# Using a transistor to amplify current

1. Oct 12, 2011

### SMD1990

Hello. As the title says, I need some help in getting a transistor to amplify a current.

I have a TIP120, whose DC gain is specified as 1000. I am applying a positive voltage of about 0.6 V to the base. The current is around 10 μA.

I have the positive side of a AA cell connected to the collector. The emittor connects to the negative side of the "battery", and to the negative of my 0.6 V signal being applied to the base.

Why am I not reading around 10 mA from the battery? Turning off the base voltage causes the reading to become zero. Still, when the signal is applied, I am only reading 20 to 30 μA.

2. Oct 12, 2011

### yungman

0.6V is only an approximation. Most likely the transistor is not even turn on. Usually it is about 0.7V. BUT BUT!!! You cannot hook up circuit like this, because the circuit is too sensitive that as soon as you raise the 0.6V towards 0.7V, the transistor will suddenly turn on and draw a lot of current and drain your battery or burn the transistor.

To play with this, you need to put a small resistor between the emitter and the negative of the battery. Say a 100ohm. So when you slowly increase the base voltage from 0.6V up, you'll see there is a voltage drop across the 100ohm emitter resistor. from that, you can calculate the collector current as the current gain you gave is 1000. The reading should be close.

this little 100ohm is called degeneration resistor that help to stabilizing the circuit.

3. Oct 12, 2011

### SMD1990

My multimeter's diode testing mode shows the drop between the base and emitter as 0.55 V.

The short-circuit current of the AA should be less than the transistor's maximum rating. Besides, should not the limited base current keep the collector current limited to well below the maximums?

If not, my reasoning is incorrect.

4. Oct 12, 2011

### yungman

The base emitter voltage change with emitter current, you cannot use that as the number because your meter drive very little current. You did say the gain is 1000, so if you put 1mA into the base, it will give you 1A at the emitter, that would be enough to pull down and drain the battery quickly. Check out some practical transistor circuits to experiment.

There is a way to ground the emitter and drive the base alone without the emitter degenerating resistor, but people use a high value resistor in series with the base to drive. Say if you solder the emitter to the negative side of the battery and collector to the positive. Then you put a 1M resistor in series with the base and drive the resistor. Then you can look at the collector vs the voltage drop across the 1M resistor and back calculate the beta of the transistor.

5. Oct 12, 2011

### jim hardy

look up datasheet for TIP120

if Beta (hfe) is 1000 it's almost certainly a Darlington which will require 1.2 volts base-emitter because there's two junctions to overcome not one.

"Darlington" is a method of connecting two transistors piggybacked.
It's so handy the manufacturers sell them in one case, as shown on this Fairchild datasheet:
www.fairchildsemi.com/ds/TI/TIP120.pdf

observe from datasheet absolute maximum base current is 0.12 amp, and a AA cell might well exceed that.
if you've ever had the battery directly across base-emitter it likely shorted one of the junctions and that could be why you read low Vbe of .55 v --- transistor wrecked.

But maybe not - Try it again at Vce of 3 volts and it might work.

but monitor and limit base current.

Also observe pinout is BCE not EBC,,,, might your 0.55 volt be a hookup error?

Last edited: Oct 12, 2011
6. Oct 13, 2011

### pantaz

Last edited by a moderator: Apr 26, 2017