Help understanding astable multivibrator

Thread Starter

gerases

Joined Oct 29, 2012
186
Hello,

I'm fairly new to electronics and can't really understand complex circuits yet. I do know the basics of the components, Ohm's low, voltage drops, etc. This is just a side hobby for me. So I've come across the astable multivibrator and I can't for the life of me understand a few things about its operation. Here's the circuit as an example:
1642776792736.png
I understand that the C1 and C2 take turns in turning on the cross connected transistors. Ie, C1 will turn on Q2 (eventually causing a turn-off of Q1) and C2 will turn on Q1 (causing a turn-off Q2). My main difficulty is in understanding the charging process of these capacitors. The way they are connected in this circuit doesn't look like anything I've seen before. Usually, I've seen the plus of the battery connected to the plus of the capacitor and the negative of the battery connected to the negative of the capacitor. The capacitor can be connected in series with a load or in parallel with a load. Regardless, its positive terminal should lead to the plus of the battery and the negative to the negative.

Here we have something unusual looking. First of all, the circuits use specifically polarized capacitors. So, looking at C2 for example, how come its negative side is connected to the positive terminal of the battery via R3? Its official right side is also connected to the positive (I guess?) when Q2 is turned off or negative when Q2 is turned on. That's so weird and so never mentioned in any descriptions of this circuit I've read. Yet you also read to always observe the polarity of the polarized capacitor. So yeah, for a newb like me, it's tough.

Next thing that is odd to me is that when we just turned on the circuit, I'm told that the voltage on both C1's and C2's negative side begins to positively charge through R2 and R3 respectively until one of them reaches the BE threshold voltage of 0.7V thus causing the ensuing events (let's forget about those for now). Ok, so as I mentioned before, it's hard to imagine the negative side of the capacitor being charged. But somehow it does work?

On top of it, the positive side of those capacitors (before any of them reach 0.7V on the negative side) reaches about the positive voltage of the battery (roughly). But how is that possible if the LEDs are not letting any current through yet and the positive sides of the capacitors are connected below the LEDs?

There's also the issue of negative voltage on the capacitors, but I'm not even going there yet.

I must be missing a lot of background on all this, but if anybody patient enough could enlighten me on the operation of this puzzling circuit, it would be so helpful.

Thanks!!
 

ericgibbs

Joined Jan 29, 2010
15,338
hi,
You say:
C1 and C2 take turns in turning on the cross connected transistors.

The timing is achieved by the charged capacitors turning Off the Transistors.

Due to unbalances within the circuit, one of the transistors will turn On, at power up, consider the polarity of the Capacitor charge of the capacitor connected to the On transistor to the Base of the Off transistor.
E
 

Alec_t

Joined Sep 17, 2013
12,732
Eric's waveforms also demonstrate an often-overlooked snag with that basic circuit when powered by a 9V battery; the transistor base gets pulled down to about -8V. That is 3V beyond the point where a normal silicon transistor suffers breakdown of its base-emitter junction. Not good for the transistor. A diode in series with the base-emitter junction can overcome that problem.
 

Thread Starter

gerases

Joined Oct 29, 2012
186
Unfortunately, the graph doesn't help understand
Does this plot of the voltages at vc1 and vb2 help you understand.?
Sorry, not really, I need to understand the polarity issues I described. It's clear that the circuit works and the voltages go up and down. My question was why.
 

dl324

Joined Mar 30, 2015
14,306
I must be missing a lot of background on all this, but if anybody patient enough could enlighten me on the operation of this puzzling circuit, it would be so helpful.
The circuit you found has a couple issues.

The first is that the BE junctions of the transistors can be reverse biased enough to cause them to breakdown. That will damage transistor beta. Without enough current gain, the circuit won't work.

The second problem is that the LED on the transistor that's off will likely be dimly lit. The circuit works better if you put the LEDs across the collector-emitter of the transistors or use N channel MOSFETs. A hack is to put a resistor in parallel with the LEDs.

The way the circuit works is that when power is applied, one of the transistors, e.g. Q2, will be on and the other off. LED2 will be on and the C2 terminal on Q2 will be near ground. Since the voltage across a capacitor can't change instantaneously, the other terminal will be at -9V and start charging towards the supply voltage. When that voltage gets to about 0.7V, Q1 will turn on, the voltage on the base of Q2 will get pulled to -9V and it will turn off.

The problem is that the BE junctions will break down at about 5V of reverse bias.
 

ericgibbs

Joined Jan 29, 2010
15,338
hi ger,
OK,
Lets try it this way.
Consider you had a 100uF cap, and you charged it to, say +10v.

Measuring with a meter one way would show +10v if you now Reverse the meter connections, now it would measure -10V

On the Astable, the Collector of the transistor that is Off allows the capacitor to charge to +10v relative to the Base of the other transistor, which is conducting

If now the Off transistor conducts, it connects the +10v charged end of the capacitor to 0V and now as the other end of the capacitor is on the Base , this Base is set to -10V relative to the 0V supply rail, and it switches Off. The -10V charge cannot flow into the Base as it is reversed biassed, so it loses its charge via the 47k which is connected to the +10v supply rail.

As the Base becomes forward Biassed due to the 47K discharging the -10V this transistor turns On and so the On/Off cycle is repeated,

E
 
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