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bcwilson_oz

Voltage Multiplier

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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.

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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:



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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.

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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

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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.

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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.

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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.

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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.

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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.

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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.

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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.

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Thanks for all your responses but PLEASE could you answer the questions Q1 - Q3 directly? (I've got Q4 now.)

Of course I can't increase the power output by multiplying the voltage, but if the multiplier is only a half-wave multiplier then I can only presume it will have a poor efficiency.

Yes I'm sure it's AC, it's a motor in reverse. (Not that that necessarily means it's AC output, but it I've tested it, it is definitely AC.)

The 6V/0.5W dynamos are hugely restrictive (you must have noticed how much more effort it takes to cycle with it engaged), in this age of high-output LEDs, I'm convinced I can make something wind-powered that can at least trickle charge my rechargeable batts or hopefully if I'm lucky just power an LED-light directly, keeping a reserve in a large capacitor for when you're stopped at traffic lights, etc.

I'm hoping for something like 10V 50mA, so that would be 1/2W.

Thanks in anticipation,
Clark.

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Hi Clark,
I still think your motor/dynamo has a DC output. An AC motor usually doesn't have the required permanent magnets to make it a generator. The alternator on a car uses battery power to make magnetism for its field coil.

Why multiply the voltage to drive a 2V LED? With a 10V output you'll be throwing away 8V which is 80% of the dynamo's power. 50mA is too much current for an ordinary LED unless you divide the current among a few LEDs.

What will limit the dynamo's current when you pedal fast into a strong wind? The LED will burn out.

How much wind resistance will a big windmill propellor cause? I bet it will have the same or more resistance as pedalling a dynamo. Try pulling an umbrella or parashoot around.

Answers to your voltage multiplier questions:
1) Sure you can use your electrolytic caps. You must connect them with the correct polarity. The voltage in your application isn't very much.
2) We have shown half-wave voltage multipliers. You are correct, an AC generator will be more efficient with a full-wave multiplier.
3) Full-wave multiple multiplier? I dunno, why not just use a transformer or a generator with a higher output voltage? Add gears or pulleys?

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Thank you, very much appreciated!

It's a brushless motor (from a pc-fan) - I will triple-check when I get a chance but my multimeter gave very spurious results when I connected it in DC-mode, but very consistent results when connected in AC-mode (of course, this was when I was blowing on the fan blades to turn it).

I have already attached the fan/motor/dynamo to my bike and seen the blades turn - it's about 4" dia (same size as a pc-fan obviously!) Yes, there'll be more resistance when I connect a load on it, but I still think it will turn, if more slowly. But no, there will be negligble extra wind-resistance from the fan/dynamo.

I need 10V+ because I have a vast array of LEDs, 3 in series x14 sets in //. They are 3.2V/30mA rated: that's 9.6V plus a bit for the drop across the resistors. This already works with a 10.5V supply (8 rechargeable NiMH AA batts). The whole setup draws c.350mA when connected.

Honestly, I do, mostly, know what I'm doing, I just can't get my head round the voltage multiplier circuit.

Many thanks again,
Clark.

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Hi Clark,
Are you serious?
You're going to power 42 LEDs (a traffic light?) with a 4" PC's fan?
Ha, I figure that to develop nearly 4W of wind resistance (which won't be neglegible) the little fan will need a wind nearly the speed of sound! You'll need to step its voltage down, not up, to get your required 1/3rd of an Amp. Maybe you could put a big venturi in front of it to funnel lots of air to it.

BTW, the PC brushless fans that I've seen operate on 12VDC.

Never mind the LEDs. Use the fan to charge a battery when you ride downhill. Then let it propel you uphill. ;D

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Please please read the postings properly.

I'm sorry, I know I'm really a guest on this site, and I do appreciate all your help, but I'm quite exasperated that you haven't read my points.

I want to trickle charge the batts from the dynamo, I DO NOT expect to get 4W of power from the dynamo. Depending on how much power I CAN generate, and I'll keep you posted when I've got it working, then maybe, just maybe, I can generate enough power to trim my batts down from AAs to AAAs without having to recharge them after every 2 hours' use. If I'm REALLY lucky - and if I can find a full wave voltage multiplier circuit that more than doubles the p.d. - then I may just drag enough power out of the dynamo to power a lesser LED array in which case I'll have a rethink.

Yes the PC fan motor started life as a 12V DC motor, but I've taken the PCB off it and I'm just taking the 'raw' AC output it generates in its new 'skeleton' mode.

I am also quite well aware that I need to regulate my eventual DC output -- the dynamo will of course provide a different p.d. depending on how fast I'm going, so I have to multiply it up maybe 4 or 5 times so that when it is - and I am - going slowly I still have the 10V I need. When I'm going faster, this multiplied voltage will be way too high, so yes, I need to step it down. I have found circuits elsewhere that show me how to tap off a constant voltage.

Is there any other unsolicited advice you'd like to give me? I only asked a few simple questions.

Regards,
Cloink.

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Hi Clark,
I am sorry to have sounded so pessimistic about your idea, but I don't think a 4" fan will provide much power at bike speeds.
When you said you needed 350mA at 10.5V to power a 42-LED traffic light, I didn't realise that your Ni-MH rechargeable battery will be supplying most of the power.

I am also sorry that you may not be aware of this additional unsolicited advice:
1) A voltage doubler (tripler, etc.) is designed for a mains voltage input, that's why its caps have such a high voltage rating. If your dynamo has a peak output voltage of only 2.0V, the voltage drops of the two diodes in the 1st half-wave voltage doubler will allow it to have an output voltage of only 2.2V to 2.6V. The 2nd doubler will have an output voltage of only 2.6V to 3.8V, etc. After these two doublers you still haven't yet doubled the voltage but you have quartered the output current. The dynamo provides 2.0V at say 0.1A (200mW) in, and you get only 2.6V at only 25mA (65mW) out. That is only 33% efficiency before deducting ripple voltage.
2) Your dynamo probably has at least 3 coils which will have outputs on different phases. I think that each coil must have its own separate voltage tripler, quadrupler, etc. before being combined at the output. It will require many diodes and capacitors.

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