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Old 11th July 2010, 07:29 AM   #1
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Default Supply voltage doubler question

In the attached circuit I've shown a voltage doubler circuit driven by a center tapped transformer ( S1,S2) . I checked the voltage across the capacitors C1 and C2 after switch on. These capacitors seem to have a reverse voltage across ( up to 15 V peak depending on component values) for one cycle followed by a lower voltage on the second cycle and then charges up correctly according to the polarity of the capacitor.

Assuming my simulator is correct , would these two reverse polarity cycles damage the capacitors progressively after multiple switch on cycles ? I added the two diodes D5 and D6 across the capacitor to limit the reverse voltage to a diode drop. Is that OK or even required ? I don't see any diodes like these being used in published circuits . Is that an indicator that they will not create a problem ?

I'll add the voltage plots to show the effect with and without the diode.
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File Type: bmp Voltage multiplier.bmp (54.7 KB, 245 views)
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Old 11th July 2010, 08:31 AM   #2
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Quote:
Originally Posted by ashok View Post
In the attached circuit I've shown a voltage doubler circuit driven by a center tapped transformer ( S1,S2) . I checked the voltage across the capacitors C1 and C2 after switch on. These capacitors seem to have a reverse voltage across ( up to 15 V peak depending on component values) for one cycle followed by a lower voltage on the second cycle and then charges up correctly according to the polarity of the capacitor.

Assuming my simulator is correct , would these two reverse polarity cycles damage the capacitors progressively after multiple switch on cycles ? I added the two diodes D5 and D6 across the capacitor to limit the reverse voltage to a diode drop. Is that OK or even required ? I don't see any diodes like these being used in published circuits . Is that an indicator that they will not create a problem ?

I'll add the voltage plots to show the effect with and without the diode.
Hi,

It's an interesting circuit. In my view however I don't think it does what you want it to do. If you use the outer transformer leads without the centter tap and a diode bridge you'll get the voltage you want.
Don't forget that voltage doublers by design have a much larger ripple than 'normal' rectifier circuits and have much higher Zout meaning that the voltage drops significantly with load.

Voltage doublers are generally only used when it is impractical to have a xformer for the full voltage and where small currents are required, e.g. in high-voltage bias circuits.

jd
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Old 11th July 2010, 09:25 AM   #3
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Hi Jd,
This is meant for low operating currents and yes the ripple is ( expectedly!) high.
I don't really plan to use this as I can get a much cleaner supply by just using additional windings. However from the operation point of view I wanted to check if my idea was correct about using those diodes. I forgot to add that the transformer already produces a +/- supply rail for the main circuit and the voltage doubler circuit has to use the center tap as gnd. The HV is being used by a another circuit that needs very low currents. There is enough extra voltage to sacrifice and clean up the ripple .

I'm attaching the voltage build up plot across one of the capacitors , C1.
Without the diode D5 and with the diode installed. Any ideas on that ?
Thanks.
Attached Images
File Type: jpg No diode.jpg (14.9 KB, 298 views)
File Type: jpg With diode.jpg (18.5 KB, 297 views)
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Last edited by ashok; 11th July 2010 at 09:30 AM.
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Old 11th July 2010, 10:44 AM   #4
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Quote:
Originally Posted by ashok View Post
Hi Jd,
This is meant for low operating currents and yes the ripple is ( expectedly!) high.
I don't really plan to use this as I can get a much cleaner supply by just using additional windings. However from the operation point of view I wanted to check if my idea was correct about using those diodes. I forgot to add that the transformer already produces a +/- supply rail for the main circuit and the voltage doubler circuit has to use the center tap as gnd. The HV is being used by a another circuit that needs very low currents. There is enough extra voltage to sacrifice and clean up the ripple .

I'm attaching the voltage build up plot across one of the capacitors , C1.
Without the diode D5 and with the diode installed. Any ideas on that ?
Thanks.
It's not a doubler, it's a halver
You get the same or more voltage from one 30V winding. You'd get the double if you use the outer windings.
I don't think D3 and D4 do anything; they are always opposed by another diode and will never conduct. Take D3 & D4 out and see if it makes a difference.

jd
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Old 11th July 2010, 11:58 AM   #5
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Quote:
Originally Posted by ashok View Post
I don't see any diodes like these being used in published circuits . Is that an indicator that they will not create a problem ?
Quote:
Originally Posted by ashok View Post
...from the operation point of view I wanted to check if my idea was correct about using those diodes.
No, it's not.

Your circuit is FUBAR.

You don't understand what you're doing. Don't fiddle around with it any more.

Just copy a conventional circuit. Connect a bridge rectifier across one winding and follow it with a cap.

Click the image to open in full size.

w

If you want to be pedantic you can build a full-wave centre-tap rectifier and save 2 diodes. Look it up...

Last edited by wakibaki; 11th July 2010 at 12:03 PM.
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Old 11th July 2010, 12:33 PM   #6
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Originally Posted by wakibaki View Post
No, it's not.

Your circuit is FUBAR.

You don't understand what you're doing. Don't fiddle around with it any more.

Just copy a conventional circuit. Connect a bridge rectifier across one winding and follow it with a cap.

Click the image to open in full size.

w

If you want to be pedantic you can build a full-wave centre-tap rectifier and save 2 diodes. Look it up...
Don't be such a scientist...

jd
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Old 11th July 2010, 12:58 PM   #7
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Hi jd

I can't help it, it just comes out that way... All the guys I learned the most from didn't pull any punches. I guess it shows

w

Last edited by wakibaki; 11th July 2010 at 01:05 PM.
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Old 11th July 2010, 01:34 PM   #8
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Hmmmm....

Let me add that apart from those two diodes D5 and D6 the circuit is functional and working for quite a long time.

In fact the circuit is borrowed from one of the threads on this forum. I have shown only one half , the other half produces the -ve doubled voltage.

Wakibiki , you probably haven't understood how this circuit works. If you spend a few minutes you can see how the voltage doubles here. It is quite simple . You also haven't apparently understood my question.

You will find that as the supply switches on , the initial current flow through the capacitor C1 will reverse charge it ( junction with diode D1 and D3 will be -ve) by an amount depending on the ratio of C1 and C3 and the current flow. On subsequent cycles the capacitor C1 will get charged such that the side connected to the junction of diodes D1 and D3 will be positive. After that the voltage across C3 keeps ramping up till it reaches close to double the peak supply voltage. Subsequently C1 will always be positive at the junction of the diodes D1 and D3 .There are some drops in the circuit and other losses and so it will never reach the full doubled voltage. Any load comes in parallel with C3 and hence the greater the load ( more mA) the lower the output voltage. The circuit has it's uses and it's limitations.

Now the "question" was , "are the diodes D5 and D6 of any use ?".

On further reflection I think the appropriate answer is that it does cut out the reverse voltage across C1 or C2 to a diode drop for the first cycle or so. But since this happens briefly only at switch on and not after the circuit is operational plus it happens for just one or two half cycles ( 20 mS), it can be dispensed with. It shouldn't have enough time to cause any problems.

There the problem is solved .
AND you've understood how the circuit works !
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Last edited by ashok; 11th July 2010 at 01:43 PM.
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Old 11th July 2010, 01:56 PM   #9
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Originally Posted by janneman View Post
It's not a doubler, it's a halver
You get the same or more voltage from one 30V winding. You'd get the double if you use the outer windings.
I don't think D3 and D4 do anything; they are always opposed by another diode and will never conduct. Take D3 & D4 out and see if it makes a difference.

jd
I think you never looked at the circuit closely. D3 and D4 are absolutely essential for operation of the circuit . Without them it wouldn't work.

Take for example D1 and D3. On the positive cycle of the incoming signal, D1 conducts and charges up C3 incrementally. On the negative cycle D1 cuts off C3 from the input circuit but current flows through D3 and C1 and charges that up incrementally with +ve side at the junction of D1 and D3. On the next positive cycle D1 opens up again and C3 sees the supply voltage plus voltage across C1 ( they are additive and in series) and hence there is again an incremental increase in voltage on C3. This continues over several cycles till it stabilises at some point, as explained earlier. C1 is also essential as it is the one that helps build up the voltage to double the incoming ac signal.

Amazing. You guys never even looked at the circuit before making those sweeping statements in your posts.
Never mind !

The attachment shows the voltage build up plot with a 30-0-30 V (peak) 50 Hz supply.
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File Type: jpg Voltage doubler-output.jpg (19.3 KB, 269 views)
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Last edited by ashok; 11th July 2010 at 02:08 PM.
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Old 11th July 2010, 02:15 PM   #10
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Originally Posted by wakibaki View Post
Hi jd

I can't help it, it just comes out that way... All the guys I learned the most from didn't pull any punches. I guess it shows

w
I know, it's a natural for me too. Doesn't make it optimal

jd
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