I'll sketch a diagram for you. I've just reformatted so no software on here yet... edit: edit 2: This one is better:
Great!!! Upon analysing the schematic and understanding it, I like it. Thanks a million. I owe you. I'll give it a try tomorrow. I hope I don't break anything, I'm down to my last MOSFET and few transistors. The shops only open on Monday again and I'm strapped for cash as well . Sputnik EDIT: Um... What would happen if +ve is greater than logic in the second diagram? Couldn't that be harmful to digital circuits? If the input is logic high and +ve is say 12v. 12v > 5v. So wouldn't current still flow? Forgive me if this is silly and stupid, I'd just would like to make sure and learn since I have never seen this sort of configuration before.
Vlogic will need to be lower than Vload for the second circuit to work. In that case, if in = Vlogic, Vbe = 0 so the transistor is turned off, no current can flow, so no damage.
There does seem to be a problem with your circuit. It won't work. After looking through an electronic book, I noticed that it is illustrated that the transistors need to be swapped. The NPN transistor must switch the high-side and the PNP transistor must switch the low-side. I don't know why it works this way only. Sputnik
that is a crude class-b amp (if I'm not mistaken) when the signal is high, it biases the top NPN on, pulling the voltage up, and when it is low, it sinks current from the bottom pnp, pulling the output low, As the current though the load increases, the voltage dropped over it would decrease the vBE of the transistor, starting to turn it back off, till it reached a balance point. . I think it would work as a voltage follower, however you would lose the vBE of the transistor. For saturating the transistor completely, and providing the full power to the load, you would need Vcc+1.2 (1.2 is a nominal vBEsat voltage) to fully turn it on. NPN goes on the bottom, because the base sinks current, and PNP goes on top because the base sources current.
The diagram you've seen isn't what we want for driving the MOSFET. If you were to use that schematic, you may as well not put those two transistors on at all. We're after the switching effect of the transistors not running them in their linear region where they'll dissipate lots of power. If you can't get my circuit to work, you've done something wrong
Paged a little further in the book to see if I could fine another diagram where the transistors were used as a push-pull configuration. Sputnik
That diagram uses 10V logic, which is enough to turn most MOSFETs fully on. Your MOSFET won't turn fully on with 5V logic, hence the PNP connected to Vload in my diagram. Swap the NPN and PNP round like in Fig 3.74, and the emitter voltage can't get any higher than Vlogic - 0.6V so you can't turn the FET.
AH YES!!! The collector of a NPN transistor will always be more than the emitter and vice versa for PNP. But I thought I'm using 12v to switch the MOSFET? Thanks for your help guys, Sputnik
I'm not sure if you dont understand what is happening in the 2nd diagram by SteveyG or not When the input is high, (pretty close to Vlogic), the first npn transistor has no base current, so it turns off, blocking the mosfet drive voltage(+ve), preventing damage to the logic circutry. The pull up resistor to +ve biases the lower NPN transistor on, by raising its base above its emitter, and turns off the upper PNP by raising the base voltage, close to the emitter voltage. That makes the mofets gate low, which turns it off. When the input is low, current flows from Vlogic through the first NPN transistor, pulls the base on second NPN low, turning it off, and lets current flow from the PNP, turning it on. pulling the mosfet up to +ve. which should be enough to bias it completely on. The first NPN transistor doesnt swith anything in the normal way that we think of it, it just blocks voltage, in a pretty clever way too, I havent seen it like that before, but that doesnt mean much,
