7
\$\begingroup\$

In an article in LT Journal of Analog Innovation, Linear Technology brags about their high-voltage operational amplifier, the LTC6090, which can operate on ±70 volts. They present an example circuit, a photodiode amplifier:

enter image description here

This is perhaps impressive; I don't know much about photodiode amplifiers.

What strikes me as odd is that, while the feedback loop operates on voltages up to 120 volts, taking advantage of the increased voltage range, they then place a 1:10 voltage divider on the output, seemingly negating the benefits.

Why not replace the 10 MΩ resistor with 1 MΩ and get rid of the output divider? Then you can use a completely average operational amplifier because the output only needs to swing between 0 and 12 volts.

In my simulations, lowering the feedback voltage gave me a number of benefits:

  • Higher bandwidth (using the same LTC6090)
  • 0.3 pF capacitor can be increased, loosening the layout requirements
  • Lower output impedance
  • A larger selection of operational amplifiers are now available
  • ...and so on

I have searched in the article and the datasheet for reasons behind this, but I could not find any. Perhaps it is obvious to people experienced with photodiode amplifiers, but since the "feedback action" makes sure that the photodiode sees the same potential difference, it should not affect its behaviour - there will be 3 volts across it, and the same current will flow.

Why is this circuit improved by a high feedback voltage?

\$\endgroup\$
1
  • 3
    \$\begingroup\$ I think, from seeing so many TIA questions for photodiodes around here, that part of the point of the circuit is to have a high impedance around the diode. 10M and fractional pF, coupled with the very low bias current of the LTC6090 seem to meet that goal. The SFH213 looks fairly typical, but I don't have the knowledge to understand much about its specifications in this application. \$\endgroup\$
    – user2943160
    Sep 9, 2016 at 3:01

1 Answer 1

Reset to default
6
\$\begingroup\$

Why not replace the 10 MΩ resistor with 1 MΩ and get rid of the output divider?

It's all about the thermal noise created by the 10 Mohm resistor. A 10 Mohm resistor at 20 degC will produce a noise voltage of 90 uV RMS across a 50 kHz bandwidth.

Compare this with a 1 Mohm resistor - it produces a noise of 28 uV across the same bandwidth at the same temperature. That isn't a 10:1 reduction in noise voltage and that's the point here.

So, if you use the 10 Mohm resistor (to give ten times the signal gain) then 10:1 attenuate afterwards using "low" value resistors, you will achieve a significant net improvement in signal to noise ratio (\$\sqrt{10}\$). There are other factors that make this not as good as it sounds (such as the noise gain of the circuit due to the photodiode's self capacitance) but, generally speaking, you will get between 5 dB and 10 dB improvment in SNR.

Here's a bit of theory about noise power produced by resistors across a given bandwidth. If you need to see real numbers use this website as a calculator: -

enter image description here

Noise power increases with resistance but, noise voltage only increases with the sqare root of resistance. Signal voltage increases with resistance.

Why is this circuit improved by a high feedback voltage?

If the target signal output level is (say) 10 Vp-p and you adopt the noise reduction approach shown then you have to produce a signal that is 100 Vp-p in order to be able to attenuate to 10 Vp-p.

\$\endgroup\$
3
  • \$\begingroup\$ +1 although I'm not sure if OP realistically doubts the (quite well known) theory--the question, it seems to me, arises from the fact that feedback resistor thermal noise isn't mentioned as a consideration, let alone the dominant one, in deriving this design. There are several other reasonable considerations apart from noise, which, as OP points out, will give different optimum designs. \$\endgroup\$
    – Oleksandr R.
    Sep 9, 2016 at 11:12
  • \$\begingroup\$ Yes, I sahll change my words given that @pipe probably does understand this. \$\endgroup\$
    – Andy aka
    Sep 9, 2016 at 11:19
  • 2
    \$\begingroup\$ I was indeed aware of resistor noise, but not this trick! Naturally I assumed that a higher resistance in the loop lead to more noise, but it did not cross my mind that you could balance the ratios and come out ahead by adding even more resistance to the equation. \$\endgroup\$
    – pipe
    Sep 9, 2016 at 12:10

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service and acknowledge that you have read and understand our privacy policy and code of conduct.

Not the answer you're looking for? Browse other questions tagged or ask your own question.