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Voltage doubler confusion

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carbonzit

Active Member
Charge up two large caps with 10v each. Connect them in series. What do you get across the two? You get 20v because the cap voltages add.
The cap is like a battery for a time, so replace the first cap on the left with charged battery and go from there for that next half cycle.

But it depends on the polarity. If you connect two charged batteries in series with opposite polarity, you get ZERO voltage (assuming equal voltages). In other words, YES, the voltages add, but if you add x and -x, you get 0. That's my problem here.
 

crutschow

Well-Known Member
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So what's happening is that even though the now-positive voltage is actually charging the capacitor in the opposite polarity to the previous negative half cycle (right?), the incoming charge "lifts" the total charge on the capacitor (from V to 2V). Is this correct?

Yes. The point is that, if the + side of the cap has no load, then you can't "charge" the cap. When you charge a cap, equal currents have to flow through both terminals, in one terminal and out the other. Thus if you raise the voltage of one side of the cap and the other side is open then the cap just acts like a battery and it's output voltage is "lifted" by the input voltage.

I'm still confused about how you can start out with a capacitor charged thus (by the negative half-cycle)

- --||-- +

then have a positive-going charge, which wants to charge the capacitor thus

+ --||-- -

and somehow have them add.

Answered above.

To help my understanding, can you answer the question I posed above? If you have a capacitor charged by a positive-going cycle applied to the left side:

---||---

what polarity is it charged at afterwards?
If the right side of the cap is grounded then the cap will be charged to the positive voltage on the left. If the right side of the cap is open, then the cap is not charged.

Make sense yet?
 

carbonzit

Active Member
If the right side of the cap is grounded then the cap will be charged to the positive voltage on the left. If the right side of the cap is open, then the cap is not charged.

Make sense yet?

Starting to. But a couple problems still:

1. You say "if the right side of the cap is open", but it isn't really, is it? After all, there's that other diode and a (possibly uncharged) capacitor downstream of it: doesn't this constitute a load?

2. Here's my problem still: so we have the first capacitor charged by the first negative half-cycle thus:

-(0V) ---||--- + 10V

So if the positive half-cycle "lifts" both sides of the capacitor equally, wouldn't we then have

+10V ---||--- + 20V

(because we added 10V to both sides)? giving a net charge of 10V, not 20V.

Obviously, it doesn't work this way. But it looks as if it works this way, from the explanations I've read.

Looking forward to having the errors in my thinking corrected.

(I appreciate your forbearance on this seemingly trivial point that everyone else takes for granted. But I find that if I don't really understand first principles, my understanding of more complex things built on them is shaky.)
 
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crutschow

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1. You say "if the right side of the cap is open", but it isn't really, is it? After all, there's that other diode and a (possibly uncharged) capacitor downstream of it: doesn't this constitute a load?

Yes, of course. I left the cap open to try to simplify the discussion about how the cap works. The load will obviously affect this charge and discharge. With a load the capacitor will discharge until it has transferred all the charge it can.

For example if it were connected to a second equal value cap C2 to ground that started at 0V, then the result with 10V applied at the left side would be that the transfer of charge would leave the first cap with 0V across it and C2 would have +10V across it.

If we repeat the process with out discharging C2 then you end up with +15V across C2. Repeating the process many times incrementally increases the voltage on C2 until it approaches 20V (thus a voltage doubler).

2. Here's my problem still: so we have the first capacitor charged by the first negative half-cycle thus:

-(0V) ---||--- + 10V

So if the positive half-cycle "lifts" both sides of the capacitor equally, wouldn't we then have

+10V ---||--- + 20V

(because we added 10V to both sides)? giving a net charge of 10V, not 20V.

Obviously, it doesn't work this way. But it looks as if it works this way, from the explanations I've read.

Yes that is correct. The net voltage on the cap is still 10V, but the output voltage is 20V.

Looking forward to having the errors in my thinking corrected.
Getting a little clearer?
 
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ronv

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Voltage doubler

I think I understand your problem. If you notice all the circuits you see like this are fed from a transformer. The voltage input needs to swing both above and below ground. So if the input is +5 to -5 the output will be +10 (minus the diode drops). If you feed it with 0 to +5 you only get +5 out not +10. Attached is a simulation with both cases so you can see the voltage across the cap.
I think what you are looking for is the charge pump from the earlier post.
 

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carbonzit

Active Member
crutschow said:
Yes that is correct. The net voltage on the cap is still 10V, but the output voltage is 20V.

Doh!

Of course. The (net) voltage on the first capacitor is only 1V. It's the second capacitor that "doubles" the voltage ("charges towards 2V", as my textbook puts it). I think I finally get it now.

And no, Ron, I'm not thinking of a charge pump: why would you say that? (Plus your explanation about the transformer was needlessly confusing.)

So I can now accept it on faith that the first capacitor charges in such a way as to aid the positive half-cycle. Thanks, especially to Carl.
 
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ronv

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I guess I was following your post on the boost converter.
 

MrAl

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But it depends on the polarity. If you connect two charged batteries in series with opposite polarity, you get ZERO voltage (assuming equal voltages). In other words, YES, the voltages add, but if you add x and -x, you get 0. That's my problem here.

Hi again,

There's only one way you can connect two capacitors that each have 10v across them and get 20v across the two. THAT is exactly what happens. The polarities are such that the two voltages sum to a higher voltage, not a lower voltage.
The first diode forces the first cap to charge up - on the left and + on the right. Now when the phase reverses, the first cap - side gets a + voltage and so it is like having the two caps in series. It's actually more like a battery in series with a cap at that point. If you charge a cap and connect it in series aiding with a battery, you get a higher voltage. Isnt that simple enough?
 
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carbonzit

Active Member
OK, let me try to further explain my understanding of what's going on, which is still apparently not 100% certain.

On the first negative half-cycle, the first capacitor gets charged thus:

**broken link removed**

On the next positive half-cycle, let's say for the sake of illustration that the diode is reversed. Therefore, the capacitor would be charged thus, right?:

**broken link removed**

Butbutbut ... notice that the signs on the capacitor are reversed. Doesn't that mean that the positive half-cycle is trying to charge the capacitor in reverse, compared to the existing charge from the negative half-cycle? It looks as if that positive half-cycle would just about cancel any charge from the negative half-cycle, based on the polarity of the charge as indicated by the + and - signs.

Do you see the source of my confusion? Do you have an answer for it?
 
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crutschow

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Most Helpful Member
Yes. That is true. But why are you trying to confuse yourself by reversing the direction of the diode? The diode is not reversed in the actual circuit so the charging from the positive cycle does not go through the diode to ground, it goes through the diode to the next cap.
 

MrAl

Well-Known Member
Most Helpful Member
OK, let me try to further explain my understanding of what's going on, which is still apparently not 100% certain.

On the first negative half-cycle, the first capacitor gets charged thus:

**broken link removed**

On the next positive half-cycle, let's say for the sake of illustration that the diode is reversed. Therefore, the capacitor would be charged thus, right?:

**broken link removed**

Butbutbut ... notice that the signs on the capacitor are reversed. Doesn't that mean that the positive half-cycle is trying to charge the capacitor in reverse, compared to the existing charge from the negative half-cycle? It looks as if that positive half-cycle would just about cancel any charge from the negative half-cycle, based on the polarity of the charge as indicated by the + and - signs.

Do you see the source of my confusion? Do you have an answer for it?


As you can see in this new drawing we have one terminal grounded so we can reference all the other voltages to that point.

When the top of the battery is negative, it charges the cap through the diode so that the cap is charged as shown. Note that the battery is 5v and the cap now has 5v across it too, but the voltages cancel out so we have 0v at the top of the diode (and the diode stops conducting).

When the phase reverses the top of the battery becomes positive now, as you can see it is +5v now in the lower figure. The cap previously had been charged to 5v with that shown polarity and so its voltage adds to the battery voltage. The result is +10v at the top of the diode. Since the diode is now reversed biased, it does not conduct. That means the cap has no path to discharge (or charge in reverse). The diode acts as a switch, allowing the cap to charge in only one direction but not the other. Note there is no current flow in the lower figure.
 

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