In this circuit you have connected the OPamp as comparator. So, you will have a much higher voltage at output than your 5V, because the output goes into saturation.


I show you how I would solve this design issue. We discuss two cases:

1. Output voltage is actually +5V, assuming you have added suited feeback components to assure this. Also, a current of 10mA shall flow through the LED, which is an adequate value for OPamps.

From this diagram


you can see, that for an output voltage of 5V the internal current limiter allows a current of up to 22mA to flow out of the OPamp. But as we want only 10mA, we choose R2 as follows:

R2 = (5V - Uled) / 10mA

So, R2 = 330Ohm seems to be a good start. Make measurements to verify it!

And, please, put away R1!


2. In this case the output shall be in saturation, as explained above. This case is a bit more difficult, because we don't know the exact output voltage of OPamp.

Nevertheless, by the help of this diagram


we can at least estimate what's going on:

When a current of 10mA is flowing a voltage of about 2.5V is dropping across the internal circuitry. But this is only true when Vcc=+-15V. At Vcc=+-10V I guess the voltage drop could be slightly higher. So, with Vcc=+-10V the output voltage will be about 10V - 2.5V = 7.5V, maybe slightly less. This leads to

R2 = (7.5V - Uled) / 10mA

So, R2 = 560Ohm seems to be a good start. Make measurements to verify it!

Again, please put away R1!


The design isn't finsihed yet. We must protect the LED, when the output goes negative! This is simple: Just connect a 1N4148 in parallel to the LED, but reversely connected, so that the reverse voltage of LED is limited to one diode's forward volatge drop, when the output of OPamp becomes negative.

Another issue is to be taken into consideration: Heat dissipation!

10mA^2 x 560Ohm = 56mW. So, a 1/4W resistor is enough.

Heat disspation of OPamp is about 3V x 10mA = 30mW, so, no problem.

Kai