Looking for pulse-rated zener.

[Winfield Hill]

Robert Baer wrote...

John Popelish wrote:
Robert Baer wrote:

One thing to consider:
Allowing current to flow after the switch opens will mean that the
magnetic field will continue to exist while it collapses, and if
current continuse to flow,the magnetic field will reverse.

(snip)

I think you are laboring under a misconception, here. The coil
inductive voltage reverses the moment the current starts to decrease
(V=L*(di/dt)), but the magnetic field polarity does not change till the
current changes direction. Once the current passes through zero and the
coil (and zener and switch) stray capacitance that is sitting at zener
voltage starts to dump current into the coil as the capacitance
discharges back toward zero volts, then the magnetic field polarity will
also pass through zero and reverse.

Err...
Take a look at the current and voltage waveforms in a flyback system.
Turn on the switch, inductive current increases in the standard R/L form.
When the switch is opened, the magnetic field starts to collapse; the
waveform across the inductor is square-ish in most practical circuits.
During that time, the current goes rapidly to zero as the flyback
voltage pulse rises; roughly remains near zero at the flattish top,

This is incorrect. The current goes "rapidly" to zero _after_
the flyback voltage pulse rises, during the flattish top...

and then goes negative as the flyback voltage pulse drops to zero.
However, the voltage would continue to decrease and go negative
(L-C oscillations), but the switch (FET) internal diode conducts,
allowing the coil current to continus to flow.
The voltage pulse seen has the *same* polarity as the supply.
In a standard flyback scheme, the negative current waveform is
mirror-image of the ramp-like charging time.
This remains to be a fairly close picture of operation waveforms,
even if the core of the inductor saturates to some extent.

John is correct, the current shouldn't reverse, excepting perhaps
for a small amount of ringing. Perhaps, when you speak of mirror
image, you're thinking of a reversal of the rate-of-change in the
flyback coil current, rather than the current polarity, per se?

In the rapid magnetic-field collapse system we're talking about,
using MOSFET or TVS avalanche to absorb the coil's energy, the
avalanching junction will stop conducting the instant the current
drops to zero (the physics of avalanche doesn't have any reverse-
recovery time). There will be a small amount of current reversal
(and ringing) due to discharging the system capacitance from the
avalanche voltage level back to the supply V, but it'll be small
compared to the 15A magnetizing current.

For example, a 34n20 FET's capacitance is 400pF at 150V, a 1.5kW
150V TVS is 100pF, and 1 meter of cable is another 100pF. The
resonant frequency with 8uH will be 2.3MHz. The energy stored
in 600pF of capacitance at 130V (assume Vs = 20V supply for the
coil) pushes the inductor to a peak reversal of i = V sqrt(C/L)
= 1.1A, and the voltage to +20V -130V = -110V after a T/2 time
of 220ns, assuming nothing else in the path. However, the FET's
intrinsic body diode and the TVS diode will prevent the voltage
from going below ground, bringing things to a stop after only
130ns, limiting the ringing to the 20V supply Vs, and the peak
reversal current to about -600mA, only 4% of the original 15A.

Since the magnitude is only 4% and it only lasts a few hundred
ns, it's not useful to characterize this as "current reversal."

A 2pi f L = 100-ohm resistor paralleled with the inductor would
nicely damp the 2.3MHz ringing, and reduce the magnitude of the
single ring as well, to say 2%. It would need to be a 3W part,
etc., unless a 0.01uF series capacitor was added to stop any DC
current. We call this R-C network a snubber.


--
Thanks,
- Win

 

[Mark Becker]

Hello -

First, I wish to thank everyone participating in this discussion. While
I'm an EE, it has been a long time since some of this has crossed my
mind.. and the technology has improved considerably. The discussion has
poked a few embers.. Thank you.

Why a zener? Why not just a fast diode to circulate the current back through the inductor?

Using the effective series resistance of the inductor does not dissipate
the energy fast enough. For 8 uH and a series resistance of something
like .050 ohms, tau is on the order of 160 microseconds.

The zener has the feature of placing an (almost) constant voltage across
the coil during field collapse. Ideally, superposition applies.. and
I(t) = I(0) - Integral(v(t), t)/L. Faster than exponential decay.

[ TVS ]

Think I looked at this when first thinking about it.. and wasn't happy
at the response time. I'll go and look again. I am also not familiar
with "avalanche mode" transient absorbers and will stare at those as
well.

The current circuit (pun not intended) uses a pair of IRFB18N50K MOSFETs
and a pair of series-connected BZW03D100 100V across the MOSFETs. Hmm..
maybe the MOSFETs should be across the coils with a series fast diode..

Another poster suggested using the MOSFET intrinsic diode as the zener
clamp. I stared at that one.. is this feasible (I'm still climbing out
of using BJTs) ? I like the idea of the snubber better.. but have to
think a little more on it.

Especially as now the researcher wants the magnetic field to enclose a
larger volume. The coils for THAT assembly work out to about 400 uH ..
and still need 15A.

Your time is *really* appreciated.

Mark

 

[John Popelish]

Robert Baer wrote:

John Popelish wrote:

Robert Baer wrote:

One thing to consider:
Allowing current to flow after the switch opens will mean that the
magnetic field will continue to exist while it collapses, and if
current continuse to flow,the magnetic field will reverse.

(snip)

I think you are laboring under a misconception, here. The coil
inductive voltage reverses the moment the current starts to decrease
(V=L*(di/dt)), but the magnetic field polarity does not change till
the current changes direction. Once the current passes through zero
and the coil (and zener and switch) stray capacitance that is sitting
at zener voltage starts to dump current into the coil as the
capacitance discharges back toward zero volts, then the magnetic field
polarity will also pass through zero and reverse.

Err...
Take a look at the current and voltage waveforms in a flyback system.
Turn on the switch, inductive current increases in the standard R/L form.

Yes. And the magnetic field is instantaneously proportional to coil
current.

When the switch is opened, the magnetic field starts to collapse; the
waveform across the inductor is square-ish in most practical circuits.
During that time, the current goes rapidly to zero as the flyback
voltage pulse rises; roughly remains near zero at the flattish top, and
then goes negative as the flyback voltage pulse drops to zero.

No argument. The switch current goes to zero first, the coil current
detours first to the coil capacitance, and to snubber, if any. Then
the voltage rises till either the capacitance (and snubber) absorb all
the energy stored in the coil, or a clamp voltage is reached. If the
latter (as the op is doing) then the current detours to the clamp.
The current continues to ramp toward zero as does the magnetic field.
As the current and magnetic field pass through zero, the voltage
starts to swing back toward that connected to the other end of the
coil (toward zero volts across the coil). But to change the voltage
across the coil requires current to charge the stray capacitance
across those nodes, so this current passes through the coil and and
this causes a reversal of the magnetic field.

Perhaps I misunderstood your last post when you said, "Allowing
current to flow after the switch opens will mean that the magnetic
field will continue to exist while it collapses, and if current
continues to flow, the magnetic field will reverse."

How could the field not continue to exist while it collapses? And the
coil current must cease and change directions before the field ceases
and changes directions.

However, the voltage would continue to decrease and go negative (L-C
oscillations), but the switch (FET) internal diode conducts, allowing
the coil current to continus to flow.
The voltage pulse seen has the *same* polarity as the supply.
In a standard flyback scheme, the negative current waveform is
mirror-image of the ramp-like charging time.
This remains to be a fairly close picture of operation waveforms, even
if the core of the inductor saturates to some extent.

My only point was that the applicable generalization for this op is
that the magnetic field is instantaneously proportional to the coil
current at all times, and the field reverses exactly when the current
reverses. If you agree with this than we are on the same page.

 

[Rich Grise]

On Sat, 18 Jun 2005 07:22:15 -0400, Mark Becker wrote:

Hello - (hope this is the right place .. pardon the cross-post)...

Crossposting is actually preferred, when it's applicable.

I have found only one maker of pulse-rated zener diodes.. and while I
was able to get a few as samples, now a small quantity is needed (<100)
and these things are only sold in bulk. I don't need 12,000 or 25,000
diodes, not for a research effort.

You don't want a Zener doide here - what you want is a TransZorb!
http://www.google.com/search?q=transzorb

They're essentially high-power zeners that are designed to switch fast
and absorb amazingly high transients.

Good Luck!
Rich

 

[Winfield Hill]

Mark Becker wrote...

First, I wish to thank everyone participating in this discussion. While
I'm an EE, it has been a long time since some of this has crossed my
mind.. and the technology has improved considerably. The discussion has
poked a few embers.. Thank you.

Why a zener? Why not just a fast diode to circulate the current back
through the inductor?

Using the effective series resistance of the inductor does not dissipate
the energy fast enough. For 8 uH and a series resistance of something
like .050 ohms, tau is on the order of 160 microseconds.

Yes, the t = L/R time constant is what's at work there. So can we
assume your coil's resistance is about R = L/tau = 0.05 ohms? This
means it should drop about 0.75V at 15A. How are you establishing
the 15A current?

The zener has the feature of placing an (almost) constant voltage across
the coil during field collapse. Ideally, superposition applies.. and
I(t) = I(0) - Integral(v(t), t)/L. Faster than exponential decay.

Yes, that's the appeal of any high-voltage coil flyback scheme.

Think I looked at this when first thinking about it.. and wasn't happy
at the response time. I'll go and look again. I am also not familiar
with "avalanche mode" transient absorbers and will stare at those as
well.

You're familiar with zener diodes, same thing... Same junction,
just added blocks of copper to absorb transients without heating.

The current circuit (pun not intended) uses a pair of IRFB18N50K MOSFETs
and a pair of series-connected BZW03D100 100V across the MOSFETs.

Well, the IRFB18N50K are 500V FETs with a rather high Ron, I'd
say, not the best choice for your 15A current.

Hmm..
maybe the MOSFETs should be across the coils with a series fast diode..

??

Another poster suggested using the MOSFET intrinsic diode as the zener
clamp. I stared at that one.. is this feasible (I'm still climbing out
of using BJTs) ?

Hey, I don't usually think of myself as just another poster....

I like the idea of the snubber better.. but have to think a little
more on it.

Yes, fine.

Especially as now the researcher wants the magnetic field to enclose a
larger volume. The coils for THAT assembly work out to about 400 uH ..
and still need 15A.

That'll be fun. Now you'll have 45mJ, getting up into slightly
more interesting territory... You can use larger TVS parts
(5kW) or use lots of 1.5kW parts in series, or you can use
this common approach, recommended by someone else here, IIRC.

.. 8A <-- coils
.. ,---UUUUU--UUUUU--- constant-current supply
.. | or supply + ballast R
.. |
.. | diode storage cap
.. +---|>|--+---||--- gnd
.. | | / zener, TVS p6ke150A etc.
.. | '-/\/\--|<|---- gnd
.. |--' /
.. ||<-, 200V FET
.. --/\/\--'|--+ fqaA34n20
.. |
.. gnd

Here you deliver the flyback energy to the capacitor, and use
the zener simply to limit the capacitor voltage and ultimately
to dissipate the energy.

Let's see, 400uH. Oops! You'll have to raise the flyback to
V = LI/t = 6kV to stop in 1us, now I'll say, that's decidedly
more interesting! Just don't kill yourself.

Eventually you may have to redesign your coil parameters, etc.,
but in the meantime perhaps you can run them separately, rather
than in series, dramatically that reducing the inductance. And
you can change to serious 1200V IGBTs rather than puny FETs...

.. 15A <-- coil (one of two)
.. ,---UUUUU--- constant-current supply
.. |
.. | diodes storage caps
.. +-|>|--|>|--+---||--||--||-- gnd
.. | | 150V TVS (8)
.. | '--|<|--|<|- - - -|<|-- gnd
.. |/V
.. || 1200V IGBT t = 15A 100uH/1.2kV = 1.25us
.. --/\/\--'|\V IRGP30B120KD etc
.. | on 40W heat sink
.. |
.. gnd

Your time is *really* appreciated.

You're very welcome.

Next your boss will decide he needs more current. :&gt
What's the magnetic field and its rapid shutoff for anyway?


--
Thanks,
- Win

 

[Lasse Langwadt Christense]

Winfield Hill wrote:
snip

A second point is that such an active-zener method isn't necessary,
because most power MOSFETs are just as happy dissipating 250mJ in
avalanche mode as in their active power regions. So there's little
reason to use them in a fashion that risks high-power oscillation.
(In the *old* days some FETs had problems with possible gate damage
due to the current pathways during avalanche. To satisfy oneself
that the FET has been designed to avoid this, and is safe to use,
look for an Eas Single-Pulse Avalanche Energy rating on the spec
sheet. Most parts have this rating, including the 34n20 types.)

I've seen an app-note, onsemi I think, that stated that
the fets could handle more power with the active clamping

And you could can get transistors with the clamping buildin
though I guess it's mostly ~40 volt FETs for automotive or
~400 volt igbts for ignition

snip

--
-Lasse

 

[Winfield Hill]

Lasse Langwadt Christensen wrote...

Winfield Hill wrote:
snip

A second point is that such an active-zener method isn't necessary,
because most power MOSFETs are just as happy dissipating 250mJ in
avalanche mode as in their active power regions. So there's little
reason to use them in a fashion that risks high-power oscillation.
(In the *old* days some FETs had problems with possible gate damage
due to the current pathways during avalanche. To satisfy oneself
that the FET has been designed to avoid this, and is safe to use,
look for an Eas Single-Pulse Avalanche Energy rating on the spec
sheet. Most parts have this rating, including the 34n20 types.)

I've seen an app-note, onsemi I think, that stated that the
fets could handle more power with the active clamping

It'd be nice if you could find the reference. It's a common piece
of urban engineering wisdom, but it doesn't hold up to theory, to
calculations using datasheets of real parts, or to measurements.

And you could can get transistors with the clamping buildin
though I guess it's mostly ~40 volt FETs for automotive or
~400 volt igbts for ignition

The predictability of an added gate-zener function may be desirable.


--
Thanks,
- Win

 

[Winfield Hill]

Winfield Hill wrote...

Lasse Langwadt Christensen wrote...

Winfield Hill wrote:
snip

A second point is that such an active-zener method isn't necessary,
because most power MOSFETs are just as happy dissipating 250mJ in
avalanche mode as in their active power regions. So there's little
reason to use them in a fashion that risks high-power oscillation.
(In the *old* days some FETs had problems with possible gate damage
due to the current pathways during avalanche. To satisfy oneself
that the FET has been designed to avoid this, and is safe to use,
look for an Eas Single-Pulse Avalanche Energy rating on the spec
sheet. Most parts have this rating, including the 34n20 types.)

I've seen an app-note, onsemi I think, that stated that the
fets could handle more power with the active clamping

It'd be nice if you could find the reference. It's a common piece
of urban engineering wisdom, but it doesn't hold up to theory, to
calculations using datasheets of real parts, or to measurements.

And you could can get transistors with the clamping buildin
though I guess it's mostly ~40 volt FETs for automotive or
~400 volt igbts for ignition

The predictability of an added gate-zener function may be desirable.

Actually, there is another possible reason. Higher inductor flyback
voltages mean shorter inductor discharge times. Dissipating a fixed
amount of energy in a shorter time means higher junction temperatures
because the heat can't diffuse as far through metal frames, and into
the heatsink, etc. So that's an argument both for lower FET clamping
voltages and for more predictable voltages.


--
Thanks,
- Win

 

[Jim Thompson]

On 20 Jun 2005 18:09:32 -0700, Winfield Hill
<hill_a@t_rowland-dotties-harvard-dot.s-edu> wrote:

[snip]

Actually, there is another possible reason. Higher inductor flyback
voltages mean shorter inductor discharge times. Dissipating a fixed
amount of energy in a shorter time means higher junction temperatures
because the heat can't diffuse as far through metal frames, and into
the heatsink, etc. So that's an argument both for lower FET clamping
voltages and for more predictable voltages.

On CMOS ASIC's it can get even trickier. Any "diode" will have a
myriad of parasitics, thus using flyback diodes is risky business.

Years ago, on an Ethernet chip that also needed relay drivers, I
developed a scheme that simply turns the driver transistor back on if
it's drain voltage exceeds VDD by one volt or whatever.

...Jim Thompson
--
| James E.Thompson, P.E. | mens |
| Analog Innovations, Inc. | et |
| Analog/Mixed-Signal ASIC's and Discrete Systems | manus |
| Phoenix, Arizona Voice480)460-2350 | |
| E-mail Address at Website Fax480)460-2142 | Brass Rat |
| http://www.analog-innovations.com | 1962 |

I love to cook with wine. Sometimes I even put it in the food.

 

[Robert Baer]

Winfield Hill wrote:

Robert Baer wrote...

John Popelish wrote:

Robert Baer wrote:


One thing to consider:
Allowing current to flow after the switch opens will mean that the
magnetic field will continue to exist while it collapses, and if
current continuse to flow,the magnetic field will reverse.

(snip)

I think you are laboring under a misconception, here. The coil
inductive voltage reverses the moment the current starts to decrease
(V=L*(di/dt)), but the magnetic field polarity does not change till the
current changes direction. Once the current passes through zero and the
coil (and zener and switch) stray capacitance that is sitting at zener
voltage starts to dump current into the coil as the capacitance
discharges back toward zero volts, then the magnetic field polarity will
also pass through zero and reverse.

Err...
Take a look at the current and voltage waveforms in a flyback system.
Turn on the switch, inductive current increases in the standard R/L form.
When the switch is opened, the magnetic field starts to collapse; the
waveform across the inductor is square-ish in most practical circuits.
During that time, the current goes rapidly to zero as the flyback
voltage pulse rises; roughly remains near zero at the flattish top,


This is incorrect. The current goes "rapidly" to zero _after_
the flyback voltage pulse rises, during the flattish top...

**** Nope; i dug out a scope picture and we are both incorrect.

Current drops rapidly toward zero, then flyback pulse rises, has
flattish top, then falls. During flyback pulse time, the current is near
or at zero. After flyback voltage gets to zero, thenone has rapid
current change to negative value, etc.

and then goes negative as the flyback voltage pulse drops to zero.
However, the voltage would continue to decrease and go negative
(L-C oscillations), but the switch (FET) internal diode conducts,
allowing the coil current to continus to flow.
The voltage pulse seen has the *same* polarity as the supply.
In a standard flyback scheme, the negative current waveform is
mirror-image of the ramp-like charging time.
This remains to be a fairly close picture of operation waveforms,
even if the core of the inductor saturates to some extent.


John is correct, the current shouldn't reverse, excepting perhaps
for a small amount of ringing. Perhaps, when you speak of mirror
image, you're thinking of a reversal of the rate-of-change in the
flyback coil current, rather than the current polarity, per se?
*** After the flyback pulse gets to zero, the inductor current rapidly

goes to a negative value and ramps toward zero (and could continue to
positive) during the FET body diode clamp time.

In the rapid magnetic-field collapse system we're talking about,
using MOSFET or TVS avalanche to absorb the coil's energy, the
avalanching junction will stop conducting the instant the current
drops to zero (the physics of avalanche doesn't have any reverse-
recovery time). There will be a small amount of current reversal
(and ringing) due to discharging the system capacitance from the
avalanche voltage level back to the supply V, but it'll be small
compared to the 15A magnetizing current.

For example, a 34n20 FET's capacitance is 400pF at 150V, a 1.5kW
150V TVS is 100pF, and 1 meter of cable is another 100pF. The
resonant frequency with 8uH will be 2.3MHz. The energy stored
in 600pF of capacitance at 130V (assume Vs = 20V supply for the
coil) pushes the inductor to a peak reversal of i = V sqrt(C/L)
= 1.1A, and the voltage to +20V -130V = -110V after a T/2 time
of 220ns, assuming nothing else in the path. However, the FET's
intrinsic body diode and the TVS diode will prevent the voltage
from going below ground, bringing things to a stop after only
130ns, limiting the ringing to the 20V supply Vs, and the peak
reversal current to about -600mA, only 4% of the original 15A.
*** And the resulting magnetic field might be unwanted. We have not

heard about that one way or the other; which is one reason i suggested
transferring the energy to another inductor or to a capacitor.
Certainly the switches that would be doing that would dissipate some
of the energy to the good.

Since the magnitude is only 4% and it only lasts a few hundred
ns, it's not useful to characterize this as "current reversal."

*** Maybe, and maybe not; depends on desired result of the magnetic

field and how picky one wants to be.

A 2pi f L = 100-ohm resistor paralleled with the inductor would
nicely damp the 2.3MHz ringing, and reduce the magnitude of the
single ring as well, to say 2%. It would need to be a 3W part,
etc., unless a 0.01uF series capacitor was added to stop any DC
current. We call this R-C network a snubber.


*** Now that is to the good!

 

[Robert Baer]

John Popelish wrote:

Robert Baer wrote:

John Popelish wrote:

Robert Baer wrote:

One thing to consider:
Allowing current to flow after the switch opens will mean that the
magnetic field will continue to exist while it collapses, and if
current continuse to flow,the magnetic field will reverse.


(snip)

I think you are laboring under a misconception, here. The coil
inductive voltage reverses the moment the current starts to decrease
(V=L*(di/dt)), but the magnetic field polarity does not change till
the current changes direction. Once the current passes through zero
and the coil (and zener and switch) stray capacitance that is sitting
at zener voltage starts to dump current into the coil as the
capacitance discharges back toward zero volts, then the magnetic
field polarity will also pass through zero and reverse.


Err...
Take a look at the current and voltage waveforms in a flyback system.
Turn on the switch, inductive current increases in the standard R/L
form.


Yes. And the magnetic field is instantaneously proportional to coil
current.

When the switch is opened, the magnetic field starts to collapse;
the waveform across the inductor is square-ish in most practical
circuits. During that time, the current goes rapidly to zero as the
flyback voltage pulse rises; roughly remains near zero at the flattish
top, and then goes negative as the flyback voltage pulse drops to zero.


No argument. The switch current goes to zero first, the coil current
detours first to the coil capacitance, and to snubber, if any. Then the
voltage rises till either the capacitance (and snubber) absorb all the
energy stored in the coil, or a clamp voltage is reached. If the latter
(as the op is doing) then the current detours to the clamp. The current
continues to ramp toward zero as does the magnetic field. As the
current and magnetic field pass through zero, the voltage starts to
swing back toward that connected to the other end of the coil (toward
zero volts across the coil). But to change the voltage across the coil
requires current to charge the stray capacitance across those nodes, so
this current passes through the coil and and this causes a reversal of
the magnetic field.

Perhaps I misunderstood your last post when you said, "Allowing current
to flow after the switch opens will mean that the magnetic field will
continue to exist while it collapses, and if current continues to flow,
the magnetic field will reverse."

How could the field not continue to exist while it collapses? And the
coil current must cease and change directions before the field ceases
and changes directions.

However, the voltage would continue to decrease and go negative (L-C
oscillations), but the switch (FET) internal diode conducts, allowing
the coil current to continus to flow.
The voltage pulse seen has the *same* polarity as the supply.
In a standard flyback scheme, the negative current waveform is
mirror-image of the ramp-like charging time.
This remains to be a fairly close picture of operation waveforms,
even if the core of the inductor saturates to some extent.


My only point was that the applicable generalization for this op is that
the magnetic field is instantaneously proportional to the coil current
at all times, and the field reverses exactly when the current reverses.
If you agree with this than we are on the same page.
** And the op has not mentioned that a reverse magnetic field is

allowable or not, so reverse curent may be a no-no.

 

[Tony Williams]

In article <d97pcc024a3@drn.newsguy.com>,
Winfield Hill <hill_a@t_rowland-dotties-harvard-dot.s-edu> wrote:

Actually, there is another possible reason. Higher inductor
flyback voltages mean shorter inductor discharge times.
Dissipating a fixed amount of energy in a shorter time means
higher junction temperatures because the heat can't diffuse as
far through metal frames, and into the heatsink, etc.

Hello Win..... you don't have to get the energy out of
the coil AND dissipate it in the same uS. Think about
transferring the coil-energy into capacitor-energy in
the uS, then discharging the capacitor at leisure.

8uH 15A
+---+---+---////////---+--<-[Const-I Source]
| | _|_ | |
| [R] \_/D1 | |
| | | | |
+ +---+ + /|\
Sw1 / | Sw2 / Disable
+ ===50nF +
| | |
--+-------+--------------+---

Assume 15A is flowing. The turn-off sequence is:-

1. SW1 goes OFF, SW2 goes ON, Disable started in the cc-source.

The 1/4 cycle resonance of 8uH and 50nF transfers the 15A
energy in the inductor to a 190V energy across the cap.
Coil current has ceased after 1uS.

2. Wait until the cc-source has Disabled, the turn SW1 back ON.
The energy in the 50nF is discharged through R, at leisure.

3. Sw1, Sw2 = ON, (holding a short across the inductor), and
cc-source Disabled is the rest position.

$. To re-energise the coil. Sw2 to OFF, Enable the cc-source.

--
Tony Williams.

 

[Jim Thompson]

On Tue, 21 Jun 2005 13:34:23 +0100, Tony Williams
<tonyw@ledelec.demon.co.uk> wrote:

In article <d97pcc024a3@drn.newsguy.com>,
Winfield Hill <hill_a@t_rowland-dotties-harvard-dot.s-edu> wrote:

Actually, there is another possible reason. Higher inductor
flyback voltages mean shorter inductor discharge times.
Dissipating a fixed amount of energy in a shorter time means
higher junction temperatures because the heat can't diffuse as
far through metal frames, and into the heatsink, etc.

Hello Win..... you don't have to get the energy out of
the coil AND dissipate it in the same uS. Think about
transferring the coil-energy into capacitor-energy in
the uS, then discharging the capacitor at leisure.

8uH 15A
+---+---+---////////---+--<-[Const-I Source]
| | _|_ | |
| [R] \_/D1 | |
| | | | |
+ +---+ + /|\
Sw1 / | Sw2 / Disable
+ ===50nF +
| | |
--+-------+--------------+---

Assume 15A is flowing. The turn-off sequence is:-

1. SW1 goes OFF, SW2 goes ON, Disable started in the cc-source.

The 1/4 cycle resonance of 8uH and 50nF transfers the 15A
energy in the inductor to a 190V energy across the cap.
Coil current has ceased after 1uS.

2. Wait until the cc-source has Disabled, the turn SW1 back ON.
The energy in the 50nF is discharged through R, at leisure.

3. Sw1, Sw2 = ON, (holding a short across the inductor), and
cc-source Disabled is the rest position.

$. To re-energise the coil. Sw2 to OFF, Enable the cc-source.

Yep, That's what I used back about 25-30 years ago, with bipolar MJ...
something or other transistors.

I also had some inductive goodie that I used as well. I can recall
getting near textbook switching load lines along each axis.

It was so effective at reducing the transistor dissipation that I
removed the heat sinks from the transistor flags. Then made the
mistake of grabbing the flags to check the temperature while it was
powered.

The technicians were delighted... almost as much as when I got knocked
off a lab stool while working on a CD ignition ;-)

...Jim Thompson
--
| James E.Thompson, P.E. | mens |
| Analog Innovations, Inc. | et |
| Analog/Mixed-Signal ASIC's and Discrete Systems | manus |
| Phoenix, Arizona Voice480)460-2350 | |
| E-mail Address at Website Fax480)460-2142 | Brass Rat |
| http://www.analog-innovations.com | 1962 |

I love to cook with wine. Sometimes I even put it in the food.

 

[John Popelish]

Robert Baer wrote:

Winfield Hill wrote:

John is correct, the current shouldn't reverse, excepting perhaps
for a small amount of ringing. Perhaps, when you speak of mirror
image, you're thinking of a reversal of the rate-of-change in the
flyback coil current, rather than the current polarity, per se?

*** After the flyback pulse gets to zero, the inductor current rapidly
goes to a negative value and ramps toward zero (and could continue to
positive) during the FET body diode clamp time.
(snip)

Does your flyback circuit have more than one winding on the coil?

 

[colin]

"Winfield Hill" <hill_a@t_rowland-dotties-harvard-dot.s-edu> wrote in
message news:d93g5602b26@drn.newsguy.com...

Daniel A. Thomas wrote...

This active-zener method works well with low-voltage power MOSFETs,
such as under 100V, but it's dangerous with high-voltage FETs, 200V
and up, because they have a bad tendency to go into RF oscillation.
This is a high-power RF oscillation at frequencies of 15 to 40MHz,
which is very difficult to damp with external parts such a ferrite
beads, gate resistors, etc. That's because the RF oscillation is
internal to the FET, employing its inductance and self capacitance.
The required linear properties occur whenever a high current flows
while the drain-source voltage is higher than 10 to 20V. The latter
condition causes the FET capacitances to drop to the levels where RF
amplification is efficient.

--
Thanks,
- Win

If the RF feedback path is wholy internal how would this affect the method
of using a higher gate drive resistance to slow the current fall to limit
the voltage to less than the breakdown voltage ?

or does the zener just add more parasitics to make the difference ?

Colin =^.^=

 

[Winfield Hill]

Robert Baer wrote...

Winfield Hill wrote:
Robert Baer wrote...
John Popelish wrote:
Robert Baer wrote:

One thing to consider:
Allowing current to flow after the switch opens will mean that the
magnetic field will continue to exist while it collapses, and if
current continuse to flow,the magnetic field will reverse.

(snip)

I think you are laboring under a misconception, here. The coil
inductive voltage reverses the moment the current starts to decrease
(V=L*(di/dt)), but the magnetic field polarity does not change till the
current changes direction. Once the current passes through zero and the
coil (and zener and switch) stray capacitance that is sitting at zener
voltage starts to dump current into the coil as the capacitance
discharges back toward zero volts, then the magnetic field polarity will
also pass through zero and reverse.

Err...
Take a look at the current and voltage waveforms in a flyback system.
Turn on the switch, inductive current increases in the standard R/L form.
When the switch is opened, the magnetic field starts to collapse; the
waveform across the inductor is square-ish in most practical circuits.
During that time, the current goes rapidly to zero as the flyback
voltage pulse rises; roughly remains near zero at the flattish top,

This is incorrect. The current goes "rapidly" to zero _after_
the flyback voltage pulse rises, during the flattish top...

**** Nope; i dug out a scope picture and we are both incorrect.
Current drops rapidly toward zero, then flyback pulse rises, has
flattish top, then falls. During flyback pulse time, the current is near
or at zero. After flyback voltage gets to zero, thenone has rapid
current change to negative value, etc.

I'm talking about the current inside the coil, are you talking
about the HOT current in a TV flyback circuit?

For example, a 34n20 FET's capacitance is 400pF at 150V, a 1.5kW
150V TVS is 100pF, and 1 meter of cable is another 100pF. The
resonant frequency with 8uH will be 2.3MHz. The energy stored
in 600pF of capacitance at 130V (assume Vs = 20V supply for the
coil) pushes the inductor to a peak reversal of i = V sqrt(C/L)
= 1.1A, and the voltage to +20V -130V = -110V after a T/2 time
of 220ns, assuming nothing else in the path. However, the FET's
intrinsic body diode and the TVS diode will prevent the voltage
from going below ground, bringing things to a stop after only
130ns, limiting the ringing to the 20V supply Vs, and the peak
reversal current to about -600mA, only 4% of the original 15A.

*** And the resulting magnetic field might be unwanted. We have not
heard about that one way or the other; which is one reason i suggested
transferring the energy to another inductor or to a capacitor.

Wouldn't help much, if at all; this reversal comes from the energy
stored in the node's capacitance, of which only 1/6 was in the TVS.


--
Thanks,
- Win

 

[Winfield Hill]

Jim Thompson wrote...

On Tue, 21 Jun 2005 Tony Williams wrote:

Winfield Hill wrote:

Actually, there is another possible reason. Higher inductor
flyback voltages mean shorter inductor discharge times.
Dissipating a fixed amount of energy in a shorter time means
higher junction temperatures because the heat can't diffuse
as far through metal frames, and into the heatsink, etc.

Hello Win..... you don't have to get the energy out of
the coil AND dissipate it in the same uS. Think about
transferring the coil-energy into capacitor-energy in
the uS, then discharging the capacitor at leisure.

8uH 15A
+---+---+---////////---+--<-[Const-I Source]
| | _|_ | |
| [R] \_/D1 | |
| | | | |
+ +---+ + /|\
Sw1 / | Sw2 / Disable
+ ===50nF +
| | |
--+-------+--------------+---

Assume 15A is flowing. The turn-off sequence is:-

1. SW1 goes OFF, SW2 goes ON, Disable started in the cc-source.

The 1/4 cycle resonance of 8uH and 50nF transfers the 15A
energy in the inductor to a 190V energy across the cap.
Coil current has ceased after 1uS.

2. Wait until the cc-source has Disabled, the turn SW1 back ON.
The energy in the 50nF is discharged through R, at leisure.

3. Sw1, Sw2 = ON, (holding a short across the inductor), and
cc-source Disabled is the rest position.

$. To re-energise the coil. Sw2 to OFF, Enable the cc-source.

Yes that's very nice.

Yep, That's what I used back about 25-30 years ago, with bipolar
MJ... something or other transistors.

I also had some inductive goodie that I used as well. I can recall
getting near textbook switching load lines along each axis.

A friend of mine used a similar configuration in his ignition
design, but with an added inductor that returned the energy to
the battery. That proved to be VERY efficient. Perhaps that's
what you did way back when. Did you keep a copy of the drawing?


--
Thanks,
- Win

 

[Jim Thompson]

On 22 Jun 2005 02:18:59 -0700, Winfield Hill
<hill_a@t_rowland-dotties-harvard-dot.s-edu> wrote:

Jim Thompson wrote...

[snip]

Yep, That's what I used back about 25-30 years ago, with bipolar
MJ... something or other transistors.

I also had some inductive goodie that I used as well. I can recall
getting near textbook switching load lines along each axis.

A friend of mine used a similar configuration in his ignition
design, but with an added inductor that returned the energy to
the battery. That proved to be VERY efficient. Perhaps that's
what you did way back when. Did you keep a copy of the drawing?

I hope. That was in my BC days (before CAD, ~1977-1980). I'll make a
trip to the archive storage (I rent an off-site, air-conditioned,
10'x10' storage "cell" that I have filled with file drawers) and take
a look.

...Jim Thompson
--
| James E.Thompson, P.E. | mens |
| Analog Innovations, Inc. | et |
| Analog/Mixed-Signal ASIC's and Discrete Systems | manus |
| Phoenix, Arizona Voice480)460-2350 | |
| E-mail Address at Website Fax480)460-2142 | Brass Rat |
| http://www.analog-innovations.com | 1962 |

I love to cook with wine. Sometimes I even put it in the food.

 

[colin]

"Tony Williams" <tonyw@ledelec.demon.co.uk> wrote in message
news:4d7e7d0b3dtonyw@ledelec.demon.co.uk...

In article <d97pcc024a3@drn.newsguy.com>,
Winfield Hill <hill_a@t_rowland-dotties-harvard-dot.s-edu> wrote:

Actually, there is another possible reason. Higher inductor
flyback voltages mean shorter inductor discharge times.
Dissipating a fixed amount of energy in a shorter time means
higher junction temperatures because the heat can't diffuse as
far through metal frames, and into the heatsink, etc.

Hello Win..... you don't have to get the energy out of
the coil AND dissipate it in the same uS. Think about
transferring the coil-energy into capacitor-energy in
the uS, then discharging the capacitor at leisure.

8uH 15A
+---+---+---////////---+--<-[Const-I Source]
| | _|_ | |
| [R] \_/D1 | |
| | | | |
+ +---+ + /|\
Sw1 / | Sw2 / Disable
+ ===50nF +
| | |
--+-------+--------------+---

Assume 15A is flowing. The turn-off sequence is:-

1. SW1 goes OFF, SW2 goes ON, Disable started in the cc-source.

The 1/4 cycle resonance of 8uH and 50nF transfers the 15A
energy in the inductor to a 190V energy across the cap.
Coil current has ceased after 1uS.

2. Wait until the cc-source has Disabled, the turn SW1 back ON.
The energy in the 50nF is discharged through R, at leisure.

3. Sw1, Sw2 = ON, (holding a short across the inductor), and
cc-source Disabled is the rest position.

$. To re-energise the coil. Sw2 to OFF, Enable the cc-source.

--
Tony Williams.

The same thing would happen with the simpler RC snubber accros a single
switch as was sugested before, but put a diode in series with the inductor
so that when the current stops the diode blocks the voltage acros the
capacaitor cuasing curent in the reverse direction through the coil.

when the switch opens the current travels through the resistor and
(discharged) capacitor, initialy creating the necessary voltage accros the
R, then when the curent drops the voltage acros the resistor falls and the
voltage accross the capacitor has risen keeping the voltage high enough all
the time the curent is flowing for a fast fall time/peak voltage, when the
switch next closes it discharges the capacitor through the resistor as it
turns on the curent to the coil.

o
|
|
C|
C| coil
C|
|
|
V diode
-
| ____
+-----|____|----+
| |
| D |
||-+ ---
||<- ---
o-----||-+ |
| S |
| |
-+---------------+-


Colin =^.^=

 

[Winfield Hill]

Jim Thompson wrote...

archive storage (I rent an off-site, air-conditioned, 10'x10'
storage "cell" that I have filled with file drawers) ...

I never could understand why Western houses eschew basements.
A full basement is a VERY useful thing to have. Get double
or half-again the floor area of the house in one quick step.


--
Thanks,
- Win