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Max Ratings L298N - V*A!=W?

Robyn

Apr 17, 2013
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Hi everyone,

I am currently looking for a DC brushed motor driver for an artist who made a rotating scculpture with LEDs. I have measured the current draw of the motor at full speed to be just under 2A at 12V. My best candidate sor far is the L298N but I cannot make sense of the absolute maximum ratings on the datasheet (https://docs.rs-online.com/af53/0900766b80b34864.pdf).

The chip is supposed to be able to handle up to 50V and 2A per channel (of which it has 2). The max ratings also indicate that the max power dissipation of the case is 25W. To me it seems that 50V*2A*2channels = 200W. Which I understand would be ridiculous. Please can someone explain what I am not getting here?

Thank you!
 
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bertus

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Hello,

First of all, the mentioned 50 Volts is the absolute maximum rating.
One should never use a device on its absolute maximum value.
The recommended maximum is 46 Volts.
The device is a switching device and will never have the full powersupply on the end stage when it is driving 2 Amp.
The power it will dissipate is likely the saturation voltage of the endstage transistors times the current in these transistors and some power for the control circuit.

Bertus
 

Robyn

Apr 17, 2013
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Thank you for your responses :)

I will read about the heat sinks right now but I guess my main issue here is my lack of understanding of the power consumption in switching devices.

Hello,

First of all, the mentioned 50 Volts is the absolute maximum rating.
One should never use a device on its absolute maximum value.
The recommended maximum is 46 Volts.
Of course! I will use 12V so I'm good on that level. I am a little more worried about the current draw as the motor draws about 1.8A at full blast which is close to the 2A max. I am considering parallelling the 2 channels as suggested on the datasheet to be safe.

The device is a switching device and will never have the full powersupply on the end stage when it is driving 2 Amp.
The power it will dissipate is likely the saturation voltage of the endstage transistors times the current in these transistors and some power for the control circuit.
Bertus
So the current rating is regardless of the supply voltage if I understand well and that's the figure I can trust. When you say "saturation voltage", do you mean at the base? Why doesn't the collector emitter current come into play? Is that because the fact that it is a switching device makes it intermittent so less overall?
 

bertus

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Hello,

The saturation voltage is the voltage accross the output transistor, driving the motor.

l298_saturation_voltage.png

Bertus
 

Robyn

Apr 17, 2013
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OMG I just got it. Because not the whole supply drops at the transistors obviously otherwise there would be nothing left to drive amps into the motor so only what is used by the transistors participates to the power dissipation. Feels good learning, thank you!
 

Audioguru

Sep 24, 2016
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If a motor draws a max current of 2A when running at full speed then it might draw 5A or 8A when it starts from a stop
Then paralleling a few drivers is needed.

If your supply is 12V and one driver produces a 2V saturation voltage loss then the motor gets only 10V and probably will not run at full speed.
 

Robyn

Apr 17, 2013
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If a motor draws a max current of 2A when running at full speed then it might draw 5A or 8A when it starts from a stop
Then paralleling a few drivers is needed.

If your supply is 12V and one driver produces a 2V saturation voltage loss then the motor gets only 10V and probably will not run at full speed.
That is a lot! Think it might be better to make a circuit with discrete components in this case.
 

Robyn

Apr 17, 2013
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Perhaps drop in a couple of mosfets on the output.
So if I do that I would have to put the gates of a P channel and a N channel on each output of the H bridge to keep the ability to reverse the direction of the motor. The source of the N to the positive and the drain of the P to ground. Is that correct? Tried to upload a diagram but iOS doesn't seem to let me.

Edit: link to image:
https://ibb.co/9bffLzb
 

Audioguru

Sep 24, 2016
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Complementary Mosfets are common in H-bridge motor reversing circuits. Your Mosfet polarity descriptions are backwards.

Here is an H-bridge circuit using complementary Mosfets. When one transistor is turned on then the motor runs forwards. When the other transistor is turned on then the motor runs backwards. When both transistors are turned off or both are turned on then the motor is shorted and braked.
 

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Robyn

Apr 17, 2013
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Complementary Mosfets are common in H-bridge motor reversing circuits. Your Mosfet polarity descriptions are backwards.

Here is an H-bridge circuit using complementary Mosfets. When one transistor is turned on then the motor runs forwards. When the other transistor is turned on then the motor runs backwards. When both transistors are turned off or both are turned on then the motor is shorted and braked.
Thank you.
So basically if I'm gonna make a whole H bridge I might as well not spend money on the L298N and just make an opamp converter to turn the PWM from the Arduino into an input for the transistors that feed into the H bridge and make sure my MOSFETS can handle 8A / use Darlington arrangements to get to that rating. Is that correct?
 
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Audioguru

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Without seeing the spec's for your motor I simply guessed how much current it draws when it starts running.
How much is the current when you slam it suddenly at full power in the other direction? Double the stalled current?
Is the braking action of the circuit not wanted?
 

Robyn

Apr 17, 2013
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Without seeing the spec's for your motor I simply guessed how much current it draws when it starts running.
How much is the current when you slam it suddenly at full power in the other direction? Double the stalled current?
Is the braking action of the circuit not wanted?
I'll pay the artist a visit on Monday to check on that. I was silly and only measured the current at full speed thinking that it was as high as it was going to get. The motor is salvaged so I don't have a data sheet.
I think I can do without breaking because the idea is to slowly increase speed then decrease it then reverse. It's a sort of led Spirograph thing. There is a belt on the linkage to the spinning wheel so it stops pretty quick with the friction.
I can't say how much I appreciate the help. Don't waste more time on me until I come back with more measurements :)
 

duke37

Jan 9, 2011
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With a simple DC motor
The voltage sets the speed.
The current sets the torque.
Adding an extra load will increase the current. Add a fuze.
 

Robyn

Apr 17, 2013
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I've measured the stall current by blocking the motor shaft and running 12v through it. The amp reading never exceeds 2.2. This is not much higher than the full speed current and I assume it is so because the full speed current that I've measured is under load (actually a fair bit of it: 3 arms in a star configuration each weighing about 100g at a length of 36cm).
I think designing a circuit to handle 4A max should keep me out of trouble.
 
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