I want to make an oscillator for my step up transformer but I can't use an ordinary one because I can't buy anything right now. I have a few components (resistors and exactly two tranzistors). So I came up with this oscillator:


simulate this circuit – Schematic created using CircuitLab

At the beginng the Q2 starts to conduct because it has voltage on his gate so it charges the capacitor. After this step the Q1 is quickly turn on which turns off Q2 but it remainslike that until C1 discharges on R1...And then it starts over again create a square wave (almost).

But when I tried to build it, it failed... Why? What is wrong with this circuit and can you explain why ? And don't tell me that there are other oscillators (I tried arduino but I fried it...) Thank you in advance..

  • \$\begingroup\$ Maybe you should instead try explaining why you think it should work (or what the expected output is) \$\endgroup\$
    – PlasmaHH
    Oct 28, 2015 at 20:25
  • 2
    \$\begingroup\$ More voltage? (should Q3 be flipped?) \$\endgroup\$ Oct 28, 2015 at 20:27
  • 1
    \$\begingroup\$ I don't see any Q1. BTW the "gate" of a BJT is called "base". \$\endgroup\$
    – Curd
    Oct 29, 2015 at 1:42

2 Answers 2


Q3 will work upside down; the hFE will be low but it will still work as a transistor (it will actually saturate at a lower voltage but that is besides the point).

Why doesn't this work? Or better still why do relaxation oscillators work - the simple answer is hysteresis. Your circuit has no hysteresis therefore it operates in a linear fashion and finds a perfectly likable equilibrium with about 0.7 volts across R2.

What would have been preferred is for Q3 to act like a trigger - once activated it stays activated until the voltage across C1/R2 has dropped (maybe) to 50%. Then the whole process will start again with C1 rapidly charging then triggering Q2 which then discharges C1/R2 thus an oscillator is created. Remember that Q3 can work upside down and the base-collector region is the new base-emitter region.

Sorry, it won't happen like that in a linear circuit - it just finds a nice equilibrium and sits there. Relation oscillators need hysteresis! Here's the archetypal op-amp relation oscillator: -

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R3 and R2 provide positive feedback and this creates hysteresis. Here's a useful oscillator to try: -

enter image description here

It's called an astable multivibrator but if you want some other options hit this link.

  • \$\begingroup\$ It may be useful to note that loading most (elementary/unbuffered) relaxation oscillators with a coil significant enough for power transfer will usually influence operation. Not central to the question, but quite likely "a next question". \$\endgroup\$
    – Asmyldof
    Oct 28, 2015 at 23:05
  • \$\begingroup\$ Please reconsider my question..I made a mistake and now it is ok. The second transistor (Q4) was upside down.... \$\endgroup\$ Oct 30, 2015 at 17:14
  • \$\begingroup\$ And please re-read my answer - Q4 being upside down is of no consequence to it not working (as I said in my answer!!!). Also please renumber Q4 back to Q3 because you've made my answer look wrong. \$\endgroup\$
    – Andy aka
    Oct 30, 2015 at 17:21

Since the voltage drop across Q2's base-to-emitter junction will always keep Q3's emitter more positive than its base and, since Q3 is an NPN, it'll never be able to turn on and your circuit will be dead in the water from power-up until the end of time.

  • \$\begingroup\$ Please look again at the circuit..I made a mistake and now it is fixed. The problem was with the second transistor (Q4)..it was upside down... \$\endgroup\$ Oct 30, 2015 at 17:16

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