Running a signal generator and signal dependent load on the same power supply

Question

I want to use a square wave generator to drive a load in the same circuit but I am confused about how to do so. Here is what my current circuit looks like:

^{simulate this circuit – Schematic created using CircuitLab}

This works fine up until I turn my power supply up to 12 volts, at which point the load can turn on. The problem is, at this point the square wave disappears. I think it's because the current draw on the diode somehow limits the current in the generator but I don't really know.

How would I "split" my power supply such that it acts like two separate supplies (one for the generator and one for the load)? Should this square wave disappearance even happen or is it just because I have built something wrong?

Answer 1

One problem I see is you have a darlington pair (Q3 and Q4) with no current limiting resistor to the LED. This could create problems for the supply or burn out the LED due to large currents through D1. Make sure there is a current limiting resistor in series with D1.

Answer 2

First of all, there is a problem with exceeding Veb = -5V which happens when the collector pulls down the Vcc charged cap negatively to start off-time ramp up.

This is easily corrected with blocking diodes and a base to ground large R to prevent leakage turn on.

This method works well from 3V up to any voltage you want.

Your LED driver has a high threshold with the large base R plus the emitter voltage required. R6/hFE^2 appears as the emitter current limiting R .

A 10K base and a load current limiting R would have been more successful to drive harder but limit output.

Answer 3

I see two problems:

1. You are powering your astable multivibrator via R1, a 10k resistor. I believe the top rail of this entire oscillator (the node joining R2, R3, R4 and R5) should be connected directly to the positive supply. I do not understand the purpose of R1 here. If you are trying to reduce the amplitude of the square wave, this would be better achieved by a potential divider at the oscillator ouput, rather than at its power supply:

2. The LED is able to short-circuit the power supply, when Q4 is on. There should be some resistor in that path (in series with the LED), to bridge up the gap between 1.6V across the LED and the 15V across the power supply, limiting the current that the LED draws, as shown above.