Need Help Please

[Tony3dd]

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.

Found put I was using the 10K resistor which is not in the current flow. All's good.

[Mr Eastwood]

hi,  I'm not very good at analogue stuff but I think 12 /  (680 + 217)   or  V /  (R1 + R3) would give you about ~13mA?

[ edit: This is where 2 DMM's come in handy! as you can measure the transistor base current vs. the collector current, and you'll see that the base only requires a very small amount of current to flow in order to allow a larger current to flow through the collector.]

[Tony3dd]

hi,  I'm not very good at analogue stuff but I think 12 /  (680 + 217)   or  V /  (R1 + R3) would give you about ~13mA?

[ edit: This is where 2 DMM's come in handy! as you can measure the transistor base current vs. the collector current, and you'll see that the base only requires a very small amount of current to flow in order to allow a larger current to flow through the collector.]

Yes, I was actually counting the 10K resistor which is actually not in the current path. That's why I was getting low mA reading. My meter actually measures it as 11.5mA's, and it calculates to 13.38.

[mariush]

You will never get an exact mA value with your multimeter, due to burden voltage.  Multimeters have a current shunt inside when measuring current, and that affects the circuit a bit.

11 mA is perfectly fine if you expected 13 mA according to your math (it may very well be the case that the circuit does 13mA with the multimeter outside the circuit)

See Dave's video in case you're interested to learn more about this issue :

http://www.youtube.com/watch?v=u2yRR4G3yTA   (from 3:22)

For the circuits you play with, it doesn't matter that you multimeter alters your circuit slightly.

Also, as a suggestion, it would help if you mention what transistor you're using. Depending on how much current you push into the base of the transistor and its hFe, it's possible to have the transistor limiting current instead of acting like a switch. It's unlikely to be the case in your example.

[Tony3dd]

You will never get an exact mA value with your multimeter, due to burden voltage.  Multimeters have a current shunt inside when measuring current, and that affects the circuit a bit.

11 mA is perfectly fine if you expected 13 mA according to your math (it may very well be the case that the circuit does 13mA with the multimeter outside the circuit)

See Dave's video in case you're interested to learn more about this issue :

http://www.youtube.com/watch?v=u2yRR4G3yTA   (from 3:22)

For the circuits you play with, it doesn't matter that you multimeter alters your circuit slightly.

Also, as a suggestion, it would help if you mention what transistor you're using. Depending on how much current you push into the base of the transistor and its hFe, it's possible to have the transistor limiting current instead of acting like a switch. It's unlikely to be the case in your example.

Thanks for the explanation. I was using a 2n2222 transistor with a 12 volt power supply.

[Mr Eastwood]

11 mA is perfectly fine if you expected 13 mA according to your math (it may very well be the case that the circuit does 13mA with the multimeter outside the circuit)

Perhaps the difference is due to the LED in the circuit;  when you bypass the LED, what does it measure?

[Tony3dd]

11 mA is perfectly fine if you expected 13 mA according to your math (it may very well be the case that the circuit does 13mA with the multimeter outside the circuit)

Perhaps the difference if due to the LED in the circuit;  when you bypass the LED, what does it measure?

Ok, I took the LED out of the circuit, and got 13.32 mA's! By my calculations it should be 13.38. That's spot on!must be the resistance in the LED then.