For the attached IR circuit what exactly is the purpose of R2 and how is its value decided?









Thanks

 

Hard to tell without the complete circuit. Do you expect that we have supernatural powers?
Is D2, D_Photo, an emitter or a receiver? It does not make sense as an emitter

 

Sorry D2 is a receiver and D3 is the emitter. The other part of the circuit cut-off is just an LED turning on and off when the transistor turns on and off. The circuit is supplied with 5V.

 

You'll get an answer when you post a more complete schematic. Otherwise, radio silence.

 

Sorry that's the full circuit

 

When D2 conducts sufficient current, transistor Q1 is turned on and LED D1 is lit.
I would determine R2 experimentally as follows:
Remove R2 and block any light from D3 to D2.
Start with R2 = 1MΩ. Reduce R2 until the LED goes off.

 

Actually, even with the circuit shown in post#1 it was perfectly clear what R2 does. It pulls the base voltage down so that the transistoris off until the photodiode conducts. I am not sure, but it does look like that diode is connected backwards.

 

Now that we know the complete circuit, we can analyze various options.
D2 in that configuration would not be able to supply base current to the transistor. If is the other way and no current flows, in the absence of illumination, then the base of the transistor will be pulled down to GND. When D2 is illuminated and current can flow from the supply through D2 and into the base to turn on the transistor the current through R2 should be smaller than the current required by the transistor. D2 will drop about 0.7V, the Vbe drop of the transistor will be another 0.7VV, and the LED and R1 will drop the remaining 3.6V as long as the Vf of the LED is not too large.

In order for this circuit to work the currents and the forward voltage drops of the diodes and the transistor need to be accounted for.

 

A photodiode is almost always reverse-biased, then light on it causes it to have leakage current.
We were not told the function of this circuit. Is it a break beam detector?
Here is a typical IR light detector circuit using a 1.1M resistor for its much higher sensitivity:

 

A photodiode is almost always reverse-biased, then light on it causes it to have leakage current.
We were not told the function of this circuit. Is it a break beam detector?
Here is a typical IR light detector circuit using a 1.1M resistor for its much higher sensitivity:

And the small current through it relative to the available base current. It is notable that the TS did not specify an actual photodiode so the properties of such a device might tend to be obscure. Is it the case that such a leakage current is sufficient to provide enough current for the emitter follower to light an LED at a forward current of say 2-10 mA?

 

Some LED devices generate quite a bit of voltage when exposed to light.The best produce over a volt from a fluorescent bench light at six inches. So that is another possibility.