mybuickskill6979 said:
okay i have a question about the mathematics.
the VCE Grows or decays in relation to the current supplied to it right? and Ib is very much decided by the VCE and ICE correct? i'm looking at the first drawing by len and the sheet for bc547b!
Vce in the active region is the sum of voltage across the resistor (which = Ic * R1) and the voltage across the LED (assumed to be 3.6 V @ 40 mA) subtracted from the supply voltage, ie. Vcc - Vled - Ic * R1.
In other words, you add the resistor and LED voltages together and subtract them from Vcc.
mybuickskill6979 said:
whats the mathematics to from .250v @100mA With a Ib of 5mA, and a Vce of 90mV@10mA with a IB= .5mA
This is Vce sat at a collector current of 10 mA and a base current of 0.5 mA (not 5 mA). A typical transistor will have a Vce sat of 90 mV under these conditions. But the worst case will have a Vce sat of 250 mV.
What this means is if you were to buy say 1000 of these transistors and measured every one at an Ic = 10 mA and Ib = 0.5 mA, the Vce sat would be spread around 90 mV; the maximum could be as much as 250 mV.
So Vce sat for one transistor may be 85 mV, another one may be 134 mV, another one may be 79 mV, etc. The average will be 90 mV.
mybuickskill6979 said:
to a Ib of 4A and a VCE of 80mV @40mA i know where the 40mA comes from but how about the 80mV?
As Audio said, you mean 4 mA, not 4A.
I took this from Figure 4 of the data sheet. This graph shows that at a collector current of 40 mA, Vce sat will be about 80 mV and Vbe sat about 0.8V.
Note that it is a log/log scale, not a linear one.