Cytron Technical Forum

Hello,
if i use the 12 v, 2A adapter then i use L293D to stepper motor(rated 1.2A). will the L293D burn?

donny WROTE:Hello,
if i use the 12 v, 2A adapter then i use L293D to stepper motor(rated 1.2A). will the L293D burn?

No, it won't.

Allen

some of my friend say it will but i need confirmation before i build the circuit..
anyone out there would like to share?The L293D spec. is 600mA.. thanks allen for you answer.

donny WROTE:some of my friend say it will but i need confirmation before i build the circuit..
anyone out there would like to share?The L293D spec. is 600mA.. thanks allen for you answer.

Sorry, I didnt check the specs of L293D. In that case sure it will toast the L293D like a fuse. Dont try it. It is just like connecting a 230V 1KW water heater which needs a current of 4.3A and the fuse is only 2.2A in series. Do you think it will blow the fuse? It has nothing to do with the power supply from Tenaga.

Allen

i see. the L293D will not sustain the 2A supply but the stepper motor can right?

donny WROTE:i see. the L293D will not sustain the 2A supply but the stepper motor can right?

Yes, there has been discussion in our facebook, posted by the admin. cannot really remember the question exactly like, but it should be:

If I have a DC motor rated at 12V, 1A max. And a power supply that can source 12V 2A. Can I connect the DC motor to that power supply directly in no load condition?

It surprise me that a lot of people give opinion, which is good.

As conclusion. Current supplied to load in this case the motor does not depend on power source when the power supply can supply more than what needed . The current draw is determine by the motor (load). For example, if you have a resistor of 1/4W, 1Kohm, can you connect to 12V power supply with 2A output? So you might have say, 2A from power supply flow through 1K Ohm, so the power is VI, correct? So the wattage is 12*2A = 24W, this will blow the resistor But in reality, nothing happen, why? Because the current flow through resistor is not determine by the power supply, is the resistor (load), current, I should be = V/R = 12V / 1K = 0.012A = 12mA. The real Power, P = V*I = 12*12mA = 0.144W < 1/4 W

Hopefully you understand.

ok.i understand already.Thanks.
all is depends on the load.since the load is rated 1.2A, therefore the motor driver should also support 1.2A.Am i right?

donny WROTE:ok.i understand already.Thanks.
all is depends on the load.since the load is rated 1.2A, therefore the motor driver should also support 1.2A.Am i right?

To be safe, the driver current capability should higher than the load current.

This is different in fuse case as highlighted by Allen. Fuse is to protect the load, while driver is to control the supply to load depending on the control