimpedance amplifier, and

$C_{\rm i} = C_{\rm j} + C_{\rm id} + C_{\rm icm}$

where $C_{\rm j}$ is the photodiode's junction capacitance, $C_{\rm id}$ is the differential input capacitance of the opamp, and $C_{\rm icm}$ is the common-mode input capacitnce of the opamp.

• Is the crossover frequency $f_{\rm c}$ of an opamp the same as its gain bandwidth product (GBP)?

From the ADA4505 datasheet; GBP = 50kHz, $C_{\rm id}$ = 2.5pF, $C_{\rm icm}$ = 4.7pF. From the photodiode datasheet $C_{\rm j}$ = 5pF.

Using a feedback resistance $R_{\rm f}$ = 100M, and substituting these values into the above equation yields a first pole frequency $f_{\rm fp}$ at 2,554Hz. However, the equivalent simualtion in LTspice appears to show a first pole frequency (peak) at 320Hz (see attached) - an order of magnitude less.

• Any ideas what I'm missing or doing wrong?
Thanks!

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2. ### Ron H AAC Fanatic!

Apr 14, 2005
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Try setting the DC on your current source to 25nA. That will set the output at 2.5V. When I did this, The peak moved to about 950 Hz. Still not as calculated, but closer.

I ran the sim with a single pole, 50kHz GBW op amp, and the peak was at 2.55kHz, as calculated.

The difference is apparently partially the result of the second pole inside the op amp.

3. ### crutschow Expert

Mar 14, 2008
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Not sure if that equation is correct. And the large peak indicates the circuit is unstable. A small feedback capacitor in parallel with the feedback resistor will eliminate that. Here is an article on calculating the proper value.

4. ### Ron H AAC Fanatic!

Apr 14, 2005
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As I said, a simulation of a single pole op amp gave the same results as the equation. AFAIK, there are no single pole op amps available in the real world.

5. ### jbriaris Thread Starter New Member

Apr 11, 2013
19
0
Great, thanks.

I've been following the theory in Jerald Graeme's book Photodiode Amplifiers, and perhaps, aspointed out, that equation is only valid for an ideal single-pole opamp - the text doesn't seem to make this clear.

Thanks for the link crutschow. Yes, it is unstable, picking a phase compensation capacitor was my next job, but doing some investigating first.

The link provides a formula for the phase compensation capacitor which differs slightly to the one offered by Graeme, i.e., $2\sqrt{2}$ cf 2 in the denominator - perhaps a slightly different phase margin in the compensation.

Since the premise of the transimpedance amplifier is to create a virtual ground at the inverting input, would you recommend using a split-supply opamp?

Also, how did you model a single-pole opamp in spice? Do you use something like this - http://www.ecircuitcenter.com/Circuits/opmodel1/opmodel1.htm

Thanks again.

6. ### jbriaris Thread Starter New Member

Apr 11, 2013
19
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Agreed. Using LTspice's default opamp with GBW = 50K.

7. ### crutschow Expert

Mar 14, 2008
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Obviously you can't go below ground at either input or output with a single-supply op amp. But a photo-diode only outputs signal in one-direction, thus if connect the diode so it puts a negative current into the (-) summing-junction, giving a positive output, then you should be OK with a single supply.

8. ### Ron H AAC Fanatic!

Apr 14, 2005
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657
Yep. You got it.