36V astable multivibrator

Thread Starter

betchern0t

Joined Dec 29, 2018
3
Hi I want to build a wig wag (us slang)/astable multivibrator (electro geek slang) to flash a 36 volt led tail light. The light has two sets of leds and designed to take 36v. I have been looking at the standard two transistor circuit


however most of these diagrammes are specced to 5v or 9v. Looking at the components - transistors especially on the activation voltage on the base pin - are limited at well below 36v. I am just wondering what I need to do to get this working with a 36v supply?

Is there a better approach for this voltage?

many thanks in advance...

Paul
 

dl324

Joined Mar 30, 2015
11,279
Welcome to AAC!

What is the battery voltage? Are you opened to other solutions, such as a 555 timer or other integrated astable?
 

Thread Starter

betchern0t

Joined Dec 29, 2018
3
Hi Dennis, the battery voltage is 36v. This is an eBike. I have no problem with a 555 version however the 555 I believe requires a 5v supply. Ideally I didn't want to provide multiple voltage supplies.

Cheers Paul
 

dl324

Joined Mar 30, 2015
11,279
A 555 timer will operate from around 5-15V. It's current demand is low enough that a zener could be used for its supply. The output can switch a MOSFET that drives the LEDs.
 

dl324

Joined Mar 30, 2015
11,279
what does that look like?
Something like this:
upload_2018-12-30_17-38-43.png
D1 = 12V 1/2W zener, Q1 an N channel MOSFET that can handle the LED current. R3=820 ohms, 1W.

High side switching would be a little more complicated.

Since this is battery operated, you might opt to use a 12V regulator to save some power dissipation. Or decrease the zener current; voltage regulation shouldn't be much of an issue.
 

AnalogKid

Joined Aug 1, 2013
8,536
To your original question - yes, that circuit will work just fine as long as you cover the details.

1. Use transistors rated for at least 60 V.

2. Add a small signal diode (1N4148, 1N914, etc.) across each base-emitter junction with the diode's anode to GND and cathode to the base. This will prevent a charged up coupling capacitor (C1, C2) from driving the base below the transistor's reverse voltage rating when the circuit changes state. Note that the diode also will make the circuit oscillate faster than the equations indicate.

ak
 
Last edited:

LesJones

Joined Jan 8, 2017
2,687
I would modify the circuit in a slightly different way. When Q1 starts to conduct the voltage at the junction of C1 and R2 will try to go down to close to -36 volts. With the diode clamping the voltage to about -0.7 volts the capacitor very rapidly until the voltage on the negative end reaches -0.7 volts. The time taken for the negative end of the capacitor to go from -0.7 volts to +0.7 volts (Which is when Q2 would switch on.) is the only time the time constant would be effected by R2. I would add the diode between the base of Q2 and the junction of C1 and R2 wth the cathode connected to the base. (The same would be done for the other side of the multivibrator.) This way the time constant should be closer to the calculated value. I don't thing the small reverse leakage current through the diode would harm the base emitter junction of the transistor.

Les.
 

DickCappels

Joined Aug 21, 2008
6,533
If you really want to use the two transistor multivibrator, be kind to the transistors by preventing reverse break-down of the emitter-base junctions. This can be done by putting a small signal (1N916) is series with the base to block reverse voltages.
 

AnalogKid

Joined Aug 1, 2013
8,536
Les and Dick make the same excellent point. I would expand on it by using a zener diode in series with the base rather than a signal diode. In this way the circuit will use a much larger voltage swing across the capacitors, decreasing their size significantly.

Also, I think you still need a diode in parallel with the base-emitter. When the base and a diode in series with it are pulled below GND by the capacitor, you have no way of guaranteeing that the majority of the reverse voltage appears across the signal diode instead of the b-e junction. So, two additional diodes per transistor, one in parallel with the base-emitter for protection and one in series with the base to decrease the size of the timing capacitors.

ak
 

crutschow

Joined Mar 14, 2008
25,265
Also, I think you still need a diode in parallel with the base-emitter. When the base and a diode in series with it are pulled below GND by the capacitor, you have no way of guaranteeing that the majority of the reverse voltage appears across the signal diode instead of the b-e junction.
That's likely not a real problem.
At the very low reverse leakage current of a typical small diode, such as the 1N4148, even if the reverse breakdown voltage across the B-E junction is esceeded, that current is too low to cause any damage to the B-E junction.
The junction will just Zener at that voltage and low current with no problem.
So all you need is a diode is series with each base.
 

AnalogKid

Joined Aug 1, 2013
8,536
A fair point, and I agree. Still, I think this is one of those things that someone without training or experience could mis-apply by remembering the conclusion without the constraints.

ak

ps. I *know* you'll be at the game to cheer on my Buckeyes. Where are your seats?
 
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