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I've got a motor driver which can take 3A. But my motor seems to draw more than 3A as peak current when it's been turned on. And I would like to get rid of that since it blew up my motor driver a few times already by now. I do not have a oscilloscope and do not know how much the peak current exactly is. And I also don't have the type of the motor (Nothing is written at it at all). All I know is that it's an 18V motor with the TC4424AVP as motor driver connected like this.


I do not have the inputs of the TC4424 connected to a PWM signal sadly.


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up vote 7 down vote accepted

The TC4422 is a MOSFET gate driver and is not really intended to drive motors.

The internal resistance is 6.3 ohms typically (could be 50% more when hot), so your current is being limited by the driver to not more than a couple amperes. You'll note that 18V/6.3 ohms is about 3A, so that's about the current you can momentarily get from it with maximum supply voltage into a short circuit (instantaneous power dissipation is 50W!). Actual current will be less because MOSFETs limit when the voltage drop gets high. Normally current flows to ground so only one switch is involved at a time.

enter image description here

I strongly suggest you get something beefier if you want to drive a relatively large motor like that.

The 3A mentioned in the large print on the datasheet is for momentary currents drawn when charging and discharging the gate of a MOSFET, it is not a continuous current rating. Note also that the resistance measurement is made at low current (10mA), which prevents significant heating (so the numbers look good) and also at the highest power supply voltage (again which makes the numbers look better than you'll see at lower supply voltages).

General rule with data sheets: The big print giveth and the small print taketh away.

Do the measurements that Brian Drummond suggested to determine your actual requirements. The driver should be able to withstand the starting current without failing, either indefinitely (stalled motor) or with some kind of protection should that situation occur.

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Thanks. I'll show your answer to my Teacher who's forcing me to use the TC4424 as motor driver ;) – Handoko Jun 18 '14 at 11:34
+1 for "General rule with data sheets: The big print giveth and the small print taketh away" :-) – m.Alin Jun 18 '14 at 12:48
@Handoko If there is an interesting response (other than sputtering), editing your question to include it might be appropriate. – Spehro Pefhany Jun 18 '14 at 13:02
Where is the 6.3 internal resistance measurement coming from? All I see is a HIGH/LOW otput resistance metric. – sherrellbc Jun 18 '14 at 16:09
@sherrellbc See where it says "sum of these two" (in red). The current must flow out of the "high" pin and into the "low" pin, so the total series resistance is the sum of the two. That is not the case when it is driving a MOSFET gate. – Spehro Pefhany Jun 18 '14 at 17:31

Measure the resistance of the motor. (Do this a few times and rotate the shaft, and take the lowest value, because the brushes can interfere with getting a good measurement).

Dividing 18V(or your actual supply voltage) by this resistance gives you the starting current; your motor driver needs to withstand that.

Say you measure 2 ohms; the starting current would be 9 Amps.

Two ways of reducing this current : run the motor off a lower voltage, or use PWM.

A series resistor is a traditional way of starting a motor, but it should be switched out of circuit as soon as possible once the motor is running. PWM during start is better and more efficient.

EDIT : given Spehro's revelation on the controller IC, the right answer might be to keep it and add external MOSFETs (it is after all a MOSFET driver!) capable of handling the current you need.

EDIT for SherrellBC's questions...

1) I am not going to recommend a way to disconnect a series resistor once started. It is a traditional "big motor" (think tramcar!) approach but there are better approaches here.

2) True the inductance limits current momentarily then the current settles to 9A. However this is NOT the steady state current - unless the motor is stalled. And if stalled - unless the motor is specially designed for such abuse - you had better disconnect the current quickly!

Then the rotor starts to rotate. And this is what you are missing : when it rotates it also acts as a generator, and the voltage it generates (called "back EMF) opposes the driving voltage. This reduces the current.

If the motor is lightly loaded (as it should be for efficiency) it will spin up to generate (say) 15V, leaving only 3V across its 3 ohm resistance, and consume 1 amp.

Load it a bit more, and it will draw more current (say 3 amps) to generate the extra torque. It slows down so that the back EMF is reduced to allow the extra current draw.

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Thanks! I'll measure the resistance. Sad to hear that I need to use PWM or lower the voltage though. :( – Handoko Jun 18 '14 at 9:40
... or rewind the motor or substitute a different one, or use a more suitable motor controller. – Brian Drummond Jun 18 '14 at 9:42
Yeah, I understand. It appears that the peak current would be 6 AMPS I've calculated. That's twice the amount my driver could take. – Handoko Jun 18 '14 at 9:44
When I add a 3 Ohm resistor on one output of the motor drive could that work? – Handoko Jun 18 '14 at 10:20
@Handoko, you must cosider the effect of voltage division that the 3 ohm resistor would induce. You must consider the winding resistance and the output resistance of the driver. – sherrellbc Jun 18 '14 at 16:11

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