I have some problems on understanding this equivalent circuit for a photodiode (picture taken from this article):

enter image description here

The author states that:

The equivalent circuit for a photodiode is shown in Fig. 1, where \$I_P\$ is light generated photocurrent, \$I_N\$ is noise current, \$R_{SH}\$ is the shunt resistance, \$R_S\$ is the series resistance, CJ is the junction capacitance, and \$R_L\$ is an external load resistance connected to the photodiode.

I have some questions about it:

  1. I do not see any biasing network. I think it has been omitted for simplicity. How would it have been connected to the photodiode terminals?

  2. I do not understand why the ideal diode has been inserted in this circuit. Which is the ideal behaviour of a photodiode? I'd say that it is simply a current source Ip, without diodes in parallel.

  3. Which is the cause of Rsh? Is it quite high or low?

  1. This circuit is just the photodiode model and a load resistance. No bias network.

  2. The ideal diode is there because this is a photodiode. Which means that if you forward bias it, a lot of compoments gets bypassed, but when you reverse bias it those components come into play.

  3. The shunt resistance is really high. It's basically the resistance across the PN junction when the PN junction doesn't want to conduct, I believe. It's the resistance that everything has, even when it really doesn't want to conduct anything.

  1. For DC analysis, start with the model of your reference:


simulate this circuit – Schematic created using CircuitLab

R_L is omitted, because it is an external component, load

To reverse bias a photodiode, you use a network like this (transimpedance amplifier):


simulate this circuit

Inverting a V_bias polarity, you can also try and use this opamp configuration to forward bias a photodiode, but notice that for V_bias close to V_oc (V_oc is a photovoltaic voltage developed in an open circuit configuration), current grows exponentially. For this reason, the forward biased photodiode (operating in a photovoltaic mode) is best examined when simply loaded by a resistor R_L.

In this passively loaded configuration, without external voltage applied, the photodiode, when lighted, is forward biased by its own photocurrent; a "self inflicted" biasing voltage depends on R_L and varies from V_oc (for R_L = ∞) to 0 V (for R_L = 0)

  1. You're right, an ideal diode is just a current source. In your model, you can safely substitute for an ideal diode a voltage controlled current source $$ I_D = I_{SAT}*(exp(qV_{bias}/(k_BT)) - 1) $$ (Shockley equation)
  2. Shunt resistance is determined by diffusion current. Diffusion current dominates in a photovoltaic mode. Shunt resistances vary from KiloOhms (Ge photodiodes) to GigaOhms (Si photodiodes). Shunt resistance produces a Johnson noise. The noise level thus produced is inversely proportional to resistance.

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