Playing around with an astable multivibrator

Question

I had a doubt regarding the astable multivibrators using BJTs.

This is a typical astable multivibrator.

I have been told that if I connect a capacitor to the output channel, there would be no difference.

I've also been told that if one would instead connect the Output 2 channel to just a load resistor, which in turn is connected to the ground, the peak voltage of the square waveform would drop, and also the duty cycle might increase (which is 50% initially, as it can be seen.)

I'm having a hard time figuring it out why this happens at all, for the resistor.

Can someone work out the formulae and try to come up with an explanation as to why this happens?

The load resistor can be assumed to be of 1k ohms.

I'm new to electronics, so it would be really helpful if someone could work out the formulae and the explanation.

Answer 1

I'll give you a partial explanation about how the circuit works, and let you work out the details from there.

Consider transistor TR1, and let's think about what happens around the time that it switches from the nonconducting state to the conducting state. Just before that switch occurs, C1 has been charging through a path that includes R1 and the B-E junction of TR2. This has a short time constant, mainly because R1 has a much lower value than that of R2 or R3. This means that C1 is pretty much fully charged -- to the supply voltage minus the B-E drop (about 0.65V) of TR2. See the diagram below.

^{simulate this circuit – Schematic created using CircuitLab}

Note that if you have a load resistor attached to the collector of TR1, it forms a voltage divider with R1, which reduces the maximum voltage on C1. The diagram below shows this situation, and the part in the dashed box on the left can be replaced with its Thévenin equivalent on the right, which better illustrates its effect on the capacitor.

^{simulate this circuit}

Next, TR1 switches on and pulls its collector to within a few hundred mV of ground. Since the voltage across C1 can't change (at least, not quickly), this means that its other end — the one connected to the base of TR2 — is pulled to a negative voltage that is nearly equal to the supply voltage. This reverse-biases the B-E junction of TR2, insuring that it is cut off.

^{simulate this circuit}

Now, C1 is discharging (charging in the other direction) through a path that includes TR1 and R3. Since this path has a much longer time constant, this takes a while, and nothing else happens until the base of TR2 reaches about 0.65 V, at which point TR2 starts to turn on. At this point, everything that I described above now starts to happen in the other half of the circuit.

In other words, right after TR1 switches on, the voltage at point A in the diagram above is about -5.15 V,¹ and slowly rises to +0.65 V.² If there's a load resistor as shown before, then the voltage at point A starts out a -2.15 V instead of -5.15 V. Clearly, it's going to take a lot less time to get to +0.65 V than in the original case.

So, from this, can you see why the presence of a load resistor, which affects the maximum voltage on the corresponding capacitor, would have an effect on the timing? Can you work out the equations and other details from that?

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¹ As @jonk points out in a comment, if the supply voltage were any higher than 6 V, this negative voltage could easily exceed the maximum reverse B-E voltage of the transistor, which would mess up this analysis by clamping the peak negative voltage.

² The relevant equation is the one for a capacitor changing its voltage from a start value to an ending value through a resistor, which is:

$$V(t) = (V_0 - V_{end}) e^{-\frac{t}{RC}} + V_{end}$$

The question we want to answer is if \$V_0\$, \$V_{end}\$, R and C are known, how long does it take for the capacitor to reach some intermediate voltage \$V_x\$? That requires solving the above equation for t, which isn't exactly straightforward.

$$V_x = (V_0 - V_{end}) e^{-\frac{t}{RC}} + V_{end}$$

$$V_x - V_{end} = (V_0 - V_{end}) e^{-\frac{t}{RC}}$$

$$\frac{V_x - V_{end}}{V_0 - V_{end}} = e^{-\frac{t}{RC}}$$

$$\ln\left(\frac{V_x - V_{end}}{V_0 - V_{end}}\right) = -\frac{t}{RC}$$

$$RC\ln\left(\frac{V_x - V_{end}}{V_0 - V_{end}}\right) = -t$$

$$-RC\ln\left(\frac{V_x - V_{end}}{V_0 - V_{end}}\right) = t$$

Since negating the logarithm of a fraction is equivalent to taking the logarithm of the fraction inverted, the final equation becomes

$$t = RC\ln\left(\frac{V_0 - V_{end}}{V_x - V_{end}}\right)$$

For the first case, without the load resistor, we have

$$t = 100k\Omega 1\mu F \ln\left(\frac{-5.15 V - 6 V}{+0.65 V - 6 V}\right) = 73.4 ms$$

For the second case, with a load resistor on either side, this becomes

$$t = 100k\Omega 1\mu F \ln\left(\frac{-2.15 V - 6 V}{+0.65 V - 6 V}\right) = 42.1 ms$$

Answer 2

The output transistor pulls the output to near 0V and the 1k collector resistor pulls the output up to 6V if there is no load. When the load is a 1k resistor then the output will go only as high as the voltage divider of the resistors making 3V.