Don't understand this sch. Please help.

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vlad777

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Hi.

This is a car electro flasher and I want to figure out how it works.
I am not sure about transistors and even if they are transistors?

I know this is much to ask but can you tell me did I get the sch right?
I think I made a mistake because I don't see how C 47uf gets charged and
also first transistor configuration makes no sense .


Many thanks.


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I suspect that the wire that you have shown as a link, from pin 49, is actually a current sense resistor.

I also think that the white components that you have shown as capacitors are actually resistors fitted upside down. Most assembly standards allow for components upside down, or even on their sides, so there is no reason to suppose that all resistors would be the right way up.
 
I suspect that the wire that you have shown as a link, from pin 49, is actually a current sense resistor.

Yes the wire is visible in the picture. I thought it was a fuse.


Unfortunately I don't have it any more. When I measured one of them I got no reading so I jumped to a conclusion they caps.
 
Why do relays in these flashers have resistor instead of snubber diode?
How does 100uF cap discharge?

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The resistor is used so that the relays turn off faster than if a snubber diode is used. If the resistor has 10 times the resistance of the coil, then the voltage surge is limited to 120 V so a suitable transistor can be chosen. The current will stop very quickly. With a diode, the current could carry on of a long time compared to the fast flash rate of a flasher if there is a blown bulb.

The 100 uF capacitor charges and discharges through the 2k resistor. It charges when Q1 is on, so the negative end is low. The relay is on, so the top of the 2k is high. When the Q1 turns off, the negative end goes high, and the relay turns off so the flasher bulbs pull the top of the 2k resistor low, and that discharges the 100 uF cap.
 
Just a couple of points to clarify. The circuit is not actually a multivibrator as such but a high-gain DC amplifier with positive feedback.
The electrolytic is actually discharged via the diode across the 1k resistor. This produces a quick discharge as the electro will have about 12v across it.
 
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