# Dual halfwave rectifier polarity inverter

1. Jun 10, 2009

### JZD

Dear honored forum members.

Thanks in advance for any information and help.

The title of this post might not be technically correct, but it incompases the issues that im working with.

My goal is to charge a 48 volt batterypack from a 12 dc volt source. I need a rather high amperage, 50 amps or more so buying a specialised dc-dc charger is very expensive.

I solved the problem by buying a cheap mass produced 12 volt dc to 110 volt ac inverter and then putting a half wave rectifyer on the back of it, giving me a good 58 volt Dc.

It works perfectly and my batteries are charging fine.

My issue is the inefficiency of only using half the sinewave.

My conceptual idea is that If I could somehow invert the polarity of the other half of the sine wave - half wave rectify that and then parallel the two. I would then be able to use both halfs of the sinewave while still having the approx 58 volt dc for charging.

I have searched and read alot, but my fundemental problem is that I dont know what the right question is to ask. Why the long explanation :)

Thanks very much for your time

John Z D

2. Jun 11, 2009

50 amps...... holy moly.

Would a bridge rectifier work?

You can try a custom made charge pump to just quadruple the voltage.

3. Jun 11, 2009

### skeptic2

What are you using to supply the 12 volts to the inverter, your car battery?

What is the rated amperage of the inverter?

How did you measure the 58 volts? Unless you used an oscilloscope, it probably isn't really 58 volts. Your peak voltage may be as high as 170 volts.

Yes a full wave bridge rectifier is the way to go but be careful. Your rectifier should be rated at somewhat more than 50 amps. The full wave rectifier will double the charging current. Please check that the 48 volt battery pack isn't getting hot.

4. Jun 11, 2009

### JZD

I might have been unclear.

The 50 amps is on the 12 volt side so the power usage is 600 watts and up.

The reason is dont use a full bridge rectifier is that it would give me over 115 volts dc out. Using the half wave rectifier cuts the voltage in half. Im looking for a method to keep the voltage in half but still use the full sinewave. Hence the polarity inverter question. If you could half wave rectify both the positive and the negative part of the sinewave from the inverter and then invert the polarity on one you should have the same potential albeit with a time difference. Them having the same potential + 58 volts approx they could be parallelled for double the current. Or atleast that it the idea.

Sorry if my intentions where unclear

5. Jun 11, 2009

### skeptic2

A 115 vac sinewave will go up to a peak voltage of about 162 V. Ordinary meters are designed to read that waveform as the DC voltage with the equivalent heating power which is known as RMS. When you measure the voltage of a half wave rectified waveform, you are measuring 115 V for half the cycle and 0 V for the other half. The meter averages the two readings and shows 58 V. The peak voltage is still 162 V (unless loaded down by the batteries you're charging).

If you use a full wave rectifier, your peak to peak voltage will be double the peak voltage or 324 V. Your meter will probably read it as 230 volts. I would not attempt to use that voltage to charge a 48 V battery pack. This means I was wrong when I said above that a full wave rectifier was the way to go.

You need to find out from the manufacturer of the battery pack what the peak charging current and voltage are and stay under that. Unfortunately finding the peak charging current is not easy. Since no current is drawn until the voltage exceeds 48 V, and above that, it will be determined by the series resistance in the circuit including the charging resistance of the battery.

Probably the easiest way is to put a one ohm resistor in series with the battery and measure the peak voltage across the resistor with an oscilloscope. That voltage would correspond to the current in amps.

6. Jun 11, 2009

### Bob S

A full-wave center tap would work, if you had a center tap on the 110 V ac, but you don't. A full-wave bridge is too much voltage out. One possibility is a step-down transformer so you could use a full-wave bridge. There really is no inefficiency in using only half the sinewave. It is sort of like using a V-8 in a VW Beetle. But there is little power wasted. There is quite a large current surge when the rectifier starts conducting; it would be nice to have an inductance between the 110 V ac and the rectifier. The inductance would be very expensive however because of the large dc current requirement. So maybe just a surge resistor will do.

7. Jun 11, 2009

### JZD

Good info regarding the peak voltage. Havent thought about that.

I have reading up on batteries and how they handle peak voltage and rms. But info seems to be scarse. If the charge current dont exeed the batterys limit, does the voltage peaks matter ?

The battery manufacturer only stipulates that it charges optimum if the charge voltage is 2 volt above pr 12 volt cell and then specify a max amperage.

From taking a car alternator apart years ago, it only have 6 diodes and no other form of smoothing (from what I could se) so wouldent that mean that car alternators also have voltage spikes ?

If the voltage spikes are a problem wouldent a simple capacitor be the easiest way to fix that ? Or am I completely off ?

thanks so much for your attention.

8. Jun 11, 2009

### Bob S

This is a very good idea. I looked up and found a full wave voltage quadrupler circuit in my ARRL handbook (1956 Edition). It is for selenium rectifiers, but I am sure it will work for silicon rectifiers. The big cost will be the capacitors. Probably the circiut should be analyzed in Spice to determine the optimum capacitor values for the required current. Unfortunately, it works only on an AC power source.

Last edited: Jun 11, 2009
9. Jun 11, 2009

### TurtleMeister

Ripple in the battery charger output is not good for the battery. And it appears that your setup is producing a lot of ripple. Here is a very detailed article about it:
http://www.cdtechno.com/custserv/pdf/2131.pdf [Broken]
Yes, but the output is fed to a voltage regulator.
It would help, but with your setup you would need a very large bank of them.

Last edited by a moderator: May 4, 2017
10. Jun 12, 2009

### Bob S

I think that the alternator regulator regulates the output voltage by controlling only the voltage that is fed back to the armature, and the raw rectified stator AC output is fed to the battery and accessories. If your alternator has 3 pairs of diodes, it is probably a 3-phase output that has less ripple than single phase. I doubt that the regulator is fast enough to remove any voltage ripple from the rectified output.

Last edited by a moderator: May 4, 2017
11. Jun 12, 2009

### TurtleMeister

Bob is correct. The low ripple output (less than 90mv) for a car alternator is not due to it's regulator but to it's 3-phase full wave rectification. There is a huge difference between the ripple output of a 3-phase full wave bridge and a single phase half wave rectifier.

12. Jun 13, 2009

### RonL

when I mess around with DC motors, I use a router speed controller plugged into the inverter, the output lead is wired to a full wave bridge rectifier (1000 volt 35 amp) the output then is DC, I just dial up to whatever speed the motor needs for a certain RPM, with a meter inline you can see exactly what voltage is going to the motor. (or whatever)

13. Jun 13, 2009

### Bob S

In this case you are feeding the same reduced voltage to both the stator and the rotor winding. It does not compare to regulating the output of the alternator.

14. Jun 13, 2009

### RonL

Guess I drifted away from the OP, your point of the step down transformer (2:1) was what I had in mind at first, the variable speed controller was how I would change the voltage to the transformer so that it's output was 48 volts, the voltage then passed through the rectifier is 48 volts DC.
If a pure sine wave is needed, then I would hook two motors together and control the speed to the unit working as a generator, convert that output through the rectifier for a smooth DC output.

I might go the long way around, but I generally get the end results I need.

Ron

15. Jun 26, 2009

### JZD

Ok, Have been doing some reading inspired by the suggestions in this thread. Especially thanks to Turtlemeister for the very informative article about ripples effect on batteries.

It is obvious that I need to smooth it out, alot. It works for now, but my batteries wont last long.

Using a couple of high wattage zenerdiodes @56 volts would that be the best and or most economical way to smooth out the ripple ?

What suggestios do you have for smoothing the ripple out ?

And does anyone have an idea how to economically utilize the other half of the sinewave?

Thanks very much. Im learning alot from this.

16. Jun 26, 2009

### Bob S

If you can find a centertap on the inverter, you cold solve the problem by using a full-wave centertap rectifier configuration. An alternative is to get a 60 amp triac-style lamp dimmer (I don't know of any) to reduce the average or rms voltage out of the inverter,
60 amps is a lot for an inductor, but if you could find an inductor rated at V=60 volts, and dI/dt = 60 amps/0.01 sec so L = 60 x 0.01/(60) = 10 mH.

17. Jun 26, 2009

### vk6kro

You are lucky it is working at all. You are basically shorting the output of your inverter into a battery bank via a diode, without any current limiting. Getting it to work at all was quite an achievement. The inverter must have some pretty effective current limiting in it.

Anything like adding a bridge rectifier will give you more voltage which you don't need. There was a post earlier that explained this in detail. I quote Skeptic2:
A 115 vac sinewave will go up to a peak voltage of about 162 V. Ordinary meters are designed to read that waveform as the DC voltage with the equivalent heating power which is known as RMS. When you measure the voltage of a half wave rectified waveform, you are measuring 115 V for half the cycle and 0 V for the other half. The meter averages the two readings and shows 58 V. The peak voltage is still 162 V (unless loaded down by the batteries you're charging).

Adding a bridge rectifier gives you exactly the polarity inversion you are asking for , but the peak voltage will be still 162 volts and the average will rise to 115 volts.
You are measuring 58 volts now, but your meter is not telling you the true story.

To save your batteries, you could try putting some 110 volt lamp globes in parallel with each other and then in series with the batteries. This will increase the charging time, but may also increase the battery life.

You could get a power transformer to give about 40 volts from the inverter, rectify that with a bridge rectifier and current limit the current into the battery. That would give you peak voltages of about 56 volts and useful charging capability.

Maybe it would be more efficient to look at the source of the power, though. If it was AC from a windmill generator, could you put a transformer and bridge rectifier there to directly charge up the 48 volt batteries?