boylesg
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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.
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.
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 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.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 found that if I reduce those resistors to 300R I can get just below 10mA out through my zenner.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.
1k and 10k then perhaps? Or even 100R and 1k?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.
How come? Too hard basket?In a real circuit, Vbe reverse breakdown that's often ignored in SPICE models may affect the circuit operation.
Wooahhh! 100mA output.....brilliant.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.
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.
Preferably you'lll make a symmetrical change and keep the duty cycle. By increasing the reistor ratio, you increase the multivibrator output voltage.I think I get it, by changing the ratio you change the duty cycle of the multivibrator to 90% or what ever.
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