[SOLVED] transistor voltage doubler

[boylesg]

After a week and a half of hair tearing frustration and agrivation I have finally done it.

I have found a discrete transistor voltage doubler that works nicely in the usual direction but that I was also able to modify and get it to work in the reverse direction, i.e. -12V to -24V

All I did was swap all the NPNs for PNPs, reversed the polarity of all electrolytic caps and diodes, swapped Vcc to Vee and it worked first time without a hitch.

Only thing I can't see is what part of the circuit does the doubling and how to change that to X1.5 or any other value.
Probably should just leave it but add a voltage divider and two output sockets - 24V/-18V and +24V/+18V......keep my options open.

After

 

[betwixt]

Re: I don't f'ing believe it!!!!!!!

Well done.

A voltage douber does just that, you get twice the input voltage minus the forward drop in the two diodes so it should be about (2 * 12) - (2 * 0.7) = 22.6V at the output. I'm afraid you can't adjust it to get less out unless you starve it of current and rely on it being overloaded. You will get slightly more out and with higher efficiency if you use fast switching diodes rathe rthan 1N4007s.

A suugestion if you want to experiment: remove R7 & R8, then connect the emitter of Q2 to the base of Q1 and the emitter of Q4 to the base of Q3. It uses two less components and should give better efficiency.

Brian.

 

[boylesg]

Re: I don't f'ing believe it!!!!!!!

Well done.

A voltage douber does just that, you get twice the input voltage minus the forward drop in the two diodes so it should be about (2 * 12) - (2 * 0.7) = 22.6V at the output. I'm afraid you can't adjust it to get less out unless you starve it of current and rely on it being overloaded. You will get slightly more out and with higher efficiency if you use fast switching diodes rathe rthan 1N4007s.

A suugestion if you want to experiment: remove R7 & R8, then connect the emitter of Q2 to the base of Q1 and the emitter of Q4 to the base of Q3. It uses two less components and should give better efficiency.

Brian.

Schotty diodes? Some suggestions for commonly used ones in this situation?

Or signal diodes? I have some recovered 1N4148s.

Perhaps that circuit simulator applet will make it clear to me as to why you can't adjust this one.

Can schmitt triggers and schmitt trigger oscillators be inverted in the same way. If not then no wonder I was having trouble with the last cicruit I was fiddling with.

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Well done.
A suugestion if you want to experiment: remove R7 & R8, then connect the emitter of Q2 to the base of Q1 and the emitter of Q4 to the base of Q3. It uses two less components and should give better efficiency.
Brian.

Tried this Brian and it causes a simulator error within a second. Decreasing those resistors to 1k cause it to take far longer to ramp up to peak voltage. Increasing their value to 100k appears to have little effect.

Swapping the 1n4007s for 1n4148s makes the rise time to peak voltage a little faster. It that what you were expecting?

 

[betwixt]

I have doubts about your simulator if it gives an error.

1N4148 will work better than 1N4007, the problem with all diodes is they take a short time to start conducting after they have had a reversed voltage across them. When you have sharp rising and falling edges in the signal, there is a brief period during the transition where they appear more like resistors and waste power rather than block the flow. The faster a diode switches, the better it is as a rectifier, in 50/60Hz sine wave applications such as transsformer power supplies they are perfectly adequate but for square-wave and higher frequencies, the recovery time becomes a serious problem.

You need to understand how your circuit works to see why the voltage isn't variable. Think of the output stages like this, both sides are identical:

1. Q3 conducts so it's collector is at almost 0V - D2 and D4 conduct and charge C2 and C5 to nearly 12V.
2. Q3 then shuts off and it's collector is pulled to 12V by R4. This makes the negative side of C2 rise to nearly 12V and because it holds charge the positive end goes to near 24V.
3. D2 is now reverse biased and doesn't conduct so the 24V doesn't leak back into the 12V rail but D4 is forward biased and transfers energy from C2 to C5 raising it's voltage close to 24V.

Brian.

 

[boylesg]

We Brian I have wondered how accurately Multisim, or any other electronics simulator for that matter, simulates real circuits.

It must be extraordinarily difficult to account for al the variables and nuances of real life solid state physics in a software simulator.

Do you have any suggestions as to adjustments of Multisim 11 'simulation settings' I might try to to make your alteration work?

What schotty diodes would you recommend in particular. I have not used them as yet and am not at all familiar with the different types available in multisim - wouldn't know where to start. I have come across a couple of heavy duty schottys in the power supplies of tv circuit boards but I doubt they would be particularly useful for this situation. Usually the voltage drop across heavy duty diodes, including schottys, is rather high compared to the ones you would probably suggest.

 

[D.A.(Tony)Stewart]

Voltage doublers rely on the fact that when you switch a cap in parallel with a load cap, you transfer charge to equalize the voltage based on V=Q/C. If the input cap is smaller it takes more time to reach the same voltage but less current. ∂v = I/C ∂t . Since both caps share the same rise time ∂t and current, I, then the change in voltage is due to the Capacitance ratio. So if the load cap was much smaller than the previous stage it could transfer most of the voltage in a single cycle, but then have low capacity for the next stage if needed. Conversely if the switched capacitor was 10x smaller as in this design, it might take more than 10 cycles to reach doubling and if the load time constant was < 10 cycles it would drop below optimal doubling. So the best designs use equal values and ensure the current drive is adequate. Normally for AC lines, simple diodes and equal caps are adequate. For better size at low currents, smaller caps may be used at higher switching frequencies so the impedance at that frequency is low enough to transfer charge quickly between cycles where load current decays the voltage. Hence this is how MAXIM designed their V+/V- bipolar RS232 power chip from 5V to +/-5V using a trippler circuit with CMOS switches and 0.01 to 0.1uF capacitors.

 

[betwixt]

To be perfectly honest, I've never used Multisim, apart from very occasional use of LTSpice I use old fashioned theory and a calculator. While simulators are undoubtedly useful in some instances, I see far too many people making fundamental design mistakes then fiddling the values in a simulation to get the results they want.

The type of diode you need depemds on several factors. As mentioned earlier speed is important when the waveform has fast transitions in it and in general, the faster the diode the better. Schottky doping is a chemical process that speeds up the recovery of a 'normal' silicon junction and it also reduces Vf (the forward voltage drop) of the diode but it also makes it more difficult to ensure a high reverse voltage rating and they tend to have slightly more reverse leakage current. The different types on the market are optimized to make best use of whichever of these parameters is most important. Because they have a lower Vf, they also dissipate less heat for a given current (W = Vf * I).

In your schematic, I would guess you would be lucky to get more than about 5mA from the 24V before it starts to significantly drop so a small signal diode like the 1N4148 should be adequate. It can switch fast enough and although not Schotky, it still has a reasonably low Vf. If you want to go for a higher rated fast diode, try the 1N5819 series. As SunnySkyguy points out, it would be better to use a higher value transfer capacitor in C1 and C2 positions, around 47uF would be best. If you want more output current, you have to make it available at the input to the voltage doubler so some circuit changes would be necessary.

Brian.

 

[BradtheRad]

Only thing I can't see is what part of the circuit does the doubling and how to change that to X1.5 or any other value.

I have a Youtube video showing how a charge-pump doubler operates.

It's an animated simulation, portraying current bundles traveling through wires, and capacitors charging and discharging.

https://youtu.be/3czj7J_FE_k

Link to my Youtube video showing an animated simulation of a bridge (Greinacher) voltage doubler.

Its concept of operation is similar to the final stage of your schematic.

https://www.youtube.com/watch?v=qTS26qfGa30

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You can feed a voltage multiplier AC or pulsed DC.

When fed AC the output voltage can rise to the peak of the sine waves. Both the positive and negative waveforms actively contribute.

With pulsed DC you get a lower net volt output. It has only one polarity waveform as the active force.

You can add stages to obtain greater multiples.

A Villard tripler with AC will give you 3x the nominal input V (more or less, depending on load).

A Villard tripler with pulsed DC will give you 1.8X (more or less).

 

[boylesg]

To be perfectly honest, I've never used Multisim, apart from very occasional use of LTSpice I use old fashioned theory and a calculator. While simulators are undoubtedly useful in some instances, I see far too many people making fundamental design mistakes then fiddling the values in a simulation to get the results they want.

The type of diode you need depemds on several factors. As mentioned earlier speed is important when the waveform has fast transitions in it and in general, the faster the diode the better. Schottky doping is a chemical process that speeds up the recovery of a 'normal' silicon junction and it also reduces Vf (the forward voltage drop) of the diode but it also makes it more difficult to ensure a high reverse voltage rating and they tend to have slightly more reverse leakage current. The different types on the market are optimized to make best use of whichever of these parameters is most important. Because they have a lower Vf, they also dissipate less heat for a given current (W = Vf * I).

In your schematic, I would guess you would be lucky to get more than about 5mA from the 24V before it starts to significantly drop so a small signal diode like the 1N4148 should be adequate. It can switch fast enough and although not Schotky, it still has a reasonably low Vf. If you want to go for a higher rated fast diode, try the 1N5819 series. As SunnySkyguy points out, it would be better to use a higher value transfer capacitor in C1 and C2 positions, around 47uF would be best. If you want more output current, you have to make it available at the input to the voltage doubler so some circuit changes would be necessary.

Brian.

I tried plying with the values of C1 and C2 but it only seemed to have a marginal effect on the output current through my zenner diode but seemed to reduce the rise time to peak voltage……in Multisim of course.

I tried proportionally reducing the value of all resistors to reduce the total resistance and hopefully get more current but it didn’t seem to work.

Which part is the voltage divider you are referring to? The 10k - BC548 collector and 1k – BC458 base? I just tried playing with the values of these and either increasing them significantly or decreasing them significantly seems to result in substantially less current through my output zenner.

What changes would you suggest if I was to try and get 50mA at the output rather than about 5mA?

 

[betwixt]

Before going any further, you must understand that this kind of design isn't used commercially because it has inherent deficiencies and will always be very inefficient. In particular, you must understand that you are trying to increase voltage, not increase power. The output power you are asking for is 24V * 0.05A = 1.2W so the input power must be significantly higher than that. Even at 100% efficiency, you need to draw 100mA from the supply.

The biggest problem is that you need to fully charge and discharge C1 and C2 once per oscillator cycle. When Q1/Q3 are conducting, there isn't a problem (assuming they are saturating) because there is a ready supply of current through D1/D2 to charge them up. When Q1/Q3 turn off, the charge in C1/C2 is transferred via D3/D4 into the output capacitor but the resistors R3 and R4 are in series with the current path and will limit the transfer rate. Effectively, every half cycle you add a 1K resistor in series with the output.

The problem is if you reduce the value of R3/R4 to eliminate the series resistance, you also waste more power in the resistors and in Q1/Q3 and risk either not saturating them or worse, actually damaging them. So you have a conflict of requirements, a high value to reduce losses but a low value to increase output current. The solution is fairly straightforward and is used commercially but it requires changes to your circuit: The objective is to pull the negative sides of C1/C2 to ground on one half cycle and pull then to 12V on the other half cycle. A passive pull-up (R3/R4) can't do this so you need to actively pull them to 12V rather than rely on the current through R3/R4 to do it. What you have to do is remove R3/R4 and replace them both with another transistor, switching alternately with Q1/Q3, so when one is off, the other is on.

Brian.

 

[boylesg]

Before going any further, you must understand that this kind of design isn't used commercially because it has inherent deficiencies and will always be very inefficient. In particular, you must understand that you are trying to increase voltage, not increase power. The output power you are asking for is 24V * 0.05A = 1.2W so the input power must be significantly higher than that. Even at 100% efficiency, you need to draw 100mA from the supply.

The biggest problem is that you need to fully charge and discharge C1 and C2 once per oscillator cycle. When Q1/Q3 are conducting, there isn't a problem (assuming they are saturating) because there is a ready supply of current through D1/D2 to charge them up. When Q1/Q3 turn off, the charge in C1/C2 is transferred via D3/D4 into the output capacitor but the resistors R3 and R4 are in series with the current path and will limit the transfer rate. Effectively, every half cycle you add a 1K resistor in series with the output.

The problem is if you reduce the value of R3/R4 to eliminate the series resistance, you also waste more power in the resistors and in Q1/Q3 and risk either not saturating them or worse, actually damaging them. So you have a conflict of requirements, a high value to reduce losses but a low value to increase output current. The solution is fairly straightforward and is used commercially but it requires changes to your circuit: The objective is to pull the negative sides of C1/C2 to ground on one half cycle and pull then to 12V on the other half cycle. A passive pull-up (R3/R4) can't do this so you need to actively pull them to 12V rather than rely on the current through R3/R4 to do it. What you have to do is remove R3/R4 and replace them both with another transistor, switching alternately with Q1/Q3, so when one is off, the other is on.

Brian.

I found that if I reduce those resistors to 300R I can get just below 10mA out through my zenner.

But after reading your second paragraph I started thinking about incorporating a PNP in there some where so to allow the capacitor to continue charging when Q1 is not conducting.

After reading your third paragraph I can see I was on the right track.

So when the multivibrator's Q2 is conducting Q1 flogs a bit of current to turn on and charge up capacitor C1. However Q1 requires the resistor to ensure its max current ratings are not exceeded, which limits the charge current to C1 possibly resulting in it not being fully charged within the cycle.

You're saying pull C1 up to 12V and pull it down to GND so that it quickly charges and discharges. That sounds rather like complementary totem pole as was trying to use to drive a mosfet gate. Is that what you are talking about here?

 

[boylesg]

So clearly this is what you are talking about with that resistor-transistor swap. With it I can get about 20mA out which is probably enough more most purposes.

But for some reason with the inverse circuit negative booster the mod doesn't work quite so well. At first I could only get about 9mA out of it but now it is giving me a simulation error as soon as I run it. I can't see anything wrong with it compared to the positive booster circuit.
Can you see anything obviously wrong with the inverse circuit?

Never mid about the above, I spotted what was wrong........one of my multivibrator diodes was around the wrong way in the inverted circuit and for some reason the ammeter below the 1000uF cap cause an immediate simulator error, but only in the inverted circuit. Any idea why it is only the inverted circuit?. Beats me! The ammeter in the same position in normal circuit works just fine.

+12V - +18V - works brilliantly


-12V to -18V - works brilliantly

 

[FvM]

There are several voltage drops adding up, the 10k to 100k resistors ratio is further reducing the available multivibrator output voltage. If you review the individual circuit voltages, you should be able to understand the problems involved with the present component dimensioning.

In a real circuit, Vbe reverse breakdown that's often ignored in SPICE models may affect the circuit operation.

 

[boylesg]

There are several voltage drops adding up, the 10k to 100k resistors ratio is further reducing the available multivibrator output voltage. If you review the individual circuit voltages, you should be able to understand the problems involved with the present component dimensioning.

1k and 10k then perhaps? Or even 100R and 1k?

In a real circuit, Vbe reverse breakdown that's often ignored in SPICE models may affect the circuit operation.

How come? Too hard basket?

This modification to this circuit brings to mind another question. If I remove the 1k resistor in the original circuit then I get a sim error, presumeably because the BC548 fries with to much current passing through CE.

Why don't you need the same current limiting when you combine it with its complement in a totem pole?

Surely you have the same problem of a current limit through both their CEs.

 

[FvM]

I was talking about resistor ratio. Changing the 10k to 2K or 1k will considerably increase the output voltage respectively available current.

I don't wanted to start a detail discussion about the BE reverse breakdown point, just mention a possible simulation compliance problem. A possible consequence is to refer to a multivibrator circuit that avoid large negative BE voltages.

 

[boylesg]

I was talking about resistor ratio. Changing the 10k to 2K or 1k will considerably increase the output voltage respectively available current.

Wooahhh! 100mA output.....brilliant.

I think I get it, by changing the ratio you change the duty cycle of the multivibrator to 90% or what ever. That gives a longer on and time for the 1000u capacitor to charge up and deliver more current during the off time.

I don't wanted to start a detail discussion about the BE reverse breakdown point, just mention a possible simulation compliance problem. A possible consequence is to refer to a multivibrator circuit that avoid large negative BE voltages.

Sounds like the too hard basket then.

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I just want to say a big THANKYOU to all of you in here who have helped me with this circuit.

Despite the initial frustrations with it, this is been a BRILLIANT learning exercise and I have ultimately enjoyed the journey.

I hope to learn a great deal more in the future from you guys!

 

[betwixt]

Thre duty cycle is best at 50%, it's a multivibrator so if one side is 90% the other is only 10%.

As an experiment, try removing Q6 & Q11, if you get half the output current it's working properly - but I bet you dont !

Brian.

 

[boylesg]

Thre duty cycle is best at 50%, it's a multivibrator so if one side is 90% the other is only 10%.

As an experiment, try removing Q6 & Q11, if you get half the output current it's working properly - but I bet you dont !

Brian.

Another source of inefficiency would be that zenner diode on the output - it carries away some of the output current in order to disipate some of the voltage.

Is there a way I could redirect that current back into the output stream but without raising the voltage above 18V?

I suppose I could try a large value resistor in series with the zenner to limit the current flow.

Update....I just tried a 10k resistor and got the out put current up to about 170mA but the voltage went up to between 19V and 20V.

Any other suggestions?

 

[FvM]

I think I get it, by changing the ratio you change the duty cycle of the multivibrator to 90% or what ever.

Preferably you'lll make a symmetrical change and keep the duty cycle. By increasing the reistor ratio, you increase the multivibrator output voltage.

If you are targetting to regulated output voltage, larger output currents and high efficiency, you should think about inductive boost converters.