Using a transistor to amplify current

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Discussion Overview

The discussion revolves around the use of a TIP120 transistor for current amplification, focusing on the setup and behavior of the circuit. Participants explore the conditions under which the transistor operates, including base-emitter voltage requirements and the implications of circuit design choices.

Discussion Character

  • Technical explanation
  • Debate/contested
  • Mathematical reasoning

Main Points Raised

  • One participant notes that the TIP120 has a DC gain of 1000 and questions why the expected collector current is not achieved with a base current of 10 μA.
  • Another participant suggests that the base-emitter voltage is likely insufficient, recommending a minimum of 0.7 V to turn on the transistor properly.
  • Concerns are raised about the circuit's sensitivity, indicating that a small increase in base voltage could lead to excessive current draw, potentially damaging the transistor or draining the battery.
  • A suggestion is made to include a degeneration resistor to stabilize the circuit and allow for better control of the collector current.
  • One participant points out that the diode testing mode of their multimeter shows a base-emitter voltage of 0.55 V, questioning whether this is adequate for operation.
  • Another participant emphasizes that the base current limitation should theoretically prevent the collector current from exceeding maximum ratings, but acknowledges that this reasoning may be flawed.
  • Discussion includes the possibility of grounding the emitter and using a high-value resistor in series with the base to drive the transistor, allowing for back-calculation of the transistor's beta.
  • A participant highlights that the TIP120 is a Darlington pair, which requires a higher base-emitter voltage due to its two junctions.
  • Concerns are raised about the absolute maximum base current specified in the datasheet, suggesting that exceeding this could damage the transistor.
  • One participant questions whether a wiring error could be responsible for the low base-emitter voltage reading.

Areas of Agreement / Disagreement

Participants express differing views on the appropriate base-emitter voltage and the implications of circuit design choices. There is no consensus on the exact cause of the low collector current readings or the best approach to resolve the issue.

Contextual Notes

Participants mention the importance of considering the transistor's specifications and the potential for damage due to incorrect wiring or excessive current. The discussion also highlights the need for careful measurement and circuit design to achieve the desired amplification.

SMD1990
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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.
 
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SMD1990 said:
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.
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.
 
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.
 
SMD1990 said:
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.

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.
 
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:

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