# 9. Two-phase squarewave oscillator

It’s one of the simplest circuits to get working but one of the more tricky to understand. This simple two-transistor oscillator has a technical name of “astable multivibrator” and provides a squarewave (of sorts) output in and out of phase.

Looking at the circuit, there are three basic states – Q1 on/Q2 off, Q2 on/Q1 off and the start-up period, which we’ll look at shortly.

This will get a bit confusing so let’s start by assuming transistor Q1 has just switched on on and Q2 is off.

First up, keep in your mind that capacitors cannot change their voltage instantly. So we’ll assume that the cap attached to R2 has 0V across it. This pulls the base of Q2 to 0V, which means it’s off. However, with Q1 on, that 0.01uF capacitor now begins to charge via R2. When the voltage at the junction of the cap and R2 (which is also the base of Q2) reaches 0.6V.

This now turns Q2 on. Because Q1 was on, that means the cap joining R3 had 0.6V across it (it can’t go any higher because the Vbe drop of Q1 won’t let it go higher than 0.6V) but Q2 switching on effectively pulls the side of the capacitor connected to R4 from 9V down to 0V. As I said before, capacitors cannot instantly change the voltage across them so what happens on the R3 side of capacitor C2 is its voltage goes from 0.6V down to -8.4V, which well and truly turns Q1 off.

At the same time, a similar thing has happened to cap C1, connected to R1 and R2. The R1 side of that cap has jumped from 0V to 5VDC and the other side jumps from 0.6V to 5.6V – well, it would if it wasn’t for the Vbe junction of Q2, which holds it to 0.6V.

So now that Q2 is on and Q1 is off, C2 now begins charging up via R3 and the Q2 collector-emitter path. But it’s starting from -8.4V and it has to reach 0.6V before it turns Q1 back on again. However, it will eventually do so, turning Q1 on, which pulls the R1 side of C1 from 9V down to 0V and the R2 side from 0.6V to -8.4V.

And around it goes. Basically, each side switches the other to the opposite state and the time delay between states is set by R2/C1 and R3/C2 combinations.

Now as you probably noticed, we started by making the assumption that Q1 was on and Q2 was off. But can you actually assume that? The answer is no.

How the circuit actually starts in practice is going to be a bit of a lottery, even with component pairs the same as shown above. What actually happens in practice is that as soon as the power is applied, there’s a race between the two RC combinations (R2/C1 and R3/C2) to see whichever one reaches 0.6V first. The one that does turns on the opposite transistor first (eg R2/C1 turns on transistor Q2 or R3/C2 turns on Q1), which forces the other transistor off. And as soon as we reach that state, the circuit gets into a groove and off it goes.

Because the transistors operate in opposite states (eg. Q1 is off when Q2 is on and vice versa), you actually get dual outputs from the two collectors that are out of phase. This can be very handy in plenty of circuits.

Of course, you could also connect a speaker to the emitter of Q1 (make it a BC337) and provided you choose the right R/C values, you’ll hear a tone coming from it.

You can also use this circuit as a crude clock or pulse generator to drive some other digital circuitry.

And if you wanted to be able to electronically control turn this on and off, lift up the +V rail ends of resistors R2 and R3 and drive them with the output of some other circuit. When the tops of those resistors are pulled to ground, the circuit stops but pull them up to the positive rail as shown, and it operates as normal.

This is actually one of three similar circuits of the “multivibrator” family. Later, we’ll look at the monostable multivibrator that produces a pulse for a set period when triggered.

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