# Transistor Multivibrator Circuit

Discussion in 'General Electronics Discussion' started by eptheta, Oct 16, 2010.

1. ### eptheta

188
0
Dec 20, 2009
I was looking up oscillators and i came across this transistor based oscillator.

What i don't actually understand is why current through the 1kOhm resistors goes through the capacitor to ground rather than to the other transistor's base.
For example: when C1s left plate is getting charged... near C2, the current goes ahead and charges C2s left plate instead of triggering Q1s base and letting C1s current go to ground..

Excuse me if this is basic electronics, I am unfamiliar with much of this subject.

Thank you.

2. ### (*steve*)¡sǝpodᴉʇuɐ ǝɥʇ ɹɐǝɥdModerator

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Jan 21, 2010
That is an astable multivibrator. There are plenty of explanations on the web.

Here is one of them.

3. ### eptheta

188
0
Dec 20, 2009
Oh, so the potential without the capacitor isn't enough to trigger the base of the transistor unless the capacitor is charged enough to provide it's own additional potential because the transistor is biased in that particular way ?
Why can't electronics work without math... Luckily its a relatively simple calculation anyway.

Thanks for not giving me a lmgtfy link, I feel silly enough.

4. ### (*steve*)¡sǝpodᴉʇuɐ ǝɥʇ ɹɐǝɥdModerator

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Jan 21, 2010
I didn't give you an lmgtfy link because you had clearly made some effort to understand it yourself.

If you've put some effort into it, and you're happy to tell me what it is that you understand, and why you think your understanding is faulty, then it would be pretty bad form of me to belittle those efforts.

The circuit is confusing, because in a perfect world it wouldn't oscillate. Indeed the explanations of its operation almost always assume that it is already oscillating. A more naive attempt at explaining it tends to fail because you don't assume one transistor is already turned on.

5. ### eptheta

188
0
Dec 20, 2009
Hi,
I've gone really far back and realized that I haven't really learned much about capacitors at all.
From what I know,
the charge on a cap at time t(with a resistor in series)= qo(1-e^-t/τ)
τ being the time constant. Same for voltage...

Now in this circuit:

It turns out that the final voltage on the cap is 2.5V when Vcc is 5V... (this is because the capacitor is parallel to the 100 ohm resistor which is at 2.5V)
What I'm not really clear about are the following:
1. In what ratio does the current split along the resistor and capacitor branch ?(same 1-e^-t/τ ? Also whats the time constant, i know of it when there is just resistance in series not parallel)
2. From the graph, why does current in the resistor start at 0 ? ( is it because it seems that the capacitor's resistance is 0 at the beginning and so relative to that the resistance's resistance is infinite ?)
3.After the capacitor is fully charged, why doesn't its potential contribute to the potential of the resistor parallel to it ? (As in discharge through the resistor while simultaneously being charged by the other ? Or is that happening already ? But if i remove the capacitor after it is charged, then the circuit stays the way it is without a reduced potential)

Sorry if these are very basic questions, but I am rather confused after staring at so many simulations.

6. ### (*steve*)¡sǝpodᴉʇuɐ ǝɥʇ ɹɐǝɥdModerator

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Jan 21, 2010
In that circuit, when things are in equilibrium (i.e. battery connected and capacitor fully charged) the voltages and currents are identical to what they would be if the capacitor was not present.

The capacitor has a voltage that rises as it charges. Initially there is 0 V across the capacitor, therefore no current flowing through the resistor in parallel with it. As the capacitor charges, the voltage across the resistor in parallel increases and obviously so does the current through it.

Think of the resistor in series charging the capacitor, and the resistor in parallel discharging it. At some point the current available to charge the capacitor becomes equal to the current available to discharge it. At that point there is no net charge flowing into or out of the capacitor.

7. ### eptheta

188
0
Dec 20, 2009
Good, That clears a few doubts, now on to this...
I assembled this circuit on a breadboard (the simulator won't let it work though for some reason)....
NOTE: the 2 NOT gates are just a buffer so that i get nice flat slopes and instant switching off of the LED instead of it lingering on.. This lets me time it better...

Vcc is about 8V... When i don't have the 2.2kOhm resistor, the LED goes off in about 4 seconds.. When i do introduce it however, it turns off in about 8 seconds.
I am guessing this is some form of biasing ? If so, how do i get the math in place? I went through several wikipedia pages and play-hookey ones too but all of them have standard assemblies for calculations.
In any case, this serves as a crude delay circuit. How am i supposed to calculate the time period ?

Thanks.

8. ### (*steve*)¡sǝpodᴉʇuɐ ǝɥʇ ɹɐǝɥdModerator

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Jan 21, 2010
There are many problems with that circuit, but the big one is that the input impedance of the gate can be considered pretty close to infinite, so no current can flow through the transistor. The path through the 2k2 resistor and the BE junction of the transistor is enough to create a high level at the input of the first inverter, this the LED is always on.

It is functionally equivalent to connecting the input of the first gate directly to the battery.

9. ### eptheta

188
0
Dec 20, 2009
Oh, that explains the simulator crashing, but strangely enough it still worked on a breadboard giving me the delays i posted earlier.

Anyway, since i can't re-create the conditions without the logic gates, I may as well ask straight away.
Without the NOT gates, the LED stays on for about 11s, until the capacitor is charged sufficiently, and then goes off(well not actually, but becomes very very dim).
Is there any way whereby i can increase this delay period to more than 11s (well without increasing the capacitance over 4700u? Say double of it or 22s ?

I'm hoping for a mathematical approach to this because that's the part where I'm really clueless. I could just fiddle around with variables on the simulator and figure something out, but I'd rather not.
Could you point me in the right direction ?

Thanks.

EDIT: While my above queries are equally relevant, I wanted to add on to what i last posted.
I realized that this is what i had actually assembled on the breadboard, which seems to make more sense.

Now that i have that 10k resistor that pulls the input of the NOT gate to ground if there is no current, it seems to please the simulator as well....
Now, while I have not tested these exact values in the circuit myself, the simulator seems to think that with the 2.5M resistor, the LED stays on for about 4000 seconds.
With the 2.5M resistor, it stays on for approx 9000 seconds.....

Now there is a definite relationship between these variables but I know not the mathematics of transistors well.... How can i put this to paper and calculate a desired time period ?

Thanks !

Last edited: Oct 17, 2010
10. ### (*steve*)¡sǝpodᴉʇuɐ ǝɥʇ ɹɐǝɥdModerator

25,491
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Jan 21, 2010
Do you really mean 4.7 milli farads (= 4700uF)

What are you trying to do. It may be better for you to explain that than just add components to a circuit.

11. ### eptheta

188
0
Dec 20, 2009
Yeah, whats wrong with 4700uF ?

I'm actually doing nothing. Just experimenting to no predetermined end.There's a lot of stuff that I see on this simulator that I never encountered in school textbook problems...
I have a breadboard, I have the components and I have the time.

12. ### (*steve*)¡sǝpodᴉʇuɐ ǝɥʇ ɹɐǝɥdModerator

25,491
2,833
Jan 21, 2010
I've seen people use "m" for micro instead of milli.

You are drawing circuits that are an unusual combination of analog and digital, and which seem to have no real function. If you're trying to achieve some end, then perhaps we can suggest a circuit. If you're putting together stuff and asking us why it doesn't work, you may not really benefit from the explanation.

For example, in your most recent circuit, the 2.5M resistor provides some bias to turn the transistor on. However, unless the transistor has a very high gain, it will turn on only very slightly. This will essentially be the case when the capacitor has charged. While the capacitor is not fully charged, the current charging it biases the transistor on, with it slowly turning almost off (to the state as described above).

When power is removed, the 2.5M resistor provides a discharge path for the capacitor, however one that is probably swamped by the leakage current of the capacitor itself.

You appear to be using inverters, I would hazard a guess that they are CMOS inverters. You are driving them with a linear signal that varies from just under 0.9 x your supply voltage down to almost 0 volts. Having 2 such inverters in series means that the second will switch quite rapidly when the input voltage passes approx 0.5 of the supply voltage.

13. ### eptheta

188
0
Dec 20, 2009
Okay, you're right, this would just get me going in circles....
The whole circuit actually stemmed from half of the astable multivibrator circuit, that is just one transistor, 2 resistors and 1 capacitor. The arrangement is pretty much the same , the capacitor in series, the resistor between the capacitor and the base etc....

Without 2 sets of capacitors/transistors/resistors, can I use the idea of a capacitor in series with a transistor's base to act as a timing mechanism and thereby generate square-waves (which is why i thought of using the inverters)

OR

Can i generate square-waves like the 555 timer does without the chip(and without Op-Amps: Most references on google are to 555s and Op amps) , and rather only with transistors resistors and capacitors ? the transistor multi-vibrator doesn't give a squarewave output does it ?.....

The wikipedia link says :"Multivibrators find applications in a variety of systems where square waves or timed intervals are required"
But I can't find any way of making the output a square wave......

Thanks.