Engineering Half wave voltage doubler

[PhysicsTest]

Homework Statement

The circuit shown is a half-wave voltage doubler. Analyze the operation of this circuit. Calculate
(a) the maximum possible voltage across each capacitor,
(b) the peak inverse voltage of each diode.
Compare this circuit with the bridge voltage doubler of Fig. 4-22. In this circuit the output voltage is negative with respect to ground. Show that if the connections to the cathode and anode of each diode are interchanged, the output voltage will be positive with respect to ground.

Relevant Equations

V = IR ohms law, Diode conducting in forward direction and Off in negative direction

The input is sine wave vi= Vmsin(wt), the problem is it is increasingly difficult to analyze these kinds of circuits for sinusoidal inputs and with positive and negative portions, most of the time i analyze in the mind without the equations and to remember the states is tough. Do we have simpler techniques to analyze these kinds of circuits?
As per the analysis during the positive cycle the C1 will follow the input waveform the D1 is forward biased and D2 is reverse biased.
During the negative portion D2 is Forward biased and D1 is reverse biased, C2 will be following the negative voltage. But recommend me a procedure where i can use it for any circuit. Is it possible?

 

[Baluncore]

But recommend me a procedure where i can use it for any circuit. Is it possible?

It is all in the way you draw or imagine the diagram. Keep positive at the top, then diodes that block will be more obvious. Draw storage capacitors vertically, with AC coupling capacitors horizontal. Follow the path through the diodes from the transformer to the output.

 

[DaveE]

These switching circuits are difficult to analyze on paper in general. Basically you would need to identify each of the states of the circuit that correspond to the switch positions (2 switches means 4 states). You also need to identify at which point the switches change. Then you are left with a bunch of transient response problems which need to have the initial and final states (capacitor voltages and inductor currents) conditions matched when the switching occurs.

There are often tricks to simplify if you only care about the steady state response. Also good techniques if the switching occurs at a high frequency compared to the circuit dynamics and the frequencies of interest, like state space averaging for SMPS. In self commutating circuits, like this with low frequency switching, these tricks usually don't work well. Otherwise, it's a big PITA that people avoid. You don't get generalized answers, you have to solve it for each different set of I/O conditions. This is a good example where simulation can be helpful, but there are issues that make those simulations difficult too. What EEs really do is examine the circuit heuristically and approximate solutions for the parameters of interest.

There are issues with the simplified model shown where idealized zero impedance circuits are shown which may create infinite solutions, like the current from one capacitor to another. While this is a very non-linear circuit, it is reminiscent of the LTI circuit rule that no loop can contain only capacitors and/or voltage sources.

In this case there is no load and all of the questions relate to the peak values, so it's set up to be easy. They are really only asking about the final state of the circuit, like how high could a voltage ever get. IRL there is a load and parasitic elements, which make the problems much harder.

If you want to know more about voltage multipliers, this is a great place to start:
https://www.voltagemultipliers.com/...1MjE3MjkkbzEkZzEkdDE3NjY1MjE3NDIkajQ3JGwwJGgw

 

[PhysicsTest]

Thank you for all the support. I have a very basic question in the circuit before i start attempting

when he shows the negative and positive symbol at the output side as above what does it indicate, is it arbitrary or some logic involved in it? Can you please clarify?

 

[DaveE]

It's so that when you say the answer is, let's say 12.7V, everyone will agree on the polarity. It is just a definition of how we should communicate, nothing more.

 

[PhysicsTest]

Ok just to re confirm my understanding

The drop across voltage is Vr= 5V, based on the output polarity o/p it is -5V, Is my understanding correct? please confirm.

 

[DaveE]

Ok just to re confirm my understanding
View attachment 368354
The drop across voltage is Vr= 5V, based on the output polarity o/p it is -5V, Is my understanding correct? please confirm.

yes, exactly right.

In circuit analysis the first step is to draw the schematic and label the voltage and current polarities, often arbitrarily. This sets a basis for avoiding sign errors in the equations you may develop later. There is no need to guess in advance the "correct" polarity.

There is a "passive sign convention" that says that the current flowing through a passive device, like your resistor, should flow from the high (most positive) voltage towards the lower voltage. But this doesn't always work and isn't necessary. I only bring it up because people will often assume that unless stated otherwise. So everyone says Ohms law is V=IR, but sometimes V=-IR, based on the definitions.

In your circuit the polarity of the voltage at the output is the opposite of the defined polarity at the other components. This is poor practice which can create confusion. But in your question it's a great illustration that the polarities are determined by the definitions. So, I would say the voltage across the resistor and source (wrt ground) is +5V but the output voltage is -5V. The discrepancy is sorted out in your KCL/KVL equations.

 

[Baluncore]

You can analyse it in two steps.
The first stage, Ls, C1 and D1, shifts the AC waveform down to generate -AC, then the second stage, D2, C2, follows the most negative voltage, producing -DC.