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DRV8840 - Motor load vs. resisitve load
This is really confusing me and I would appreciate the advice of the E2E community!
Supply voltage = 15V
Orange: voltage wrt ground on load terminal 1
Green: voltage wrt ground on load terminal 2
Red = Orange - Green
When I drive an unloaded motor (16705) in slow decay mode using the DRV8840 at 25% duty cycle and PWM frequency of 32.5kHz, no current sense, I get the following response:
Great, but when I drive a wire-wound resistor of 4 ohms using the same amplifier, I get this:
How come the load terminal 2 (green) is behaving so strangely? Shouldn't be be close to ground like in my motor test?
To compare a motor with a resistor is not a fair comparison. Motors are inductive in nature and when you apply a voltage to them you will see current changing slowly as determined by the power supply voltage, motor inductance and motor BACK EMF. Since current can not change abruptly, you should see the voltage on the lower side not that much affected.
Now you do the same with a resistor, but in this case current will in fact change abruptly! In your case you are putting 15V across a 4 ohm resistor, so you should get 3.75A instantaneously.
At first I thought the low side FET's RDSon was causing the 5V rise, but this doesn't make much sense as this resistance is rather low. The max is 0.160 ohms and at 3.75A you should only see 600 mV, which is less than a tenth of the voltage we are seeing. I do believe, however, this explains the wave shape. Notice it goes into higher voltage at first and then it decreases to the 4-5V line. Why? Well, because FETs take time to fully turn on and their resistances decrease accordingly. So it is easy to imagine that at first the FET was more resistive, forming a voltage divider in where the low side resistance is comparable with the 4 ohm load.
But I still wonder, where is the 4-5V "settling" coming from?
So my next question is: how good is this GND plane? We often think of GND as being zero potential and somehow we also assume it is zero resistance. However, this is a relative matter. As you put so much current, the resistance on your ground plane's resistance starts to become significant. This is why when we switch lots of current in a very short period of time we suffer from ground bounce. At this point in time, your GND is no longer at 0V as we though initially, which is why you must be seeing the extra 4V or so.
This is of course quite conjectural as we do not know anything else about what is on your board, what kind of parasitic effects are being caused by the small inductive elements on the wound resistor, or its internal parasitic capacitance. All of these components will also play a role in making the signals rather different than when driving a full fledged inductive load. A good experiment is to add a 1-10 uH inductor in series with the resistor and see what happens. Chances are it will look cleaner.
Hope the info helps. Best regards,
Wow. Thank you so much for that wonderful explanation. I am sorry it took me so long to reply. The article got moved around and I lost track of my question thread.
I will do some tests with the DRV8840 to try to test for the things you commented on. I will let you know how it goes! Will probably need your advice again.
Thanks again. Really appreciate it.
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