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# Current problems of TIP122

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I was testing some things on my protoboard to learn a little about transistors. I made this:

The batteries are two AA 1.5V in series. The resistor is a 1Kohm. The transistor in anpn darlington.

I uploaded a code that just put Pin9 as OUTPUT and HIGH, so I can test the currents on the circuit.

1.Current beetwen the Pin9 and the Resistor: 1,32mA.

2.Current beetwen the Baterry(+) and the Colector: 0,43A.

3.Current beetwen the Emissor and the Ground: 0,32A.

I can`t understand this currents. I was expecting 3mA ((5-2)/1000) on the 1st case and 3A (1000(gain)*3mA) on the 2nd and 3rd.

Am I doing something wrong? Is this circuit going to damage my Arduino, since there is 0,32A going to the Ground Pin?

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You may wish to look at:

### Using the TIP120 & TIP125 Darlington Transistors with Arduino

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Why are you using the TIP122 darlington to short the 3V battery? The collector and emitter currents will be whatever the shorted battery current will be which might be 8A if the cells are new or be a few mA if they are old. The current gain of a TIP122 is a range from 1000 (the minimum) for some of them to maybe 10,000 or more for others. The Arduino output voltage at 3mA will be less than 5V, maybe 4V so the base current of the TIP122 might be 2mA.

Your current meter has a voltage drop so it measures currents low. Maybe the Arduino pulses the Currents and your meter shows the lower average currents.

The high current is not going into the Arduino, instead it is going from the battery into the TIP122.

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