With reference to the question from post #10 which I did not notice yesterday. Here is what I was suggesting.
The 2.2K resistor will limit the LED current to about 9 mA. The two diodes D1 & D2 are to limit the voltage drop across R1 to about 1.4 volts and must be rated for the full load current. R2 is to limit the base emitter current of the PNP transistor to less than that rating of the transistor. So if we want to limit it to 1 mA then we have 1.4 volts across R1 less Vbe of the transistor (say 0.6 volts) This gives us 0.8 volts. So we could use an 820 ohm or 1K for R2. To calculate R1 you need to supply the following information. What is the minimum current you define for detecting the load being connected ? What is the input current to the regulator when there is no load on the regulator output ? What is the maximum load current ? (That is for the current rating of the diodes) This design will not work if the no load current of the regulator is several mA and you want to be able to detect a load of only a few uA . As you have now changed your original specification from no allowable voltage drop to 0.8 volts. Your original idea could probably be made to work if you could obtain a germanium (Rather than silicon.) transistor as these only have about 0.2 volts (From memory.) Vbe to turn them on. I do not think germanium transistors have been manufactured for many years.
The method suggested by gophert in post #18 is a better solution than the one I suggested on the regulator input.
Les.
The 2.2K resistor will limit the LED current to about 9 mA. The two diodes D1 & D2 are to limit the voltage drop across R1 to about 1.4 volts and must be rated for the full load current. R2 is to limit the base emitter current of the PNP transistor to less than that rating of the transistor. So if we want to limit it to 1 mA then we have 1.4 volts across R1 less Vbe of the transistor (say 0.6 volts) This gives us 0.8 volts. So we could use an 820 ohm or 1K for R2. To calculate R1 you need to supply the following information. What is the minimum current you define for detecting the load being connected ? What is the input current to the regulator when there is no load on the regulator output ? What is the maximum load current ? (That is for the current rating of the diodes) This design will not work if the no load current of the regulator is several mA and you want to be able to detect a load of only a few uA . As you have now changed your original specification from no allowable voltage drop to 0.8 volts. Your original idea could probably be made to work if you could obtain a germanium (Rather than silicon.) transistor as these only have about 0.2 volts (From memory.) Vbe to turn them on. I do not think germanium transistors have been manufactured for many years.
The method suggested by gophert in post #18 is a better solution than the one I suggested on the regulator input.
Les.
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Oops I see that it just was suggested...
