# BJT Astable Multivibrator

I have the basic knowledge about multivibrators, but I did not understand the working principles of the circuit schematic provided down below.

I have done the basic research about how it works, but I could not understand this particular BJT Astable Multivibrator circuit. Could someone please explain why 470 ohm and 10K ohm resistors are used and why the outputs of the circuit are not the emitter legs of the transistor but the collectors of the transistors.

And the most important question is : I put LEDs to the outputs. Does that changes the duty cycle of the Multivibrator (or frequency)?

The emitters of NPN bipolars are often directly connected to ground for various reasons. For one thing, this makes the circuit analysis easy. The NPN transistor starts conducting when there is more than about 0.7 volts base-emitter voltage. At that point the base-emitter current starts to flow, causing also the collector-emitter current to flow. As the emitter is connected to ground, you can see that this threshold voltage of 0.7 volts is 0.7 volts from the ground.

Consider T2. When a transistor's base has at least 0.7 volts, then the collector is drawing current, causing the collector voltage to be near zero volts. At that point, the LED D2 is not light. The other state is that the transistor is not conducting: at that point the LED is light and there is about 2 volts at OUT2. C2 charges to 1.3 volts : the left plate will have 0.7 volts (the threshold voltage of T1) and the right plate has 2 volts.

Now, let's say that T2 starts conducting. OUT2 will drop from 2 volts to 0 volts. But there is still 1.3 volts charged into the capacitor. This means that the left plate of C2 suddenly drops to -1.3 volts, causing T1 to stop conducting. C2 starts to discharge, C1 starts to charge.... and round and round we go.

• Wow Sir, quickest answer ever! You are great at this! Thank you for the detailed answer! Commented Dec 30, 2014 at 20:36

This would be a good one to take in slow-motion:

1. The 10k resistors want to pull the bases to the positive supply, kinda like springs, but the caps are kinda like syrup - it'll get there, but not immediately. As soon as one gets high enough to turn on its transistor, the opposite one gets jerked low through the cap and forces its transistor off.
2. After the jerk low, the voltage starts to rise again, slowly, and eventually gets high enough to turn its transistor on. This jerks the first one low, and the process repeats.

The 470 resistors are required to pull up the opposite sides of the caps so that the jerk low actually does something. Using the spring analogy some more, they need to be really stiff compared to the 10k's so that we can ignore them during that part of the analysis. And that point is also convenient to tap off and feed something else.

In your case, you put an LED directly across the transistor so as to use the existing 470 pullup for double duty. It'll probably work, but it limits how far the voltage can rise at that point to the forward voltage of the LED, which limits how far the jerk can go. If the circuit still works with that little voltage swing, then you're probably good.

As for the frequency, it has to do with how long it takes the RC circuit to get from where it got jerked down to, to the base-emitter voltage of the transistors. This is not a simple RC cutoff, as in signal filtering, nor is it a trivial relation to an RC time-constant. You can either experiment with different values until you like it, or you can derive the math a couple of different ways and then solve that.