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I just watched Bullzoid/Actually Hardcore Overclocking's video on how VRMs actually work, and I have a question.

From wikichip

First, in my understanding, the diodes in the VRMs are included to stop the inductor from having a massive voltage build-up. Normally this could cause destruction of the CPU, so the diodes prevents an open circuit in case the two transistors are opened at the same time.

From physics EM, I know the current in the inductors tries to stay constant because changes in the induced magnetic flux are resisted due to Lenz's law.

However, I'm not sure why in an open-circuit configuration with both transistors off at the same time, the charged inductor's voltage would continue to rise to an extremally high voltage. The video says that it is due to the inductor's stored current being suddenly dumped out because of the rapidly collapsing magnetic field, but could someone clarify intuitively why the inductor's voltage only spikes with an open circuit, as opposed to the "clean" and safe discharging in a closed-circuit configuration with the diode?

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    \$\begingroup\$ note that this diode is implicit when using a Mosfet for the lower switch and a discrete diode may not be necessary \$\endgroup\$
    – tobalt
    Oct 2, 2022 at 1:15
  • \$\begingroup\$ To the OP --> What is a VRM? \$\endgroup\$
    – Andy aka
    Oct 2, 2022 at 7:50
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    \$\begingroup\$ VRM is a throw back to 90’s PCs where the voltage regulators for the cpu were sometimes a module, thus ‘Voltage Regulation Module’. The term somehow persists in reference to the voltage regulator for a cpu/gpu. Probably much like ‘chipset’ where the motherboard had a collection of chips to do the housekeeping. Now, a singular chip is called a ‘chipset’. \$\endgroup\$
    – Kartman
    Oct 2, 2022 at 12:00

3 Answers 3

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Let me walk you through it.

Assume the high side switch is on and current is flowing from PSU through choke to CPU.

Then, the high side switch is turned off, because, well it has to turn on and off to maintain the correct voltage at the CPU.

So a very short time AFTER the high side switch is turned off, the low side switch is turned on.

During that brief dead time when both transistors are off, the voltage at A will spike in the negative direction. Without the diode, it would shoot far below ground and probably destroy the high side switch.

But with the diode, the voltage at A is constrained to one diode drop below ground which is no problem for the high-side mosfet. The current through the diode will be approximately the same as the load current.

You might wonder why we have to turn the low-side on after the high side is turned off. Well, that dead time is needed because we can't risk having high and low side on at the same time (that would be a short circuit of the PSU). The dead time might be something like 100 to 300 nanoseconds. It is not much but it is very important.

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The diode in the diagram prevents the voltage from rising.

If you don't have a diode, such as with just battery and a coil, if you connect the battery to coil, current starts flowing, and when you disconnect battery, the circuit is open but the inductor would stil need to pass current, which is impossible. Thus inductor voltage rises in theory up to infinity to keep the current flowing. And that's why e.g. relay coils need diodes so switching transistors don't burn out or mechanical switches degrade from the arcing.

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Skip to the TL;DR further down, for the quick answer.

To illustrate inductor behaviour, and see why the voltage across an inductor can reach ridiculous, damaging levels, I'd like to use a switching setup for which it's easier to visualise current paths:

schematic

simulate this circuit – Schematic created using CircuitLab

In this arrangement, when SW1 is closed, SW2 is open and vice versa. I use voltage-controlled switches, under control of a 1kHz square signal to switch them on and off in anti-phase.

When SW1 is closed, the right-hand half of the circuit, consisting of only R2 is isolated, passes no current, and plays no role. In the second half of each cycle, SW1 opens, removing V1 and R1 from consideration, and SW2 closes so that R2 is now the only participating element, the only path for current. Inductor current is therefore routed alternately around the left loop and then the right, and so on.

While SW1 is closed, voltage source V1 is applied (via R1) across the inductor, and current will rise in that loop, initially very quickly, but settling eventually at some maximum \$I_{MAX}\$ determined by R1 and V1. This is really a very basic LR configuration, characterised by an exponential current curve:

$$ I = I_{MAX}(1-e^{-\frac{t}{\tau_1}}) $$

\$\tau_1\$ is the time constant associated with this exponential rise, and for a simple LR circuit like this, it is calculated as:

$$ \tau_1 = \frac{L_1}{R_1}= \frac{100\mu H}{1\Omega} = 100\mu s$$

This is the time it takes to reach about 63% of final value (10A in this case, 63% of which is 6.3A), so when we plot this current later, check for yourself that \$I\approx 6.3A\$ at a time 100μs following the start of the rise of current.

Eventually, when current is no longer changing, inductor L1 behaves like a short circuit, and the entire loop effectively becomes voltage source V1 connected directly across R1. When that happens, current is at a maximum, and \$I_{MAX}\$ will conform with Ohm's law:

$$ I_{MAX} = \frac{V_1}{R_1} = \frac{10V}{1\Omega } = 10A $$

The next graph shows our expected rise of current \$I\$ around the left loop, in a clockwise direction (downwards through L1; this is an important point), when SW1 is closed:

enter image description here

Now examine behaviour when SW1 is switched off, and SW2 is closed, introducing R2 and removing the entire the left-hand loop. By now, inductor current \$I\$ has already reached its maximum of 10A. The inductor has no regard for the sudden change of physical circumstances surrounding it, and since its behaviour is to oppose changes in current, that's exactly what it will do; continue to push 10A through whatever it's connected to.

Before we consider any voltages, let's continue our assessment of current, and find out what we should expect in the coming microseconds. Now we have different LR loop, with different time constant:

$$ \tau_2 = \frac{L_1}{R_2} = \frac{100\mu H}{2\Omega} = 50\mu s $$

Without a voltage source to deliver energy to the system, it is the inductor's own energy (stored in its magnetic field) which will drive current. That energy will deplete over time, causing an exponential decay of current:

$$ I = I_{MAX}e^{-\frac{t}{\tau_2}} $$

The decay is two times quicker than the rise, due to there being twice the resistance in the loop, and we should expect 63% of the decay to have occurred 50μs from the start of its fall from 10A:

enter image description here

Those two curves represent expected current \$I\$ during each phase (first and second halves of a complete cycle). Plotting two entire cycles together will give us:

enter image description here

Until now we have focussed on inductor current. Now it's time to consider voltages.


During the first phase, where a voltage source is used to "charge" the inductor (the word "charge" is not really appropriate, but I don't have a better one). The largest voltage across the inductor was the voltage of the source, initially 10V, diminishing eventually to zero. Current eventually reaches some maximum, which we calculated to be 10A.

In the second "discharge" phase, the inductor initially tries to maintain those 10A of current flow, but here's the thing: the external system it's connected to has a very different impedance now, and the inductor will develop whatever voltage is necessary to maintain 10A, regardless of what it's connected to.

We can predict the voltage that will appear across the inductor, the voltage necessary to pass 10A of current through a resistor of 2Ω, using Ohm's law:

$$ V = IR = 10A \times 2\Omega = 20V $$

That's the Eureka moment; we have obtained 20V from a system that has only one voltage source of 10V. There's one more thing to consider, direction of current, and polarity of voltage.

From the graphs of current above, it's clear that it never changed direction. It is always downwards through inductor L1. This makes sense if we remember that the inductor opposes change in current. If current ever suddenly changed direction, this would be a gross violation of that principle.

Given that current is always downwards through L1, then it follows that during the discharge phase, it must be flowing anti-clockwise around the right-hand loop, upwards through R2. Current always enters a resistor at the end with higher potential, and emerges from the lower potential end. This implies that node X must be negative.

At the instant the switches changed state, and the inductor went from being energised via R1 to delivering energy to R2, the voltage across the inductor does actually flip polarity, as well as magnitude. We can see this if we plot the potential at X:

enter image description here

The big takeaway is that voltage across the inductor during "charging" will be whatever the source V1 produces (eventually falling to zero), and the eventual current through the inductor will be capped by whatever impedance is present around the charging loop. However, when the switches flip, current will stay momentarily at that maximum, and the voltage across the inductor will be whatever it takes to push that current through whatever load is now present across the inductor, and with opposite polarity.

You can obtain whatever voltage you desire across R2, at that instant of switch-over. Knowing that the inductor will push 10A through it, just choose a resistance that will develop your required voltage with 10A through it, according to Ohm's law. Let's aim for 100V:

$$ R = \frac{V}{I} = \frac{100V}{10A} = 10\Omega $$

schematic

simulate this circuit

enter image description here

I won't address what happens when you connect a capacitor, or diodes, across the inductor in phase 2, but I will talk about what happens when we connect nothing there.


TL;DR

When an inductor is passing a current, and then that current path is suddenly broken, it does what it always does, keeps that current flowing. However, there's nothing there to conduct it!

What happens is the voltage across the inductor suddenly flips polarity and rises to whatever potential is necessary to push that current through something, whether that be its own winding insulation, or the air between its end terminals. One or the other will break down first, and become conductive under the stresses of such a strong electric field, and you'll no doubt see some nice sparks.

The answer to your question is that the inductor will produce however many thousands of volts is necessary to cause the materials around it to ionise, break down, and conduct. That's why you see such huge potentials appear across the inductor when its current path is suddenly interrupted.

The magnetic field's energy must be dissipated somewhere, as the field collapses, which might be coupled magnetically to other conductive loops in the vicinity, or capacitively to any other conductive materials nearby.

A lot of damage can occur, if you fail to provide a sufficiently low impedance path for that current to flow around. Frequently, we could uss just a diode. In the case of DC-DC converters, we dump as much of that energy as possible into a capacitor.

In my above examples, the recipient of that energy was a simple resistor, and by controlling its value, we were able to control the magnitude of potential "spike" that occurs when inductor current experiences a sudden change in path impedance. I hope I have demonstrated that the voltage produced by an inductor will depend entirely upon the current it was passing and the impedance of the "new" loop that is formed, when the "old" loop is broken. In the absence of any explicitly provided current path, with deliberately controlled and known impedance, the only remaining current path will be through whatever parasitic impedances are unlucky enough to find themselves in the vicinity, impedances which will be very high, and which will fall victim to an uncontrolled and indeterminately enormous voltage spike.

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