transistors
I need to do some switching and I want to turn off both +ve and -ve(GND)
I can use 1 NPN and 1 PNP transistor. But I'm wondering if I can use 2 NPN transistors only, I have some confusion about NPN. I've been reading for long time about transistors and now this is how I can use NPN.
But can I use NPN transistor like below schematic too ?
or only PNP can do this ?
However, if VB = 5v, and Vc is 5V, wouldn't that violate the condition that Vb > Vc? Ve < Vb checks out, because Ve is on the GND side of the circuit.
If \$V_B = 5\:\mathrm V\$, your transistor is probably on fire. Why? Because the base-emitter junction is a silicon diode, and current rises very rapidly after about 0.65V.
I think the mistake you are making is considering the voltages not at the transistor, but at the components attached to the transistor.
simulate this circuit – Schematic created using CircuitLab
\$V_B\$ is the voltage at the base of the transistor, not the voltage at the microcontroller output (\$V_{in}\$). \$V_C\$ is the voltage at the collector, not the power supply (\$V_{cc}\$). In this circuit, \$V_B\$ will be 0V (when off) or about 0.65V (when on). \$V_C\$ will be 5V (when off) or about 0.2V (when on). Remember that when current flows through a resistor, there will also be a voltage across that resistor, by Ohm's law.
I would use a relay for this. A relay provides electrical isolation between the controlling circuit and the controlled circuit. You would not need to know anything about the controlled circuit: whether either side is grounded, polarity, voltage, AC/DC, etc., as long as the relay contacts are rated for the voltage and current they will need to handle.
Best Answer
As far as the second circuit goes in the original question, using a NPN transistor as a "high-side" switch is generally not a good idea because you would have to bring the base voltage all the way up to Vdd to turn the transistor on, which would generally be a problem, for example, if Vdd is +12v and you are trying to drive it from a microcontroller output that can source only 3.3v or 5v.
Instead, the accepted way is to use a PNP transistor. Resistor R1 ties the base to Vdd and keeps the transistor off by default, so the microcontroller only needs to ground the base to turn it on. (Note this is the opposite of how the microcontroller usually operates; 0 means on and 1 means off). However you sill have the situation where Vdd can appear at the output pin of the microcontroller, which most can not tolerate above their supply voltage.
So a better scheme is to use both an NPN transistor and PNP transistor together, like this:
The PNP part of the circuit operates as before as a high-side switch, except now instead of the microcontroller directly grounding the base of the PNP transistor, instead it turns the NPN transistor on which in turns grounds the base of the PNP transistor. This isolates the Vdd voltage from the microcontroller.
Note this does invert the sense of the microcontroller again; so once again it is zero to turn the load ff, and 1 to turn the load on.