WAM - Question on ch #9

RonM

I just finished the WAM book. Having no background in electronics did cause some pain at times but I got help here and from other books.·On to my own projects...

But I do have a question on chapter 9 (transistors). The tutorials show 2 100k ohm resistors in parallel between the IO pin and the transistor collector. Can someone tell me how (if I was designing this circuit) I would know that I need 2 100k ohm resistors in parallel here? The chapter starts with a 10k analog pot and replaces it with a digital pot and uses the sames resistors for both.

Again, if I was designing this circuit, how would I determine that these resistors were necessary? Reasoning behind it?


Thanks!

Mike Green

First of all, you don't need two 100K resistors in parallel. They're used because they're on hand (in the kit). What you need is about a 50K resistor which this gives you (put two identical resistors in parallel and you get 1/2 their value ... there's a general formula for parallel resistors ... Rt = 1 / (1 / R1) + (1 / R2) + ... + (1 / Rn)).

The idea is that, if the slider on the pot is connected to +5V, the pot is effectively not there. You have 5V going through the 50K resistor into the base of the transistor and out the emitter to ground (it's slightly more complicated than that, but you can ignore the fine print). The base-emitter junction of the transistor acts like a diode and drops about 0.6V across it leaving 4.3V across the resistor. Ohms law says E = IR or I = E/R. That's I = 4.3 / 50000 = 86 microamps. The transistor used has a current gain of about 100, maybe as high as 300 which yields a collector current of about 8.6 milliamps, maybe as much as 25ma. This won't burn out the LED, yet will provide a reasonable brightness and the brightness (and collector current) can be reduced to zero by changing the setting of the pot (from +5V towards zero).

As you can gather, you should know the formulas for series resistors (Rt = R1 + R2 + ... + Rn) and parallel resistors, for Ohm's law, and for power (P = EI).

Post Edited (Mike Green) : 4/1/2007 9:55:07 PM GMT

RonM

Mike,

Thanks for the response but I'm afraid your answer leaves me with another question. I'm teaching myself so forgive the apparently simple questions.

Where did you get that the drop off was 0.6V? I'm looking at the AD5220 spec and do not see a forward voltage note anywhere.


·

Mike Green

The voltage drop (0.6V) is the drop across any silicon diode which includes the base-emitter junction of a silicon junction transistor (the 2N3904 is a typical one).

I'm not actually referring to the version of the circuit with the digital pot, but digital pots are not silicon junction transistors. They're made from CMOS (Complementary metal oxide semiconductor) FETs (Field effect transistors) which are very different.

If you want to find out about these terms, look them up on the Wikipedia. Do an internet search with "wiki" and the term you want like "wiki cmos".