A 2N2700 is an N-MOSFET but you are showing a P-MOSFET.Back to inverted logic then, but it's a better alternative, I guess.
Which is it?
A 2N2700 is an N-MOSFET but you are showing a P-MOSFET.Back to inverted logic then, but it's a better alternative, I guess.
What the.... ?A 2N2700 is an N-MOSFET but you are showing a P-MOSFET.
Which is it?
I know... but it's an internal logic signal to a more sophisticated controller (see post #1). In fact, I'm thinking that using a simple opto could be simpler. But I'd also have to use a transistor and a resistor to drive the opto. So maybe I'll just stick to this design, judging from its smaller size and component count.Seems like a lot of work to drive a 3V signal at 2 mA.
Ahhhh... it's practically a totem-pole push-pull switch... As I mentioned before, I need to start thinking about those chips as more than just mosfet drivers. Very nice, thanks for the suggestion!Of course, you could replace the whole lot with one of these...
That's what I call a double invert circuit. I have posted it many times.shame on me
Thanks... I think I get the general principle now. One transistor is driving down the gate of the other one. Simple as that!Here is an example of using an NPN in a common base configuration especially for 8051 type open drain outputs:
Correctimundo! Because the base of Q1 is connected to the +5V supply, the collector is pulled down towards 4.3V (= +5V - 0.7V), or is allowed to float up to the +12V supply rail allowing Q2 to turn off. You can adjust the current that the microprocessor is required to sink by adjusting the values of the three resistors.Thanks... I think I get the general principle now. One transistor is driving down the gate of the other one. Simple as that!
And in your circuit, the MCU is draining current through Q1's emitter and not sourcing any.