How would I wire a light-emitting device (not necessarily an LED) in the following schematic to indicate the input (High to run) is high? Would it be in parallel with R4? Before or after R3?
Thanks, Russell McMahon, for making the schematic for me.
Electronic – MCU Electronic Power Switch
lightmicrocontroller
Related Solutions
There are not good answers to these questions because LEDs are intended for emitting light, and as such the parameters you need to answer your questions are not specified.
A LED reverse biased as a light sensor is a current source proportional to the light level. Being a current source, it have very high impedance (a perfect current source has infinite impedance). The response time is proportional to the resistance of the node times the capacitance. Since the capacitance is parasitic, it is hard to guess and will depend a lot on the particular LED and on layout. The resistance is the deliberate resistance R1 in parallel with any leakage resistance and the resistance of the LED being a imperfect current source. Other then R1, these are again hard to guess. 20 MΩ is so high that leakage can be a important factor. Even dirt on the board and ambient humidity matters at that impedance.
As for how to determine the voltage, that again must be done experimentally. Unless you have a unusual LED that is intended also for reverse operation, you're not going to get a spec. Test a few and leave lots of room for device variation.
I would use a considerably lower resistance with some amplification. The lower resistance will decrease the response time and make things more predictable by making the leakage resistance small enough in comparison to not matter. You are currently getting ouputs from 150 mV to 5 V with 20 MΩ. With 2 MΩ instead, those voltages will be 15 mV to 500 mV, which is still big enough for plenty of opamps to amplify reliably and should be low enough to make leakage ignorable. It may still be too slow, in which case you can use lower resistance still with better amplification.
Another point is that if your supply is large enough to get 5V on R1, then you may be applying too much reverse voltage to the LED in low light conditions. Check the LED datasheet (this usually is specified) and make sure you're not exceeding the reverse voltage limit. A lower resistance will let you use lower reverse bias voltage.
Based on the circuit you provide, you could just add a diode in series right after the switch (S1) (cathode connected to the switch) and them you could use an input to detect if the switch was pressed again, if so, turn off PB3.
The zener diode protects the PIC input from the voltage coming from the power supply.
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Best Answer
There are various places.
The next step is 'after' Q2, but that has a switch in parellel, so your light will light up if that is pressed too. You could extend the circuit with a resistor or a few diodes to prevent this.
Next is between the collector of Q1 and ground. Like the collector of Q1, it is also activated when the switch is pressed (and prevent this parallel activation is a bit more involved), so maybe that is not what you want.
One additional note: if you use a LED on any of those places, don't forget the series resistor.