I have a Honda EU2000i gasoline generator that has a battery charging DC outlet that I would like to use to power a 12volt inline fuel pump. Unfortunately the DC output is unregulated and puts out 28 volts at no load and high rpm. It drops to 14 volts only at 8 amps. The fuel pump needs only
2 amps at 12 volts.
I tried a MC7812 regulator but it wouldn't produce the surge current to start the pump. Short of building a more complicated and powerful voltage regulator, can any of you suggest a simpler approach to limiting the voltage to the pump?
P.S. I have already burned out one pump by hooking it to this DC output.
Simplest way. Attach a lead-acid battery to the generator, and run the pump from this. Alternatively, you might pop down to the library, and get "The Art of Electronics", which describes a number of high power regulators, or look in the 7812 datasheet for suggestions on how to boost power.
the simplest solution would be to add a high curent npn transistor as an emiter folower after the voltage regulator say maybe a 2n3055 on a heatsink, this would lower the output voltage a bit but would provide more current, the output curent would be limited by the gain of the transistor x the output curent of the regulator.
if the lower output voltage is a problem you could put a diode in the ground lead of the regulator or use a pnp pass transistor driven by the input of the regulator.
It may be a good idea to check the DC output with an oscilloscope while it is loaded. See how much ripple you have. Assuming there is not a large amount of ripple, and assuming that the fuel pump doesn't turn itself on and off automatically, and that there are no other loads on the (nominally) DC output, I think your best bet might be to just put some diodes in series with the fuel pump. Make sure they can handle the 2A current. They will get warm, and you may need a heat sink.
Probably you should first measure the actual Voltage at the "12 V" output when it is supplying 2 Amps. Then you can look at the diode datasheet to see what the forward Voltage drop is at 2 Amps, and this will allow you to select the right number of diodes. Of course you will double-check the calculation when you implement the circuit.
Another option might be to parallel several 7812's. Be sure to use a current-sharing resistor for each one. Given the information you have presented so far, it is impossible to give an accurate estimate of the power dissipation for the 7812's, but this needs to be determined so you can figure out how much heat sink you will need.
Note that I've never done this but it should work. (Famous last words.)
You might also complain to Honda. A "12V" output which puts out 28 V when open is pretty bogus, IMO. On the other hand, we have the same generator where I work, and it is otherwise wonderful. It is quiet, fuel-efficient, and provides very clean AC power. We don't use the 12V output, and we never even looked at it with a VOM, so we didn't know about this problem.
If the generator provides 8 amps into a 12 volt battery and the electric motor takes 2 amps out then there will be a 6 amp charging current into the battery as long as the generator is running. That will make for a very short battery lifespan.
A current limiter between the generator and the battery set to provide just a little bit more than the nominal motor running current will "float" the battery and still allow it to provide the starting surge.
A fairly small battery is all that is required. One of those sealed gel-cells from a small UPS should do the job.
View in fixed font. You can limit current without using a voltage regulator. Just off the top of my head: in out ______________ ____________/\\/\\/\\/\\______________ | \\____/ _|__ | |___/\\/\\/\\/\\___|_______/ \\____________|
Hook this up to the positive terminal of your DC supply. Both transistors are NPN with arrows pointing to the right. The transistor on the left is a power transistor like a 2N3055. The lefthand resistor initially biases the power transistor ON and you have current. The transistor on the right gets biased on when current rises to a point where voltage drop across the righthand resistor to exceeds about two thirds of a volt. When that happens the righthand transistor pulls the power transistor base down and biases the power transistor off. Of course it does not cut off the current to zero; current balances out at your chosen max.
Your ASCII is beyond bad. The problem is that you typed it up in some proportional font, which is not the standard practice for ASCII art. Next time, use courier or a similar font to draw it, and advise the user to use courier or similar to view it.
Also, the OP wants a Voltage regulator, essentially, and you are giving him a current regulator...
Here's one random idea that uses only passive parts:
Assuming there's no way to regulate the generator speed, then you need to dissipate some energy somewhere. So, something's going to get hot. Hmm, what can we use? Well, two things that have interestingly nonlinear v/i curves, that don't cost much, and that can deal with heat are light bulbs and diodes.
How about a 45W 12V headlight bulb in series, and a string of 20 silicon diodes (rated 3A) in parallel with the pump? When the pump is running, and drawing 2A, that won't be enough to turn the headlight bulb on so it'll be low resistance. But if the voltage goes above 14V (the 20 diodes, times nominal 0.7V; adjust by changing number of diodes), then the diodes will conduct and drop the voltage down. But they won't have to drop the full 6A, because as the current goes up the bulb will turn on, introducing series resistance.
Just a crazy idea, haven't really thought it through. Maybe the current-limiting effect of the bulb will get in the way of the pump starting. If so, maybe you could parallel it with a capacitor.
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But personally, I'd grab a couple junk box components and build the following. Note that the 2N3055 might be dissipating a good 10 or 20W, so give it a good heat sink. The unmarked transistor could be anything that can handle 500mA peak and about 100mA continuous with 10V across it - something like a TIP31 would be dandy. The TIP31 will be dissipating about
1/20th the power of the 2N3055, so you might want to mount it on the heat sink too. The two ordinary diodes in series with the Zener are to make up for the BE voltage drop of the transistors - or, you could just use a 14V Zener, if you have one around.
It is not quite as simple as an MC7812, but a simple switcher chip, the LM2678, with a boost cap, an input filter cap and an RC low pass filter for the voltage feedback from the motor (the motor performs the task of both inductor and load resistance) would make a cool running circuit.
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Figure 2 shows how to set the output voltage with a divider. Since you only have access to the pulsed output voltage (with the inductance of the motor smoothing the current), you have to add a capacitor to ground at the feedback pin to turn the pulse into a triangle of ripple riding on the average output voltage. The RC time constant of this cap and the equivalent resistance at the center of the divider sets the operating frequency. I would use a 3 amp 40 volt Schottky diode like a 1N5822.
The 28 to 14V at 8 amps suggests a generator impedance of 14/8=1.8 ohms. If you plan on 4-5V across the combination boost+regulator, then a series dropping resistor of (28-12-5)V/2A-1.75V=3.9 ohms works with power dissipation 3.9*2^2=16W. The pump turn on surge can be handled by another boost that momentarily bypasses the 3.9 ohm resistors. So something like this is cheap and uncomplicated if you can find a nice
50W heatsink block to bolt everything onto: View in a fixed-width font such as Courier.
If you use some sort of electronic switch (relay or transistor) between the generator and the battery, you can connect the generator to the battery to charge it and then run off the charged battery for a while.
The switch or relay could be made to go open for, lets say, 1 minute any time the battery voltage exceeds 13.9V.
This has the advantage that very little power is lost in the switch.
It does mean that you need a bigger battery but there is no large heat sink needed.
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