small amp

See what happens if, rather than simply leaving the base lead open, add another resistor, say 4.7K to 10K, from the base of the 2N3904 to ground. I think what is happening is that the collector-emitter leakage (with the base open) of the 2N3904 is sinking enough current from the base of the 2N2907 to cause what you're seeing. The 2N2907 can have a beta of 300, so it wouldn't take much, especially if it's a sensitive-coil relay.

A 4.7K to 10K from the base of the 2N2907 to the +12 rail wouldn't hurt, either.

Good Luck! Rich

HI all

I have made circut B at this scheme.

it somewhat works, and when there is current on base of transistor 1 (from the left) the output is 9 volts.. witch is great.

BUT, when there is no base on trans 1 , there is a output of 4 volts.. what causes this ?

it makes the relay hit when there is base current, but not release when the base current ends :(

Thanks for any help

Chris

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Chris,

If no base current means an open input, the floating base does not block the collector emitter current. Not reliably. The base should be tied to ground to make the transistor block. Like in all digital circuits an open input always should be avoided. If the current to the base of the transistor is disconnected by a switch or a plug you can introduce a pulldown resistor between de base of the first transistor and ground. it's value depends on the available input voltage. I guess 22k for the upper and 47k for the lower circuit may be a good starting point.

petrus bitbyter