Polarity reversed coil - how to protect FETs from back EMF?

Thread Starter

hazyj

Joined Dec 14, 2020
29
Hello. My first time here and I'm sure I'll make at least a few mistakes. I've searched but can't find an answer to my question so I'm hoping I can find it this way. My EE knowledge is pretty rusty as you'll soon see - thanks for your patience as you read on ...

The diagram attached shows a single coil to be pulsed by n & p FETs. Each FET has 100V d-s breakdown and < 10 mOhm Rds. Vgs is +/-2V for each. There is no possibility of both FETs ever switching on at the same time.

coil1.png

The coil is ~ 50mH & Vdd may be as low as 12V and as high as 50V as I experiment with pulse widths, duty cycles and field strengths. I realize back EMF may be significant, but I'm not sure how to protect the FETs. Current demand per pulse will be way below FET capability.

I realize this circuit will need a lot of work and caps added to lessen ringing/noise, but my concern for now is to protect the FETs. I also realize I'm a bit challenged with regard to correct wiring of T2. I ran out of time - sorry about the mess. The ESP32 is a 3.3V device.

Last point which I'm sure some will ask me regards pulse widths and duty cycle. The hope is to create pulses at widths between 0.1ms and 2ms. Duty cycle flexible depending on noise, heat, ringing and of effects from back EMF. The ideal case which I don't expect to achieve: 0.5ms pulses that aren't a terrible mess (some ringing/noise will be acceptable). Current approx 0.5A. Duty cycle 0.3 max. Just the ideal case. The final result can be far off from this and I'll be happy.

Many thanks for your help. I'll be surprised if my diagram doesn't have horrible flaws. I'm just getting back into this so please go easy on me!
 

Deleted member 440916

Joined Dec 31, 1969
0
I dont understand your circuit very well, why are the batteries not referenced to ground ?
In any case the FETS have inverse diodes internally between the source and drain so those diodes will prevent any excess back emf from exceeding the rail voltages
 

Thread Starter

hazyj

Joined Dec 14, 2020
29
I dont understand your circuit very well, why are the batteries not referenced to ground ?
In any case the FETS have inverse diodes internally between the source and drain so those diodes will prevent any excess back emf from exceeding the rail voltages
Really sorry about that! I'm fixing the diagram now wrt. batteries. Regarding inverse diodes, I didn't think they'd be enough given large emf. So you're meaning is that back emf won't ever be so large as to breakdown the diodes. Correct?
 

crutschow

Joined Mar 14, 2008
27,463
The circuit is incorrect.
The top P-MOSFET is connected with a positive voltage to the drain, which is incorrect biasing.
Did you intend for the drain to be connected to the load, the same as the bottom N-MOSFET?
 

Thread Starter

hazyj

Joined Dec 14, 2020
29
The circuit is incorrect.
The top P-MOSFET is connected with a positive voltage to the drain, which is incorrect biasing.
Did you intend for the drain to be connected to the load, the same as the bottom N-MOSFET?
Whoops. Circuit is correct but diagram isn't. pFET is upside down. I'd never wire it up this way - I just don't know the symbols. Looks like I've really botched this question because of bad diagramming skills. Sorry ... wasted most of the day trying to get Tina-Ti working on my Mac and it's been a disaster. 4 hours absolutely wasted.

I'll fix the diagram now but I guess no way to replace it. Just keep embedding more images. Any idea why that's the policy here?
 

michael8

Joined Jan 11, 2015
155
inductor about 50 mH (40.8 mH?), vdd 12 to 60 volts, pulses widths 0.1ms to 2ms, current about 0.5A

Just considering the inductor, Vdd using E = L * di/dt => di/dt = E/L

for 12 volts 12/40.8e-3 -> 294 A/second, so the current will reach 0.5A in 1.7 mS
for 60 volts 60/40.8e-3 -> 1470 A/second, so the current will reach 0.5A in about 340 uS

OK, you've reached your target current. Now what?

If you leave the mosfet on you will exceed it your target current.

If you turn the mosfet off the internal mosfet diodes will clamp the voltage on the inductor to about +Vdd or -Vdd
and the current will ramp down.
 

Thread Starter

hazyj

Joined Dec 14, 2020
29
inductor about 50 mH (40.8 mH?), vdd 12 to 60 volts, pulses widths 0.1ms to 2ms, current about 0.5A

Just considering the inductor, Vdd using E = L * di/dt => di/dt = E/L

for 12 volts 12/40.8e-3 -> 294 A/second, so the current will reach 0.5A in 1.7 mS
for 60 volts 60/40.8e-3 -> 1470 A/second, so the current will reach 0.5A in about 340 uS

OK, you've reached your target current. Now what?

If you leave the mosfet on you will exceed it your target current.

If you turn the mosfet off the internal mosfet diodes will clamp the voltage on the inductor to about +Vdd or -Vdd
and the current will ramp down.
Hi Michael. Thanks for slogging through all my imprecision. From here on I hope to be much more accurate with details.

Regarding current, I didn't know how much info to include in the post. 0.5A was my max estimate for a coil @ 140 ohms and max voltage which has yet to be determined. While I'll be experimenting with max values to determine tolerances, eventually I expect to settle for Vdd between 24V and 32V so a max current ~ 0.23A. Further, I don't need to reach max current per pulse.
The primary goal (for this week anyhow) is to produce nice looking pulses. I'm concerned that back emf will get in my way toward that goal.
 

Thread Starter

hazyj

Joined Dec 14, 2020
29
From your input I'm less concerned about the FETs now and instead wondering how to quickly attenuate back emf. I've shunted the coil with a 2.2k resistor here. Not sure if a diode of any sort would make sense. Thoughts?

6A1BBB85-1B4D-4DF5-B1C7-20112C2CF78A.jpeg
 

michael8

Joined Jan 11, 2015
155
Oh, 140 ohms is a lot more than I was thinking. At 12 volts
the maximum current is only 85 mA, at 60 volts 420 mA. You won't
get to 0.5A.

As far as "back EMF", look at the current flows.

When the n-mosfet turns off the current which was flowing through the coil
will continue flowing. It will raise the drains of both mosfets until
it's greater than +Vdd and the internal diode in the p-mosfet conducts.
That will clamp the voltage to one diode drop above +Vdd.

It sounds like your problem is going to be getting to your 0.5A
current.
 

Thread Starter

hazyj

Joined Dec 14, 2020
29
Oh, 140 ohms is a lot more than I was thinking. At 12 volts
the maximum current is only 85 mA, at 60 volts 420 mA. You won't
get to 0.5A.

As far as "back EMF", look at the current flows.

When the n-mosfet turns off the current which was flowing through the coil
will continue flowing. It will raise the drains of both mosfets until
it's greater than +Vdd and the internal diode in the p-mosfet conducts.
That will clamp the voltage to one diode drop above +Vdd.

It sounds like your problem is going to be getting to your 0.5A
current.
As I said, 0.5A was an estimate intended to keep from bogging anyone down with details. "Current approx 0.5A" and "0.5A was my max estimate for a coil @ 140 ohms" was what I wrote and I stand by it as a very good approximation for where I'm headed with this. I don't see why anyone would need for me to write "0.43A" - it changes nothing for such a general question because breakdown is 100V per FET.

In any case, I also continued with "I'm less concerned about the FETs now and instead wondering how to quickly attenuate back emf. I've shunted the coil with a 2.2k resistor here. Not sure if a diode of any sort would make sense. Thoughts?"

Also provided a new schematic. Anyhow, I think I have a handle on this. Just needed to think it over for a few hours and get back into an EE mindset. It's been awhile, clearly.

Thanks!!
 
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Thread Starter

hazyj

Joined Dec 14, 2020
29
When the n-mosfet turns off the current which was flowing through the coil
will continue flowing.
I doubt that's what you meant to write. Maybe you mean that *if* induced EMF due to the quick cutoff exceeds Vdd (-Vdd for nFET) then current will keep flowing. But the current doesn't raise the drains - the EMF does that and as a result current flows only if drain exceeds Vdd (-Vdd for nFET).
 

michael8

Joined Jan 11, 2015
155
No that is what I meant. There is no "*if* induced EMF".

My 2 cent inductor model is that the current in an inductor doesn't
change in an instant, instead the voltage across the inductor changes
to whatever it takes to continue the previous current flow.

I think this is a better model than some mysterious back EMF. Since you
know the current you can calculate the resultant voltage required to
maintain it (for an instant, the current will be decreasing as the energy
comes out of the inductor).

This doesn't specify where the current flows -- you need to figure
that out. It could flow into charging capacitances without much voltage
rise and into resistances like your 2.2K shunt resistor. But the
voltage across the inductor still needs to rise enough to continue
the current flow. I'm a firm believer in conservation of energy
and the energy in the "charged" inductor has to go somewhere if
the current in the inductor is going to change.

I don't think the 2.2K resistor makes much difference, it only takes
about 5 mA of 85 mA at 12 volts and and 27 mA of 420 mA at 60 volts
so most of the current is going to flow to +/-Vdd via the mosfet
diodes. This doesn't take very long as the 140 ohm coil resistance
is a major energy sink.
 

Thread Starter

hazyj

Joined Dec 14, 2020
29
No that is what I meant. There is no "*if* induced EMF".

My 2 cent inductor model is that the current in an inductor doesn't
change in an instant, instead the voltage across the inductor changes
to whatever it takes to continue the previous current flow.

I think this is a better model than some mysterious back EMF. Since you
know the current you can calculate the resultant voltage required to
maintain it (for an instant, the current will be decreasing as the energy
comes out of the inductor).

This doesn't specify where the current flows -- you need to figure
that out. It could flow into charging capacitances without much voltage
rise and into resistances like your 2.2K shunt resistor. But the
voltage across the inductor still needs to rise enough to continue
the current flow. I'm a firm believer in conservation of energy
and the energy in the "charged" inductor has to go somewhere if
the current in the inductor is going to change.

I don't think the 2.2K resistor makes much difference, it only takes
about 5 mA of 85 mA at 12 volts and and 27 mA of 420 mA at 60 volts
so most of the current is going to flow to +/-Vdd via the mosfet
diodes. This doesn't take very long as the 140 ohm coil resistance
is a major energy sink.
Appreciate the willingness to continue with this. First, let me be clear that I've seen plenty of emf spikes on my scopes over the years, so I know what you're getting at, and this doesn't have to be only about theory. Second, spikes occur because an attempt to quickly cutoff current results in large rate of change of current in the direction opposite to flow. The fact that cutoff begins at all means (for nFET case) |Vgs| is below threshold and |Vds| and Rds are increasing rapidly. Again, Rds is increasing rapidly. You cannot have much current flow continue unless and until one of two things happen: Rds comes back down rapidly or if |EMF-Vdd| > 0. You seem to be saying that Rds comes back down to a low value even with |Vgs| ~ 0V?

That doesn't make sense, so I assume you're making the point that for this very short burst I'll see a condition where |EMF-Vdd| > 0. There is no guarantee that this will happen though, hence my very firmly worded statement "*if* induced EMF due to the quick cutoff exceeds Vdd (-Vdd for nFET) then current will keep flowing". This is what we're trying to prevent isn't it? In short, how to keep |EMF-Vdd| < 0. If successful current will not keep flowing. Not through the FET or its diode anyhow. Unless you mean a very small current through the enormous Rds.

I agree my shunt resistor doesn't help much. A shunt cap makes sense though, because as emf starts to spike at cutoff the cap works to lower its value by essentially shorting that end of the coil to ground during that spiking. So I believe what I need is an RC shunt but one that doesn't cause ringing. Back to energy conservation: for most of the pulse width I'm interested in I can consider rates of change to be on the order of 2000 - 5000hz. Goal is for energy to be almost entirely magnetic in this domain. So back of enveloping it ... electric field energy can dominate within the cap above 10,000hz. So isn't it a simple matter to create an RC shunt with XC << XL above 10,000hz?
 

Thread Starter

hazyj

Joined Dec 14, 2020
29
Let me create a schematic appropriate to this last point. I'll crunch some numbers and throw out an appropriate RC shunt and see where the conversation goes after that.
 

michael8

Joined Jan 11, 2015
155
> spikes occur because an attempt to quickly cutoff current results in
> large rate of change of current in the direction opposite to flow

Rather spikes occur due to the curent which was flowing at the instant
before still flows after that instant. The spike is just the voltage
required for the same current flow to continue.

> The fact that cutoff begins at all means (for nFET case) |Vgs|
> is below threshold and |Vds| and Rds are increasing rapidly. Again,
> Rds is increasing rapidly. You cannot have much current flow continue
> unless and until one of two things happen: Rds comes back down rapidly
> or if |EMF-Vdd| > 0. You seem to be saying that Rds comes back down to
> a low value even with |Vgs| ~ 0V?

No, my model is only for the inductor, I'm assuming the nFET is an
instantaneous switch with zero switching time and zero resistance.
The nFETs, do however, have an internal diode across them and when
one switches off the inductor current can continue through the *other*
nFETs diode to whichever Vdd it's source is connected. This clamps
the voltage across the inductor to Vdd plus the diode drop.

Do you have ltspice?
 

Thread Starter

hazyj

Joined Dec 14, 2020
29
Doesn't this 6nF cap acting as a shunt help considerably?

8FCC0F05-E25D-4520-B407-4FC68F0A5DE1.jpeg

The shunt should conduct most of the current at cutoff but for most of coil rise time (eg. approx 0.2-1.0ms) it does not conduct much at all.
 
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Thread Starter

hazyj

Joined Dec 14, 2020
29
Michael, I may not be an EE but I do have a physics PhD, and I'm sorry but i need to correct some of this...

spikes occur due to the current which was flowing at the instant
before still flows after that instant. The spike is just the voltage
required for the same current flow to continue.
Absolutely not! The spike is the response to changing magnetic field! That field changes because the current is *rapidly* being cutoff. When that happens a large electric field is produced. This is the EMF which then moves current. I think the reason you think the current wants to "continue to flow" is because the rate of change of current is opposite to the flow (must be since current is lessening since Rds is increasing). Since the rate of change is opposite the current an EMF is induced which moves current in the same direction it was already moving.

No, my model is only for the inductor, I'm assuming the nFET is an
instantaneous switch with zero switching time and zero resistance.
The nFETs, do however, have an internal diode across them and when
one switches off the inductor current can continue through the *other*
nFETs diode to whichever Vdd it's source is connected. This clamps
the voltage across the inductor to Vdd plus the diode drop.
As I have made clear this point is true only if |EMF - Vdd| > 0. Until that occurs the diode has nothing to do.

Regarding Itspice, is it free? I'm using Tina-TI Spice because I've had it for years, but I'll move to something else if it's better and free. I guess Fritzing isn't free?
 
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