Why would a photodiode with an op-amp be very linear?

Question

I read here mentioned

The configuration in Figure 1a creates a very linear response in terms of the irradiance of the light source.

Would this configuration of photodiode not be linear? Why?

The model of photodiode model should be following picture:

Voltage across photodiode should be \$V = I_{PD} \cdot R_{Load}\$ (ignore dark current)

V of above formula seems to very linear to Id and Rload

Why does a photodiode need a transimpedance amplifier for linearity?

The opamp is holding the bias voltage constant, which is more linear than the resistor where the bia voltage depends on load. — user1850479, Feb 19 '23 at 14:04
@user1850479 The load would usually be something high-impedance anyway, like an ADC/comparator input. The real disadvantage of the simple divider is that the bias voltage depends on the current through the photo-diode itself. — Dmitry Grigoryev, Feb 20 '23 at 14:55
Sorry yes, I mean depends on the loading from the photodiode. — user1850479, Feb 20 '23 at 14:59

score 11 · Accepted Answer · answered Feb 19 '23 at 11:27

11

Why does a photodiode need a transimpedance amplifier for linearity?

It's a simple case of if the bias voltage across the photodiode changes, then the photo-current for a given illumination will also change: -

Original image from here.

So, if you have a photodiode in series with a resistor and, the photo-current increased, then the voltage across the resistor must also increase and, inevitably the voltage bias across the photodiode must decrease. This creates a more non-linear relationship.

An op-amp trans-impedance-amplifier (TIA) holds the voltage across the photodiode constant.

There are also other benefits of a TIA.

answered Feb 19 '23 at 11:27

Andy aka

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That's *energy conservation*. I'm not sure if I understand correctly: When the photodiode absorb the energy, Vout rise of course. Since my circuit contain 2 only component(photodiode and resistor) and *energy conservation* -> current of photodiode and resistor must be the same -> Voltage of photodiode decline. – curlywei Feb 19 '23 at 14:10
In the op-amp(TIA) case, photodiode connect between V- of op-amp and GND(or reverse bias), since virtual ground, voltage across photodiode is V- - GND(or reverse bias), it's constant(linear). – curlywei Feb 19 '23 at 14:18
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@curlywei correct – Andy aka Feb 19 '23 at 18:02

DKNguyen · Answer 2 · 2023-02-19T10:22:53.773

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Voltage across photodiode should be $$V = I_D \cdot R_Load$$

This is wrong. The entirety of \$I_d\$ does not flows straight to \$R_load\$. It gets split up to also flow through the junction capacitance and shunt resistance.

\$I_d\$ itself is also of limited power in that it has limits as a current source and can't produce any arbitrary voltage to maintain its output current. It gets loaded down.

edited Feb 19 '23 at 10:22

answered Feb 19 '23 at 10:17

DKNguyen

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score 2 · Answer 3 · answered Feb 19 '23 at 20:38

Your circuit should be close to linear if the voltage across the load resistor is moderate. The voltage across the diode changes, but that doesn't have a strong effect on the current unless it gets so small (< 1V) that significant forward current flows. However, the depletion depth of the diode varies with bias: photocharge generated outside the depletion region is poorly collected, so that influences the current. This is most pronounced for infrared, where the radiation may penetrate the entire volume of the diode.

So, for fussy photometry use a transimpedance amp. For less fussy applications, a load resistor may be OK.

score 1 · Answer 4 · answered Feb 19 '23 at 10:35

For DC, your formula works.

But bear in mind that in your voltage output setting, the output voltage varies based on photocurrent, so based on photocurrent, there will be different DC bias over the photodiode, and the generated photocurrent just might depend on the voltage bias over it.

And, like mentioned, the load resistance and thus output voltage varies, and so does the internal capacitances need current to get charged to new voltage values.

With the photocurrent mode, the op-amp keeps a fixed bias voltage over the photodiode. Thus the DC bias does not vary based on photocurrent. Also the internal capacitances are kept relatively constant voltage as the photocurrent can't change the voltage of capacitances that much, as there is only the internal series resistance.

The photocurrent mode is just faster an better, at least for high speed data communications.

Why would a photodiode with an op-amp be very linear?

4 Answers4