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

enter image description here

The batteries are two AA 1.5V in series. The resistor is a 1Kohm. The transistor in a TIP122 npn 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?

  • \$\begingroup\$ 600 pixels wide is about the biggest image that can usefully displayed on the site. Making your image 2500x1500 pixels just means that everyone looking at it will have longer to wait downloading it. Could you provide a smaller version? \$\endgroup\$
    – The Photon
    Jan 20, 2013 at 17:40
  • \$\begingroup\$ Sorry, I changed the links. Now is better. \$\endgroup\$
    – Rafael
    Jan 20, 2013 at 17:42
  • \$\begingroup\$ @Zebonaut you beat me by 5 seconds! \$\endgroup\$ Jan 20, 2013 at 17:47
  • \$\begingroup\$ @Madmanguruman being withing some seconds for editing in the image is one thing. Seeing three very similar answers withing minutes is another... SE is amazing. \$\endgroup\$
    – zebonaut
    Jan 20, 2013 at 20:43

3 Answers 3


You have a minimum current gain given as 1000 in the datasheet, so this means 1mA into the base should result in at least 1A from collector to emitter, assuming the supply can provide this.

The calculation for current into the base (assuming the Arduino pin outputs 5V high) and we take the maximum base-emitter voltage from the datasheet is:

(Vpin - Vbe) / Rbase = (5V - 2.5V) / 1kΩ = 2.5mA;

It looks like your multimeter has quite a high burden voltage (too high a high resistance used for current shunt, so you get a voltage drop across it which affects things) hence the readings being out on the base current and the difference between the supply-collector reading and emitter-ground reading (which should be practically the same - the emitter-ground reading only has the base current added to it, which is tiny compared to Ic)

The darlington transistor has a high saturation voltage (higher than a normal transistor), so a higher voltage supply is preferable for reasonable results, and gain also drops at saturation. In any case, controlling the current in this way is not very practical, since the gain can vary widely between parts, with temperature, etc. Try adding another battery or two, adding a small value collector resistor (capable of the wattage it needs to handle to control max current accurately), and lowering the base resistor to around 200Ω.

If you want to learn about the base current vs gain and saturation relationship, try using a higher supply voltage you know is capable of the current you are testing with, adding a potentiometer (wired as a rheostat) at the base, a small value resistor to give you a known maximum current, and plotting the base current vs collector voltage/current. If you do the calculation you should be able to see how the gain varies and drops approaching saturation. You should get plots similar to the datasheets.
Doing the above in SPICE is also another option if you don't have enough test equipment to make things easy.

  • \$\begingroup\$ Thanks, I think I understand what you told. I still have a few doubts: 1. What supply should I use? 9V is enough? If I put 6 AA batteries is series, it should give me 9V, but what is the maximum current I can draw from it? Is there any problem to Arduino if I use it`s ground as the grounf of the circuit, with a current of about 1.5A flowing through the circuit? \$\endgroup\$
    – Rafael
    Jan 20, 2013 at 19:11
  • \$\begingroup\$ @Rafael - Yes, these things can be hard to diagnose from afar, there are quite a few possibilities for what you are seeing. You definitely need to test the battery voltage when it's under load, as it could be sagging quite a lot, which combined with the things I mention above could be causing the lower collector current. \$\endgroup\$
    – Oli Glaser
    Jan 20, 2013 at 20:14
  • \$\begingroup\$ 1. Ideally use a bench supply, or e.g. a good quality wall wart supply that has a reliable output current. 9V is enough, yes. With batteries it's impossible to say what the maximum current you can draw is, since it depends on type, quality, temperature, etc - you can only guess at a rough figure - If you use batteres, ideally you would only draw say, less than half the rated capacity to make sure the voltage does not sag too much. There is no problem for your Arduino with the current flowing through the ground of your circuit, as long as the cables are rated to handle it. \$\endgroup\$
    – Oli Glaser
    Jan 20, 2013 at 20:19
  • \$\begingroup\$ Going from your diagram, the transistor current will not flow through the Arduino, only through the transistor, breadboard and power cables. 1.5A may be quite large for your breadboard - check what it is rated to handle. Also, note that there will be voltage drops along the cables/breadboard, so use thick cables and makes the spacing less on the breadboard (try and imagine the current loop and make it as small as possible) \$\endgroup\$
    – Oli Glaser
    Jan 20, 2013 at 20:22
  • \$\begingroup\$ Ok. Thanks a lot for your answers. I will get an wall supply and try other measures. One last question about this transistor: If I connect the Battery(+) to the Emissor and Ground to Colector, what should happen (with and without current on the base)? Remembering that this TIP 122 has a built-in Diode. \$\endgroup\$
    – Rafael
    Jan 20, 2013 at 21:34

Your circuit is basically this:


When you turn on the transistor, it basically shorts out the battery. I bet your batteries are nearly dead, and I bet that if you measure the battery voltage when the transistor is on, it's about 2.52V, not 3V. Why? because you measured the base current at 1.32mA. By Ohm's law, 1.32mA through a \$1k\Omega\$ resistor means the voltage across the resistor must be \$1.32mA\cdot 1k\Omega = 1.32V\$. The two base-emitter junctions of T1 add another 1.2V: \$1.32V + 1.2V = 2.52V\$

R1 and R2 represent the internal resistance of the batteries. I've used \$200m\Omega\$ here, which is an estimate for a fresh battery. As the batteries deplete their stored chemical energy, this resistance becomes greater. The consequence is that when they must pump more current, their voltage decreases more.

It's a bit odd that you are measuring more collector current than emitter current. I bet you measured the collector current first, and by the time you measured the emitter current, the batteries were more dead, and this is why you measured less emitter current.

On a practical note, the current gain (\$h_{fe}\$) of transistors varies, a lot. Even two transistors of the same model can have very different current gains. A well designed circuit should thus be relatively insensitive to this parameter. If you wanted to make 1A flow through the transistor, you should not try to find the right base current to make that happen, but rather limit the current through some other means, then calculate the necessary base current, then give it a bit more base current to be sure the transistor is "saturated", that is, as "on" as it can be. Thus the voltage between the collector and the emitter of the transistor will be at its minimum.

Here's a more reasonable circuit for experimenting with transistors:


Here you see the collector current is limited by R4. Of course, you could power an LED directly from your microcontroller. A transistor like this becomes more useful when your load is big enough that you can't do that. Imagine it's a headlight, instead of an LED.

  • \$\begingroup\$ I think it's also possible the burden voltage of the meter lowered the effective base voltage, hence the discrepancy. \$\endgroup\$
    – Oli Glaser
    Jan 20, 2013 at 17:59
  • \$\begingroup\$ I measured the battery and is 2.94V. When I short circuit then, the current is about 1.25A. \$\endgroup\$
    – Rafael
    Jan 20, 2013 at 18:02
  • \$\begingroup\$ @Rafael: 2.94 when this circuit is on, or when its off? \$\endgroup\$
    – Phil Frost
    Jan 20, 2013 at 18:03
  • \$\begingroup\$ 2.94V measuring directly trough the poles. \$\endgroup\$
    – Rafael
    Jan 20, 2013 at 18:04
  • \$\begingroup\$ @Rafael: I forgot to add the extra 0.6V to the battery voltage for the 2nd base-emitter junction in the Darlington pair. See edits. \$\endgroup\$
    – Phil Frost
    Jan 20, 2013 at 18:19

The circuit looks like your transistor has a connection from C to +BAT and from E to -BAT(GND).

When you turn your transistor on, it shorts the battery. This is not a good way to test a transistor.

The current you are measuring is probably just the max. value the battery can deliver. The voltage of the battery drops once it is shorted. You seem to be using a supply voltage of 5  in your calculation, but the picture shows 2 * 1,5 V = 3 V as the battery supply.

Also, the current gain of a transistor has a very (!) wide tolerance. The TIP122 has min. 1000, max = ?. It is not a reliable parameter to work with, and a good design almost always tries to get along with a wide variation in current gain.

  • \$\begingroup\$ But considering that the current on the base is 1,32mA, shouldn`t the current on the Colector/Emissor be 1A instead of 0,32/0,43A? And why the current on the base is 1,32mA instead of 3mA? \$\endgroup\$
    – Rafael
    Jan 20, 2013 at 17:54

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