Voltage Multiplier

[bcwilson_oz1]

???
I am looking for the basic voltage multiplier circuit. Diodes & Caps. I have lost my basic circuits notebook and I need valuse etc.

Thanks in anticipation.

Brian
Canberra Australia

 

[mixos1]

The circuit and some typical part values are the next:

 

[bcwilson_oz1]

Thanks

I was about 1 - 2mF which was close.


thanks for your help

Brian

Canberra Australia

 

[bcwilson_oz1]

The circuit and some typical part values are the next:

The first CCT is OK but the second wil provide a 1.8V output from virtually any AC input. The caps are in parallel and all the diodes will do is clip the voltage.

 

[mixos1]

This happens if you measure the voltage across the final diode.

If you measure the voltage at the point where the last capacitor and diode are connected (upper right corner of the circuit) refering to the ground then the voltage you measure is the multiplied one.

 

[bcwilson_oz1]

Thanks

I will try that and advise. It certainly looks better

Brian Wilson
Canberra Australia

 

[mixos1]

By clicking here you can download a circuit that is an ion generator and consists of a voltage multiplier circuit.

This is a usage of an voltage multiplier circuit

Also this is another design of the voltage multiplier:

 

[Kevin Weddle]

Does the final capacitor in a voltage multiplier determine the current that is supplied to the load? I think the other capacitors can be smaller, which results in a reduced output voltage.

 

[MP1]

Current does not go up and down in a path. It remains the same.
Perhaps I do not understand the question?
MP

 

[Kevin Weddle]

The load is not across the capacitor. The final capacitor charges to the peak of the cycle. The other capacitors keep charged to a voltage and pass the AC. When the cycle continues the diode is reversed and the final capacitor delivers to the load.

 

[Ldanielrosa1]

All of the capacitors in a voltage multiplier contribute to the impedance. Until I read otherwise I'll go with my belief that they work best when all of the capacitors are of the same value.

 

[MP1]

Here are some good links on voltage multipliers:

http://www.tpub.com/neets/book7/27m.htm

http://braeg.piranho.com/e_s/2theory/2_6_1_voltage_multiplier.htm

Yes, all of the capacitors should be the same value. The circuit depends upon the direction of voltage from the rectifiers charging the capacitors, which depend upon the charging rates to be the same. The voltage multiplier could become unstable with different capacitor values.

MP

 

[hotwaterwizard2]

Let's compare a Capacitor to something Common say a Water Filter in a Swimming Pool.
As the flow of water goes into the filter it is slowed down by the filter cartridge. The Water Flow is like Current in electricity and the Water Pressure is like Voltage.The speed at which it is slowed down depends on how many microns the filter is. The filter also removes unwanted impurities in the water . The unwanted impurities in Electricity are voltage spikes, unwanted AC voltages. Ect.The amount it removes also depends on the microns of the filter. If the holes in the filter are big then the water will not come out as clean. With a Capacitor the bigger holes are a lower microfarad. So a larger microfarad would be the same as a finer Micron Filter cartridge in the Pool Filter.
In the pool filter the smaller the holes are the higher the pump pressure remains.,
The same can be said about a Capacitor. The voltage will stay higher for a longer time with a large capacitor.

 

[Cloink]

Just looking for a bit of help, I'm not an electronics whizz, remember some from school & uni, so assume I'm an idiot please!

I'm trying to make the voltage multiplier circuit, but nowhere does it say whether ceramic caps must be used or whether electrolytic ones are ok. (Just to be absolutely sure I'm not using terms that mean the wrong thing, ceramic caps charge whichever way you connect them - yes? - whereas electrolytic ones only charge 'the right way round' - yes?)

Also, another site I found has a voltage doubler circuit but it's connected differently (as far as I can tell) and outputs (rippled-)DC not AC and seems to only charge the caps in the same direction whichever half of the AC cycle you're on. I'd rather assumed that the multiplier on this page is AC output. Which has put me in rather more confusion than before, not less!

Help very much appreciated.

 

[Kevin Weddle]

The output is not AC. The output is pulsating DC. This is because there is a DC voltage offset becuase of the DC charging point on the sine wave. Here is the voltage relationship.
20 40
10 30
10 -10 | | 0 | | 20
diode diode diode
0 | | 20



The -10 never materializes because it's limited to .7 volts by the first diode. But only until the positive cycle clamps the first capacitor at 10 volts.Notice how the 20 sets the minimum voltage for the next stage. This is a solid DC that will try and discharge to maintain the voltage on the last diode,but you can use a larger capacitor here to maintain a solid DC.

 

[Cloink]

Ooo-kaaay.... Sorry, that's not really helped.

Q1
Can I use the electrolytic caps I've already got, or do I need to go get ceramic ones?

The other site I mention is
http://www.tpub.com/neets/book7/27m.htm
Towards the bottom of that page it's got "Figure 4-49. - Full-wave voltage doubler" - after having talked about half wave doublers & triplers.

Q2
Is the multiplier circuit given on this site a half-wave or full-wave?

I want to get as much power as possible as I'm generating AC from a wind-powered dynamo on my bike and the output isn't going to be too great to start off with.

Q3
If it's a half-wave multiplier, can you tell me how to do a full-wave tripler/ quadrupler/ etc? The above site only shows a doubler, but I need to multiply by more than 2.

Q4
You're saying to minimise ripple I can just use a whopping great cap as the final one rather than including a cap in // with the load?

Elsewhere in this thread, someone says you must use all the same size caps else it may become unstable.

Many thanks,
Clark Pearson.

 

[audioguru2]

I want to get as much power as possible as I'm generating AC from a wind-powered dynamo on my bike and the output isn't going to be too great to start off with.

Hi Clark,
A voltage multiplier will not give more power output than the dynamo already gives. It increases the voltage output by rectifying AC and adding the capacitors' voltages, and its output current is reduced by the same mulipication amount. Just like in an AC transformer, power (voltage times current) in, equals power out. A voltage multiplier and a transformer actually give less output power than the input power because of losses.

What do you want to power with the increased voltage? If your dynamo gives 6V at 0.5A (a really big windmill propeller and your bike is going very fast) then its power output is 3W. If you multiply the voltage 10 times to 60V, the output current will be less than 50mA, which isn't enough for anything.

Are you certain that your dynamo has an AC output? If it has a DC output like the one on my bike, these voltage multipliers won't work.

 

[Kevin Weddle]

I don't know what he means by unstable. Maybe the voltage is so high that it shakes the circuit unstable. I don't know. But yes you can use a whopping big capacitor to reduce the ripple and obtain a very fine DC. After all it is just a rectifier and capacitor.

 

[audioguru2]

If the output cap is too big it will take a long time to fully-charge from the smaller charge that is "dumped" into it from the smaller doubling caps on each half-cycle of the AC. Look at the staircase signal in the posted article.
Fill a bucket with another same-size bucket or use a teaspoon to fill another teaspoon. But don't fill a bucket with a teaspoon unless you have lots of time to do it.

If an output cap that is too big is suddenly discharged then allowed to charge again, it will take another long time for it to fully-charge. Maybe that is why it is called unstable.