How to understand this oscillating circuit made of two transistors and R C?

This is an oscillating circuit which is made of two transistors and some R C.

As below schematic shows, during the power on the capacitors, C1 and C2 are charged through R1 and R2.

After about 0.5s the C1 is fully charged to 24*(1/(1+0.2))V = 20V and after about 10s the C2 is charged to about 20V+0.34V=20.34V.

The voltage difference between E and B of Q1 makes the Q1 PNP transistor to work, the current flow through the R5 resistor then makes the Q2 NPN turn on. after then the current flow to the LED D2, then the D2 turns on and emit light.

But quickly as soon as almost 3-5ms the energy in C1 and C2 capacitors will be exhausted which then makes the process go back to the very beginning.

Then the capacitors C1 and C2 are recharged.

The above process is the basic description of the circuit. I simulate the circuit by using the Tina V9 software and the result is as the pic. shows.

My questions are:

  1. How to calculate the suitable value of the R and C which can make the oscillating time equal to 5s?
  2. How does the circuit work which very precisely makes the circuit to oscillating?
  3. What are the roles of the two transistors and how they work?


simulate this circuit – Schematic created using CircuitLab



1 Answer 1


The principle of this oscillator is similar to other "relaxation oscillators" and oscillators that use a Unijunction Transistor, see this article.

The two transistors form a self-latching circuit, when the PNP conducts, it keeps the NPN in conductive state and that keeps the PNP in a conducting state. To get from the non-conducting state to the conducting (latched) state, the PNP transistor's base-emitter junction needs to enter breakdown. This generally happens at a voltage between 7 to 10 V. That breakdown will make current flow into the base of the NPN which will make it switch on. That then switches on the PNP as well. That will discharge C2, partly through the LED D2 which will light up.

When C2 has been discharged enough the current through the PNP and NPN drops so low that they will switch off again, releasing the latch. Then C2 can start charging again until the PNP enters breakdown and the cycle repeats.

Getting precise timing from this circuit is a matter of "trial and error", this isn't a very precise circuit as for example the breakdown voltage of the PNP is highly unpredictable, it varies even between the same model transistors. So you just build the circuit and adjust the values of R2 and C2 to get the timing you want.

If you want better predictability use a timer chip like the NE555 and even that isn't an accurate timer. To reach the next level of timing accuracy you will need a crystal as a time base in combination with a logic circuit or (much simpler) a microcontroller with a crystal as its timebase.

  • \$\begingroup\$ As I simulate this circuit, I found that the current flow into the LED consists of two parts, firstly from the C2 and when the voltage of C2 drops a little then the C1 began to discharge. That is to say in the simulation pic. the AM1 ≈ IC1+IC2.The two parts current made each of the two capacitors discharge completely. \$\endgroup\$
    – lukeluck
    Feb 26, 2020 at 0:43
  • \$\begingroup\$ Your share web really helps me a lot with this circuit. A more question. Compared with the web, what's the function of the R5 10M Ohm resistor? I found if there is no thus a resistor the oscillation can not be produced at least from the simulation's view \$\endgroup\$
    – lukeluck
    Feb 27, 2020 at 9:44
  • \$\begingroup\$ My guess is that R5 helps when the transistors need to be switched off. Without R5 some charge can get "stuck" at the base of Q2, with R5 in place that charge can get away so that Q2 is properly switched off. \$\endgroup\$ Feb 27, 2020 at 9:53

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