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Will this circuit work?

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Again someone asking the in a million different threads. Do you want a second/third opinion on each thread?
 
Word is a lousy software for schematic drawings. Use a proper software such as Eagle or PCB123.

In fact MS Word sucks even as a word processor...

thanks
aa
 
No that circuit will not work, there is no power anywhere in it. All wires lead to ground!

(Word as a cad program) :p
 
It looks like it will work. There is certainly some delay in the design and it's one of those clever circuits that you have to try and see for yourself.
We are talking about milliseconds delay and you need to see how long it takes to charge the 0.1u and transfer this rise to the base of Q2. Q2 is an emitter follower. It all depends if you have supply-voltage (and current) at the point labeled "to flash."
 
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I replied on one of his many other threads. It is from the net. It is powered from leakage of the hot shoe.
 
thanks sir



It looks like it will work. There is certainly some delay in the design and it's one of those clever circuits that you have to try and see for yourself.
We are talking about milliseconds delay and you need to see how long it takes to charge the 0.1u and transfer this rise to the base of Q2. Q2 is an emitter follower. It all depends if you have supply-voltage (and current) at the point labeled "to flash."
 
sir i want to delay current few more milisecond.
plzz tell me which value to change and how much ?
plzzz
tell me

It looks like it will work. There is certainly some delay in the design and it's one of those clever circuits that you have to try and see for yourself.
We are talking about milliseconds delay and you need to see how long it takes to charge the 0.1u and transfer this rise to the base of Q2. Q2 is an emitter follower. It all depends if you have supply-voltage (and current) at the point labeled "to flash."
 
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