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i want to use the TIP122 darlington transistor with my atmega8 to control a 3A 12V geared DC motor. i found that the current gain is 1000 for tip122 from its datasheet. so the base current for 3A output is 3mA(am i wrong?). So i used a 1.6k resistor to limit the current from atmega8. I breadboarded the circuit and found that i get only 4v emitter voltage even though i the supply voltage is 12.4V. why is this so? i removed the resistor and again i get the same voltage( base current from atmega8 without resistor is 40mA). i simulated the circuit in Proteus and still the same result. So something is wrong? Please help me with this.

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    \$\begingroup\$ ALWAYS supply a circuit diagram. Always. Steven is probably right about what you are doing but it is not possible to be sure. Answering questions like "why do I only get 4 volts" is hard to answer well when you have not supplied a circuit. Note for a bas eresistor the current should be AT LEAST as much as needed. Severa, times more is usually a good idea. \$\endgroup\$
    – Russell McMahon
    Oct 29, 2011 at 13:45

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TIP122 datasheet is here

Fig 2 of the data sheet shows that Vbe is typically 1.75V "saturated" when Ic = 3A. Using Vbe = 2V is safer. Using 2.5V does no harm.

Fig 2 shows that Vce = 1V typical at 3A. So the transistor will dissipate Power = V x I = 1V x 3A = 3 Watt. If you use eg a TO220 package you will need a modest heatsink. If you use a surface mount package you may need to check heat sinking issues.

enter image description here

Your circuit should look like the diagram below. Note that the motor connects from 12V to collector.

enter image description here

Fig 1 in datasheet shows that gain improves wuth IC upto about 3 and that a gain of 4000 may be expected. Do not trust it :-).
Designing with a gain of 1000 will work well.

For Ic = 3a you need 3 mA drive at gain = 1000 as you noted. If we assume a 5V supply then R1 is such as to allow 3 mA to flow when Vbe = 2V (see above)
R = V/I = (5-2) /0.003 A = 1000 ohms.

If the process ior was running on 3V3 say then
R1 = V/I = (3.3-2) / 0.003 A = 430 ohms =say 390 ohms

Worst case if you allowed Vbe = 2.5V and drove it with a 3V3 processor pin then
R = V/I = (3.3-2.5)/0.003 = 266 ohm = say 270 ohms.

So R1 may have a 4:1 variation depending on what Vbe you decide to choose* and what processor supply voltage is. In fact 1000 ohms would probably work O in all cases and 470 ohms is probably a good all round compromise. (* I said "what Vbe you decide to choose" but in fact the transistor does the choosing - actual Vbe will depend on the transistor in the given circumstances and we choose a Vbe for design purposes based on what the datasheets tell us. ).


A motor is inductive. When you turn it off the motor current cannot styop instantaneously and MUST go somewhere. Diode D1 gives it somewhere to go.
Without D1 you will get a LARGE inductive driven voltage spike. The voltage will rise until the motor current finds somewhere to go ! :-). This can be fatally bad for the transistor and for other electronics, ALWAYS include a D1 equivalent on such cases.

To be safe D1 should be rated at motor current. In practice this may not be needed depending on how you are driving the motor. If you are using fast PWM then D1 should be a fast diode BUR in almost all cases a standard power diode will do OK. 1 x 1N400x will probably do. 2 o3 3 in parallel will be better. A single power diode rated at3A or more is better still - but 1N400x are cheap and more usually available.


No no no !!!; You mentioned driving the transistor base directly with no resistor. This is very bad practice as it very probably violates the processor's datasheet specifications. Once you do that anything can happen and anything might. In some cases (not in this one) you may also destroy a driven device doing that. Circuit components should always sbe designed !.

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What you seem to have built is an emitter follower, where the emitter follows the base voltage, minus the base-emitter voltage. This configuration is also known as common collector. This \$V_{BE}\$ will be around 1.4V since the Darlington gives you two PN junctions between base and emitter. The reason the emitter voltage doesn't get higher: if it would there wouldn't be enough voltage drop across the base-emitter junction to cause any base current, and hence no collector current either. That's why the emitter voltage is stuck at about 4V.
Like Leon suggested, place the motor on the collector side, and connect the emitter to ground, so that you get a common emitter configuration. You also need a base resistor. The datasheet says that \$V_{BE}\$ can be as high as 2.5V, so

\$ R = \dfrac{5V - 2.5V}{3mA} = 833\Omega \$

That's a maximum value, I would pick a 680\$\Omega\$; the 1.6k\$\Omega\$ is too high to give you 3mA of base current. Don't forget to place a diode in parallel with the motor, anode to the transistor's collector.

edit
Something important I forgot to mention: at 3A the transistor may have a \$V_{CE(SAT)}\$ of 2V, which gives you a 6W power dissipation. That's a lot and requires a sizable heatsink.
Russell also calculates the dissipation, but arrives at other values because he works with typical values, whereas I calculated for maximum values. You always should calculate for worst case. A heatsink you picked for 3W dissipation will be insufficient in case the dissipation is 6W, which may happen. That's why it is mentioned as maximum value.

A better alternative for the Darlington is a MOSFET with a low \$R_{DS(ON)}\$, like the FDC653N. Thanks to its low \$R_{DS(ON)}\$ of 0.055\$\Omega\$ it doesn't need a heatsink if it's mounted on sufficiently large copper. You may find MOSFETs with even better specs.


Further reading
More on common collector
More on common emitter

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Connect the motor between the collector and the supply, and ground the emitter.

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  • \$\begingroup\$ thanks that will really work... any hint about why the emitter voltage is 4 volts only? thanks in advance \$\endgroup\$
    – user420528
    Oct 29, 2011 at 12:28
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Somebody thought that this circuit is an emitter follower. I don't think so. There is no load circuit in the emitter and in fact the emitter is to be grounded. It is just a plain vanilla saturated switch in common collector mode.

You are driving the base from a microprocessor or similar that only can supply small numbers of mA and the load is A. So using that formula that was suggested is correct. Take the output of the micro and subtract the Base-Emitter Voltage expected in the TIP device. That difference is what will determine the potential current in the base of the TIP. To get the value of the resistor needed, divide the voltage difference by the drive current you want and just apply ohms law. But since the TIP has a guaranteed current gain of 1000, you should assume the base current is 1/1000 of the load current. So if the load current is 1 A, then the minimum base current would be 1 mA. If the Collector current is expected to hit 3 A, then the minimum base current would need to be 3 mA. But as folks mentioned, we typically overdrive the base a little bit. If the base current required is 1 mA, I would supply 2 mA and so on. So for example, if your micro supplies 5 Volts output for a high and you are trying to drive a 3 A load, then assuming a Base-Emitter voltage of 2 V, then (5-2)/(3/1000) = 1000 ohms. But to ensure that the TIP part is overdriven, use half that (500 ohms) to guarantee twice the current.

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