Photodetector Simulator Circuit

Discussion in 'The Projects Forum' started by moot, Sep 28, 2013.

  1. moot

    Thread Starter Member

    Sep 20, 2009
    46
    1
    I would like to make a small, stand-alone circuit that simulates a photodetector. This circuit would receive a voltage input, and convert it into a current in a manner that imitates a photodetector. This would allow me to test other circuits that receive signals directly from a photodetector from the comfort of my electronics bench.

    The photodetector I would like to simulate is a Thorlabs DET36A.

    We generally use a simple preamp (photodiode_preamp.pdf) to convert the photodiode current to a voltage. I did a quick test to see how much light is incident on the photodiode in a typical experiment, and what the resulting current from the detector and voltage from the preamp are. Results of that:

    • 800 μW of 780 nm light incident on photodiode
    • 100 kΩ potentiometer set to only 2.2 kΩ
    • current from DET36A is 0.235 mA
    • voltage after preamp is -0.5 V
    A pseudo-circuit idea of what I would like to do is shown in photodiode_simulator_outline.pdf. Such an arrangement would allow me to use a function generator with, say, a 1 V signal amplitude to analyze the AC response of my other circuits.

    According to the spec sheet for the DET36A, we can model it with a circuit similiar to photodiode_eq.pdf. Note that the DET36A has negligible series resistance, 1 GΩ shunt resistance, and 40 pF diode capacitance.

    My idea comes from a circuit in The Art of Electronics (Horowitz & Hill, 2nd Ed., pg 181). My adapted schematic is shown in photodiode_simulator_schematic.pdf. This is the circuit I'm thinking of building. A few notes:

    • the input would connect to the function generator
    • the output would connect to the preamp
    • the emitter-side resistor value of 56.2 kΩ was chosen to convert 1 V to about 0.25 mA
    Does this seem like the right idea?
     
    Last edited: Sep 28, 2013
  2. #12

    Expert

    Nov 30, 2010
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    The polarity of the amplifier in the photodiode_simulator_schematic.pdf is backwards and the LF356 will not work because it can not output higher than Vcc minus 2 or 3 volts, so it can not turn the transistor off.

    Try this:
     
  3. moot

    Thread Starter Member

    Sep 20, 2009
    46
    1
    Hmm, could you maybe elaborate? My very basic understanding of op amps tells me you're right, in that for an open-loop configuration, the op amp swings low whenever the (+) input is more negative than the (-) input. Here the (-) input is near 15V (depending on the load and whether the transistor is on), so any signal applied to (+) that is less than ~15V would result in an op amp output near the negative supply, -15V. Since this puts the pnp's base voltage always below the emitter voltage by a fair bit (supposing inputs voltages in the 0-1V range), the pnp will never turn off (as you say).

    But H&H has this circuit in their book! Maybe they mean for us to use an op-amp that can operate with its inputs near the positive supply voltage (and signals in that vicinity, too). They don't say anything like that. They do provide some fancier circuits, though...

    Isn't this just a basic noninverting amplifier (with a gain of 5)? I thought this would act as a voltage source (as well as a buffer, with perhaps the ability to source more current than the signal input to (+)).

    Since the photodetector acts as a current source, wouldn't I need a different op-amp design - one that provides constant current regardless of the load?

    (Thanks, by the way; I really appreciate your reply.)
     
  4. #12

    Expert

    Nov 30, 2010
    16,321
    6,818
    1) In the original circuit, even if you did use an op-amp that could output at the positive rail, the 1 volt input would never be higher than 15 volts so the op-amp would never lower its output voltage and turn on the transistor.

    2)The circuit you provided, if it did work, would try to provide 15 volts through a 56.2 k resistor. If the load resistor isn't very large, that would be about 1/4 milliamp. It isn't a current generator in the first place. It's a pseudo current generator. That is, a high voltage through a large resistor causes a pretty stable current as long as the load resistance isn't high compared to the source resistance. The circuit I drew provides 4.9 volts through a 19.6 K resistor which is exactly 1/4 milliamp. They both do the same thing, only mine works. If you want to change the gain to 12 and use a 48K resistor, you are welcome to do that, or you could change to a chip that can output 15 volts with a 15 volt supply and use a 60K resistor. It still won't be an ideal current generator.

    3) The third option is to design a real current generator, but I seem to be having a senior moment and I can't remember that one right now.

    Anybody else have a current generator? 0volts = 0 current, 1 volt = 250 uamps. Output goes positive into a grounded load.
     
    Last edited: Sep 28, 2013
  5. moot

    Thread Starter Member

    Sep 20, 2009
    46
    1
    Here's another. Just wired it up. Works great for positive inputs at frequencies of DC-10kHz, but then acquires a DC offset* for higher frequencies. Not sure why yet.

    (Note, also from H&H, pg 257, "circuit ideas", with note: "Howland-style current source for transconductance voltage-current control circuits (1μA to 1mA)".)

    I actually need a bandwidth of 1kHz-200kHz, so I might need to keep looking (if I don't figure out what's limiting this)...

    *EDIT: this DC offset can be attributed to bandwidth limitations of the V-to-I converter; the input test voltage is a 0-1V amplitude sine wave, i.e., a 0.5 V amplitude sine wave with a 0.5 V DC offset. filter out the AC and you get a DC offset only.

    *EDIT 2: Okay, did a little more testing. I tried different load circuits, and turned up the frequency until the -3dB point to find the approximate cutoff frequency. Results for different circuits:

    • plain old 10 kΩ resistor: 250 kHz
    • "photodiode_preamp.pdf" with R = 3 kΩ: 110 kHz
    • "photodiode_preamp.pdf" with R = 1 kΩ: 1.2 MHz
    So the load makes a huge difference with respect to the bandwidth of this circuit. I'm not really sure what to do yet about that... since I was hoping to use the "photodiode_preamp.pdf" I-to-V converter circuit with a gain resistor setting of about R~50kΩ (+/- a 20k). I just want to convert that 1/4 mA into a decent voltage to work with.... say 3-10 V...

    *EDIT 3: More testing. I made a point of adjusting the function gen's peak-to-peak voltage (into the V-to-I converter) until I measured a current of 0.20 mA into the I-to-V "preamp". This changed things. Apparently, I had been testing it in a range where the current was quite small (~0.03 mA).

    When the current was increased to 0.20 mA, with the feedback resistor in the "preamp" set to ~10k, I get a peak-to-peak output voltage of about 5.8 V, which is great. There is very slight attenuation around 80 kHz, and the 3 dB cutoff frequency is a few hundred kHz, which is fine.

    I'm satisfied with this for the time being.
     
    Last edited: Sep 29, 2013
  6. moot

    Thread Starter Member

    Sep 20, 2009
    46
    1
    My circuit appears to be working. I thought I would share some of my findings.

    The voltage-to-current converter from my last post works fine. I've added a parallel cap and resistor to the output to better mimic a typical photodetector (schematic attached). I typically tested it with a sine wave input of about 2 Vpp.

    I've also attached the front-end of my main circuit that connects to the photodetector. It includes the current-to-voltage converter, some filtering, and an 5x inverting amplifier.

    The bandwidth issues I had before can be traced back to the current-to-voltage converter. There is an excellent application note from Burr-Brown about photodiode monitoring with op amps found here: http://www.ti.com/lit/an/sboa035/sboa035.pdf.

    My application doesn't restrict me to one op amp, so I've used two to get around my bandwidth issues. Generally, increasing the feedback resistor on the I-to-V op amp results in a decrease in bandwidth. Therefore, I've set the feedback resistor to 1 kΩ, which keeps the cutoff frequency in the few 100 kHz or higher. (There is a noise trade-off, though, as discussed in the app note.)

    Although not necessary for low feedback resistor values (such as 1kΩ), I added a matching resistor between the noninverting input and ground, along with a parallel cap, to compensate for possible thermal DC voltage drift. More important was the feedback capacitor (100 pF). With no feedback cap, I saw +5 dB resonance near 1.2 MHz, and op amp instability (oscillations). I thought a 1 pF or 10 pF feedback cap would be enough to compensate for the capacitance of the photodetector circuit, but they weren't. Not until I up'd it to 100 pF did I get a nice roll-off in the MHz range and stable behavior everywhere.

    I've attached some hastily taken frequency sweeps looking at two points in the second circuit. -3 dB points are shown with vertical gray lines. "Point A" is immediately after the initial current-to-voltage converter, and before the LC filtering. The I-to-V converter appears to operate well for DC to about 2 MHz. "Point B" is the output, terminated at 50 Ω. Anything below about 20 kHz and above 1.2 MHz is filtered out, before the signal is amplified 5x.
     
  7. upopads

    Active Member

    Dec 18, 2007
    42
    0
    Interesting, I really like your photo detector circuit. Too bad DET36A is so expensive
     
  8. moot

    Thread Starter Member

    Sep 20, 2009
    46
    1
    You can get an equivalent diode from Thorlabs or Hamamatsu (or elsewhere) for under $20 and bias it yourself.

    The FDS100, for instance. The spec sheet has a recommended circuit you can copy (battery + low pass filter).
     
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