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[JohnH]

How do voltage multiplier circuits work? I’ve come across them before but I just don’t get how you can get something like 90v from a 15v supply...

[DrNomis_44]

How do voltage multiplier circuits work? I’ve come across them before but I just don’t get how you can get something like 90v Dom a 15v supply...

It's pretty simple really, when the output of the 60kHz oscillator goes high, each of the 1N4007 diodes will conduct and charge up each of the 47uF/100V caps, since there are five of the 47uF/100V caps, each of them will charge up to about a fifth of the output voltage, when the output of the 60kHz oscillator goes low, each of the 1N4007 diodes will stop conducting, the cap on the end of the 10k resistor will then charge up to the sum of the charges on all the other 47uF/100V caps, this happens at the rate of the 60kHz oscillator.

With five 47uF/100V caps, the output voltage will be about +45V DC, to get +90V DC out of the circuit you just add another five 47uF/100V caps and 1N4007 diodes, I've actually tried doing it in practice and it does work.

Marcel could probably elaborate more on how the circuit works.

[Marcel]

Simplest explanation for multiplier circuits is it's a bit like voltage stacking.
We all know how with a diode and a cap we can convert AC into DC, well, if we do it multiple times in (theoretical) parallel and then put all the resultant DC voltages in series you can achieve higher DC voltages from lower AC voltages.

The mathematics and the trickery pokery of reality are a lot more complex but that only really concerns the designers... for us we just need to be able to recognise the basic circuit and to know what voltage we can expect at the output... Usually you can recognise it as an AC power signal of some sort going to a bunch of diodes (often 3 or more) in series with appropriately voltage'd caps (usually tied to chassis/earth) connected between every diode with a higher voltage needing 'load' at the end.

One thing to remember is only voltage is multiplied...Power remains the same.... Discounting any losses in the circuit If we have 12v at 1A at the input to the multiplier then if we multiply up to 24V we can only consume 0.5A, and at 48V we will have 0.25A at our disposal, and if we multiply up to 96V then the maximum theoretical current we can draw is only 0.125A (12V x 1A = 96V x 0.125A = 12W assuming nil losses). In reality due to losses and the effects of 'loading' the multiplier to 96V will probably only supply a maximum of about 0.1A before drastically dropping of in the higher voltage it can supply.

[DrNomis_44]

Simplest explanation for multiplier circuits is it's a bit like voltage stacking.
We all know how with a diode and a cap we can convert AC into DC, well, if we do it multiple times in (theoretical) parallel and then put all the resultant DC voltages in series you can achieve higher DC voltages from lower AC voltages.

The mathematics and the trickery pokery of reality are a lot more complex but that only really concerns the designers... for us we just need to be able to recognise the basic circuit and to know what voltage we can expect at the output... Usually you can recognise it as an AC power signal of some sort going to a bunch of diodes (often 3 or more) in series with appropriately voltage'd caps (usually tied to chassis/earth) connected between every diode with a higher voltage needing 'load' at the end.

One thing to remember is only voltage is multiplied...Power remains the same.... Discounting any losses in the circuit If we have 12v at 1A at the input to the multiplier then if we multiply up to 24V we can only consume 0.5A, and at 48V we will have 0.25A at our disposal, and if we multiply up to 96V then the maximum theoretical current we can draw is only 0.125A (12V x 1A = 96V x 0.125A = 12W assuming nil losses). In reality due to losses and the effects of 'loading' the multiplier to 96V will probably only supply a maximum of about 0.1A before drastically dropping of in the higher voltage it can supply.

Yep, but since the typical 12AX7 Triode gain-stage circuit will only draw maybe up to 1mA of current at best from the HT supply, a limit of .1A is probably nothing to worry about, so, to power a circuit that uses one 12AX7 valve, using the Voltage Multiplier circuit to generate the +HT supply, we would only need a regulated power supply to generate +12.6V DC for the heater, and +15V DC for the Voltage Multiplier.

That means we could use something like a 16V AC/1.25A plugpack adaptor to power the regulated power supply circuitry together with a basic Bridge-Rectifier and filter cap to rectify the 16V AC to raw DC.

[Marcel]

For the sake of understanding I wanted to keep the numbers simple there Doc..

Did you look at the Valvecaster pdf..???

[Marcel]

Cheers for that Marcel, I downloaded a copy of that pdf, I've experimented with the Valvecaster circuit in the past but wasn't able to get it working at it's best, maybe the valves I was using were getting a bit worn out, the circuit tended to be a bit hummy, I'm thinking of building a circuit that uses two 12AX7 valves, but with a higher HT supply that's produced by a square-wave oscillator made from a CD4049 IC that drives a Voltage Multiplier circuit consisting of some 1N4007 (or similar) diodes and some 1uF/100V, or 2u2/100V (can even be some 47uF/100V caps too) electrolytic caps, I found that using the circuit I can easily generate up to something like +90V from a 15V DC supply.


Here's the Voltage Multiplier circuit:

Attachment 24457


The 10k resistor on the output of the Voltage Multiplier is there just to limit the output current a bit, the square-wave oscillator consisting of the CD4049 IC, 33k resistors, and 220pF cap runs at about 60kHz.

The CD4049 IC has six Inverters in it, three of them are used to form the 60kHz square-wave oscillator and the other three are used as inverting-buffers so that there is enough current drive to drive the Voltage Multiplier section.

I definitely need to get to it and do some breadboarding soon.

If it were me I'd be increasing the size of the cap on the oscillator circuit to lower the frequency to just out of earshot range. The charge caps in the multiplier will work and hold charge better at 20kHz than at 60kHz.
There won't be much power coming out of the IC, so if there is a voltage sag issue on the tube's plate resistor then I'd consider using the output of the IC to drive a transistor switcher so that more power (wattage) can get put into the multiplier so that the multiplied voltage is "stronger"....

[DrNomis_44]

For the sake of understanding I wanted to keep the numbers simple there Doc..

Did you look at the Valvecaster pdf..???

Yep, had a look at it and it looks like it's not that hard to build at all, I seem to remember building it a couple of years or so ago.


The way Matsumin, the designer of the valvecaster circuit, goes about implementing a gain control is a bit unusual, I had to mod my valvecaster build so that the gain control worked a bit better cause when it was built as per the circuit diagram it sounded a bit mis-biased, or it could have been due to the valve I was using at the time, I'll have another go at building the valvecaster circuit and will make sure I start a build-thread for it, will also do some signal tracing with a signal-generator and scope (will post some screen shots too) once I've got it working properly.

[DrNomis_44]

If it were me I'd be increasing the size of the cap on the oscillator circuit to lower the frequency to just out of earshot range. The charge caps in the multiplier will work and hold charge better at 20kHz than at 60kHz.
There won't be much power coming out of the IC, so if there is a voltage sag issue on the tube's plate resistor then I'd consider using the output of the IC to drive a transistor switcher so that more power (wattage) can get put into the multiplier so that the multiplied voltage is "stronger"....

So maybe increase the 220pF cap to maybe something like 1nF, or 2n2?

[JohnH]

Simplest explanation for multiplier circuits is it's a bit like voltage stacking.
We all know how with a diode and a cap we can convert AC into DC, well, if we do it multiple times in (theoretical) parallel and then put all the resultant DC voltages in series you can achieve higher DC voltages from lower AC voltages.

The mathematics and the trickery pokery of reality are a lot more complex but that only really concerns the designers... for us we just need to be able to recognise the basic circuit and to know what voltage we can expect at the output... Usually you can recognise it as an AC power signal of some sort going to a bunch of diodes (often 3 or more) in series with appropriately voltage'd caps (usually tied to chassis/earth) connected between every diode with a higher voltage needing 'load' at the end.

One thing to remember is only voltage is multiplied...Power remains the same.... Discounting any losses in the circuit If we have 12v at 1A at the input to the multiplier then if we multiply up to 24V we can only consume 0.5A, and at 48V we will have 0.25A at our disposal, and if we multiply up to 96V then the maximum theoretical current we can draw is only 0.125A (12V x 1A = 96V x 0.125A = 12W assuming nil losses). In reality due to losses and the effects of 'loading' the multiplier to 96V will probably only supply a maximum of about 0.1A before drastically dropping of in the higher voltage it can supply.

Ok, that makes some sense to me. I still have to think about it a bit I think...

[DrNomis_44]

Ok, that makes some sense to me. I still have to think about it a bit I think...

No worries, take your time cause it is a bit hard to understand at first, I went through the same learning-process when I first started out in electronics.