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Photodiode Amplification

  1. Oct 29, 2014 #1
    I am using (attempting to at least) a photodiode to count fringes from a Michelson interferometer. Here is the amplification circuit I am using:


    Coming out of the op-amp into the comparator I get about 4.2V in a well lit room, 2.17V in total darkness, and 2.33V when the brightest part of the interferometer fringes are hitting it. That range 2.33-2.17 is my problem. I am having a hard time getting the reference voltage of the comparator at a spot where I get a count on every fringe, but don't get extra counts. Is there something I can do in the circuit to make the voltage range between bright fringes and dark fringes bigger? I can't get any more intensity out of the laser, and I have the spot the smallest I can trust it.

    EDIT: I missed something. I ended up using the comparator to make sure my voltage into the debouncer and pulse counter went high enough and low enough to make counts. Notice that the lowest voltage I got coming out of the op-amp was around 2V. I was under the impression that for this configuration, when there is total darkness (or I pull that photodiode out of the circuit), that the output of the op-amp should go to 0, since the non-inverting input is sent to ground. What's the deal with that?
    Last edited: Oct 29, 2014
  2. jcsd
  3. Oct 29, 2014 #2


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  4. Oct 29, 2014 #3


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    but the + input isn't going to ground, is it

    you have it lifted above gnd potential by the 4M7 resistor
    ( you probably realise the idea of that resistor is to balance out the input bias current.
    It is normal, as in this circuit, to make it the same value as the feedback resistor)

    I also seem to recall from other discussions that if you use a 741 op-amp on a single rail supply,
    you wont get the output below 2V. You would need a + - rail supply to get to 0V on the output

    EDIT: further thinking .... you could try using the offset null capabilities of the 741 to zero the output when there is no input voltage

    Last edited: Oct 29, 2014
  5. Nov 2, 2014 #4
    First up, the 741 needs a negative supply voltage as well as a positive. First, I'd add a -5V supply for the op amp.
    Next, you may consider swapping the amp for a more modern device, I'd recommend the OP07C as it is similar with the exception of having a smaller offset and bias voltage. If you go with the OP07C, I'd do away with the resistor between the +input to ground as it is a detriment when using these bias-compensated op-amps.

    Next, you've probably noticed some overshoot and even ringing on the output of the op amp. This is caused by the combination of large feedback resistor and too much input capacitance on the -input.
    Start by reducing the gain of the trans-impedance amp. Don't worry as you can easily recovery the gain with a second op-amp stage. How much gain? I'd aim for about 220 k-Ohm. By following this with a second op-amp set for a gain of about 22, you'll have roughly the gain of the original circuit, but a much easier design on the input. When designing the second stage, it's best to keep the feedback resistor between 5k-20k to ensure that stray capacitance on the -input doesn't cause issues.

    Finally, you have to "tweek" the input amplifier to compensate for the capacitance of the diode, any cabling, and the input of the transimpedance op amp. If you don't need the circuit to settle quickly (i.e within 50us), the easiest answer is to place a 47pf capacitor in parallel with the 220k feedback resistor. If you're not using a large area photo-diode and significant cabling, this will ensure you have a stable input.

    If you want to adjust the transiimpedance amp for fastest response, blink an LED in front of the photo diode at about a 10 kHz rate, and monitor the output of the op amp with a scope. Then vary the value of the feedback capacitor until you get you fastest response without overshoot.

    Now, you should have a nice representation of the light coming out of the second op amp with only a few mv of offset when dark.

    Some additional cautions are in order:
    - Tie the photo diode to the ground rather than the input of an op amp with a resistor to ground. Why? Any stray signals the photo diode become coupled to should be clamped to ground - not confusing the amplifier.
    - If you are building this on a solder-less breadboard, there is more capacitance hanging about than might seem and some of that is to whatever the board is setting on. That's why I use metal-backed boards and always ground the metal plate. It's also why I sometimes bend the -input lead up into the air and solder the components to it. This helps reduce capacitance loading and leakage currents, especially on transimpedance amplifiers.

    There's more to this circuit as we deal with the comparitor, but I'll leave you with this for now as It's my lunch time :-)

    Best Regards,

  6. Nov 2, 2014 #5
    Back from lunch (and dinner).

    The 311 comparitor is a good choice. However, since the dark signal level is around zero volts and the op-amp may transition to negative voltages during power up / down, it's probably best to power the 311 from both +5V and -5V, so that it operates consistent with the output of the second op-amp.

    Also, the comparitor is an open-collector type, which means it needs an external source of current on the output. I'd suggest a 4.7k-Ohm resistor to +5 volts. Thus the comparitor's unloaded output will swing from about -5V to about +5V.

    Next, the comparitor needs a bit of positive feedback to keep it from oscillating as the input signal passes it's threshold. I'd accomplish this by disconnecting the +input of the comparitor from the output of the second op amp and adding a 2.2k-Ohm series resistor between the two. Then add a 4.7 Meg-Ohm resistor between the output of the comparitor and it's +input. This should cause the threshold voltage to vary by about +/- 2.2 mv as the signal passes the threshold. An additional 1pF capacitor between the output of the comparitor and +input will also help overcome the parasitic capacitance on the +input. As long as the signal coming in is cleanly rising and falling and your reference is steady, this will provide a clean, jitter-free transition coming from the comparitor.

    Since the comparitor reacts only as good as the reference, it's best not to use the +5V supply for your reference. This is actually a general rule when designing precision circuitry - use a reference for your reference, not a power supply. I'd suggest an LM4040-2.5 shunt reference with a .1uF ceramic and 22uF electrolytic capacitors in parallel. Tie it's cathode to ground and pull the anode up via a 2.2k-Ohm resistor and you'll have a clean, stable 2.5V reference for as much as 1.5ma.

    As long as you're not fighting background lighting, I suspect you'll find that you'll find a much easier to detect waveform coming out of this circuit. However, It's unlikely that you'll find it's crossing the 2.5V reference. If were, I would reduce the gain on the op amps until your signal doesn't exceed 2V. This way, you can be sure of adjusting to it with the reference voltage you have.

    Unless you're signal level is much higher than it's DC offset, getting the potentiometer to work properly may be an issue. 10 turn pots make precision adjustments easier, and adding 1% resistors in series with the pot can help by reducing it's operating range. For example, if you have a 200mv - 300mv signal, you would want the threshold set to 250mV. If the pot were attached to the +2.5V reference, this adjustment may prove difficult because a small turn makes a big difference. However, if you grounded a 1K pot and supplied voltage to it through a 2.0k-Ohm resistor, the adjustment range would be 0-.5 volts, which is easier to adjust into .250 volts.

    So far, if everything is working, you will have a way to look at the fringes going by on the scope, but not drive to communicate to anything useful, such as a counter. This is where the analog designer's only digital friend comes in handy - CMOS. CMOS IC's and analog were made for each other. For example, you want a 0-5V or a 0-3.3V logic output. The 74HC14 hex inverting Schmitt trigger will operate from 3.3V or 5V and can be tricked into accepting the -5V to 5V level from the comparitor as valid with a simple resistor. Simply place a series 100K resistor between the output of the comparitor and the input of one of the inverting Schmitt triggers and the internal ESD circuit of the trigger will clamp the input voltage to a safe level. Run that trigger through a second trigger, and you have a non-inverted version of the signal. Just be sure to tie the inputs of all the unused triggers to either power or ground, because they're not happy floating.

    I hope this helps. Don't forget the basics like placing capacitors from power to ground here and there.

    Best Regards,

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