The signal required to control the PNP transistor can also be reversed from NPN transistors
The calculations for base current along with the base resistor are identical to those outlined in Component seven for NPN transistors except the polarities are reversed.
One additional factor you need to be careful with PNP higher facet switches may be the voltage utilised to drive the load. Commonly it really is very best to make use of exactly the same voltage to drive the load which is utilised to power the microcontroller. Take into account the subsequent.
Suppose the load voltage is +12V along with the microcontroller is operating at 5 volts. Ignore R2. R2 would typically possess a worth higher adequate to get little impact and ignoring it tends to make the calculations that follow easier. Assume that P0 is higher, at 5V, and R1 is one thousand ohms. Base current could be calculated by
Ib = (Vcc – Vp0 – Vbesat)/R1 =( 12 – 5 – .seven)/1000 = .0063A = six.3ma
Except if the transistor has exceptionally affordable gain, it’ll be turned on despite the fact that with P0 higher it must be turned off. Even worse yet, the microcontroller pin P0 is going to be viewing extra than 5V, which greatly exceeds the typical limitation of the microcontroller supply voltage additionally 0.3V. The microcontroller would most likely be broken in this particular scenario. Component 12 discusses techniques of operating a PNP transistor when the load must be pushed with a greater voltage than the microcontroller.
In the event the transistor is working, then you may have gotten a worth of about 0.5 – 0.seven volts for 2 of the 6 checks. The remainder of the checks should certainly have said that there is infinite resistance among the legs tested. When you get extra than two beneficial results, or you receive much less than 2 beneficial results, then the transistor is fried.
What is extra, the 2 checks that return a beneficial outcome may have a normal leg. ie: should you check a BC550 transistor, you may obtain a outcome for testing legs 2 to 1, and in addition 2 to three.
The normal leg may be the base of the transistor.
In the event the normal leg conducts with the beneficial lead of the multimeter linked to it, then it really is a NPN assortment.
In the event the normal leg conducts with the damaging lead of the multimeter linked to it, then it really is a PNP assortment.
So for our BC550, the normal leg is leg 2, and due to the fact the red probe was on Leg 2 when it performed, then it really is an NPN transistor.
So how would you determine which legs are the collector and emitter? This really is a bit trickier, but still not that tough. You are going to need the remainder of the components talked about before. You might want to build this circuit:
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