Can I control 2 transistors w 1 output from a SR ?

[DriftZ]

Meaning,

1 output from a ' 595 Shift register connected to the base of 2 transistors;

595 pin 1_______B-C-E_____Load1

............      |____B-C-E_____Load2

Will this work ?

or.. how can I calculate to make it work ?

greetz

[Duggle]

hi,

each base should have it's own current limiting resistor (I assume youre driving BC547 or similar) value 1k to 10k typical.

regards

[DriftZ]

Hmmm.... actually the situation is more like this:

595 pin 1(1)_______B-C-E_____Load1___R___pin 2(0)

....................|................|___Load2___R___pin 3(0)

....................|____B-C-E_____Load3___R___pin 4(0)

......................................|___Load4___R___pin 5(0)

B-C-E = BC639 (1A)

Load 1+2 = .7A (a bunch of leds infact :) )

Load 3+4 = .7A (yes, more  leds infact :) )

Do I still need a resistor at the base ?

Thanks

(I really wish I studied electronics  :-/ 8) )

[TK.]

Yes, you definitely want to add a resistor to the base to limit the I_BE current. I_CE doesn't really matter in this case.

You can possibly connect 20 or more transistors to every pin

Best Regards, Thorsten.

[DriftZ]

Okay, many thanks

Is there any way to calculate what value of resistor I should use ?

greetz

[TK.]

Since the transistors are not used as an NF/HF amplifier, but as a simple driver (in other words: an overloaded amplifier) here, no special calculations are required. Just take 1k and it will work

Best Regards, Thorsten.