Hi everyone, I am currently learning about TIP125 PNP transistor control with arduino and have a quick question which is puzzling me at the moment.
So I know that NPN transistors can generally be connected directly to an arduino output pin like the picture below, as it is easy enough to control with 5V and 0V from the arduino and the small amount of current.
However PNP transistors require an NPN transistor as well if the source voltage is above 5V. Like the picture below:
In terms of operation I assume when the NPN transistor is on there is 0V at the base of the PNP transistor and when the NPN transistor is off there is 10.8V at the base of the PNP transistor.
So my question is why is it that there can be 10.8V (after voltage drop across transistor of course) at the base of the PNP TIP125 when the datasheet says the maximum Base-Emitter voltage can only be a potential difference of 5V? Also whatever the answer, can the same answer be applied to MOSFETS that also only need 5V?
Alexisa:
In terms of operation I assume when the NPN transistor is on there is 0V at the base of the PNP transistor and when the NPN transistor is off there is 10.8V at the base of the PNP transistor.
That's a wrong assumption There is still a resistor between the base and the NPN collector and that will drop most of the voltage. The base-emitter voltage of a transistor is almost fixed (around 0,6 - 0,7V) like a diode. That 10,8V is when the PNP is ON. The TIP125 is a darlington so it's especially two PNP's together. So 2 x 0,6V lower then 12V makes 10,8V. When it's off it will almost be 12V on the base of the PNP. Most circuits like to use a little pull up op the PNP base to make sure it's fully off and not turned on a tiny bit because of leak current.
Alexisa:
So my question is why is it that there can be 10.8V (after voltage drop across transistor of course) at the base of the PNP TIP125 when the datasheet says the maximum Base-Emitter voltage can only be a potential difference of 5V?
Although that voltage is here when it's on, that observation is a problem with understanding voltage Voltage is ALWAYS measure across something. When we talk about "the voltage at the base" we implicitly assume the other side to be at GND. But we are free to choose where that GND is! So yes, the voltage at the base is 10,8V in reference to GND. But it's emitter is connected at the 12V line (again, with reference to GND) so the voltage across the base emitter is 10,8V - 12V = -1,2V. That's why the datasheet also mentions a negative voltage
In real life you also have that with height. If you call the floor 0 (or the ground ) and you place a block of 5cm on top of it you say the block is 5cm. But if you place the same block on top of a 1m high table you still say the block is just 5cm (the distance across the block). Although the total height of the block is at 105cm
Alexisa:
Also whatever the answer, can the same answer be applied to MOSFETS that also only need 5V?
Mosfets are a bit different. Where BJT's (what you refer to as transistor, but a MSOFET is also a transistor ) are current controlled devices, MOSFETs are voltage controlled devices. So the gate source voltage isn't fixed. You need to apply a certain amount of voltage across the gate and source to turn it on. Without load that's the threshold voltage. And if you have a load and you want to fully turn it on you probably need to apply more and the graphs will show you want voltage at least you need to apply. And yes, they also have a max voltage across gate and source. But again, that's from gate to source (gate referenced to source) and NOT referenced to GND.
Yet again lets quash the myth that the MOSFET threshold voltage is where it turns on, it isn't.
Its the voltage below which it is considered fully off...
A MOSFET is on when the channel is fully formed. That requires 3 times the threshold voltage on
the gate for typical power devices.
Ok thanks for clearing up some of my knowledge. So what is the significance of the -5V max rating from the datasheet for the base-emitter voltage? Because what I'm gathering there's going to effectively be 0v or 12v on the base and that the base-emitter voltage is effectively always -1.2v for a darlington transistor?
In regards to an npn(2n2222 for example), can I apply something over 5V from base through to ground with an appropriate resistor in place (since a transistor is current driven rather than voltage driven)?
Also interesting about mosfets needing 3x the threshold voltage, is this fixed? As in always 3x or dependant on the load? Roughly how would you calculate that?
Thanks for the help as well, definitely makes things clearer
Alexisa:
Because what I'm gathering there's going to effectively be 0v or 12v on the base
What the reference of the base to circuit ground is doesn't matter. The transistor can only see voltages between it's own legs.
Just like a diode, if you put current through it (in forward) you get a more or less fixed voltage across it. But it doesn't stop you from connecting it to a voltage and go over that voltage. Same for the transistor, as long as you drive the base with current it will have that 1,2V across the base and emitter. But you can connect a voltage to it of 5V and kill it.
Alexisa:
and that the base-emitter voltage is effectively always -1.2v for a darlington transistor?
If you drive it with a current like you should then yes, for a darlington it's always going to be around (-)1,2V for the base emitter voltage.
Alexisa:
In regards to an npn, can I apply something over 5V from base through to ground?
On most, no. But you shouldn't in the first place. Drive it with a current. That's why you connect the base (in most cases) with a resistor. A resistor is a crude but fine current source
Alexisa:
Also interesting about mosfets needing 3x the threshold voltage, is this fixed? As in always 3x or dependant on the load? Roughly how would you calculate that?
Bit of a discussion but I say depends on the load. But more voltage doesn't hurt (as long as you stay under the max gate source voltage and you want to switch). Look at the datasheet graphs to see what VGS you need for what ID to keep the mosfet in linear region.
Alexisa:
Ok thanks for clearing up some of my knowledge. So what is the significance of the -5V max rating from the datasheet for the base-emitter voltage?
The maximum reverse bias voltage the base-emitter junction can withstand. If you use the transistor properly, it's highly unlikely you'll even need to be concerned about that.
If you do reverse bias the emitter junction (not usually possible in a switching circuit), its a very
easy way to trash the transistor. Normally this is only an issue with analog circuitry where the emitter
isn't connected to the relevant rail. You sometimes see a reverse diode between E and B terminals
in a circuit for this reason.
One last thing in regards to mosfets, I am looking at this low vgs p-channel mosfet I found (NDP6020P) https://cdn.sparkfun.com/datasheets/Components/General%20IC/NDP6020P.pdf
It states that the maximum Vgs is +/- 8V, does this mean if I have a 12v supply to the source and use arduino's 5V on the gate, then it will switch on. As the gate - source = 5V-12V = -7V which is just under the max rating. I also notice is says max Vgs is + or -, but the VgsTH is distinctly only negative, why is that?
That's correct. But how do you plan to turn it off then? If you make the Arduino pin LOW the gate source will be 12V - 0V = 12V => killed mosfet.
For a P-channel you need to pull the gate lower then the source to turn it on. But noting stops you from applying a positive voltage (instead of 0V) to turn it off. So in both ways, negative to turn it on and positive to turn it off you may not exceed +/-8V.
Alexisa:
I also notice is says max Vgs is + or -, but the VgsTH is distinctly only negative, why is that?
Because the threshold voltage is a specific value, which is of course negative for a p-channel just
like it is positive for an n-channel is its Vgs (voltage of gate relative to source).
The voltage limit for the gate is the breakdown voltage for the gate oxide layer, which is only a
few nanometres thick, and being a breakdown voltage it is polarity independent, its due to
the properties of silicon dioxide.
There is still a resistor between the base and the NPN collector and that will drop most of the voltage. The base-emitter voltage of a transistor is almost fixed (around 0,6 - 0,7V) like a diode. That 10,8V is when the PNP is ON. The TIP125 is a darlington so it's especially two PNP's together. So 2 x 0,6V lower then 12V makes 10,8V. When it's off it will almost be 12V on the base of the PNP. Most circuits like to use a little pull up op the PNP base to make sure it's fully off and not turned on a tiny bit because of leak current.
