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I recently purchased a DRV8412 H-Bridge motor driver. It acts like a transformer, decreasing voltage and increasing current. I knew this before I purchased it, and I get that the power is the same therefore, as the voltage decreases the current increases.

I have no idea however how this works. Intuitively, I get how a normal transformer works with the magnetic flux, but what are the working principles of this transformation?

Edit: In my circuit the current drawn by the DRV8412 is less than half of the output. The output goes straight to a Peltier.

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  • \$\begingroup\$ A warm welcome to the site. The DRV8412 doesn't do anything like act as a transformer. It's just a switching circuit. What made you think it did? Please edit your question and greatly expand it, explaining in detail what you have seen and what conclusions you had reached. I imagine you've misunderstood its role in a circuit you've seen. Lots of detail, please. Again, welcome. \$\endgroup\$
    – TonyM
    Sep 2, 2021 at 16:34
  • \$\begingroup\$ @Alhydra: It would be good to link to where you read about the DRV8412 decreasing voltage and increasing current. As a motor driver, it doesn't really do that. It is possible, though, that a circuit using the DRV8412 could do what you have described. It is also possible that someone, somewhere was talking out of his hat. \$\endgroup\$
    – JRE
    Sep 2, 2021 at 16:38
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    \$\begingroup\$ This is very similar to a DC-DC converter more so than a transformer. It is not just the H bridge but the H-bridge in combination with the battery and capacitors on VBAT and the inductive motor winding itself that give rise to this DC-DC conversion effect. Look at it like a buck converter. \$\endgroup\$
    – user57037
    Sep 2, 2021 at 16:43
  • \$\begingroup\$ Please look up a buck converter. A full bridge is just a buck with extra steps. \$\endgroup\$
    – winny
    Sep 2, 2021 at 18:04

3 Answers 3

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Think first of a step-down DCDC converter. At its output, it consists of two switches (one to supply; one to ground) and an inductor.

When an inductor is carrying a current, it stores energy. When the supply switch is on, then current flows from the supply, through the switch, the inductor and into the load. During this time, the current in the inductor increases. Current (and power) is consumed from the input supply. Power is delivered the load, and energy stored in the inductor increases (because its current is increasing).

Now, when the supply switch opens, and the ground switch closes, the inductor current continues to flow; current flows 'up' from ground, through the inductor and into the load. During this time, the inductor current decreases.

Notice that during that 2nd period, current (energy) is still flowing into the load, and no power is being consumed from the input supply. The energy now being delivered into the load comes from the stored energy in the inductor.

If the switching frequency is high enough, the inductor current will look like a DC current with a triangular ripple superimposed on it. Generally a capacitor at the output will minimize the effects of this ripple component on the load.

Now you have a situation where current is only consumed from the power supply for a fraction of the cycle time, yet current is delivered continuously (with some minor ripple superimposed) to the load. This is the transformer-like behavior you mention.

An H-bridge is similar, with the motor's windings acting as the inductor.

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A Boost regulator using an H-bridge acts like a power transformer, not the bridge itself. This is due to the ratio of switched storage inductance and stored capacitance ratios. So that Vin/Iout ~ Vout/Iin * efficiency as you described.

In fact, the PWM reduces current on the average at all times compared with fully ON.

By pulsing a switch to a resistor it's average resistance goes up by the inverse of duty cycle. It's something like that for average reactances but more complex with saturation effects and slew rate.

Since start surge current is defined the DC resistance of the motor or DCR, this can be 10x lower than the AC impedance at rated load thus 10x more current (+/-20%). and even 100x greater than no load with low friction.

So the job of PWM is to enable you to soft start acceleration to 1/10th of full to enable not exceeding the rated current and over stressing your ESC driver. Otherwise , you can choose an ESC or bridge that is 10x your motor rated current and put the pedal to the metal.

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Yes, power (voltage, current) conversion systems work similar way.

Transformer:

Input_power = Input_voltage x Input_current.
Input <- coupled by magnetics -> Output
Output_power = (1 - loss) * Input_power = Ouput_voltage x Output_current

Output_current = Output_power / Output_voltage
= (Input_power / Input_current) x (1 - loss)

For a motor-driver you are using, the "coupling" happens by "stages" of voltage and current conversions using Inductors, Capacitors, Switching devices, and "controllers", in place of a transformer.

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