Saturday, May 31, 2008

Driving stepper motor using TIP122

If your stepper motor has a high current rating then its better to use TIP120/TIP121/TIP122 for driving your stepper motor. (TIP122 is most commonly available in Ritchie street).
The TIP122 is silicon epitaxial-Base NPN power transistor in monolithic Darlington configuration mounted in TO-220 plastic package. It intended for use in power linear and switching applications.

About Darlington pair:

Darlington transistors or Darlington pair is two transistors connected together so that the current amplified by the first is amplified further by the second transistor. The overall current gain is equal to the two individual gains multiplied together.
A darlington pair behaves like a single transistor with a very hihg current gain. It has three leads (B,C and E) which is equivalent to the leads of a standard individual transistor. To turn on there must be 0.7V across both the base-emitter junctions which are connected in series inside the Darlington pair, therefore it requires 1.4V to turn on.

Circuit Description :

The driver circuit must withstand the voltage and current required by the stepper motor. The stepper motor which i used required 12volts and 1.5A to provide good torque, so i selected using TIP122. Driver for each wire include a TIP122, a 1k ohm resistor and a diode. The resistors are used for limiting the current and the diodes are used to avoid back EMF. The circuit is shown below,

The common terminal of both the winding are shorted and connected to motor supply. When logic 0 input is provided to the base of TIP122, the corresponding motor will remain floating as the impedance between collector and emitter of TIP122 is very high. So no current flows through that motor winding. When logic 1 input is provided to the base of the TIP122, its collector and emitter get shorted as a result the motor wire will be grounded resulting in current flow through the corresponding coil.

I used the PCB displayed above to drive two stepper motors. While driving the stepper motor having high current rating, large amount of heat will be generated hence to dissipate the generated heat you must use an Heat sink. Only by providing heat sink the TIP can drive maximum current.


solitarys said...

can you drive a dc motor with same circuit

Keerthi said...

Thank you very much for the circuit. It was really helpful.

arvinda said...

hey could please provide information how the TIP122 works as a switch plzzzzzzzz

Shailesh Ajmera said...

Hi,I used the same circuit shown above to run a unipolar stepper of 1.8 deg step rated at 12V/0.6 Amp per phase. I used ATMega8L microcontroller to give signals after every 4 seconds. However the voltage output of mc is 0.68 V when i use 1k resistor between base of TIP122 n mc output, voltage at base of TIP122 being 0.58V. Thus the motor doesnt run. For 10kohm resistor, i get the voltages as 1.8V and 0.8V, and for 4.7k,I get 0.8 and 0.7V. When open-circuit, mc output signal amplitude is 5V...
Pls help me by telling the appropriate value for the resistor... I am not able to identify the problem...I need to submit my project in 5 days.
pls reply at Thanks in advance...

mify said...

it's very good. thank you