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Topic: Transistor issue (Read 5576 times) previous topic - next topic


with the 240 ohm resistor we are driving at about 4.3V/240 = ~18ma

That's not how that chart works.

You take your desired Ic, divide it by the 10x or 50x curve to get the Ib (100ma - 20ma). From that, you calculate your base resistor for 5v drive -> 43ohm - 200ohm.

Alternatively, your Ib with a 220ohm base resistor is 18ma, yielding a Ic of 18ma * 10x - 18ma * 50x, or 180ma to 800ma.

For switching applications, you should simply forget about hFE.


Dec 11, 2012, 09:40 pm Last Edit: Dec 11, 2012, 09:57 pm by BillO Reason: 1

For switching applications, you should simply forget about hFE.

Look, it's not that cut an dried.  They are only showing two lines, IC/IB  = 10 and 50.  There a an infinite number of possible lines you could draw in there, and they behave in a fully predictable manner.  Just because they only show 2 lines does not mean that other lines do not exist.  Or that if you wander off the space between those two lines by a tiny bit, the transistor will behave in a hugely different fashion.  A line for a ratio of 55 would be just a tiny bit higher than the line showing the ratio of 50.

My calculations are right.  The transistor, will be in saturation, VCE will be approximately 0.055V with IC @ 1amp, and it will be at less than 10 degrees above ambient temperature.

You are obviously set on this, so I will not try to convince you further.  This argument is just starting to get inane.  If you want to prove me wrong, get a ZTX851, drive the base from 5V through a 240 ohm resistor, allow 1 amp IC (5V through 5 ohms, perhaps) and take some measurements.  If it blows up, or even get's luke-warm, I'll play for the transistor and send you $5 for your trouble. Okay?

Other than that, let's agree to disagree.
Facts just don't care if you ignore them.

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