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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?

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  • \$\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
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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

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

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  • \$\begingroup\$ Isn't the junction capacitor in series with R1 here? \$\endgroup\$ – zud Feb 3 '16 at 17:05
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    \$\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

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