Forums » General Electronics Discussion

Hello. I am learning about transistors and am using the tab electronics guide to electricity and electronics book. So far it is pretty good. I am an electrical/electronics teacher in an automotive trade school so I have a fair understanding of automotive electronics but am looking to learn more. so my question surrounds figure 6-2 on pg 156 there is a bi-polar transistor (NPN) with a beta of 100, and it has an Re of 1K Ohms, and 1.7V at the base, and an Rc of 10K Ohms with 30V at the collector you forward bias the base/emitter junction and drop .7V in the transistor so you effectively have 1V going through the 1K resistor which gives you 1mA of current this I understand, but the book states that the 1mA is the SUM of the base emitter current AND the emitter/collector current. so I have 2 questions surrounding this question #1, if you calculate the amperage flowing through Re by applying Ohms law to the base/collector circuit, how does the emitter/collector circuit get the credit for 90% of the current that is measured at RE? Question #2, they say in the book that out of the 1mA that is measured at Re, 990 micro amps come from the emitter/collector and 99 micro amps come from the base emitter. How do they apply math to get those numbers, it would seem to me that if you start at 1mA then they took the original 1mA and divided it by the beta, subtracted that amount from the original 1mA and that was the amount supplied by the emitter collector and then divided THAT number by 100 and that was the amount supplied by the emitter base. I tried that with different numbers If I had .023A calculated through the emitter/base at a beta of 100 then .02277A is from the emitter/collector and .0002277A is from the emitter base (for a total of .0229977A) does this sound correct to you?

Thanks for any help you can give here

actually, just question 1 as i have tried my math and it is solid. thanks again

if they still exhist you might check out a "ARRL" handbook