Hopefully a simple question. I'm still not comfortable reading the datasheets without an advisor. If somebody could please advise me, I would be grateful.

I have a pump which requires 12VDC, 1.4A.

I want to power the pump with a wall adapter (what some folks call a "wall wart") that gives 20VDC 1.0A.

Now, there is a regulator called an LM7812, that will give a regulated 12VDC output. The headline on the datasheet (from Fairchild) describes this component as a "1A regulator". But later, the datasheet claims a maximum 2.2A peak current.

If I put 20VDC 1.0A into an LM7812, do I get 12VDC 1.6A out?

Thanks for your help.

Originally Posted by nyoo

If I put 20VDC 1.0A into an LM7812, do I get 12VDC 1.6A out?

No, you'll get 12V @ 1.0A output with 8 watts of heat dissipated in the 7812.
The easiest solution is to find a 12V 2A wall wart.

Originally Posted by nyoo

I have a pump which requires 12VDC, 1.4A.
I want to power the pump with a wall adapter (what some folks call a "wall wart") that gives 20VDC 1.0A.
Might burn up both the pump and the adaptor

Now, there is a regulator called an LM7812, that will give a regulated 12VDC output. The headline on the datasheet (from Fairchild) describes this component as a "1A regulator". But later, the datasheet claims a maximum 2.2A peak current.

If I put 20VDC 1.0A into an LM7812, do I get 12VDC 1.6A out?
Not continuously.

But from http://www.hosfelt.com/
you could buy an

"adaptor", which is really a
12 VCT 3 Amp (6-0-6) Power Transformer
for $8.99, it's part number is 56-192

and four
Diodes, 1N5401, at $0.12 ea.

and build yourself an unregulated supply that should be able to power this pump.

Hosfelt has wall transformers but they don't go up to 1.4 A.

You could also get two of their

12 VDC @ 800 mA Adapter $US 3.25
Mfg. - DV-1280
2-1/4" x 1-7/8" x 2-1/8"
Input: 120 VAC, 60Hz, 16 watt
Output: 12 VDC @ 800mA
The enclosure and the 6ft cord are
black. Center is positive for polarity.
Right angle coaxial jack.
OD - .215", ID - .082"
part number 56-922

and wire up some kind of current sharing arrangement with two 0.5 ohm 1/2 w resistors, but it's not very elegant.

The difference between the 1A and 2.2A figures is mainly that 1A is the guaranteed minimum that it will deliver while 2.2A is not. It isn't clear but I suspect that the 2.2A peak is measured for a short duration, not steady state. You cannot rely on that 2.2A at all times.

This series of regulator works a lot like a smart series resistor. By this, I mean that all it does is put just enough resistance in series with the power supply line to drop the voltage from whatever you put on the input to 12 volts. By smart resistance, I simply mean that the regulator varies this resistance so that it maintains 12 volts at the output regardless of the current drawn through it.

If you use this regulator, you will find that the amount of current you can get out of it in your application will depend on how hot it gets. The amount of power turned into heat in the regulator is equal to the current passing through it multiplied by the voltage drop across the input to output path. If your wallwart is indeed able to push out 20V at 1.4 amps (more on that later) then the regulator input voltage will be around 20 volts and the output will be 12 volts. So the regulator must dissipate 8 x 1.4 = 11.2 watts.

Now, to figure out how this will affect the regulator, we must look up the Thermal Resistance Junction -Air and I find the value there to be 65 degC/watt. This means that the junction temperature will go up by 65 degrees C for every watt the device must dissipate if you attach no heatsink. In your case the temperature rise will be 11.2 x 65 plus the ambient temperature, which is way way above 125 deg C which is the maximum allowable for this device's junction. So what will happen is that the internal thermal shutdown circuit will turn off the regulator's output because it will get smoking hot. The solution to this problem would be to attach a substantial heatsink. Let's say, for example, that you bolted the regulator IC to this heatsink:
http://www.aavidthermalloy.com/cgi-b...=530002b02500g

This heatsink has a rating of 2.6 degC per watt when relying on natural convection (open air). To get the junction to ambient thermal resistance we add this value to the LM7812 Thermal Resistance Junction-Case which is 5 degC per watt. So the net is 7.6 degC per watt. Your junction temperature will get up to 110 deg C (85 degC plus ambient of 25). This is still below the absolute max rating of the regulator and so it should work OK. The heatsink will get quite hot to the touch.

The last problem is that wall wart. Usually DC wallwarts are not regulated and they have some internal resistance, so they are usually rated for voltage at their rated load. This means that your wall wart will probably deliver more than 20 volts when you are pulling less than 1.0 amps and less than 20 volts when you are pulling more than 1.0 amps. This may help keep your LM7812 a little bit cooler than our prediction above. But at that amount of current, the wallwart will likely overheat. This is bad.

Originally Posted by nyoo

I want to power the pump with a wall adapter (what some folks call a "wall wart") that gives 20VDC 1.0A.

What does it say on the adaptor name plate?

Are you sure it can provide enough current?

Is 20V what you measured with no load connected?

It's possible that the output of the adaptor will drop to 12V when fully loaded so no regulator is required.

What's the 6-0-6 mean?
I took it to mean it's 12v center-tapped.
The four diodes you mention, are you thinking of a bridge rectifier?
Yes, a full wave bridge.

sketch out for me how I would join the two adapters' output with two 0.5 ohm 1/2 w resistors?

If you don't use the 0.5 Ω resistors and just put both adaptors in parallel, + to + and - to -, you'd get 1.42 A @ 12.2 v into your motor (see Thevenin voltage and resistance assumptions underlined below).
Try it without the resistors. If you get 12-13v across your motor you'll probably get a satisfactory motor lifetime.

Otherwise, the positive output lead of each 12v adaptor goes to one end of a 1/2 Ω resistor. The other end of one resistor connects to the other end of the other resistor and connects to the + terminal of the pump, and the adaptor negative leads connect to the pump negative terminal.

Assuming each +12v @ 800 mA supply is actually a +14v supply with an internal 2.5 Ω resistance, you should then get 1.39 A @ 12v into your 12v/1.4 A = 8.6 Ω pump motor.

You can measure the voltage drop across each 1/2 ohm resistor to verify equal current sharing.

the 1.4A is the rated pump running current.

Usually it will need a few times more than that to start running.

Originally Posted by nyoo

Willbe, I measured the no-load voltage of the 12VDC wall adapters. It's wanders about 14VDC. Does that meet your assumption "Assuming each +12v @ 800 mA supply is actually a +14v"?

Yes; see below

eblc1388, you must be correct that the initial start-up of the pump will require more than 1.4A current. Does that mean the proposed solution -- two 800mA wall adapters with their positive leads tied together -- will not work?
see below

Or is there some compensating "wall adapter magic" for the start-up load? Could I use three wall adapters, tied together as Willbe suggests? Could I add a capacitor for the start-up event?
see below

So Voc = 14v and Vload @ 0.8 A = 12v. (14-12)/0.8 = 2.5 Ω internal Thevenin equivalent resistance for each wall adaptor.

Let's say the pump motor resistance during startup is 2 Ω, about 5x lower than the normal apparent resistance [due to the back emf generated by a running motor].
You can confirm the startup resistance with an ohmmeter. Rotate the pump shaft and take a reading with the shaft stationary, repeat this a few times, and average the ohm value.

Two of these wall adaptors in parallel look like a 14v supply in series with 2.5/2 = a 1.25 Ω resistor.
So the startup current draw will be 14/(1.25 + 2) = 4.3 A and the voltage across the pump will be 2 Ω x 4.3 A = 8.6 v. The pump may start under these conditions.
If the pump does start I'd check how close you are to not starting by putting a low value resistor in series with the motor. 10' of #30 or 1' of #40 AWG is 1 ohm.
Or, see if it will start when using only one adaptor.

If you wanted the 4 A startup surge supplied by a capacitor, assuming a 1v drop during startup giving 11 volts across the motor [assuming the pump motor has its own on-off switch] and a 1 second startup period, with C = I*(ΔT/ΔV), this is C = 4*(1/1) = 4 Farads.
It's probably cheaper to parallel several wall adaptors.

If you ran the pump on one adaptor, once running the pump would draw 14/(8.6 + 2.5) = 1.26 A and have 1.26*8.6 = 10.8 v across it.
There would be (1.26/0.8)^2 = 2.5x the normal heat dissipation in the adaptor, so the adaptor wouldn't last very long.
Startup current would be 14/(2 + 2.5) = 3.1 A and startup voltage would be 3.1 x 2 = 6.2 v. The pump might not start with this low voltage.

Willbe, thank you for the detailed explanation of how the low value resistor is for limiting one (or the other) adapter's share of the current. I think the fault is on my side; I somehow wasn't able to hear correctly what you were sketching.

It seems like the Ohm's Law math handles eblc1388's concern: assuming 2 ohms, 14/(1.25 + 2) = 6.2 A. Therefore, two adapters in parallel will handle the surge on start-up of a 1.4 A pump. The specification page for this particular pump states that the pump requires only a 2-amp fuse. Does that confirm it?

You could PWM the motor at 60% duty cycle.

The adaptor can probably handle 1.4A surgers at this duty cycle so it wouldn't be a problem.

Originally Posted by nyoo

I'll take it that the 2-amp fuse in the pump manufacturer's spec sheet means the two-adapter solution will survive the start-up current. Can someone please confirm?
Two adaptors will handle 1.6 A steady state current which is >1.4 A. I doubt that the short-duration startup current will hurt the adaptors.

Is it a slow blow or fast blow fuse? If you can find trip curves for the fuses you can approximate the startup current. Generally a 2 A fast blow fuse can handle 4 A for one second.

If you all have the strength to go back to the point where there was
- an "elegant" solution from which I diverted you, and
- a "not elegant" solution which we pursued....

At my local Radio Shack I see, at hosfelt price's, a 12.6VAC 3A power transformer. After the discussion about start-up current for the pump motor, is that elegant power transformer still a viable solution?

The transformer has 3 (black, black, and yellow) wires facing the 115VAC mains, and 2 (black and yellow) wires facing the 12.6VAC. Also for sale was a 4A 400V full-wave bridge rectifier for about $2; however, it only has 4 connection points. Do I connect the yellow mains wire to earth/ground, and there's no polarity in the others?
The + - markings are the DC output of the bridge; the ~ marking is the AC input.
I wouldn't give Radio Shack a single penny.

At http://www.kpsec.freeuk.com/powersup.htm, they mention a smoothing capacitor.

Using their calculation, I get 2uF. Does that sound right?
I don't think you need one for a motor. It will raise the DC voltage across the motor somewhat.
This solution will cost. The two-adapter solution will cost me nothing (bless the inventor of yard sales).

Is there an upside to this power transformer solution?
The internal resistance of the transformer might be 1/2 that of the two adaptors.
Thanks again.

Measure and see how the pump behaves with one, two and three adaptors at the limits of what will be demanded of the pump. Hopefully you can then predict what the pump will do with a heftier transformer.