EEVblog Electronics Community Forum
Electronics => Beginners => Topic started by: tony3d on October 08, 2013, 01:51:24 pm
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I'm on experiment 10 in the Make Electronics book. I understand what's going on with the transistor being used as a switch, but I'm confused by the mA readings I'm getting. There are suppose to be 3 resistors in the circuit, I'm using 4 (two in series a 217, and 463 ohm) to get the required 680 ohms. There is then another 9,870ohm, and a 217 ohm. for a total resistance of 10767 ohms. Using Ohms law, that would come out to about .0011 amps or 1.11 mA's correct? My circuit actually is drawing about 11.06 mA's. Is that because the transistor is actually amplifying the current? This is all based on 12 volt input.
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Draw the schematic (or just scan it from the book) and we will be able to tell you.
Don't forget to draw in the ammeter and the values of the resistors you are using.
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You've counted the resistor on the base in your calculations. It's not actually part of the current path. What you should be concerned with are the resistors from 12V to the collector, and the one from the emitter to the load. If you add those two, you get 860 ohms, and at 12V that's 13.9 mA, which is close enough to the 11 that you're seeing.
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Draw the schematic (or just scan it from the book) and we will be able to tell you.
Don't forget to draw in the ammeter and the values of the resistors you are using.
Here you go.
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You've counted the resistor on the base in your calculations. It's not actually part of the current path. What you should be concerned with are the resistors from 12V to the collector, and the one from the emitter to the load. If you add those two, you get 860 ohms, and at 12V that's 13.9 mA, which is close enough to the 11 that you're seeing.
Ok, I see. So I need to be looking at the two 217's, and the 463. That's 897. That gives me 13.38 mA's. So, this is because the voltage at the base just opens the path through the other 3 resistors connecting the collector with the emitter, and the 9,870 ohm resistor is just there to control how much current the base gets? So why am I getting a difference of 2.32 mA's? Is it because of the LED?
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You've counted the resistor on the base in your calculations. It's not actually part of the current path. What you should be concerned with are the resistors from 12V to the collector, and the one from the emitter to the load. If you add those two, you get 860 ohms, and at 12V that's 13.9 mA, which is close enough to the 11 that you're seeing.
Ok, I see. So I need to be looking at the two 217, and the 463. That's 897. That gives me 13.38 mA's. So, this is because the voltage at the base just opens the path through the other 3 resistors connecting the collector with the emitter, and the 9,870 ohm resistor is just there to control how much current the base gets?
Exactly.
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You've counted the resistor on the base in your calculations. It's not actually part of the current path. What you should be concerned with are the resistors from 12V to the collector, and the one from the emitter to the load. If you add those two, you get 860 ohms, and at 12V that's 13.9 mA, which is close enough to the 11 that you're seeing.
Ok, I see. So I need to be looking at the two 217, and the 463. That's 897. That gives me 13.38 mA's. So, this is because the voltage at the base just opens the path through the other 3 resistors connecting the collector with the emitter, and the 9,870 ohm resistor is just there to control how much current the base gets?
Exactly.
Wow! That is so cool.
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You've counted the resistor on the base in your calculations. It's not actually part of the current path. What you should be concerned with are the resistors from 12V to the collector, and the one from the emitter to the load. If you add those two, you get 860 ohms, and at 12V that's 13.9 mA, which is close enough to the 11 that you're seeing.
Ok, I see. So I need to be looking at the two 217, and the 463. That's 897. That gives me 13.38 mA's. So, this is because the voltage at the base just opens the path through the other 3 resistors connecting the collector with the emitter, and the 9,870 ohm resistor is just there to control how much current the base gets?
Exactly.
Wow! That is so cool.
If you think about electricity as water flowing through a pipe, a transistor is basically a valve. The more current you apply to the base, the more 'open' the valve is. If you put a larger resistor there, you should see the current go down. I think in the next experiment, you replace that resistor with your finger. The resistance across your skin is several megaohms. See how much current you get out with that big a resistance going to the base.