Hi All, I am designing a driver circuit for a solenoid of 2 ohms & 2mH. What i need is that when i connect the solenoid to my driver, the effective voltage availbale across the solenoid should not exceed 6 to
6.5 Volts ( basically to limit the current). I am plannig to operate this solenoid in 20 Hz frequency of 50% duty cycle. Almost i designed the 20Hz square.oscillator with MOSFET or Transistor as Driver. The voltage input at the drain or at the collector vary from 6V to 15V. So to keep the current constant i have kept the gate voltage or the Base current constant to the drivers constant. The problem i am facing now is that the driver transistor is getting heated enormously at high voltage regions like 10 to 15 Volts. I understand that by this type of circuit the dissipation at the higher voltage range will be {(15-6)volts * (6/2)amps} = 27 watts. Is there any way to limit this current at the higher voltage ranges without dissipating much power. I do understand that by means of series inducance or some other energy storage element, we can control this. but the size would be larger. if this is the way can any one give the calculation deatils of determininig this inductance value or any other best solutoin for this problem
Hi, Nagarajan. You have two problems here. First, you're asking the transistor or MOSFET to handle a lot of heat, and also the inductive kick of the solenoid when you're turning it off may be exceeding the maximum Vceo, causing serious problems and additional heat. The classic (and easiest) way to handle the first problem is to use a series power resistor, like this (view in fixed font or M$ Notepad):
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.
This circuit has the disadvantage of wasting over 86% of the power you're using, but you're doing that anyway with your setup. The advantage is that it will impress the full 14V across the solenoid, causing the current to ramp up more quickly (also like your constant current setup). Since there will be an exponential rate of decay, it might be a little slower than your constant current setup getting up to speed.
If you're trying to operate the solenoid at its fastest, both the constant current setup and this L-R setup are relying on quick coil discharge. 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.
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). If inductive kick is a problem, it might be better to use a couple of big TVS diodes in series with a 3 amp diode to allow the kick to dissipate more quickly, while still keeping the transistor/MOSFET voltage below dangerous levels.
You'd have to provide more information to get better than a guess, but I'd think two Littelfuse 5KP12s in series might do the job. This will limit the max collector voltage to 15V + 25V = 39V, easy for most power darlingtons. They'll certainly handle the pulse current for a one-shot.
To test this, bring up the rep rate slowly, and feel the TVS diode cases for temperature rise as the rate gets closer to 20Hz. If it remains below about 70C steady-state, you should be OK. If not, go to
4 ea. 5KP6.0s in series. This will dissipate the same energy over 4 diode packages, so that should be OK.
I hope this has been of help. Good luck with your project.
Chris, one thing I wasn't sure of, but seemed to read into the OP's description, was that the power rail can be anything from about 6.5V to 15V. I may have misunderstood that, but that's how it came across to me. If that is right, the problem is that one cannot presume to dump most of the dissipation into a resistor because one cannot be sure of enough headroom for the passive device. (The OP's writing says that there is a problem in the "high voltage regions like 10 to
15 Volts" and this appears to confirm my reading -- though I cannot be sure.) I think that this means that active devices need to carry all of the dissipation, if the system doesn't using switching to develop a nicer 6V rail.
you need to PWM (pulse width modulate) it. that would work well for a solenoid since its inductive. at the on cycle you should have a regulated simple PWM osc to drive the power tranny with a feed back regulator.. a resistor in series with a cap going to ground across the load should give you a smooth output that you can use as a reference for the regulator. they do have PWM regulators that allows you to feed back from the main output to monitor and regulate. below is an example./
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Real Programmers Do things like this.
http://webpages.charter.net/jamie_5
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)
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.
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.
Thank you Kirwan. You have understood my problem correctly. i have also added my comments to Chris veiw palese check that and give your comment on my view to this problem.
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):
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):
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