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schematic

simulate this circuit – Schematic created using CircuitLab

Using two 3.6 volt cells, I'd like to drive 3A through an electromagnet that is some wound magnet wire with a series resistance of about 0.25 ohms.

Experimenting with a bench-top power supply shows that 3A through the coil is enough to pick up the bit of metal I want to pick up. DC voltage permanently magnifies the coil so I'd like to run AC through it. How would the inductance of the coil affect the circuit?

Initially, I thought to generate a 0.75V square or sine wave then buffer it with some power amplifier at unity gain. Thinking about it more, I figured I could do it with an H-bridge.

Is there anything I'm missing here. Why wouldn't this work, these seem like a fragment of a DC-AC inverter circuit.

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  • \$\begingroup\$ The inductance of the coil would be nice to know. Why you are doing this (what results are you trying to achieve) would also be nice to know. What I have is "electromagnet" made with some magnet wire. Why 3 amps, for example? Or how do you expect to limit it to that? You could write more when asking if there's anything you are missing. Sadly, I've no crystal ball pathway into your brain. \$\endgroup\$ Commented Mar 28, 2023 at 0:54
  • \$\begingroup\$ Some of the answers to this question might be helpful: electronics.stackexchange.com/questions/77549/… \$\endgroup\$
    – PStechPaul
    Commented Mar 28, 2023 at 4:34
  • \$\begingroup\$ ([DC] permanently magnifies the coil (by all means, use precious metal conductors! Alas, magnification will be due to increased temperature, and reversible.)) With a core material with low remanence/residual magnetisation, you may not need to bother and use (limited) DC current. (Searching for use of resonant circuits for demagnetisation.) \$\endgroup\$
    – greybeard
    Commented Mar 28, 2023 at 11:33

1 Answer 1

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Each side of your bridge has MOSFETs operating as source-followers, meaning that the transistors' sources can't get anywhere near the power supply potentials even with their gates at the extremes of 0V and 7.2V.

With M1's gate at +7.2V, it's source is always \$V_{GS(TH)}\approx 4V\$ or so below that, for a maximum of 3.2V. M4 has the same problem, but inverted; its source cannot fall below \$0V + V_{GS(TH)} = +4.0V\$.

As you have it, I doubt that any appreciable current would pass through that coil.

One way to fix this is to connect the transistors as common-source, but that introduces a whole slew of other issues to address, not the least of which is shoot-through:

schematic

simulate this circuit – Schematic created using CircuitLab

It's possible that shoot-through won't be a problem here, since as gate potential passes through the half-way point (+3.6V), MOSFETs M1 and M4 are not switched on simultaneously. That said, there's no guarantee, and you might have to employ dead-time, or some other method to mitigate shoot-through.

Assuming that the H-bridge problems are resolved, you still have coil inductance to contend with. The effect of inductance is to slow the rising and falling of current. If you switch too frequently, you may never allow enough time for coil current to reach 3A, in either direction.

The rate at which current rises in the coil is inversely proportional the resistance in series with it, which will include its own resistance and the resistance of the MOSFETs' channels.

If any two MOSFETs passing current through the coil have channel resistances \$R_{DS(ON)}=0.2\Omega\$, then the total series resistance will be \$0.2 + 0.2 + 0.25 = 0.65\Omega\$. This has two effects, the maximum current that can flow, and the time it takes to reach that maximum.

\$I_{MAX} = \frac{7.2V}{0.65\Omega} = 11A\$

The time constant is \$\tau = \frac{L}{R}\$, and will require you to measure the inductance \$L\$ of your coil. This constant tells you how long it takes to reach about \$\frac{2}{3}\$ of maximum current, and a common rule of thumb is that \$5\times \tau\$ is the time it takes to get within 1% of maximum.

For example, assuming \$L=1mH\$ and \$R=0.65\Omega\$, then

$$ \tau = \frac{1mH}{0.65\Omega} = 1.5ms $$

Now, if you're sure of the coil's inductance, and you are driving your MOSFETS properly, then you could probably estimate the time it takes for coil current to reach close to maximum in one direction, and then switch current direction for that same period. I estimate that for a coil of 1mH, you should be switching faster than 1kHz, to obtain an average current of about 3A in each direction, but this approach is very dependent on component tolerances, and since inductance will not be constant (it's picking up and dropping objects, which will affect that), this approach to control is fraught with inaccuracy.

Maybe it's better to limit maximum current with an additional resistance. While we're at it, let's protect the MOSFETs from inductive back-EMF when the coil is suddenly disconnected (as would happen if you implement dead-time):

schematic

simulate this circuit

Now maximum coil current is limited to:

$$ I_{MAX} = \frac{7.2}{0.2 + 0.2 + 0.25 + 1.7} = 3A $$

The time it takes for coil current to reach 1% of that target is:

$$ \tau _{1\%} = 5\times \frac{1mH}{(0.2 + 0.2 + 0.25 + 1.7)\Omega} \approx 2ms $$

That is, each time you switch current direction, you must wait 2ms for current to reach a maximum of 3A, and then wait again to keep it there for a while, then switch direction and repeat.

Again, all of this assumes certain values for each element, and you'll have to experiment and measure, and read the datasheets for actual values.

In any case, your switching frequency will have to account for all these rise and fall times. For a coil of 1mH, it seems that 100Hz would be OK.

Now consider the power dissipated in each element.

The transistors will dissipate \$P_M=I^2R=3^2\times 0.2 = 1.8W\$. They'll need heatsinks, and perhaps a fan. Or you could use MOSFETS with much lower \$R_{DS(ON)}\$ than the IRF530 and IRF9530. (Edit: I just realised, they would dissipate half that amount, since they are on for only half the time)

Resistor R will dissipate \$P_R=3^2\times 1.7 = 15W\$. That's a lot. Maybe spread the power between four 6.8Ω, 5W resistors in parallel. Still, this is a difficult one to deal with, and is probably your most limiting factor.

Coil L will also get warm, \$P_L=3^2\times 0.25=2.3W\$. That's likely to be OK, since it's just a big lump of metal and wire.

That's enough to get your teeth into. There are other ways to control inductor current without having a large resistor to limit current, but they would involve monitoring inductor current and switching current direction when it reaches a certain threshold. That relies on the slow rise of current, which turns out to be useful, after all. This would be the "switch mode" approach, but it will need a closed-loop control system.

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