Airsoft Forums UK

guess not....

Seeing as this is an AIRSOFT forum, not an electronics one...

what are you trying to work out ?

well, airsoft does have some form of electronics and mosfets are basically transistors...

and baz, trying to work out the emitter voltage which the I have no idea have to find out the collector voltage. I have the base voltage, and that's it. Has that fried your brain yet?

What other information do you have ? type of transistor NPN, PNP ?

npn with a current gain of 1000, Vcc is 6v. There is also a 5k ohm resistor on top of a 10k ohm resistor in series and a line in the middle going to the base of the transistor. Basically a voltage divider circuit. And after the emitter, there is a 10k ohm resistor there.

is the base voltage you gave 2V ?

have not gave

well, Vout is 4V definitely. And the required voltage to turn on the transistor is 0.7v. So i'm guessing base voltage is 4 - 0.7 which mean 3.3v... i don't even know if that's right.

what is the question ? what do you need to quantify to answer the question ?

here is a picture of the circuit.

http://i1172.photobucket.com/albums/r561/ak47frizzle/DSC00639_zps9d9d15b5.jpg

trying to find out emitter voltage.... been stuck on it for ages

If VCC is 6V, then the voltage at the base will be 4V initially due to the potential divider, falling to 2V as soon as the transistor switches on.

As soon as this happens, assuming there is no resistor in the collector leg, then the full 6V will appear across the 10K emitter leg resistor.

so voltage across the emitter resistor and therefore at the emitter w.r.t the negative rail is 6V as the transistor will be turned hard on (saturated) by 2-4V at VBE

sorry that is 2-4V is at base w.r.t negative rail. VBE will be 0.7V. However the answer you need is the emitter voltage which is 6V. Do you follow why ?

sorry not 2V, I mean 3V at base w.r.t negative rail.

I can see why but why doesn't the base voltage add on to 6? Because that what I have been stuck on. If Ve = Vc + VB, then if Vc = 6, then voltage would magically be created.... I have no idea why this is not the case.

if you consider the moment when the voltage is switched on at the supply, the potential divider of the 5k and the 10k will put an initial 4V at the base.

The transistor will conduct through its base-emitter (from the 6V rail through the 5k resistor across the b/e and to the negative rail.