1

We are given this schematic diagram, that our teacher called a high voltage power supply, then test the voltage of each capacitor.

Schematic diagram

Why is the voltage at C1 half of the voltage at C2 and is the wave at C1 still a sine wave?

winny
  • 13,064
  • 6
  • 46
  • 63
Jerico Bayod
  • 13
  • 2
  • 3
    Homework questions are OK, but only if you show your own effort to solve them. – K H Feb 08 '20 at 05:55
  • Draw a table with column headings of the voltages at Vin top (=C1 left), C1 right, C2 top. Draw rows with labels. Vin = 0, Vin = max initial, Vinmax after discharge period, Vin = max negative initial, Vin min after discharge period. || Repeat the line names again below the qst set. Now fill in the table voltages. eg For row 0 the values are (or course) 0, 0, 0. Tell us what you discover. – Russell McMahon Feb 08 '20 at 06:23
  • What's the impedance of C1? So what will its output *try* to do? When does the grounded diode conduct? When does the series diode conduct? What is the effect on the voltage across C1 when these diode conduction events happen? What is the effect on C2's voltage. Come back when you've thought about these questions. – Neil_UK Feb 08 '20 at 07:29
  • Hmmm. What sucks is, I **know** exactly what this circuit does because I've built a bunch of them. I just can't explain it in a way that doesn't give away all the things you are supposed to learn from it. – JRE Feb 08 '20 at 08:24
  • Did you measure (test) the voltages, or are you supposed to figure it out mathematically? If you measured it, did you use a voltmeter or an oscilloscope? – JRE Feb 08 '20 at 09:47
  • If you add further stages to it you can make voltage tripler and so on... – Bhuvnesh Feb 08 '20 at 16:48
  • @Bhuvnesh: Nope. You can't make a tripler using more stages of that circuit. You can stack multiple stages together, but it will never do a multiple of three. – JRE Feb 08 '20 at 18:57
  • @JRE It depends on across which two terminals you measure the potential difference.... And from my circuit analysis there is indeed potential difference of 3 time input voltage between two specific points! This results from the fact that the potential diff of first capacitor is the same as that of supply voltage and all other caps have twice potential difference than that of input. Hence output of 5, 7, 9...times the input potential is too possible. And it's fun(dangerous too) as you increase the stages further. I once made a 5kV output from a multiplier circuit. – Bhuvnesh Feb 09 '20 at 09:25
  • @Bhuvnesh: The circuit is a voltage **doubler.** No matter how many of them you stack together, there will always be a factor of two. 2, 4, 6, 8, 10, etc. are all possible. A factor of 3 is not. – JRE Feb 09 '20 at 09:58

2 Answers2

7

Split the circuit into two parts:

1.

enter image description here

2.

enter image description here

Each of those will function by itself.

Figure out what 1 does (that's the tricky one.)

Then imagine feeding the output of 1 into the input of 2.

Figure out what the waveform looks like at the output of section 1.

Once you have that, you'll know the answers to the two questions you posed.

Both sections are common circuits that have names. I'm sure you'll recognize the second section.

The first goes by a couple of names. I'm not going to give you the names because the point of the exercise is for you to figure out how it works. If you know the names, you could look it up instead of figuring it out.

The whole thing together also has a name. Again, though, if I told you the name it'd spoil the point of the assignment.

JRE
  • 67,678
  • 8
  • 104
  • 179
1

Looking at this simple diode circuit and answers below the question, I remember the famous Einstein's thought: “If you can't explain it to a six year old, you don't understand it yourself.”

To meet this great challenge - "to explain it to a six year old", I have fabricated the story below, in which, without any special terms and verbal cliches, I have tried to reveal the simple idea behind this well-known circuit solution. I have used my favorite approach to explain electronic circuits by means of simpler equivalent electrical circuits and elements... because this is a concept… and concepts are best revealed by conceptual electrical circuits consisting of voltage/current sources, resistors, etc. We use them in electronics since they are more primitive and comprehensible. This “scenario” is so real that it can even be implemented in the lab with real electrical elements. Here is my story…

Imagine we are (together with Robinson Crusoe:) on a desert island... and we need to activate a transmitter to send SOS. It should be powered by 12V... but unfortunately, we have only one 1.5V battery (cell). If we had at least eight cells (Fig. 1)...

Fig.1

Fig. 1. A set of eight separate 1.5 V cells

… we would connect them in series, in the same direction, to obtain 8 x 1.5 = 12 V (Fig.2)... but we have not them...

Фиг. 2

Фиг. 2. A battery of eight 1.5 V cells in series would be a remedy

Fortunately, it turns out we have a bag of capacitors… at least, seven (Fig. 3).

Фиг. 3

Fig. 3. A set of seven separate capacitors, in addition to the battery cell, could really help us

We know capacitors can “copy” the source voltage. So, if we connect them in parallel to the voltage source, all they will charge to its voltage of 1.5 V (Fig. 4).... and we will have seven 1.5 V “batteries”.

Фиг. 4

Fig. 4. A 1.5 V battery consisting of 1.5 V cell and seven 1.5 V charged capacitors in parallel

Now only remains to disconnect and quickly connect them in series to the voltage source - Fig. 5. Their voltages will be summed and added to the source voltage… and we will obtain the desired 1.5 + 7 x 1.5 = 12 V.

Фиг. 5

Fig. 5. A 12 V battery consisting of 1.5 V cell and seven 1.5 V charged capacitors in series

So, the basic idea is to use a set of capacitors as floating rechargeable sources. First we connect them in parallel to the voltage source to charge them; then we reconnect them in series to add their voltages to the source voltage (simply, we charge them in parallel and discharge in series). In this arrangement, only the voltage source is grounded, the other "sources" are floating. Thus one grounded “battery” and many floating capacitors make “higher voltage source”. The capacitors must be floating, in order to be able to connect them both in parallel and series... and rechargeable, in order to "copy" the voltage of the source.

There is only one problem - capacitors gradually discharge (especially if there is a load connected) and we have to charge them periodically. So we have to reconnect them periodically from series to parallel connection to refresh them. Let’s implement this idea into the simple voltage doubler discussed in this question.

Let's first assume a DC input voltage source V connected with its positive terminal to ground. It is drawn in Fig. 6a in a more unusual way - below the zero voltage line (ground) and mirrored since its voltage is negative. The capacitor C1 is connected in parallel to the source and also drawn below the ground since it is charged to negative voltage.

Fig. 6

Fig. 6. The capacitor C1 is charged in parallel to the source (a); then it produces opposite voltage in series (b)... however, the result is zero voltage

Now we have to disconnect C1 and connect it in series to the source... and we note that we can do both by only one action - disconnecting C1 from the ground (Fig. 6b). There is only a "little" problem - the two voltages are with opposite polarities... and the resulting output voltage is zero. So, we have to reverse one of them... and it can be implemented by replacing the DC input voltage source by an AC one - Fig. 7 (let's only temporarily represent it as two DC sources with opposite polarities in the two halve-waves).

During the negative half-way (Fig. 7a), the input voltage goes down below zero and a switch S1 (for now only outlined in the picture) connects the capacitor C1 in parallel to the voltage source. C1 is charged to its negative voltage -V.

Fig 7

Fig. 7. The capacitor C1 is charged in parallel to the source (a); then it producies voltage in the same direction in series (b)... the result is doubled voltage 2V (c)

During the positive half-way (Fig. 7b), the input voltage goes up under zero. S1 disconnects C1 from the ground so it turns out to be included in series with the input voltage source. Meantime, the polarity of the input source is reversed. Thus both sources - the input source and the capacitor "source" C1, produce voltages with the same polarity. So they are summed and, as the second switch S2 is turned on, the resulting voltage 2V is applied to C2 that is charged.

It now remains only to replace the conceptual switches S1 and S2 with real diode switches D1 and D2... and combine the separate input voltage sources into one AC voltage source... to obtain the real circuit of a diode voltage doubler.

Fig. 8

Fig. 8. A real circuit of a diode voltage doubler in action

It would be interesting to develop this idea into a voltage multiplier...

Circuit fantasist
  • 13,593
  • 1
  • 17
  • 48