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I'm currently designing a circuit to convert an infrared signal (e.g. from an IR transmitter or a TV etc. remote control) and turning it into an audio signal. The first stage of the preamplifier is shown below. enter image description here

I've worked out the resistor values for R1, R4, R5, R6, and R7. What I don't understand is what the peak voltage will be at VOUT.

According to the SFH213's datasheet, the photocurrent is 135 microamperes. How do I use this to calculate the peak output voltage?

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    \$\begingroup\$ That's the current when illuminated with 1000 lux 2865K incandescent light. Not very representative of what a remote will produce in normal operation. \$\endgroup\$
    – Spehro 'speff' Pefhany
    Commented Jun 14, 2018 at 3:07
  • \$\begingroup\$ Then, what would be a more realistic measure of current? And what would be the peak Vout? \$\endgroup\$
    – user117592
    Commented Jun 14, 2018 at 3:14
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    \$\begingroup\$ @Jack has gone through some of the calculations (we don't really know the distance or how far off-axis the receiver and sender are so it might be optimistic or pessimistic, but given the info it's a good answer +1.) The voltage gain of the amplifier is a bit less than R7/R5 so the gain is about about -100. Keep in mind that typical remotes work at ~38kHz which may be considered audio in frequency band but it is not audible. Receivers also usually use synchronous demodulation to prevent room light from saturating the PD, which will occur at relatively modest light levels (quite a dark room!) \$\endgroup\$
    – Spehro 'speff' Pefhany
    Commented Jun 14, 2018 at 4:19
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    \$\begingroup\$ That is because the DC bias of the PD is affected by R3, but the sensitivity to AC is set by R3 || R1 || R4. \$\endgroup\$
    – Spehro 'speff' Pefhany
    Commented Jun 14, 2018 at 4:25
  • \$\begingroup\$ Over what temperature range must this work? That base-bias looks a bit crucial. Adaptive biasing ==== replace R4 with two series resistors tied to the Collector, values 4.7K, with a large capacitor at the midpoint, to shunt away most of the negative feedback. \$\endgroup\$
    – analogsystemsrf
    Commented Jun 18, 2018 at 4:49

1 Answer 1

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If you look at the datasheet for the SFH213 you will notice the current is specified at a particular intensity for the incoming optical signal.

You will also notice that there is a current specified as dark current.

Assuming that your diode only sees the incoming signal you could assume that it provides only the dark current when the sender LED is OFF.
So Dark current is about 1 nA.

You have not specified your LED sender, so it's impossible to speculate on what the incoming signal power level might be.

For example you might use something like a MT5900-IR LED which could have about 45 mW radiant power spread over a 20 deg cone at 100 mA drive current.

The amount of power that reaches the PIN diode active area may only be 1000'th of this total power. Obviously the amount of signal power will vary depending on how far the LED is from the PIN diode, but let us assume for the example that the inbound signal power is <45 uW/cm^2.

You can now use the graph on the SFH213 datasheet to see what current might flow with this signal.

enter image description here

So the signal current will vary in this example from around 1 nA to a peak of about 4 uA.

The rest of the circuit is tricky ...since you have a 1M Ohm load resistor you'd expect to get about 4 V of signal if that was the only load, but the input impedance of the transistor stage will drag this down. You would actually get only about 10 mV or so input to the transistor amplifier, and the output would therefore be around 1 V peak to peak.

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