Hello AAC forum

A few years back a simple transistor flasher PCB from Velleman

was put on the back of a pin


and powered with coin batteries: CR 2032.


But the circuit was designed for 9 volts so I rigged
the battery holder so it would take three 3 volt
coins in serial. Worked great.

But my coin stack battery holder rig gave out,

So I was thinking about refactoring the setup
to use one 3 volt coin as the battery holder is designed.
I have an extra MK102
PCB. But am not sure how to change the component
values to work with less voltage.

I think reducing R3 and R4 from 10k ohms to
330 ohms will do it.

Do I need to change the capacitors also?

Thanks.

Allen in Dallas

 

Unless you reduce R1 and R2 the LEDs would be much dimmer. Although reducing R3 and R4 would help slightly, they are much less important than RV1 and RV2. Reducing VR1 and VR2, however, would drastically reduce the oscillator period unless you have a corresponding increase in C1 and C2 to compensate. In summary, you would need to change everything except the transistors!

 

Only R1 and R2, need to be changed, don't forget that LEDs need approx 2V to work so on a 3V system that's a 1 V drop across the series resistor and transistors C/E junctions, so as the battery gets to say 2.5V it may stop working.

Why don't you use a 12V Type 23A battery instead?

 

Only R1 and R2, need to be changed

Reducing R1 and R2 to maintain the same LED current is fine, but RV1 and RV2 would also need reducing to provide enough base current to switch the transistors reliably.

 

A 3V battery soon drops to 2V.
Only a red LED is about 2V. Other colors are as much as 3.6V and the series resistor also needs some voltage.
A 12V battery voltage is so high that it causes the circuit to make more heat than make light. So use a 6V battery instead.