# Transistor with emitter resistor calculation

I can't seem to get over this but this seems to be the basics in electronics. I have an NPN darlington transistor TIP122 and have built to simple test circuit around it with Rb = 150 ohm, Rc = 80 ohm and Re = 6.8 ohm. My base supply Vbb = 3.3V and collector supply Vcc = 20V.

I have used the equation from this site as mentioned for emitter biasing.

Ie = (Vbb - Vbe)/(Rb/beta + Re)

However I am getting different values theoretically and practically.

With Re = 6.8 I get theoretical Ie = 273mA and practical Ie = 222mA then With Re = 6.8||6.8 I get theoretical Ie = 535mA and practical Ie = 228ma.

I don't understand why could this be happening, how does the emitter resistor work in the transistor circuit... can you help me out?

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The simplified equations only work when you observe some basic assumptions. The most important being that the transistor is operating in it's linear region.

Leave out the transitor for a moment. How much current would you get through Rc if it was connected directly between the 20V supply and ground?

Then add in Re. You should end up with 20/(80 + 6.8) = 230 mA. This will be the maximum current that can possibly flow through the transistor. Given that there will be a small voltage drop across the transistor and errors from the resistor tolerance, that is very close to the measured value of 222mA.

Depending on what you are trying to achieve you may need to reduce the collector resistor or reduce the targeted collector current.

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That would be 20 / 80 and would mean about 250mA of current, right? –  Kevin Boyd Apr 2 '10 at 3:22
Yep, I've added a bit more to the answer that should help explain it. –  Clint Lawrence Apr 2 '10 at 3:57
Actually I was trying to test out a constant current source however I found that my assumptions were not supported by the practical readings that I later took. –  Kevin Boyd Apr 2 '10 at 4:21