[email protected] wrote...
Thanks to all who responded to my post. I think that I'm going to give
Winfield's suggested circuit for rapid switching a shot, but want to be
sure that I understand his comments regarding the zeners. In my
application, I am more concerned with the rapidity with which the field
dies down than I am with how quickly it starts up (although I certainly
don't want startup to be slow) and I know that simply applying my 24V
power source yields an acceptable startup time. So, as I understand
it, to get the quickest shutoff time, I should be using zeners with a
breakdown voltage just a hair higher than my supply (24V) rather than
ones with say a 50V rating. Am I correct here?
Here's my circuit,
.. H-bridge, with provision for rapid inductive-load shutoff.
.. _______________________________________________
.. | | zener zener | |
.. XXX '-|>|--|<|--, ,--|>|--|<|-' XXX
.. | | | |
.. '------|>|------+ LOAD +------|<|------' XXX = switch,
.. +---o o---+ e.g., a MOSFET
.. ,---------------+ +---------------,
.. | |
.. XXX XXX
.. _|____________________________________________|
My circuit can be considerably simplified by keeping the bottom
FET on during the flyback discharge. You can drive the FETs with
a bridge driver IC like Intersil's HIP4080A. This part allows
you to turn off the upper FETs while keeping the bottom one on.
http://www.intersil.com/cda/deviceinfo/0,1477,HIP4080A,0.html
I'm also curious as to how long the flyback voltage might take
to settle in this circuit. Are we talking on the order of a few
milliseconds here? How would one go about calculating the
required time?
During normal operation of the electromagnet the 8A current is
developed across the 3.3-ohm magnet resistance. When the XXX
switches are opened the inductive flyback reverses the voltage
across the magnet terminals, to the supply plus two diodes plus
a zener. The coil's inductance begins its current fall at the
dI/dt = V/L rate we're so familiar with. Ignoring the coil's
resistance we can say the time will be t = L I/V. So clearly
you want V to be as high as is reasonable. To calculate some
typical times we'd need to know your magnet's inductance.
Actually, I liked Tony's circuit the best. It can be simpler,
and has the advantage you can simply reverse the H-bridge
switches, rather than opening the first pair for a discharge
time, before closing the other pair, as my circuit required.
http://groups.google.com/group/sci.electronics.design/msg/efec70381aaaac1c
.. D1 Vs, allowed to rise to Vpk.
.. 12.5V------|>|---+----+----------+----+------+
.. | | | | |
.. + _|_ _|_ + |
.. Sw1/ /_\ D D /_\ /Sw2 |
.. + | | + |
.. | | | | +|
.. +----+--//////--+----+ ===C
.. | | R+L | | |
.. + _|_ _|_ + |
.. Sw3/ /_\ D D /_\ /Sw4 |
.. + | | + |
.. | | | | |
.. 0v---------------+----+----------+----+------+
When you reverse the set of closed switches (MOSFETs), say 2,3
instead of 1,4 the current flow through the new FETs is in the
reverse direction. However, unless the drop across the FET's
Rds(on) exceeded 0.7 volts (and generally you'd select them so
it wouldn't) in the normal drain-to-source direction, the drop
in the reverse direction would be about the same, and the diodes
shown in the drawing wouldn't conduct. This means the diodes
don't play a special role, except during the short open time as
the bridge is reversed. Power MOSFETs have intrinsic diodes as
a part of their structure, so we don't need to explicitly add
diodes if we use MOSFET switches. Tony's circuit becomes:
.. D1 allowed to rise to Vpk <= Vzener
.. V+ ----|>|---+------------------+--------+-----,
.. | | | |
.. |--' '--| | |
.. |<-, Q1 Q2 ,->| | |
.. |--+ +--| | \_|_
.. | | +| C /_\
.. +------//////------+ === | TVS
.. | L + Rs | | | units
.. |--' '--| | \_|_
.. |--, Q3 Q4 ,->| | /_\
.. |--+ +--| | |
.. | | | |
.. 0v-----------+------------------+--------+-----'
Ignoring Rs, and equating L and C energies, Vpk = I sqrt L/C,
or the TVS voltages, whichever is lower.
For rapid discharge only, with no energy return, I'd probably
eliminate the capacitor and rely on the TVS power zeners alone.
This way the discharge is a fast ramp, rather than a half-sine.
Any way you do it, the result is more simple than my circuit.
If you want, the energy in the magnet can be returned to the
power supply's output capacitor.
.. TVS zeners rises to Vzener + Vs
.. V+ -+--o--|>|--|>|---+-------+----------+--------,
.. | | | |
.. | |--' '--| |
.. | |<-, Q1 Q2 ,->| |
.. | |--+ +--| |
.. +| Cs | --> Im | | C2
.. === +------//////------+ ===
.. | | L + Rs | |
.. | |--' '--| |
.. | |--, Q3 Q4 ,->| |
.. | |--+ +--| |
.. | | | |
.. 0v--+--o-------------+------------------+--------'
You might want to use a small C2 to limit the flyback risetime,
calculate dV/dt = Im / C2, and reduce RFI, etc.
Be careful, Cs in the power supply has to be large enough to
absorb the magnet's inductive energy without soaring too much.
If Cs absorbed all of the inductor's energy (ignoring the TVS)
its voltage after flyback would be Vs' = sqrt (Vs^2 + LC I^2).
Hmm, you can add your own external electrolytic cap to be safe.
In that case add diode D1 and get the best of both worlds, fast
dI/dt ramp, plus some energy storage and return to the magnet
in the other direction, saving power and speeding up reversal.
.. V+ D1 TVS zeners rises to Vzener + Vs + dVcap
.. o--|>|-+--|>|--|>|---+------------------+--------,
.. | | | |
.. | |--' '--| |
.. | |<-, Q1 Q2 ,->| |
.. | |--+ +--| |
.. +| C1 | | | C2
.. === +------//////------+ ===
.. | | L + Rs | |
.. | |--' '--| |
.. | |--, Q3 Q4 ,->| |
.. | |--+ +--| |
.. 0V | | | |
.. o------+-------------+------------------+--------'
One big difference between Tony's circuit and mine, my MOSFET's
voltage ratings are low, just safely above the supply voltage,
whereas for Tony's scheme the FETs need to be rated at a much
higher voltage, safely above the maximum flyback. However, with
your magnet's modest 8A draw, there are plenty of good inexpensive
high-voltage FETs. Lots of attractive 800V power MOSFETs.
There are good high-voltage MOSFET driver ICs, like IR's 600V
ICs, allowing you to use impressively-high flyback voltages.
http://www.irf.com/product-info/cic/fsgatedriverics.html
http://www.irf.com/product-info/datasheets/data/ir2111.pdf
These are only $2.58 each, in stock at DigiKey.
http://www.digikey.com/scripts/DkSearch/dksus.dll?Detail?Ref=120907&Row=68496&Site=US
(fix any link wraparound) I've used the IR2113, which is rated
for 2A gate drive, but the IR2111 is fine. Its higher sinking
than sourcing gate-drive capability is appealing.
That just leaves the issue of the TVS diodes, used to maximize
the discharge ramp. These have to absorb the magnet's energy.
TVS parts have internal copper slugs for this purpose. Note I
showed parts in series. The largest practical size may be the
5kW units, which can absorb 5J of energy in one ms (less if the
time is shorter, scaled by the square root of time). You can
purchase low-cost automotive types, e.g., the 5kp26A parts,
which breakdown at about 30 volts for 8 amps, and wire them in
series for a higher voltage. E.g., 15 for 450 volts. They'd
handle 75J, which would be enough if your inductance less than
L = 2E/I^2 = 2.3H. If it's more, you can use paralleled stacks
of TVS diodes.
OK, that's it, I've said enough.
Thanks again (and thanks very much for the Art of Electronics,
Winfield!),
Hey, thanks, you're very welcome, Graham.