Full Bridge Voltage Doubler
MONO AND STEREO HIGH-END AUDIO MAGAZINE
Ever had a situation, where you simply "need" more volts out of your secondary winding, more than it actually delivers?
Let's say, you have an existing transformer, with about half of the voltage that is actually needed.
The first natural thought, idea, which would instantly come to mind, is a ... voltage doubler.
But ... the first natural objection, that would instantly come to mind is .... well, OK, ....
but the "basic" version of a voltage doubler is subject to certain limitations and does indeed have some "issues" associated with it, ones that would need be addressed, if to be made useful.
This following circuit idea was inspired by a question from a friend of mine, who was making corrections and modifications within a factory grade CD player product
{ ... and we all **DO** know what can be expected from "factory grade" sound, don't we ? :) }
So, this guy approaches me one day with his "CD modification project" and starts to jabber about these super fine, extra high quality, ultra high bandwidth, Burr-Brown style operational amplifiers of make something or other, with which he wishes to replace his present analog output buffer op-amps within his device.
So as to get him off my back, I say: "Big Deal. Go for it, Tiger !".
But then he goes on and continues jabbering about that these specific op-amp chips can be powered off from a fairly wide range of supply voltages, but actually, they start to "sing", when powered from a higher, rather than lower, range of supply voltages. That he wants to use them for the analog output stage, but unfortunately, they perform best when powered from a +12V and -12V symmetrical power supply source. Preferably even higher. Unfortunately, within his CD, he only has but a +7,5V and a -7,5V power supply line. So he will most probably need to put in an extra transformer, but there is no room for that …
Hmmm … this is starting to get interesting. Does he indeed need an extra transformer?
Since he has a +7,5V and a -7,5V line, that means that most probably he has a symmetrical AC secondary winding within the CD player’s power supply transformer.
(probably in the whereabouts of 7VAC, give or take). The only “problem” is that the value of the voltage on the DC lines is too low.
My wild guess that it is about a "half" of what is needed. So guess what !
We shall multiply it by a factor of two.
How do we accomplish this ?
Easy. We use a voltage doubler.
But … not a “simple” one. Life would be too easy, if we do not complicate it.
It needs to be a "beefed up" version of a voltage doubler.
More than one, actually.
OK, so here is what I came up with ...
A). Assuming that the PS transformer in that device is the "worst possible scenario", meaning a simple, single secondary winding with a low voltage ( like the ones that are beloved by those bean counters, who say that the purpose of any and all audio equipment production is "to make profit" and not "to play music"), we can come up with the following idea. It is basically a grouping of two independent voltage doublers. One is working so as to produce a “positive” voltage, the other one is wired “in the other direction” – so as to produce a “negative” voltage. Both are referenced to a common signal ground. This solution unfortunately provides us with what we could call similar to a “half cycle” rectification, as opposed to “full cycle” (full “bridge” ??) rectification, with ripple at a frequency of 50Hz.
But then he goes on and continues jabbering about that these specific op-amp chips can be powered off from a fairly wide range of supply voltages, but actually, they start to "sing", when powered from a higher, rather than lower, range of supply voltages. That he wants to use them for the analog output stage, but unfortunately, they perform best when powered from a +12V and -12V symmetrical power supply source. Preferably even higher. Unfortunately, within his CD, he only has but a +7,5V and a -7,5V power supply line. So he will most probably need to put in an extra transformer, but there is no room for that …
Hmmm … this is starting to get interesting. Does he indeed need an extra transformer?
Since he has a +7,5V and a -7,5V line, that means that most probably he has a symmetrical AC secondary winding within the CD player’s power supply transformer.
(probably in the whereabouts of 7VAC, give or take). The only “problem” is that the value of the voltage on the DC lines is too low.
My wild guess that it is about a "half" of what is needed. So guess what !
We shall multiply it by a factor of two.
How do we accomplish this ?
Easy. We use a voltage doubler.
But … not a “simple” one. Life would be too easy, if we do not complicate it.
It needs to be a "beefed up" version of a voltage doubler.
More than one, actually.
OK, so here is what I came up with ...
A). Assuming that the PS transformer in that device is the "worst possible scenario", meaning a simple, single secondary winding with a low voltage ( like the ones that are beloved by those bean counters, who say that the purpose of any and all audio equipment production is "to make profit" and not "to play music"), we can come up with the following idea. It is basically a grouping of two independent voltage doublers. One is working so as to produce a “positive” voltage, the other one is wired “in the other direction” – so as to produce a “negative” voltage. Both are referenced to a common signal ground. This solution unfortunately provides us with what we could call similar to a “half cycle” rectification, as opposed to “full cycle” (full “bridge” ??) rectification, with ripple at a frequency of 50Hz.
B). Assuming that we are in luck, and that indeed the bean counters were "generous" and have provided us with a power supply transformer that is center tapped and with symmetrical outputs, we can come up with a yet slightly more complex scenario, and mainly a doubling of that I mentioned above, whereby the “mirror” copy is connected to the "other output" lead of the transformer and works in “anti-phase” to the first one. All four doublers are referenced to a common signal ground. This solution happily provides us with something similar to what we could call a “full cycle” rectification, or a full “bridge” rectification, with ripple at a frequency of 100Hz.
(C) 2014, zjj_wwa, of http://hiend-audio.com
Looking at the circuit above, I can not help myself recalling a few jokes about squint eyed people, which I shall not quote here, so as not to offend anybody. But what I am getting at here, is that after a few cans of beer, you tend to see things ... double.
And after a few more cans of beer, you tend to see double of whatever you seen previously. And then, ....
Voila! A a new circuit idea comes to be!
For those of you, who are in the “dark” as to the workings of the contraption that I depicted above, please kindly make two initial observations:
a). it "consists" of four similarly looking building blocks.
b). each of the building blocks is very simple, consisting of two diodes and two capacitors.
So, let us zoom in on such a building block. Check out this very "basic" building block of a voltage doubler, as can be found in many an electronics manual,
with a positive voltage on it's output:
Here is how this basic building block, this voltage doubler circuit, works:
Imagine that you connect the transformer to the mains power supply. Now, Imagine the first half cycle, with the current flowing “upwards” though the transformer.
What results is that the left diode, D2, starts conducting and we have a current loop that loads, or fills up, the left capacitor, C1, with a charge.
At the peak of the sine wave, the capacitor with “catch” a charge that resembles the voltage of +V_peak, as referenced to commons.
Note that the right diode and right capacitor have nothing to do, since the diode is polarized in reverse,
so the C3 capacitor simply “holds” whatever voltage / charge it had stored within previously ( … whatever it may be) …
Now, imagine that the next half cycle occurs, with the current flowing “downwards” though the transformer.
What results is that the diode to the left, D2, is polarized in reverse and hence ..... cannot conduct.
Thus the capacitor C1 shall “hold” the charge that it "received" during the previous half sine wave cycle, a charge corresponding to a +V-peak voltage on it's pins.
But please also note, that in reference to ground, due to the change of the direction of the half cycle of the sine wave, now we have a situation,
where the “upper end” of right diode, D2, is polarized with a voltage of MINUS V_peak, as referenced to commons, which, just as before,
is "ankered" to the bottom pin of the transformer's secondary winding.
So now, we have …. a situation.
A situation, where the series connected pair, constituted of the right diode D1 and the serially connected right capacitor C3, that this circuit “sees” a voltage difference:
PLUS V_peak -- from the side of C1, the capacitor that still holds the charge that it gained during the previous half cycle, and, on the other side, the
MINUS V_peak -- on C3, a voltage potential which is currently being applied from the transformer secondary.
So, the {D1,C3} circuit now sees a voltage difference equal to:
+ V_peak – ( – V_peak ) = + V_peak + V_peak = 2 x V_peak
Hence, during this half sine wave cycle, D1 starts conducting and charges up the capacitor C3 with close to all of the charge that was held and available within capacitor C1.
After a few such repetitive cycles of the AC sine wave, the capacitor C3 will build up and hold a voltage that is “on average” of a value that is a double of the voltage that can be encountered on C1 (both referenced to commons). I say “on average”, as there is still a fair amount of ripple on the output of this doubler circuit, or more specifically, on C3.
Ripple that needs to be taken care of, cleaned up by extra filtering, referenced to commons.
Please also note one very important thing. The charging up of the C3 capacitor takes place only every “other” (i.e.: negative) half sine wave. During the “positive” half sine wave, we see a “replenishing” of the charge / voltage of C1, so that the next upcoming “negative” half sine wave, with the help of this “fill-it-up” charged C1, can further “fill up” the C3. Such a behavior of the doubler circuit is very similar to ... a single diode rectifier, which charges up, fills up, the following capacitor, but only during the positive peaks of a half wave, or in other words, during every “other” half cycle. What I am getting at here is that this is but only a “half cycle” rectifier-doubler, i.e. one that has poor ripple performance, due to the time gap between each and every “filling up” of C3. The ripple frequency shall be 50Hz (we can do better than that).
So, having said all of the above, I hope you now get the idea of how the basic functional unit works, the basic “voltage doubler”.
…..
Now that you have an understanding of the "basic" voltage doubler concept, now please “replicate” this idea, (use your "double vision", or lacking the aforementioned, a few cans of beer...). Replicate the idea, but in such a manner, so that the replica produces a “negative” voltage (i.e. simply: turn around the directions of all the diodes and the polarity of all the elko capacitors). What you shall obtain in result is this ( please note: all of the stuff is "reversed" here, in terms of polarity):
So, now you have a "+" version, as well as a "-" version thereof.
Now, merge them together (visually, you may need to imagine the the minus circuit "flipped" horizontally).
The two circuits are joined by their "commons". The above schematic presents the same, half cycle voltage doubling rectifier, but implemented for both the “+” and “-” symmetrical output voltage line. Having said that, it is still a lousy performer, as it still has half-wave kind of ripple, with a frequency of 50Hz.
….
Now, once you did that, “duplicate” the whole doubled-up set as above, so that you get a total of four such voltage doublers. What you obtain as a result is two times the previous, i.e. ….
Each of these (literally, cut, copy and pasted, but two times ...) have:
a). a “commons” reference input,
b). a “hot” input, and each of them also has a symmetrical set of output voltages, i.e.
c). “+” and
d). “-”, both of these output voltages referenced to commons.
Now, imagine that you do not have two separate secondary windings, but a center-tapped winding at your disposal.
Connect the respective input “commons” of both circuits as above with the “center tap” of the symmetrical winding.
Then, connect the hot input leads of each of the doublers to the respective corner taps of your symmetrical secondary winding.
After that, simply join together the “pluses” and the “minuses” on the output side, so that they work in tandem, together,
but in ***** ANTI-PHASE *****,
providing the equivalent of a “full bridge” effect and hence a smoother output voltage, with less ripple,
with at least two of the four available C3′s being filled up on each and every half cycle.
Why do I say "anti-phase"? Well, look at the side endings of the symmetrical secondary winding.
As it is referenced to commons via the center tap, whichever direction you draw an "arrow" through the core, whatever voltage you get on one of the taps, positive or negative in reference to ground, you are bound to have an "opposite" sign voltage from the "other" side of the winding. By a simple process of doubling up the symmetrical 'half-bridge' voltage doublers,
we in essence "create" a full-bridge doubler.
Having said all of the above, please bear in mind that it you shall always do better by simply using a .... “correct” transformer, with a “correct” value of the target AC voltage,
with appropriate, “correct” rectification, and such a solution shall always perform BETTER than this setup. Less parts, less capacitors, etc.
But when the going get's tough, the Tough get ... going. The only times that you will want to use this setup is at times when you *MUST* use an existing transformer, with it’s existing secondary winding, one that provides half the voltage what is actually required or needed.
Once again, …
Cheers,
Ziggy,
http://hiend-audio.com
zjj_wwa@hiend-audio.com
Matej Isak. Mono and Stereo ultra high end audio magazine. All rights reserved. 2006-2014. www.monoandstereo.com. ..:: None of the original written text, pictures, that were taken by me, links or my original files can be re-printed or used in any way without prior permission! ::..
Imagine that you connect the transformer to the mains power supply. Now, Imagine the first half cycle, with the current flowing “upwards” though the transformer.
What results is that the left diode, D2, starts conducting and we have a current loop that loads, or fills up, the left capacitor, C1, with a charge.
At the peak of the sine wave, the capacitor with “catch” a charge that resembles the voltage of +V_peak, as referenced to commons.
Note that the right diode and right capacitor have nothing to do, since the diode is polarized in reverse,
so the C3 capacitor simply “holds” whatever voltage / charge it had stored within previously ( … whatever it may be) …
Now, imagine that the next half cycle occurs, with the current flowing “downwards” though the transformer.
What results is that the diode to the left, D2, is polarized in reverse and hence ..... cannot conduct.
Thus the capacitor C1 shall “hold” the charge that it "received" during the previous half sine wave cycle, a charge corresponding to a +V-peak voltage on it's pins.
But please also note, that in reference to ground, due to the change of the direction of the half cycle of the sine wave, now we have a situation,
where the “upper end” of right diode, D2, is polarized with a voltage of MINUS V_peak, as referenced to commons, which, just as before,
is "ankered" to the bottom pin of the transformer's secondary winding.
So now, we have …. a situation.
A situation, where the series connected pair, constituted of the right diode D1 and the serially connected right capacitor C3, that this circuit “sees” a voltage difference:
PLUS V_peak -- from the side of C1, the capacitor that still holds the charge that it gained during the previous half cycle, and, on the other side, the
MINUS V_peak -- on C3, a voltage potential which is currently being applied from the transformer secondary.
So, the {D1,C3} circuit now sees a voltage difference equal to:
+ V_peak – ( – V_peak ) = + V_peak + V_peak = 2 x V_peak
Hence, during this half sine wave cycle, D1 starts conducting and charges up the capacitor C3 with close to all of the charge that was held and available within capacitor C1.
After a few such repetitive cycles of the AC sine wave, the capacitor C3 will build up and hold a voltage that is “on average” of a value that is a double of the voltage that can be encountered on C1 (both referenced to commons). I say “on average”, as there is still a fair amount of ripple on the output of this doubler circuit, or more specifically, on C3.
Ripple that needs to be taken care of, cleaned up by extra filtering, referenced to commons.
Please also note one very important thing. The charging up of the C3 capacitor takes place only every “other” (i.e.: negative) half sine wave. During the “positive” half sine wave, we see a “replenishing” of the charge / voltage of C1, so that the next upcoming “negative” half sine wave, with the help of this “fill-it-up” charged C1, can further “fill up” the C3. Such a behavior of the doubler circuit is very similar to ... a single diode rectifier, which charges up, fills up, the following capacitor, but only during the positive peaks of a half wave, or in other words, during every “other” half cycle. What I am getting at here is that this is but only a “half cycle” rectifier-doubler, i.e. one that has poor ripple performance, due to the time gap between each and every “filling up” of C3. The ripple frequency shall be 50Hz (we can do better than that).
So, having said all of the above, I hope you now get the idea of how the basic functional unit works, the basic “voltage doubler”.
…..
Now that you have an understanding of the "basic" voltage doubler concept, now please “replicate” this idea, (use your "double vision", or lacking the aforementioned, a few cans of beer...). Replicate the idea, but in such a manner, so that the replica produces a “negative” voltage (i.e. simply: turn around the directions of all the diodes and the polarity of all the elko capacitors). What you shall obtain in result is this ( please note: all of the stuff is "reversed" here, in terms of polarity):
So, now you have a "+" version, as well as a "-" version thereof.
Now, merge them together (visually, you may need to imagine the the minus circuit "flipped" horizontally).
The two circuits are joined by their "commons". The above schematic presents the same, half cycle voltage doubling rectifier, but implemented for both the “+” and “-” symmetrical output voltage line. Having said that, it is still a lousy performer, as it still has half-wave kind of ripple, with a frequency of 50Hz.
….
Now, once you did that, “duplicate” the whole doubled-up set as above, so that you get a total of four such voltage doublers. What you obtain as a result is two times the previous, i.e. ….
Each of these (literally, cut, copy and pasted, but two times ...) have:
a). a “commons” reference input,
b). a “hot” input, and each of them also has a symmetrical set of output voltages, i.e.
c). “+” and
d). “-”, both of these output voltages referenced to commons.
Now, imagine that you do not have two separate secondary windings, but a center-tapped winding at your disposal.
Connect the respective input “commons” of both circuits as above with the “center tap” of the symmetrical winding.
Then, connect the hot input leads of each of the doublers to the respective corner taps of your symmetrical secondary winding.
After that, simply join together the “pluses” and the “minuses” on the output side, so that they work in tandem, together,
but in ***** ANTI-PHASE *****,
providing the equivalent of a “full bridge” effect and hence a smoother output voltage, with less ripple,
with at least two of the four available C3′s being filled up on each and every half cycle.
Why do I say "anti-phase"? Well, look at the side endings of the symmetrical secondary winding.
As it is referenced to commons via the center tap, whichever direction you draw an "arrow" through the core, whatever voltage you get on one of the taps, positive or negative in reference to ground, you are bound to have an "opposite" sign voltage from the "other" side of the winding. By a simple process of doubling up the symmetrical 'half-bridge' voltage doublers,
we in essence "create" a full-bridge doubler.
Having said all of the above, please bear in mind that it you shall always do better by simply using a .... “correct” transformer, with a “correct” value of the target AC voltage,
with appropriate, “correct” rectification, and such a solution shall always perform BETTER than this setup. Less parts, less capacitors, etc.
But when the going get's tough, the Tough get ... going. The only times that you will want to use this setup is at times when you *MUST* use an existing transformer, with it’s existing secondary winding, one that provides half the voltage what is actually required or needed.
Once again, …
Cheers,
Ziggy,
http://hiend-audio.com
zjj_wwa@hiend-audio.com
Matej Isak. Mono and Stereo ultra high end audio magazine. All rights reserved. 2006-2014. www.monoandstereo.com. ..:: None of the original written text, pictures, that were taken by me, links or my original files can be re-printed or used in any way without prior permission! ::..