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I'm trying to get this variation on the classic circuit to work. Pardon the awkward VCC label. The voltage is 4.5 volts, though these batteries are only supplying 4 at the moment. I have an LED and a 330 resistor on the output an an indicator.

The goal is to get the third transistor fully saturated to switch on and off as much current as it can safely handle while pushing a minimum of current through the oscillator to minimize its own drain, particularly through where it dumps straight to ground. This way, I could use this as a module to hook up arbitrary loads (within reason) to switch on and off.

No matter how I play around with various resistances and capacitor sizes, but the effect is always the same. The indicator is always just on. A few times, I would see it blink about two times and then go steady. Checking the math on how quickly it should oscillate given the values here, this shouldn't be a case of it going faster than the eye can perceive.

When I look up schematics where people have this circuit tapped for output, they usually take output from the same net as the capacitor's positive terminal. This seems strange to me and makes me worry that the design below is then somehow wrong.

What puzzles me is that just before bread-boarding this one up I did the same circuit, but without the boosting transistor, putting the output of both transistors each through an LED, using 3v supply, and some different resistor and capacitor values, and have something that works beautifully.

How should I go about debugging this and where could I have gone wrong?

  • 1
    \$\begingroup\$ Please add designators to your schematic so we can talk about Q1, Q2 and Q3 instead of "the left BJT", "the middle BJT" and "the right BJT". \$\endgroup\$
    – The Photon
    Commented Mar 14, 2020 at 2:24
  • \$\begingroup\$ Try putting a diode in the emitter path of the left transistor. Your circuit may be off-balanced too much by the double emitter diodes on the right. Oh, and put the load to the collector of the rightmost transistor, not the emitter. Same reason. \$\endgroup\$
    – Janka
    Commented Mar 14, 2020 at 2:26
  • \$\begingroup\$ @Janka How do Q1 and Q3 interact if I put the base of Q3 to Q1's collector instead? \$\endgroup\$ Commented Mar 14, 2020 at 3:47
  • \$\begingroup\$ Also, it works! \$\endgroup\$ Commented Mar 14, 2020 at 3:53
  • \$\begingroup\$ I'd be the most concerned about the batteries. When you say they should be 4.5V but give 4V only by now you're just saying they might have a high and unpredictable internal resistance. This can easily interact enough with your circuit to prevent correct operations. Get a good power supply first and try again \$\endgroup\$
    – carloc
    Commented Mar 14, 2020 at 6:31

2 Answers 2


I would add a resistor between base of q3 (q1e) and negative rail (q2e). Then another larger resistor to limit base current in q3.

This change will unbalance the astable but you could balance it back by adding on q1e the same resistor as the q2e.

The load needs interfaced between positive rail and q3c.

As for resistor values, I would start with base current on q3, giving me base resistor, which leads to q1e resistor min 10 times smaller than base resistor on q3.also q3e is tie to negative rail. Hope this helps. Hope this helps.


without substantial current in R2, Q2 will not be turned off


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