Transimpedance Amplifier Photodiode

I have a 660nm LED powered in a separate circuit and am trying to build a photo diode using a transimpedance amplifier but I am extremely confused. I read all the articles on this site about them.

The photodiode is most sensitive to 660nm as well.

The circuit below seems to be a common one for converting a current into a voltage. I read I can choose a large value such as 1M ohm for Rf but I am confused where the datasheet for my photodiode comes into play. Also why is there no voltage source for the op amp? Just the op amps power alone?

Light current means its most saturated current and dark is the current when there is no light correct?

Any help would be great! Photodiodes are much harder than I expected. • That opamp symbol implicitly assumes there are two other pins, one for Vcc, and the other for Vss (or GND). Always remember they are there, even when they are not included in the picture. Apr 28 '16 at 6:27
• Try leaving a link that works - don't you guys ever double check what you write? Apr 28 '16 at 7:36
• Sorry I forgot digikey links turn into search results. Updated it. Apr 28 '16 at 12:32
– 2i3r
Feb 16 '21 at 10:08

Bottom line: what are you trying to detect - that governs how much or how little of the data sheet you need to use: - • Short circuit current - what it says on the tin - how much current you could expect to see under certain lighting conditions.
• Dark current - how much current you might expect to see in dark conditions. This will roughly double (or more) every 10degC rise in temperature and will add an offset to the output of your TIA. For instance a dark current of 10 nA at ambient might be more like 40 nA at 45degC and this through a 1Mohm resistor produces an output error of 40 mV.
• Shunt resistance - I expect it's useful to know it is rather large in a lot of applications but some slow speed (and sensitive) applications might not be able to live with 500 Mohm.
• Junction capacitance - books can be written about this and the interaction with noise in TIAs in higher speed applications. Basically, if you need high speed and decent sensitivity junction capacitance is usually the limiting factor.
• Spectral range is fairly self explanatory
• Breakdown voltage - if using the device with reverse bias (to cut-down on junction capacitance at the expense of dark current) then you need to know the limits of the device.
• Noise equivalent power - it's basically telling you that the output noise produced when light is not hitting the device is equivalent to a certain power density.
• Response time - it's usually all down to the junction capacitance and the load resistor so 18 pF and a 1 kohm load give a CR time of 18 ns. That is a 63% change as per the exponential curve: - However, what the data says is that 80% is accomplished in 50 ns and you'd expect 80% (as per the curve above) to be done within 1.5 time constants (27 ns) but, clearly there are other factors at play to make it 50 ns. Anyway it's going to help you understand how fast you can expect the output to change and is important if using it for receiving data in that sort of application.

• But why is the response time specified in nanosiemens? (I know, you didn't write the datasheet) Apr 28 '16 at 16:28
• @ThePhoton LOL you tease! Apr 28 '16 at 16:40

The datasheet of the photodiode contains many useful bits of information such as:

1. Max thermal dissipation (limits the power incident on the photodiode)
2. Dark current (this is the signal that is present in the absence of light)
3. Junction capacitance variation with reverse voltage
4. Responsivity (A/W) at various wavelengths.
5. Breakdown voltage

The datasheet of the op-amp also contains useful info such as:

1. Gain-Bandwidth product
2. Minimum stable gain (in voltage to voltage mode)
3. Bias current (some high-GBWP op-amps draw significant current at their inputs)

A good way to model photodiodes is as a current source (whose current depends on the incident light x responsivity) in parallel with a resistance (which represents the dark current) in parallel with a capacitance (the junction capacitance). The junction capacitance decreases with increasing reverse bias.

A lot of these things can be learned by looking at Application Notes (even if you have little background in electrical engineering, although you should know how op-amps work). Here's a list of my favorite resources:

If all you need is low noise, things are easy. Reduce the bandwidth by hanging a huge cap either across the feedback or at the output, and the noise will be averaged out. Also remember that if you are detecting a laser beam, there will also be shot noise, so you may want to detect as much power as you can and use a large photodiode to dissipate all that heat.

If you need a lot of bandwidth, things are a bit more tricky. The bandwidth depends on the junction capacitance, on the resistance and the GBWP of the op-amp. The junction capacitance depends on the type of photodiode (Silicon, InGaAs, GaP...), the size of the chip (more area means more capacitance) and the reverse voltage. So you would want the smallest photodiode you can deal with without damaging it due to the poor heat dissipation. Or maybe you can think of putting a Peltier cooler or something.

The National Semiconductor presentation above discusses various ways of trying to keep the bandwidth high. There's also some info on Thorlabs' website.

• Thanks alot! Very useful. I am having trouble calculating a capacitor in parallel with my feedback resistor though. The voltage output is noisy and I believe a capacitor will fix this. I am using a 1M ohm resistor. Apr 29 '16 at 6:56
• The capacitor in parallel with the feedback resistance governs the bandwidth. Figure out what bandwidth you desire. Then, you could use the formulas in one of the Application Notes I linked (I think the one from OSI). If you want the highest possible bandwidth while ensuring the stability of the amplifier, the way I have compensated it is to keep adding capacitance until it gets stabilized. 1206 SMD capacitors can be stacked up when you prototype. If you're prototyping on a bredboard, remember that the bredboard introduces lots of parasitic capacitance, as much as a few pF. May 20 '16 at 0:35

I read I can choose a large value such as 1M ohm for Rf

Sounds like you haven't understood why/how to choose $R_F$, so here's a brief explanation. If you do the DC analysis of this circuit, you can find that:

$$R_T = \frac{V_o}{I_p} = R_f \frac{A}{A+1}$$

where A is the gain of your amplifier (it could be an opamp, but could also be a transistor amplifier), and $R_T$ is the transimpedance gain (in ohms, because it's converting current to voltage). In fact, with an opamp, $A$ is very big (adding +1 doesn't change it much), so the equation is approximately

$$R_T \approx R_f$$

This means that you know how much the current is amplified as a voltage. If you have\ $1\mu A$ of current input and $R_T = 10000$, you will have 10mV on the output.

So to explain the resistor value, bigger resistor = more gain, so bigger output voltage for the same input current. If you choose 1M, then the output voltage will be one million times the input current number. The drawback to a bigger resistor is that you will also have more inherent noise due to the resistor.

I am ignoring capacitances here: if you're working at high speeds, you'll need to think about a 1st- or 2nd-order model. A big resistor will mean that your bandwidth is more limited (so the TIA is slower).

If you want to get into engineering-oriented analysis, I suggest Razavi, Design of Integrated Circuits for Optical Communications, section 4.3. (Yep, we use the same idea in high-speed fibre optic communications, except we use fast, small transistors rather than slow op-amps.)

but I am confused where the datasheet for my photodiode comes into play.

We saw above that to know the output of the circuit, you need to know the input current. What's providing the input current? The photodiode, and light hitting the photodiode.

So you need to figure out how much optical power is hitting the photodiode. The datasheet should have a responsivity $\mathfrak{R}$ (in A/W), which tells you how well the photodiode turns optical power into a current ($I = \mathfrak{R}\times P_{optical}$). I can't point anything out because your link doesn't work.

For fibre optics, if you know your coupling loss into the photodiode you can figure this stuff out. If this is just a photodiode in the LED-like packages, and you're shooting a laser or an IR LED at it without precise alignments, you probably want to experimentally characterise what the light-->current-->voltage gain is and adjust your resistor $R_F$ accordingly.

Side-note: You may want to reverse bias the photodiode (at least with the ones I've worked with, but those are fibre optics ones). Right now the circuit has the non-inverting opamp input connected to ground, so the inverting input will also be at 0V (when zero input current), and the diode has zero volts of bias. It might be good to put a positive voltage on the non-inverting input to give it a reverse bias—datasheet may specify recommendations or the voltages it was tested at.

Also why is there no voltage source for the op amp? Just the op amps power alone?

An opamp necessarily needs voltage rails. However, when discussing amplifier circuits, we omit them for clarity but they are implicitly there.

Light current means its most saturated current and dark is the current when there is no light correct?

Yes. It's important to remember that a DC dark current exists and will cause a DC offset on the output.

EDIT

Now that the original question has a corrected datasheet link: This figure, from the datasheet, shows the responsivity as a function of wavelength. If the diode receives 0.01W of light at 850nm wavelength, it will pass a current of 0.005A = 5mA. 0.01W of light at 450nm will generate around 1mA of current.

You may care about speed, in which case the junction capacitance will matter to you (remember, R-C networks form filters---in this case, the junction capacitance + $R_F$ ends up forming a low-pass pole—you can approximate it by applying Miller's theorem to $R_F$). At this point, the gain-bandwidth of the opamp will also matter.
• @Andyaka That was a typo—for a bigger input current. Fixed and clarified $R_T$, thanks for pointing it out. Apr 28 '16 at 14:41