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Driving a 12V load from 5V supply

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NJ Roadmap

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Hey, I'll try and keep this simple and hope that someone can help.

I'm using a PIC to drive a 12V 210mA valve (a solenoid irrigation valve if anyone's interested). The PIC is powered by a 5VDC 300mA regulated mains adaptor. Now I've planned and half-implemented a DC-DC converter for the valve, but am waiting on a 120uH inductor to complete the supporting circuit for the converter. The inductor has a 300mA rating so it should be able to drive the valve. Currently the DC-DC converter's input ends are connected to the 5V 300mA supply but ideally I'd like it to be connected to a pin on the PIC so it can be directly driven.

The DC-DC converter is a LT1173, and at the inductor and capacitor ratings for it's supporting circuit, it *should* supply 12V at 90mA minimum as quoted by the datasheet, the link to which is below:

https://www.electro-tech-online.com/custompdfs/2006/11/63348.pdf
(See page 11 for the configuration table and look under 5V to 12V at 90mA minimum output..I have used a 120uH L and 100uF C as stated in the table)

The thing I don't get is the 90mA minimum output current from the LT1173 in the 5V->12V configuration. What does that mean, and will it be able to drive a 210mA load in that configuration? Also how will a 5V 300mA supply limit the 12V output in terms of current?

And finally (yes I know this post wasn't short!) what other approach could be taken to drive a 12V 210mA load from a 5V logic supply?

Thanks in advance, I look forward to some informative replies!
 
Are you sure that's a DC valve?
I think all the irrigation solenoids I recall are of the AC type. They won't work on DC.
 
Well 12V at 210mA is 2.52W, and 5V at 300mA is only 1.5W - so you don't have enough power to feed the solenoid, even assuming it's DC and not AC. Switching it from a PIC is simple enough, just use a switching transistor.
 
Also, the minimum guaranteed output of the LT1173 is only 90mA at 12V. Even if you had hundreds of them to test, I doubt that you will find one that provides 210mA. Most are probably 120mA. It is micropower which means very low power.
 
Here's how I would do it. use a 12VAC 300mA (or higher) wall wart. bridge rectifier, filter caps and 7805 to power the PIC. Then use a sensitive TRIAC to control AC power to the soleniod. Take the TRIAC power from the wall wart directly, before the bridge.

If it's a DC solenoid, use a logic MOSFET and take the power after the rectifier.
 
Yep I just realised the 2.52W vs 1.5W from the supply problem. I'm going to use a 12VDC supply instead (yes, the valve is 12VDC I specifically asked for one:) ). It should be ok to step it down to 5V using the LT1173 right? I like the ideas you're sharing, and thanks for that, but I would prefer to reuse the LT1173 instead of going with a completely different setup. I don't have any heavy loads on the 5V line - just a PIC, LCD and a couple of 3mA sensors. The LT1173 can give 300mA min. at 5V, so it should be good?

I wish I'd realised the fundamental flaw earlier. I haven't been thinking straight!
 
Also your valve will probably have a hold open current/voltage and open current/voltage; the latter will be larger than the former. Suppose your valve will open at 12V 210mA, but it probably won't close until the current drops to 105mA.

To save power you can turn the valve on, then pulse it at as low as 25% duty cycle at a high frequency (>50kHz), the freewheel diode will keep the current flowing inbetween the pulses and you'll save a huge amout of power.
 
I like that idea. I'm going to experiment and find out what the hold open current/voltage is. So you're basically saying that it doesn't need to be supplied with a constant 12V, but can be switched at HF, and the frequency has to be determined. Also, whats a duty cycle? :p

On a different note, I resoldered everything today with the LT1173 in a 5V->12V config (not tested yet, waiting for farnell to deliver a 220uH inductor) only to ALSO realise that a 7805 sitting around in a box in front of me could do the job just as well!

The only question now is whether a solenoid valve driven by a 12V 500mA supply could possibly interfere with the PIC powered off a 7805 which is driven by the same 12V supply?

Thanks for the responses. I look forward to more!
 
NJ Roadmap said:
I like that idea. I'm going to experiment and find out what the hold open current/voltage is. So you're basically saying that it doesn't need to be supplied with a constant 12V, but can be switched at HF, and the frequency has to be determined. Also, whats a duty cycle? :p

You'll have to experiment a bit with this, the frequency and duty cycle will depend on the coil inductance and resistance. Remember an inductor tries to keep the current the same so when you switch it off the current will keep flowing through the diode which will decay exponentially. When the inductor is turned back on the the current will increase again but if it's not on for long enough then it won't go back up to the maximum value. The idea is you keep the average current as low as possible but ensure it doesn't go below the minimum hold on level or your valve will close or even vibrate. This is how a switching regulator works, except yout're using the solenoid as the coil in the switching regulator. Adding a 100nF and 10:mu:F capacitor across with supply to the solenoid will also help and use a schottky diode in reverse parallel with the solenoid to minimise switching losses.

On a different note, I resoldered everything today with the LT1173 in a 5V->12V config (not tested yet, waiting for farnell to deliver a 220uH inductor) only to ALSO realise that a 7805 sitting around in a box in front of me could do the job just as well!
Sorry I must be mistaken, you're still not intending to use the steupup regulator with your 5V 300mA supply are you? I thought we already decided that this won't work.

The only question now is whether a solenoid valve driven by a 12V 500mA supply could possibly interfere with the PIC powered off a 7805 which is driven by the same 12V supply?
No it won't, providing they both shate the same 0V connection.
 
Sorry my bad I meant LT1173 in a 12V->5V config! I ordered a 12V 500mA supply.

Currently it looks like this:

12V 500mA adaptor (12V) -> LT1173 -> (5V) to 7805 reg -> (5V) rail for PIC, LCD, small sensors

I think I'll take the LT1173 out of the equation, but I am going to try the LT1173 in the new 12V->5V config just to see if it works.

Thanks for the info. I'll be sure to remember what's on here and implement it. I'll let you guys know when I get it working.

Cheers.
 
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