Transistor DC Parameters

Transistor DC Parameters

There are some important equations we need to look at first. Recall that Kirchoffs Current Law (KCL) states that the sum of all currents entering a node (a point) must equal the sum of all currents leaving the node. By taking a look at Fig 3 we can see then that for both the NPN and PNP transistors:

Ie = Ic + Ib (1) i.e. Current flowing into the transistor (Ic and Ib) equals current flowing out of the transistor (Ie) for the NPN, and Current in (Ie) equals current out (Ic and Ib) for the PNP.

There is a parameter called b (Beta) for every transistor, which is a constant. The value of b for transistors is normally between 50 – 500. Equation 2 states that the collector current is b times bigger than the base current. Hence b is simply a ratio between collector and base current. Recall that the base current is relatively small and the collector current is relatively large.

Ic = b.Ib (2).

For a transistor with a b =100 and Ic=1mA, then from equation 2, Ib = 10uA. Run through this in your head to make sure…

We can now substitute equation 2 into equation 1: The highlight shows the substitution

Ie = b.Ib + Ib, which simplifies to:

Ie = (b+1)Ib. (3)

We can now substitute equation (2) into equation (3) to obtain:

Ib = Ic/b (equation 2 rearranged)

Ie = (b+1).Ic/b , and rearrange to obtain:

Ic=b/(b+1).Ie

We now define a new parameter a (alpha) where

a=b/(b+1) (4)

Hence:

Ic = a.Ie. (5)

And that's it. I highly recommend you go through the mathematics yourself and verify every step that I have done. Only after you do this will you fully understand.

In summary you should definitely try to remember the first two following equations as they crop up all the time. It's also handy to remember the third one.

Ic = b.Ib
Ie = (b+1)Ib
a = b/b+1

E.g. For the transistor with b = 100, a = 100/(100+1) = 0.99
Hence from equation 5 you can see that Ic » Ie. This is true for all transistors with high b. Now take a look at the NPN transistor in Figure 3 again. The reason Ic is only approximately equal to Ie is because of the small base current that adds in to make Ie just a little bigger.

ELABORADO POR: 

NERWIN MORA 

C.I 17557095 

EES 

SECCION 1