So one multiplier is loaded by one load, and both multipliers are also loaded by the second load?
Back to your original goal, do you really need to simulate the multiplier, or is it ok to just take the combined output power at whatever voltage you aim for and lump into one secondary that is loaded such that the load draws that total amount?

 

So one multiplier is loaded by one load, and both multipliers are also loaded by the second load?
Back to your original goal, do you really need to simulate the multiplier, or is it ok to just take the combined output power at whatever voltage you aim for and lump into one secondary that is loaded such that the load draws that total amount?

Yes exactly, that way both loads are connected to each other but they are at different voltages. I think it will be fine to lump them together so that the total load current is drawn, but that’s what I am unsure about performing. Do I add the two currents to get the total current flow and divide by the highest output voltage? Or would I divide the total power by the total current, to get a voltage and therefore a resistance that is representative both of the outputs combined? Hope it makes sense, I am scratching my head at this one really.

 

In load 1 you got V1 and I1, that gives you P1. In load 2 you got V2 and I2, giving you P2. Add the two powers together to get Ptot.
Recalculate the load resistance that is across both multipliers so that you draw the same Ptot from that voltage alone.

 

Not sure I follow the change. Am I still missing something?

 

In load 1 you got V1 and I1, that gives you P1. In load 2 you got V2 and I2, giving you P2. Add the two powers together to get Ptot.
Recalculate the load resistance that is across both multipliers so that you draw the same Ptot from that voltage alone.

I have done this and now my simulations are running much more like I expect, thank you. I have a further question though. When I did these calculations, the equivalent output voltage is less than the highest output voltage. Therefore, how would one decide the transformers turn ratio to emulate the entire stacked, multiplier output? I have included my calculations. The values for V, I, and P are arbitrary and just for example.

 

Not sure I follow the change. Am I still missing something?

View attachment 181335

I don't think this is right, Tony. The load resistors should be connected across both the capacitors in the multiplier. The only difference from a normal full wave voltage multiplier is that two of them are connected in series typically through current limiting resistor. One of the loads is connected across 1 multiplier, whereas the other load is connected across both of them. This way, 2x the rectified voltage appears on R1, and 4x the rectified voltage appears over R2. Here is a single multiplier, just imagine connecting two in series at the output and that's what I have.

 

I think we are missing something that Tony is trying to point out.. D1, D2, C1, and C2 could have an output of 1000V across C2 and the other half could have an output of 10V across C3 and the max voltage across Resistor V01 would still be less then 10V due to resistor V02 Voltage drop. D1, D2, C1, and C2 need a return path to have any effect to the output.

 

@mlharizona First - welcome to AAC. Second, thanks for pointing out some stuff I didn't see. Nevertheless, it's not quite what I was focusing on. At first I couldn't see a current path whatsoever for Sa (Secondary A). But an even bigger issue is that we have no voltage, no current and no load information. It's impossible to give an answer a question that is devoid of all detail. Nevertheless, welcome. Feel free to comment or to ask questions. The people here are always willing to help beginners. They realize that even they were beginners once. For some of us that was a very very long time ago. So there's a lot of wisdom and assistance here for those who earnestly seek to learn.

 

Hello.

What would be the process for combining two outputs from a DC/DC converter into a single output? I am trying to design a resonant converter, which requires information regarding the load resistance. However, there are two load resistances in my application. I have seen something similar done but the two outputs were not connected to each other - they were two separate outputs and therefore current did not flow between them. In my application, I have one main output, which has two current flows - one to ground (which represents losses to the body of the device) and one to the other output (which represents the useful current transfer in the device).

I have attached a schematic for illustration. I have been scratching my head at this for a while and it makes circuit analysis and converter design all the more difficult.

Petkan:
Your circuit shows transformer with two (presumably equal) secondaries, each rectified by "doubler" rectifier.
It is not clear what you want. If the secondaries are identical the voltages will match and there would be no problem joining them. Even if the voltages do not match - by joining you will get the higher of the two. With two secondaries, the use of doubler rectifier is questionable. Join the two secondaries (center tapped secondary). With two 4 diodes (in respect of the tap) you will get rectified whatever one winding gives , with 4 diodes you can get twice - why bother with doublers? Same 4 diodes, only one cap, full load handling. Define more precisely what you want.


If someone could help that would be appreciated.

Thanks, J

View attachment 181166

 

I have returned, and I see that some more questions have arrived. The unknown voltages and currents only affect the component values and ratings, not the actual circuit. The one thing that has not been mentioned is the insulation ratings in the transformer. That may be a big challenge if the voltages are "high." But the same applies if it is a system providing 5 volts and 12 volts for some hybrid circuit with digital and analog sections, or a high power tube type transmitter with the lower voltage stages at 300 volts, and the high powered final amplifier tubes using 1500 volts..

 

I understand what you are saying. However if you use voltage doublers AND stack the secondaries, then the voltage at the output will be 4 times that of the transformer voltage. In your suggestion the rectified output DC will be only 2 times that of the transformer voltage. People are misinterpreting my question (this is my fault as it is unclear). I was asking how to combine two load resistors into a single load resistor so that I could analyse the effect of the two loads easier.

What I am trying to achieve - use a transformer with a small step up to 1.5kV, then have two secondaries each feeding a voltage multiplier (3kVDC each) which are connected in series to form the final 6kV output. A second output is connected across the first multiplier only to get 3kV.

 

I understand what you are saying. However if you use voltage doublers AND stack the secondaries, then the voltage at the output will be 4 times that of the transformer voltage. In your suggestion the rectified output DC will be only 2 times that of the transformer voltage. People are misinterpreting my question (this is my fault as it is unclear). I was asking how to combine two load resistors into a single load resistor so that I could analyse the effect of the two loads easier.

What I am trying to achieve - use a transformer with a small step up to 1.5kV, then have two secondaries each feeding a voltage multiplier (3kVDC each) which are connected in series to form the final 6kV output. A second output is connected across the first multiplier only to get 3kV.

Petkan: Look at the HV section of microwaves. They use a secondary for HV and another for the filament.
The HV is derived from single winding. Old CRT TVs, oscilloscopes etc used not doublers, but multipliers i.e. higher number of diode/capacitor pairs. The same idea behind the doubler could be made 3 times, for times etc. Bear in mind that the two windings will still have to withstand very high voltages one in respect of another. This is the reason microwaves use a single HV winding

 

Petkan: Look at the HV section of microwaves. They use a secondary for HV and another for the filament.
The HV is derived from single winding. Old CRT TVs, oscilloscopes etc used not doublers, but multipliers i.e. higher number of diode/capacitor pairs. The same idea behind the doubler could be made 3 times, for times etc. Bear in mind that the two windings will still have to withstand very high voltages one in respect of another. This is the reason microwaves use a single HV winding

What voltage do these go up to though? My max output is 6kV. I usually see long strings of Cockcroft Walton multipliers in supplies that are above 10kV and sometimes even above 100kV, for example, where really low output currents exist. I don’t have very low output current relatively.

Also the regulation and voltage ripple of these types of multipliers is terrible as the number of stages is added - ripple is quite important for me!