Astable Multivibrator Circuit - Page 1
Forum section: electronics / beginners

author: skinny
author_rank: Contributor
timestamp: December 22, 2018, 03:31:41 am
content: Hi Everyone,I was just about to put together a little lesson to talk about a simple astable multivibrator circuit that I am sure many of your have seen. The best example I've found of the circuit can be seen in the simulation found here:http://www.falstad.com/circuit/e-multivib-a.htmlEvery time I've seen this circuit it's been explained to me that when one transistor charges up to the threshold voltage for the junction, current begins to flow from collector to emitter causing its corresponding cap to discharge. This flow kills the voltage at the base of the opposite transistor and the discharged cap now starts to charge in reverse to build the voltage back up at the base of that opposite transistor. What I never knew until today was that the voltage at that base does not go to zero volts when the switch occurs. It goes well into the negative voltage range (as seen in the simulation). I experienced this while bouncing around the circuit with my oscope.For the life of me I cannot convince myself as to why the voltage would go into the negative. My formal electronics training was not so good and I feel like now is one of those times it's beginning to show. I've tried simplifying things experimentally to root out an explanation but I've had no luck so far. Thanks for any help you can provide.-Skinny

author: jmelson
author_rank: Super Contributor
timestamp: December 22, 2018, 04:33:52 am
content: Quote from: skinny on December 22, 2018, 03:31:41 amThe capacitor is charged up to ~ 4.3 V, and when the collector goes close to ground on the side that just switches on, the capacitor side connected to the base is now driven to about -4.3 V for a tiny moment. That gets that transistor to turn off.Jon

author: skinny
author_rank: Contributor
timestamp: December 22, 2018, 04:41:51 am
content: I follow you. But why -4.3V? Why not zero?

author: IanB
author_rank: Super Contributor
timestamp: December 22, 2018, 04:55:36 am
content: Quote from: skinny on December 22, 2018, 04:41:51 amThink of a capacitor as similar to a battery. It generates a voltage between its terminals. So if you lower the voltage at the (+) terminal to zero, then the voltage on the (-) terminal will be less than zero by whatever the voltage on the capacitor is. If the capacitor is charged to 4.3 V, then when the (+) side is zero the (-) side will be 4.3 V lower, or -4.3 V.

author: spec
author_rank: Frequent Contributor
timestamp: December 22, 2018, 05:44:19 am
content: Quote from: IanB on December 22, 2018, 04:55:36 amGreat description.Just to add some supplementary info, capacitors are actually used as batteries to power memory, clocks, and even some power devices.

author: skinny
author_rank: Contributor
timestamp: December 22, 2018, 05:10:43 pm
content: Thanks for the help guys. I set up a little mini experiment to prove what you guys were saying. I physically manipulated the ground on both sides of the capacitor. I never thought about looking at it from the perspective of the ground reference changing. It makes perfect sense. Thanks again for your help.