# Hef on a Cen-tech multimeter

#### dl324

Joined Mar 30, 2015
12,212
You were a lot closer with this idea. The inductor CAN force flow out of a common point because energy was stored in the magnetic field of the inductor as the current ramped up when the switch was first closed. That makes the inductor act like a tiny generator or battery at the moment the switch opens. The energy is already in the magnetic package. That's what is trying to resolve to its quiescent state.
OK. Something for me to study in my copious amounts of free time. Maybe I always had this doubt in my mind when working on switching regulators. I'm about to design one from scratch for fun and this may come up and force me to study it again...

#12 and WBahn, thanks for the mental stimulation. I knew I'd never amount to much in the analog world. That's why I switched to software, process technology, and microprocessor design...

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#### #12

Joined Nov 30, 2010
18,216
One of the glaring points about energy in an inductor is in the switching regulator. When you connect Vcc on the left side of an inductor and ground the right side of the inductor, current flows to ground. During that time the voltage across the inductor is almost completely equal to Vcc. Voltage times current is power, and that power is being stored in the magnetic field of the inductor. It isn't being wasted to ground. Individual electrons are flowing to ground, but the energy they carried is not.

When the grounding switch is turned off, the energy stored in the magnetic field is what forces current to flow to the output. That energy can be a much higher voltage than Vcc because the energy stored was 1/2 LI^2 and the resulting energy is 1/2 CV^2. If the capacitor is small, the voltage quickly becomes more than Vcc. I have a blog about this conversion and a jig to measure it.

I hope you enjoy the journey you are about to travel. I did. It opened my mind up to energy flow instead of electron flow. Electron flow was my Bible for years but inductors seem to violate that theory by storing energy. When the grounding switch opens, the first electrons are long gone, but their energy is still there in the magnetic field.

@dl324

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#### crutschow

Joined Mar 14, 2008
25,983
When the transistor opens, the inductance will attempt to keep the current moving in the same direction (from top to bottom of the inductor).
This causes the voltage to drop at the transistor inductor interface until the voltage goes about 0.7V negative, which forward biases the diode.
The inductor current now flows through the diode from the bottom of the inductor to the top of the inductor until the energy in the inductance has dissipated.
Just follow the inductor current and the voltages will follow from that.

#### WBahn

Joined Mar 31, 2012
26,398
non sequitur
Quite the opposite -- the whole phenomenon being discussed is due to the fact that, for an inductor, energy storage is proportional to current (squared) and since energy must be continuous, so must the current. That is why the voltage will spike when you open circuit an inductor -- because the inductor will produce whatever voltage is required to keep the same current flowing. So the best way to view what happens is to make the change to the circuit (shutting of the transistor, for instance) and then impose the constraint that the current in the inductor is the same as it was before the change and then ask what voltage will appear where in order to force that to become true. With the diode across the coil as shown, the inductor only has to produce about 0.7 V to keep the current in the inductor continuous.

#### WBahn

Joined Mar 31, 2012
26,398
Earth ground isn't an abstract concept. It's a 0V potential. Current that goes there can't come out unless it flows to a more negative potential. The other end of the inductor is at a positive potential while the field is collapsing.
The whole notion of "0 V potential" is a human construct. In many instances we define the 0 V potential to be an infinite distance away (just as we often define gravitational potential the same way).

The best way to think of Earth ground is probably to consider it an effectively infinite supply of available charge that can be dumped into or pulled out of without changing its potential in any meaningful way -- of course, certain phenomena, such as lightening strikes, push the bounds of that description pretty hard, but what we are talking about here doesn't.

#### WBahn

Joined Mar 31, 2012
26,398
That is not a foregone conclusion if the snubber diode works as WBahn and #12 are indicating.
Since he's talking about seeing "the R of the inductor", I think the smoke he is referring to has to do with the R being too low and the DC current handling capacity of the transistor being exceeded. But I'm reading between the lines, so I could be wrong.

#### WBahn

Joined Mar 31, 2012
26,398
I don't follow the analogy. Earth ground is at zero potential and I don't see how the inductor could force current to flow out of zero potential to a higher potential.
It doesn't. The inductor will produce whatever voltage is needed in order to keep the current continuous in the inductor. If that voltage is negative, then that voltage is negative.

But let's examine your confusion at an even more basic level.

How is the battery able to "force current to flow out of zero potential to a higher potential?

#### Roderick Young

Joined Feb 22, 2015
408
OK. Something for me to study in my copious amounts of free time. Maybe I always had this doubt in my mind when working on switching regulators. I'm about to design one from scratch for fun and this may come up and force me to study it again...

#12 and WBahn, thanks for the mental stimulation. I knew I'd never amount to much in the analog world. That's why I switched to software, process technology, and microprocessor design...
Another way to think about it, if you want to mentally think of ground as always 0 volts, is that the top of the inductor goes to a negative voltage when the transistor stops conducting. Without the diode, extremely negative, possibly hurting the transistor. One source of confusion may be that this circuit is set up the opposite way from what we often see with an NPN or NFET transistor, and the coil connected to positive supply.

#### #12

Joined Nov 30, 2010
18,216
@dl324

I don't want you to think we're dog-piling you. It's just that you have shown enough intelligence to learn this and we're all trying to help because it's worth the effort for somebody like you. Examine the, "several ways to say the same thing" and work with the one that seems to make the best sense to you.

#### dl324

Joined Mar 30, 2015
12,212
I don't want you to think we're dog-piling you.
As long as the posts are on topic and don't get personal, I'm fine with it
It's just that you have shown enough intelligence to learn this and we're all trying to help because it's worth the effort for somebody like you. Examine the, "several ways to say the same thing" and work with the one that seems to make the best sense to you.
I intend to fill in this gap in my knowledge; either with help from members who understand it, or can explain it in a way I can understand. Or on my own...

I intend to read each post in detail and respond in kind. It may just take some time because, even though I'm retired, I do have other things to do

I must have been nodding off the day they covered inductor back EMF

#### dl324

Joined Mar 30, 2015
12,212
Quite the opposite
The post I replied non sequitur to was one that stated the obvious about capacitor voltage and inductor current. We weren't discussing capacitor voltage change, nor were we questioning whether current was continuing to flow in the inductor after the switch was opened. Therefore my claim of non sequitur. It did not follow the train of thought in the conversation up to that point and added no value to the discussion.

#### dl324

Joined Mar 30, 2015
12,212
Perhaps a moderator can split this discussion from the OP's thread. It's been hijacked big time.

#### WBahn

Joined Mar 31, 2012
26,398
Perhaps a moderator can split this discussion from the OP's thread. It's been hijacked big time.
I thought about it, but the discussion is very relevant to the circuit that the TS posted. If the TS would like the thread split apart, then I will be happy to do so. The TS can also pretty easily pull it back in the direction they would prefer it to go.

#### KeepItSimpleStupid

Joined Mar 4, 2014
4,370
the OP said:
That is not a foregone conclusion if the snubber diode works as WBahn and #12 are indicating.
It's totally Irrelevant to the snubber discussion

It comes from the equation of v(t)= L di/dt; for the voltage across an inductor. If I changed instantaneously, v would be infinite. Note, no R in a circuit containing a voltage source and an ideal inductor. In the real world i would be limited by R.

#### dl324

Joined Mar 30, 2015
12,212
When the transistor opens, the inductance will attempt to keep the current moving in the same direction (from top to bottom of the inductor).
This causes the voltage to drop at the transistor inductor interface until the voltage goes about 0.7V negative, which forward biases the diode.
The inductor current now flows through the diode from the bottom of the inductor to the top of the inductor until the energy in the inductance has dissipated.
Just follow the inductor current and the voltages will follow from that.
This seems to be what is preventing me from comprehending what's happening I don't see how the voltage at the diode cathode is getting negative enough to forward bias the diode.

A voltage drop across the inductor caused by current continuing to flow after the switch is opened will not cause the diode to become forward biased. On the contrary, it causes the diode to become even more reverse biased.

EDIT: in this picture I'm showing the voltage drop that causes the diode to conduct, but violates KVL.

I know where the current is flowing, but the voltages don't conform to KVL; but I know they must.

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#### WBahn

Joined Mar 31, 2012
26,398
It's totally Irrelevant to the snubber discussion

It comes from the equation of v(t)= L di/dt; for the voltage across an inductor. If I changed instantaneously, v would be infinite. Note, no R in a circuit containing a voltage source and an ideal inductor. In the real world i would be limited by R.
The current would be limited by R, but the voltage when a sudden change in current is attempted would not. It's the snubber diode that limits that.

#### WBahn

Joined Mar 31, 2012
26,398
This seems to be what is preventing me from comprehending what's happening I don't see how the voltage at the diode cathode is getting negative enough to forward bias the diode.

A voltage drop across the inductor caused by current continuing to flow after the switch is opened will not cause the diode to become forward biased. On the contrary, it causes the diode to become even more reverse biased.
View attachment 93725
I know where the current is flowing, but the voltages don't conform to KVL; but I know they must.
Let's go back and revisit the basics.

The constitutive equation for an inductor is

$$v_L(t) \: = \: L \frac{di(t)}{dt}$$

If i_L in the diagram is positive when the switch is opened, do you agree that it will be decreasing after the switch is opened?

If it is decreasing, do you agree that

$$\frac{di(t)}{dt} \: < \: 0$$

If so, then v_L(t) is negative and, in this case, clamped by the forward voltage drop of the diode.

#### dl324

Joined Mar 30, 2015
12,212
I found a physics lecture from Rutgers that seems to have cleared it up for me.

The formula to calculate back EMF has a negative sign because the current is decreasing:
$$\small V_{emf}=-L\frac{di}{dt}$$

I must have been distracted when the negative sign was introduced for the collapsing field case.

Now having or not having ground or earth ground is immaterial because the inductor generated a negative voltage so KVL is obeyed.

Did anyone else learn something today? Be truthful

#### dl324

Joined Mar 30, 2015
12,212
If so, then v_L(t) is negative and, in this case, clamped by the forward voltage drop of the diode.
I agree. This will make my upcoming switching regulator design (for the fun of it) less black magical...

#### dl324

Joined Mar 30, 2015
12,212
is that the top of the inductor goes to a negative voltage when the transistor stops conducting.
Thanks Roderick. You learned it better than I did. I had completely forgotten that the voltage across an inductor changed polarity with the current was decreasing. Makes me wonder how much other "stuff" I've forgotten or never really learned