# Transformer - Voltage doubler circuit

Thread Starter

#### jacob_c

Joined Oct 13, 2020
13
I'm working on my PSPICE project regrading diodes.

The question is about transform input source (10kHz, 22V AC) to output 1V AC and connect it to voltage doubler.

So what I have done so far is attached. I uploaded this problem before and many of you helped me but I have problems that I cannot understand yet.

1.
On the first simulation result, the curve looks like sin curve is input voltage of voltage doubler(the one came from transformer) and the other curve is the voltage of vertical capacitor V_out.

Theoretically, the V_out must be increase to the extent of half of the increased input voltage since D2 will be off and D1 is on.

Thus the V_out must be around 400mV approximately but it is about 200mV. I guess it is due to the D1N914 diode but I have no idea how should I analyze this effect. The final value of V_out is approximately 1.6V not 2V(which comes from theory)

2.
As I learend, the V_out must stay constant for some range but what I can observe is decrease of V_out during the period where it must stay constant.(30us~90us for example). It may due to discharging of capacitor but how come it can discharge even there aren't any path for discharging.

3.
The question is about asking the effect of inductance and capacitance on circuit.

What I think is increasing L would make large final convergence of V_out since it leads to higher V_in.

But I have no idea of capacitance.

As I increased it, SPICE gives me lower final V_out and relatively curve of V_out becomes more 'flat'.
Why the capacitance change both value and the shape of curve?

'sim2' is the simulation result when both capacitance is changed to 10pF. V_out looks like sine wave.

Even any book/source to find this answer will be appreciated.

#### MrAl

Joined Jun 17, 2014
8,063
I'm working on my PSPICE project regrading diodes.

The question is about transform input source (10kHz, 22V AC) to output 1V AC and connect it to voltage doubler.

So what I have done so far is attached. I uploaded this problem before and many of you helped me but I have problems that I cannot understand yet.

1.
On the first simulation result, the curve looks like sin curve is input voltage of voltage doubler(the one came from transformer) and the other curve is the voltage of vertical capacitor V_out.

Theoretically, the V_out must be increase to the extent of half of the increased input voltage since D2 will be off and D1 is on.

Thus the V_out must be around 400mV approximately but it is about 200mV. I guess it is due to the D1N914 diode but I have no idea how should I analyze this effect. The final value of V_out is approximately 1.6V not 2V(which comes from theory)

2.
As I learend, the V_out must stay constant for some range but what I can observe is decrease of V_out during the period where it must stay constant.(30us~90us for example). It may due to discharging of capacitor but how come it can discharge even there aren't any path for discharging.

3.
The question is about asking the effect of inductance and capacitance on circuit.

What I think is increasing L would make large final convergence of V_out since it leads to higher V_in.

But I have no idea of capacitance.

As I increased it, SPICE gives me lower final V_out and relatively curve of V_out becomes more 'flat'.
Why the capacitance change both value and the shape of curve?

'sim2' is the simulation result when both capacitance is changed to 10pF. V_out looks like sine wave.

Even any book/source to find this answer will be appreciated.

Hello,

What is it that you are missing?
1vac is quite small to work with diodes because the diode forward voltage is roughly 0.7 volts so this 1vac transformer output seems kind of low.
The peak is around 1.4 volts so we loose half the voltage just because of the diode.
You can calculate the transformer output voltage based on the two inductances, but the series capacitor will drop some voltage also.

What is it you have to do for this problem?

Thread Starter

#### jacob_c

Joined Oct 13, 2020
13
Hello,

What is it that you are missing?
1vac is quite small to work with diodes because the diode forward voltage is roughly 0.7 volts so this 1vac transformer output seems kind of low.
The peak is around 1.4 volts so we loose half the voltage just because of the diode.
You can calculate the transformer output voltage based on the two inductances, but the series capacitor will drop some voltage also.

What is it you have to do for this problem?
I agree with you. I tried VSIN - resistor - D1N914 series circuit and it turns out that it had 900mV across the diode.
So that's one of the point that I can't get it. Just like you said, not much voltage would be left if the input is 1VAC. But the SPICE simulation gives me the voltage across capacitor is charged up to 400mV at first. How can this happen?

#### MrAl

Joined Jun 17, 2014
8,063
I agree with you. I tried VSIN - resistor - D1N914 series circuit and it turns out that it had 900mV across the diode.
So that's one of the point that I can't get it. Just like you said, not much voltage would be left if the input is 1VAC. But the SPICE simulation gives me the voltage across capacitor is charged up to 400mV at first. How can this happen?
Well 1vac sounds so low i wonder why anyone would want to do this, i guess just an exercise.

With 1vac that is most likely RMS so the peak is around 1.4 and once the cap charges up if it can stay charged for 1/2 cycle then when the voltage reverses the output on the right of the cap will be around 2.8v. After the last diode drops maybe 0.7v then the output would be around 2v or so.
So maybe there is a method to the madness.

In Spice, you could look at the first and second cap voltages to see what is happening with them. There is no load (like a resistor) so maybe it will actually reach 2v or so. You should try it and look at the two voltages.

The funny thing about these circuits is the load usually has to be very very light or else the caps dont stay charged long enough to keep a reasonable DC output voltage.

#### crutschow

Joined Mar 14, 2008
26,732
If you use Schottky diodes (e.g BAT54) which have a lower forward-drop, then your output DC voltage will be higher.

#### MrAl

Joined Jun 17, 2014
8,063
I'm working on my PSPICE project regrading diodes.

The question is about transform input source (10kHz, 22V AC) to output 1V AC and connect it to voltage doubler.

So what I have done so far is attached. I uploaded this problem before and many of you helped me but I have problems that I cannot understand yet.

1.
On the first simulation result, the curve looks like sin curve is input voltage of voltage doubler(the one came from transformer) and the other curve is the voltage of vertical capacitor V_out.

Theoretically, the V_out must be increase to the extent of half of the increased input voltage since D2 will be off and D1 is on.

Thus the V_out must be around 400mV approximately but it is about 200mV. I guess it is due to the D1N914 diode but I have no idea how should I analyze this effect. The final value of V_out is approximately 1.6V not 2V(which comes from theory)

2.
As I learend, the V_out must stay constant for some range but what I can observe is decrease of V_out during the period where it must stay constant.(30us~90us for example). It may due to discharging of capacitor but how come it can discharge even there aren't any path for discharging.

3.
The question is about asking the effect of inductance and capacitance on circuit.

What I think is increasing L would make large final convergence of V_out since it leads to higher V_in.

But I have no idea of capacitance.

As I increased it, SPICE gives me lower final V_out and relatively curve of V_out becomes more 'flat'.
Why the capacitance change both value and the shape of curve?

'sim2' is the simulation result when both capacitance is changed to 10pF. V_out looks like sine wave.

Even any book/source to find this answer will be appreciated.
I just reread your post and i see that you have a question of the voltage output.
What could be happening is two fold.
First, there is some input resistance (100 Ohms) and that limits the charging of the first capacitor as well as the second.
Second, the diodes to have some equivalent reverse conductance as well as capacitance which most likely is not insignificant considering the two storage caps are only 10pf.

What you could do is change the 100 Ohm to maybe 10 Ohm and note the effect on the output voltage. You could do other things like this in order to try to determine what is causing the voltage loss.

Did you post the LT Spice file yet? I'd be happy to take a closer look if you feel like posting that.
I'll wait on a purely theoretical analysis.

Thread Starter

#### jacob_c

Joined Oct 13, 2020
13
If you use Schottky diodes (e.g BAT54) which have a lower forward-drop, then your output DC voltage will be higher.
Well that can be the way but I don't have any choice for changing diode by the problem condition

Thread Starter

#### jacob_c

Joined Oct 13, 2020
13
I just reread your post and i see that you have a question of the voltage output.
What could be happening is two fold.
First, there is some input resistance (100 Ohms) and that limits the charging of the first capacitor as well as the second.
Second, the diodes to have some equivalent reverse conductance as well as capacitance which most likely is not insignificant considering the two storage caps are only 10pf.

What you could do is change the 100 Ohm to maybe 10 Ohm and note the effect on the output voltage. You could do other things like this in order to try to determine what is causing the voltage loss.

Did you post the LT Spice file yet? I'd be happy to take a closer look if you feel like posting that.
I'll wait on a purely theoretical analysis.

You mean that if we see the equivalent circuit of that transformer, which will be similar to RC circuit, the value of R affects the charging time so that makes lower V_out?

I tried with 10 Ohms but V_out does not varied that much. And unfortunately, what I can change is L and C due to the problem condition. Thus I need to figure out how L and C play roles in that circuit. I can observe it by changing their value but I cannot 'analyze' why changing of C changes final V_out and the shape of V_out.

Actually I'm using PSPICE so I'm not sure you can use it but I'll attach the project file of it.

Anyway thank you sincerely.

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#### MrAl

Joined Jun 17, 2014
8,063
Hi,

No i dont use that i use LT Spice mostly and a lot of members here use that so it's good to create simulations in that environment i think.

But as long as your output pumps up with each cycle i think it is ok.

The transformer L values usually does matter, but not when the coupling is exactly equal to 1.
In that case, the L values only dictate the turns ratio.

There is also the reverse equivalent resistance of the diodes that may effect the output.

From your simulation graph though it looks like at least the circuit is working.

#### MrAl

Joined Jun 17, 2014
8,063
Hello again,

I just checked the capacitance of the 1N914 diodes and it can be as high as 4pf. That is surely comparative to the 10pf capacitors being used and thus that will affect the results over results using diodes that do not have any capacitance. The Spice model i used for the diode has 1.5pf capacitance which is still significant compared to 10pf.
There is also probably some leakage in the two 10pf caps which would contribute to a reduction of the output voltage somewhat. Cap ESR could also be a factor.

The resistance i was talking about with the transformer is the equivalent secondary resistance that you can get by reflecting the primary resistance to the secondary. You get that by dividing by the square of the turns ratio. That simplifies the problem by removing the transformer and replacing it with a voltage source that has that reflected resistance and a voltage equal to the input divided by the turns ratio. For example, if the input was 100v and the turns ratio was 10 step down, then the secondary voltage would be 100/10 which equals 10 volts. So you would end up with a voltage source of 10v in series with the reflected resistance and that simplifies the circuit not only making it easier to calculate the results but also makes it easier to understand.