I have a reverse biased photodiode as shown below.

enter image description here

where R1 = 100kΩ. The output is quite low around 50mV. I would like to replace R1 with 1MΩ to get 10x the output. My concern is, if it will have any disadvantage? Does it have any effect on the rise time?

  • \$\begingroup\$ It'll affect the fall time. \$\endgroup\$ – Brian Drummond Feb 3 '16 at 16:51
  • \$\begingroup\$ Could you explain a bit? \$\endgroup\$ – zud Feb 3 '16 at 17:05
  • \$\begingroup\$ Could you add an opamp to the circuit? \$\endgroup\$ – Voltage Spike Feb 3 '16 at 17:41

Yes, it will slow down the response (assuming the load presented by your sense circuit is still "high" compared to 1 MΩ). As far as AC circuits go, the junction capacitance of the photodiode is in parallel with the resistor, forming a low pass filter.

Increasing the resistance may also increase the noise, as the Johnson noise of the resistor goes as

$$\bar{v_n^2} = 4 k_B T R$$

|improve this answer|||||
  • \$\begingroup\$ Isn't the junction capacitor in series with R1 here? \$\endgroup\$ – zud Feb 3 '16 at 17:05
  • 1
    \$\begingroup\$ Not in terms of what signal gets to whatever's connected to the output. Remember that a power supply is an ac ground. \$\endgroup\$ – The Photon Feb 3 '16 at 17:07

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

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