I have (almost) 100ma available in a circuit. I want to drive a 12V, 1A solenoid that opens a lock. I'm thinking about using a DC/DC converter and charging up an electrolytic cap. I'm just guessing here, because I haven't done any such calculation for a long time, but it seems like I'd need quite a cap to store enough energy.
There are obviously some unanswered questions here, such as how long the relay needs to be open, etc., which needs to be studied once I have a working circuit. Also, swapping out the solenoid isn't an option.
I'd appreciate any links to a cheap/simple DC/DC. Also, if anybody's crossed this bridge before, I'm curious as to how you handled it.
You'll need a pretty large capacitor to supply 12W for any length of time. Delta-V = Ton* I/C, so for a 5V drop (say 17V to 12V), at 1A, and 100msec, you'd need 20,000uF
The energy to charge the cap is C *V^2/2, so to charge it will take about 3J, which at 100mA/5V is less than 10 seconds, assuming fairly good efficiency.
Sounds do-able, if you have the room for a ~20,000uF cap (maybe 25 x
50mm for a 25V unit), and if the actuation time is within reason.
Best regards, Spehro Pefhany
--
"it's the network..." "The Journey is the reward"
speff@interlog.com Info for manufacturers: http://www.trexon.com
Embedded software/hardware/analog Info for designers: http://www.speff.com
If one of the super caps isn't enough or turns out to be too large in size maybe you could provide a little NiMH or other battery that gets charged up. That way you may even have enough juice for a short sequence of repeated activations. For the charger you could use a little switcher around the LM3478 or similar. Those are really small and can operate around 500KHz to save space on the inductor.
Hmmm..... Shit. Algebra. OK, I get: C = Ton * I / Delta-V. A couple more points:
- Less voltage is required to keep a solenoid activated than what is required to activate the solenoid; 6V is a decent assumption for now.
- I don't want to apply too much voltage to the solenoid at first because that'll shorten its life. I'm going to go with 25% overvoltage and cross my fingers.
- The 1A rating is for 12V. I'm going to make some big assumptions here. First, since this is a DC circuit, I'm going to ASSuME that the solenoid is like a resistor. Also, and this is a big one, I'm going to assume a linear discharge of the cap. So, 12 ohms, DeltaV = 15 - 6 = 9v, gives me an avg. V of 10.5, over 12 ohms.......... Iavg = 833mA.
- Just a guess here, but give 'em 10 sec to open the door once its unlatched.
C = 8 * .833/9 = 740000 uF. Hmpf. That's a lot of space and $ for caps. Did I work this out correctly?
I don't see how I fits into this. Can you explain please?
I suggest that you look at the archives of the US Patent Office archives from the last century, since they issued a lot of patents to perpetual motion machines :-).
If you look at the solenoid and capacitor as a T=RC time constant and remembering that the voltage drops to 1/e (about 37 %) in the time constant T time. Thus the voltage should be 16.3 V at start of discharge to drop to 6.0 V at 1 T.
If we assume that the solenoid represents a 12 ohm resistor, then C=0,833 F = 833000 uF. A slightly smaller capacitor will suffice, since at the beginning of the discharge, the inductance of the solenoid will dominate and a lower discharge current is drawn in the beginning. You have to find out the inductance of your solenoid to calculate the LC and LR time constants more accurately.
To allow for voltage drops in the switch between the capacitor and solenoid, the capacitor needs to be charged to at least 17 V and with
0.833 F the full capacitor charge is 120 J, which at 5 V x 0.1 A requires 240 s or 4 minutes to charge at 100 % efficiency. Thus, the door can be opened only 360 times during 24 hours.
Using a NiMH or other batteries does not help a lot, since you are still going to be able to open the door less than 360 times each day due to the storage losses, but it will help the rush hour situation.
You should upgrade your power supply or consult the US. Pat. Office for perpetual motion machines.
One thing that will reduce the power consumption is to monitor the door switch and disconnect the solenoid once the door has been slightly opened, thus, the capacitor is not fully discharged.
However, typically an electronic lock usually also contains a buzzer to inform when the solenoid is active, so that the person knows when to open the door. This is important, if it takes more than about 50 ms from requesting entry until the solenoid is operated. Thus, you must also include the buzzer current in your calculations.
I'm hoping it doesn't take 'em 10 seconds to open the drawer. I'll use a piezo to let them know when the latch is open.
There ain't enough money or space for .8F worth of caps. I'm thinking, _maybe_, I can add 10,000 or so. I have to poke around on paper (a form of antique computing), but what I may do is combine a cap with a battery.
And how do you know how much would be "too much", for that 12V rated solenoid?
I seriously doubt that assumption is tenable. You're talking about not just some odd solenoid, after all, but a veritable magnet, i.e. there's an iron core inside that coil, boosting the inductance by a strong factor. That's about as non-ohmic as a solenoid can become. You're talking about a large inductance coupled to a seriously large capacitance here --- the behaviour of the result will be anything but resistor-like.
That's *completely* untenable, especially since you say further down that you won't be able to have a truly monstrously large supercap in there. The only way to get anywhere near linear behaviour would be to use a capacitor that is considerably over-dimensioned for the task, so it discharges almost not at all during the action. If you want to keep your cap small, you'll fall way down the exp(-t/tau) curve, and it'll be far from linear.
So, if you want advice, here's one: get yourself either a much stronger supply, or a less demanding solenoid.
--
Hans-Bernhard Broeker (broeker@physik.rwth-aachen.de)
Even if all the snow were burnt, ashes would remain.
The cheapest option may be to use a battery pack, possibly high-capacity AAA or AA batteries to make up the 12V. The batteries should have no problem supplying the current you need. Then use a 5 to 15 volt DC-DC converter, and trickle charge the batteries. A small relay would also be needed to trigger the solenoid.
You need to keep it activated rather than just pulse it? Ooops.. I don't think this is going to work with a capacitor of reasonable size. Maybe start thinking about a battery pack (eg. NiMH). This might have some other unpleasant consequences.
For 8 seconds (of course this is for constant current (linearized problem, so it's inaccurate for large voltage swings), but it's in roughly the right ball park).
That's the available energy flow to charge the capacitor up so that it can operate the door lock for 0.1 second. If you need more energy storage it will take longer to charge the reservoir. Maybe hours for a battery with a simple charger circuit.
Best regards, Spehro Pefhany
--
"it's the network..." "The Journey is the reward"
speff@interlog.com Info for manufacturers: http://www.trexon.com
Embedded software/hardware/analog Info for designers: http://www.speff.com
You realize that in common with most other solenoid-y type devices, the current required to pull it open is much greater than the current required to HOLD it open? You might only need 1A for a few dozen milliseconds, then 100mA to hold it.
He speaks the truth, though I have no documentation for you. I fiddled around with solenoids more than I'd care to admit, and the real trick is to get the slug moving fast. After that, it's all downhill, current-wise.
The relay can withstand 24V for 9S. I think Vhold of 6v is still reasonable. So there are two calculations:
1) Initial opening of the relay. Click -- can't take more than 100mS. Allow for a DeltaV of 9V, calculates out to 9200uF
2) From there, a hold current of 150mA (on the safe side), 9V swing, hold the lock open for 9 seconds.... gee, only 150,000uF.
What I can do is make the user wait before I open the hatch. The computer will say, "OK, Open the hatch Cap'n", then there will be a few second pause so that they can get their hand into place. Then I can engaged the solenoid.
I've calculated about as much as I can calculate here. Its time to hit the bench and do some experimentation. I've been thinking ahead on this problem so its going to take me a few days to get to the circuit. I'll post back.
By the way, re. the battery idea, I like it a lot. I can't go with it, though, for a couple reasons. First, those little batteries with their charging circuits appear to be quite a bit more expensive than caps. Also, this thing is expected to last for years and battery changes really aren't acceptable.
A pack of ten AA size NiCd cells would no doubt have a sufficiently low internal resistance to drive the solenoid at 1 A.
Since only 100 mA is available at 5V, about 30 mA would be available to trickle charge the batteries at 15 V. With a discharge current of
1000 mA, the charge time is 30 times longer than discharge time.
If the true battery capacitance is 1000 mAh, the solenoid could be operated for a full hour. If each lock activation is 10 s long, the battery charge is sufficient to operate the lock 360 times. It would then take more than 30 hours to fully charge the battery. With realistic efficiencies, the lock could be operated 200-250 times each day.
I think a wait of about 1 minute is all that can be tolerated, if that. It is not uncommon for four or five people to be lined up ready to use the device. In looking at some of the potential time constants, the only way to pull this off might be with a few banks of caps. I really had my heart set on just pulling power 100mA from the USB, but it doesn't look as though that's going to work. Wall wart, wall wart, here we come...
If you actually pulled your finger out and did some measurement of the required solenoid holding current you would know wouldn't you.
I have a 1A 12v spring loaded solenoid which unlocks a latching mechanism. It doesn't need to hold so I just dump a capacitor across it.
The spring provides about 10N force. I just tested it on the bench and once 'pulled' the solenoid continues to overcome the spring until the current is reduced to about 80mA. If you used a switch mode controller your puny supply would be able to put double that through the solenoid continuously.
Also why 100mA? Is your research into USB ports of the same quality? Standard USB ports are rated at 0.5A (and somewhat less than 5v at the end of a cable). If you are really using 400mA in the rest of the system then perhaps you should look for economies there.
I wanted a better understanding of the problem before I got started. I also have to build a DC/DC converter, which is something else I've never done. Jeeze.
Anyhow, I hope to have parts tomorrow.
Puny?! Don't get me started with the bumper sticker quotes 8-)
How does a switch mode power supply provide more P at the output than the input?
I've got the Rev 2.0 spec from April of 2000. Do you have something newer? Are you suggesting that each unit ship with a powered hub? That's expensive!
:-0
I wish.
Thanks for your input. I really don't know much about "switch mode power supplies" & couldn't find a quick definition (& its too late to go reading specs). Is that just another way to say "switching power supply"?
ElectronDepot website is not affiliated with any of the manufacturers or service providers discussed here.
All logos and trade names are the property of their respective owners.