Re: constant current driver


nag wrote:
If your coil is 2 ohms, and you want two amps, 50% duty cycle, you'll
need a 12 ohm, 25 watt resistor to do the job. I'd assume you'd use a
darlington power transistor here, unless you've got a lot of current to
drive the transistor base or plan on using a MOSFET.


Thank you Chris for this lot of information. But i had a doubt here.
how will that series resistor will limit the voltage across the coil to
6 to 6.5 volts itself independent of the input voltage. Any way the
voltage available across the coil will proportinately increase with the
input voltage. Also the solenoid will not get 6 volts across it till
the input voltage reach 18 volts( considering 2 amps solenoid current ,
drop across the resitor will be 6*2 =12 V)

But the inductive kick of the solenoid will try to increase
the voltage at the collector of the transistor/MOSFET to beyond the
Vceo(max), which will smoke it. This is your second problem.

I am having one idea to reduce this inductive kick of voltage. since
the coil inducatnce is 2 mH, the maximum current wil be allowed is 3
amps, the voltage rise due to turn off would be V = L* ( dI/dt)=
(2/1000)*( 3/120us) = 50Volts,. here i have considered that the turn
off time is 120 microseconds. I hope this can be achieved through the
delayed base current drive. This 50V reverse voltage will be affordable
i think.
Chris give your comment on this.

You should definitely avoid using a diode across the coil to catch the
inductive kick, because that slows down your solenoid too much (again,
20Hz is pretty fast for a solenoid -- you might want to double-check
with the manufacturer to see if that rate's even possible).

I have calculated the inductive fly back energy. The maximum energy
would be = (1/2)* L * I * I = (1/2)*(2/1000) * 3* 3 = 9 milli joules.
Even if i provide the fly back diode alone, the time taken to die those
energy may calculated like this. For 50 Volts reverse voltage the fly
back current may be 50/2 = 25 amps, the time taken will be I * I * R *
t = 9 mJ,

t = (9/1000) * /( 25 * 25 * 2) = 7.2 us. For worst case also
considering the exponential decay the time taken may be 5*7.2 = 36 us
which is very small when compared to the 20 Hz operating frequency (
1/20/2 = 25 ms).
I do not wwhether these calculations are really suitable or not. Just i
have applied the theory and formula here.
Chris please give your comment on this.

Regards,
Nagarajan.

Hi, Nagarajan. Sorry I misunderstood your first post. I'm not too
sure on a couple of things:

* Rereading your first post, I would guess you want 3 amps constant
current through your 2 ohm solenoid coil? Or 2 amps?

* What is your supply voltage maximum and minimum?

* 50 volts of inductive spike (on top of the supply voltage maximum)
seems a little small for this high of an inductance. Have you actually
measured this?

* Are you sure this solenoid can operate at 20Hz? Most larger
solenoids and relays are limited to slower speeds. The coil power has
to be balanced against the spring force, and with a longer
pull-in/release, it can get to be physically impossible to make it go
that fast. After a certain frequency, you just get a kind of low
frequency growl (and of course, coil heating).

If you're interested in a minimum power solution, the PWM idea
mentioned by another response sounds just about right. However, if
your minimum power supply voltage is greater than about 9V, and your
maximum is less than 30 or so, and if you're not too picky about
exactly 3 amps (this is a solenoid, after all), you might want to try
something simple like this (view in fixed font or M$ Notepad):

|
| VCC VCC
| VCC + +
| + | |
| | 1N5401 - C|
| | _______ ^ C|SV1
| | | | | C|
| o-|LM340T5|-o---o-------o---o TVS V |
| +| |_______|+| | | | /-/ |
| --- | --- .-. | | | |
| --- | --- | | .--o---o-. TVS V |
| | | | | | | 8 4 | /-/ |
| === === === '-' | | | |
| GND GND GND | | | '----o
| o----o7 | |
| | | | .--o
| .-. | 555 | | |
| | | | | ___ D D |/ |
| | | .-o6 3o---o-|___|->|-->|-o-| |
| '-' | | | | 47 | |> |/
| | | | | '------|<------' -| Q1
| o--o-o2 | D |> TIP101
| | | 1 5 | |
| --- '--o---o-' .---o
| --- | | |
| | | | |
| === === .-. .-.
| GND GND 2 X 1.5 ohm | | | |
| | | | |
| '-' '-'
| | |
| === ===
| GND GND
| D = 1N4001
(created by AACircuit v1.28.6 beta 04/19/05 www.tech-chat.de)

You can see you're powering a 555 with a regulated 5VDC. The output
(you can make this pretty close to a 20Hz square wave with proper
component selection) is dropped by two diodes to give 3.6V or so being
fed to the base of a darlington transistor. This will be
voltage-driven (at least once the coil current starts drawing), so you
should see about 2.2V or so on the emitter. This will correspond to 3
amps being driven through the 0.75 ohm emitter resistor (two 1.5 ohm 3
watt in parallel). Since nearly all the emitter current is collector
current, you've got a rough and ready 3 amp current source. When the
555 goes low, the diode pulling current from the base ensures a better
turn-off than just a weak pulldown. Slow turnoff is fatal here.

Of course, you'll have to tweak this a bit (I would think you should
check to see if current is too high first), and also you should do your
best to heat sink the darlington for two reasons: you'll probably be
dissipating a lot of heat, and Vbe will go down as transistor temp goes
up, which will increase emitter voltage and current. Check the steady
state transistor temperature and emitter current before you put it in
the can. If your average power supply voltage is 20V, you'll have 36
peak watts and 18 average watts dissipation in the TO-220 package IC.
If you want, you can go with a more robust TO-3 NPN darlington like my
all-time favorite NPN darlington, the dreaded MJ11016D. 18 watts is
nothing for a TO-3 package with a good heat sink.

If you want more precise control of current, you may want to put the
emitter voltage in the feedback loop of an op amp configured as
non-inverting-unity gain.

You are, I'm sure, aware that placing a diode across the inductor will
dramatically increase the time required for the energy to dissipate and
the coil to release the plunger. But if you use the back-biased diode
with the 24V of TVS, the release time shouldn't be much greater than
what it would be without any inductive kick protection. Other methods
of catching the inductive kick might add needless complexity here. If
it doesn't work with the back-biased diode and the TVS, it almost
certainly wouldn't work with anything else (or nothing, for that
matter). And exceeding Vceo can be fatal to transistors.

Hope this is of use. If not, could you please inform us of the power
supply voltage, the precise current requirement, and any other
information necesary.

Cheers
Chris

.

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