Long Time, No See. Back again from some far away business travel.
Today I will write about the power supply for an alternative op-amp analog output scenario, to be implemented within a specific CD player tuning project.
A friend of mine recently came up with the problem that he has these “fast” opamps, that he wants to use for the analog output stage, but unfortunately they perform best when powered from a +12V and -12V symmetrical power supply source. Preferably even higher, more.
Unfortunately, he only has a +7,5V and a -7,5V power supply line within the CD. So he will most probably need to put in an extra transformer, but there is no room for that …
Hmmm … 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 on the CD’s power supply transformer. The only “problem” is that the value of the voltage is too low.
I 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.
A). Assuming that the PS transformer is the worst possible scenario, meaning a simple, single secondary winding of a low voltage, 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 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 power supply transformer is center tapped and symmetrical, we can come up with a yet slightly more complex scenario, and mainly a doubling of that I presented 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 ground. This solution happily provides us with what we could call similar to a “full cycle” rectification, or a full “bridge” rectification, with ripple at a frequency of 100Hz.
(C) 2014, zjj_wwa, of hiend-audio.com.
P.S. … for those of you, who are still in the “dark” on this …. check out this very basic building block of a voltage doubler, with a positive voltage on the output.
Here is how this basic building stone circuit works:
Imagine the half cycle with the current flowing “upwards” though the transformer. What results is that one of the diodes starts conducting and we have a current loop that loads, or fills up, the capacitor C1 with charge. At the peak of the sine wave, the capacitor with “catch” a charge that resembles the voltage of +V_peak) with reference to ground. Note that the second diode and second capacitor have nothing to do, since the diode is polarized in reverse, so it simply “holds” the voltage of capacitor C3 ( … whatever it may be) …
Now, imagine the next half cycle occurs, with the current flowing “downwards” though the transformer. What results is that the diode D2 is polarized in reverse and cannot conduct. The capacitor C1 shall hence “hold” the +V-peak charge from the previous cycle. But please also note, that in reference to ground, due to the change of cycle of the sine wave, now we have a situation where the “upper end” of diode D2 is polarized with a voltage of MINUS V_peak, related to our ground.
So now, we have …. a situation.
A situation, where the pair D1 and serially connected C3 “see” a voltage difference: PLUS V_peak from the side of C1, the one 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 series path now sees a voltage difference equal to:
+ V_peak – ( – V_peak ) = + V_peak + V_peak = 2 x V_peak
Hence, during this half 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 repetitive cycles of the sine wave, the capacitor C3 will hold a voltage that is “on average” double the value of the voltage that is to 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, ripple that needs to be taken care of, cleaned up by extra filtering, referenced to ground.
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 upcomming “negative” half sine wave, with the help of this “fill-it-up” charged C1, can further “fill up” the C3. Such a behaviour 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, “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:
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 has a “commons” referenced input, a “hot” input, and each of them has a symmetrical set of output voltages, i.e. “+” and “-”, both 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.
The thing with this circuit is that the doublers are attached to but only a half of the transformer winding, so that the net value added is … limited. But the value starts to cumulate, if you make a voltage tripler, or a voltage quadrupler. In such a manner, you can actually get 1,5 times more than normally possible from such a center tapped transformer, 2x more, or yet even more.
Having said all of the above, please bear in mind that a “correct” transformer, with a “correct” value of the target AC voltage, with appropriate “normal” rectification, shall always perform BETTER than this setup. Less parts, less capacitors, etc.
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, …