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I am using a transimpedance amplifier with a BP104 Silicon PIN photodiode to receive incoming pulses from an infrared LED with a frequency around 100 kHz. Here is the relevant part of my schematic:

enter image description here

I have not included specific values for the feedback network, as it occurs for all values I have tested. Now here is the issue: this receiver only works beyond a certain distance between the transmitting LED and the photodiode; this distance is dependent on the gain, which is the value of the feedback resistor. If I have it too close to the IR LED then the output just saturates to my supply voltage (+10V). Any ideas on why this this happening? I know that it is not simply the capacitor charging up, as it also occurs without that. My best guess is that when I have the LED too close I am generating too much current, which is trying to pull up the output voltage beyond its maximum limit, thus causing it to rail and affecting it internally. I am having trouble convincing myself of this though, because my transmission is a 50% duty cycle square wave, so during half of the period there is no signal at all, and yet the output continues to hold at a steady +10V. Is such an intense amount of irradiation on the photodiode causing a current to persist through the LOW portions of the signal? This it the first time that I am using a transimpedance amplifier in a serious application, so this may be obvious. Many thanks in advance for any insight or tips.

Edit: Here is a plot showing the resistance values that I have tested. More specifically, it shows my maximum transmitted distance vs. the resistance value (gain). The blue curve is my theory prediction, and the red points are my data.

enter image description here

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  • \$\begingroup\$ What resistance and capacitance values have you tried? \$\endgroup\$ – helloworld922 Apr 21 '14 at 7:51
  • \$\begingroup\$ @helloworld922 I have edited my post to include a plot that might be useful. A capacitance of 2pF seems to work well, except for high gains, at which having no capacitor seems to be fine. I need some rest, but will respond in a few hours if you reply. \$\endgroup\$ – dsm Apr 21 '14 at 7:56
  • \$\begingroup\$ To avoid saturation issues ,have you seen IrDA design? vishay.com/docs/82633/tfbs4711.pdf \$\endgroup\$ – user38637 Apr 21 '14 at 10:58
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You're overloading the amplifier. Recovery from overload can be very fast, but 1 millisecond is not uncommon, and tens of ms is possible even with multi-MHz GBW op-amps.

There are amplifiers such as the OPA380, which has 100ns overload recovery.

I don't see a number for the LF357, you can measure it, as suggested in this classic application note AN356 from Analog Devices.

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  • \$\begingroup\$ Perfect, thank you very much for the response. Do you have any personal recommendations on how to circumvent this issue without changing the op-amp? I see in that application note (which is great) in Figure 10 that there is a "clamp circuit" for this issue, which I will build if my lab has the components, however what would you recommend if they do not? Thanks again. \$\endgroup\$ – dsm Apr 21 '14 at 15:26
  • \$\begingroup\$ The clamp circuit is a reasonable approach, although it's not so simple, I think the end result is that the FD333 diodes end up effectively in parallel with the PD rather than the feedback resistor, which is much better, and the zener capacitance drops out of the equation (the 1N4148 capacitance is a load on the output). Personally, I'd buy the $6 op-amp. The low leakage diodes are going to be hard to find datasheet.octopart.com/FDH300A-Fairchild-datasheet-28922.pdf You could try it with all 1N4148s and see if it's good enough for you, or use diode-connected JFETs for the diodes. \$\endgroup\$ – Spehro Pefhany Apr 21 '14 at 17:46
  • \$\begingroup\$ I see that the maximum output for the OPA380 is less than 10V, which is what I need. Is there another op-amp with a small recovery time that you would suggest? Thanks in advance \$\endgroup\$ – dsm Apr 22 '14 at 21:04
  • \$\begingroup\$ Maybe AD8067. Limiting amplifiers are nice, but not many that are high voltage. \$\endgroup\$ – Spehro Pefhany Apr 22 '14 at 23:15
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Spehro is correct. The "obvious" way to fix the problem is to reduce the value of the feedback resistor. This, of course, will hurt your maximum range.

The second "obvious" way to fix it is to put a zener of, let's say, 7.5 volts across the feedback resistor. Unhappily, this won't work, as such zeners have reverse capacitances of ~ 50 pf, and this will kill your frequency response at megohm resistors such as you use. I'd suggest a string of 4 or more 1N4148 diodes, anode at the opamp output. You can sum the knee voltages to whatever value you like, and the series connection will reduce the total capacitance. This will almost certainly mean you'll need to get rid of your feedback capacitor.

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  • \$\begingroup\$ Thanks for the response! I am a bit confused on why the feedback diodes would help me out (I am new to electronics). Is the idea similar to the clamp circuit (Fig. 10) in the application note that Spehro linked to? I.e., does it prevent the output from saturating? If so, how are the diodes acting to make that happen? Thanks again, and I would upvote you if I had enough reputation points. \$\endgroup\$ – dsm Apr 21 '14 at 15:31
  • \$\begingroup\$ Yes, it's like Fig, 10, but simpler. The problem with your circuit is that, when the input current gets too high, the feedback current can't keep up, because the opamp can only produce a limited output voltage range. With one or more diodes in the feedback loop, after some point (the knee voltage of the diodes, ~0.7 v / diode), the diode current goes up very rapidly and allows the current at the input to equalize. This means that the output is no longer linear wrt the input, but as long as the combined knee voltages are high enough, that's not a problem. \$\endgroup\$ – WhatRoughBeast Apr 21 '14 at 16:14
  • \$\begingroup\$ See, for instance, nxp.com/documents/data_sheet/1N4148_1N4448.pdf, Fig 3 for the voltage/current of a diode. As you can see, for voltages less than 0.6 volts, the current is essentially zero, but by the time it's up to .8 volts, it's in the tens of mA. Using regular signal diodes this way gives a very soft limit - that is, the output stops being linear long before the limit is reached, but like I say, that's not a problem in your application. \$\endgroup\$ – WhatRoughBeast Apr 21 '14 at 16:27
  • \$\begingroup\$ Sorry for the delayed response, I have had a very busy day; hopefully you haven't lost interest. I apologize to be a burden, but I am very confused about your suggestion. I have made a schematic of what I believe you are saying, is this the right idea? \$\endgroup\$ – dsm Apr 22 '14 at 4:10

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