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#11
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K7ITM wrote:
With regard to circuit simulation: have a look for LTSpice on the Linear Technology website. Or if you don't mind text-based netlists and classical Spice, see http://www.rfglobalnet.com/IndustryS...hResults.aspx? keyword=Spice&TabIndex=4&image1.x=0&image1.y=0 for some other sources. With regards the circuit, it's fairly easy to do some mental arithmetic to see if there is an advantage or not over the two-diode, two-caps, floating-transformer "full wave voltage doubler" circuit. In the old circuit, there are, if you will, four distinct time periods in each cycle (though two of them are electrically equivalent, and the other two are symmetrical). One is where the upper cap is charging, and the load current may be considered to be through the two caps in series -- or may be considered to be through the lower cap and the transformer and upper diode. Following that is a period where the transformer voltage is too low to charge the upper cap and too high (not negative enough) to charge the lower cap. Then the load current flows through the two caps in series. Following that, the transformer charges the lower cap, and after that the transformer voltage is again too low to forward bias either diode. So the net voltage across the two caps increases twice each cycle, once when the upper cap is charging and once when the lower cap is charging. The output (load) ripple fundamental frequency is twice the line frequency. You can get a reasonable idea of the ripple voltage if you assume the charging takes place in a tiny fraction of the total cycle time; in fact, that will be an upper bound on the ripple voltage. For simple analysis, assume a 1Hz input so there are two charging pulses per second (one per cap), and assume two 1-farad caps and a 1-amp load. Then the voltage across each cap between the charging pulses sags at 1 volt per second, and the output sags at 2 volts per second, but resets twice per second, so the ripple is 1 volt peak to peak. In the new circuit, you need to say something about the capacitor values. Presumably the two "input" caps are the same value, but may differ from the output cap. But what happens if you say that you are going to use the same total energy-storage ability as you had in the simpler circuit? Then the output cap might be, say, 0.25 farads, and the two input caps might be 0.5 farads each, since the output cap must be rated at twice the voltage of the caps in the original circuit. Now...can you figure out an allocation of the caps that gives you even the same performance as the original circuit, let alone a better one? And is it worth using more caps and more diodes? Especially for low voltage use, the original circuit has a distinct advantage of having only one diode drop during charging. You can also analyze transformer utilization and I think you'll find that the "improved" circuit doesn't really offer any advantage, if you set it up to give the same output ripple. Any other viewpoints, backed by some analysis or simulation? Cheers, Tom David J Windisch wrote: Build s test ckt with a filament xfmr and *small* caps and a "heavy" load, and look at the oupt ripple with a scope. Or build it virtually. OT: Which are the other good, (affordable!?) ckt sims? Anyone have a transferable version of Electronics Workbench he'd like to sell? 73, Dave, N3HE Cincinnati OH Looks like about 100V ripple. Here it is for LTSpice... Version 4 SHEET 1 880 680 WIRE -16 64 -96 64 WIRE -96 64 -96 208 WIRE -96 352 -16 352 WIRE -128 224 -128 208 WIRE -128 208 -96 208 WIRE -96 208 -96 352 WIRE 48 64 112 64 WIRE 176 64 208 64 WIRE 320 64 416 64 WIRE 416 64 416 352 WIRE 416 352 320 352 WIRE 256 352 240 352 WIRE 112 352 48 352 WIRE 112 160 48 160 WIRE 48 160 48 64 WIRE 112 256 48 256 WIRE 48 256 48 352 WIRE 176 256 208 256 WIRE 208 256 208 64 WIRE 208 64 256 64 WIRE 176 160 240 160 WIRE 240 160 240 352 WIRE 240 352 176 352 WIRE 416 400 416 352 WIRE 416 496 416 464 WIRE 416 352 528 352 WIRE 528 352 528 384 WIRE 528 464 528 496 WIRE -272 160 -272 0 WIRE 48 64 48 0 WIRE 48 0 -272 0 WIRE 48 352 48 416 WIRE 48 416 -272 416 WIRE -272 240 -272 288 WIRE -272 368 -272 416 FLAG -128 224 0 FLAG 416 496 0 FLAG 528 496 0 SYMBOL diode -16 80 R270 WINDOW 0 32 32 VTop 0 WINDOW 3 0 32 VBottom 0 SYMATTR InstName D1 SYMATTR Value MUR460 SYMBOL diode -16 368 R270 WINDOW 0 32 32 VTop 0 WINDOW 3 0 32 VBottom 0 SYMATTR InstName D2 SYMATTR Value MUR460 SYMBOL diode 256 80 R270 WINDOW 0 32 32 VTop 0 WINDOW 3 0 32 VBottom 0 SYMATTR InstName D3 SYMATTR Value MUR460 SYMBOL diode 256 368 R270 WINDOW 0 32 32 VTop 0 WINDOW 3 0 32 VBottom 0 SYMATTR InstName D4 SYMATTR Value MUR460 SYMBOL cap 176 48 R90 WINDOW 0 0 32 VBottom 0 WINDOW 3 32 32 VTop 0 SYMATTR InstName C1 SYMATTR Value 200µ SYMBOL cap 176 336 R90 WINDOW 0 0 32 VBottom 0 WINDOW 3 32 32 VTop 0 SYMATTR InstName C2 SYMATTR Value 200µ SYMBOL diode 112 176 R270 WINDOW 0 32 32 VTop 0 WINDOW 3 0 32 VBottom 0 SYMATTR InstName D5 SYMATTR Value MUR460 SYMBOL diode 112 272 R270 WINDOW 0 32 32 VTop 0 WINDOW 3 0 32 VBottom 0 SYMATTR InstName D6 SYMATTR Value MUR460 SYMBOL cap 400 400 R0 SYMATTR InstName C3 SYMATTR Value 100µ SYMBOL res 512 368 R0 SYMATTR InstName R1 SYMATTR Value 175 SYMBOL voltage -272 144 R0 WINDOW 3 -81 305 Left 0 WINDOW 123 0 0 Left 0 WINDOW 39 0 0 Left 0 SYMATTR InstName V1 SYMATTR Value SINE(0 250 60) SYMBOL res -288 272 R0 SYMATTR InstName R2 SYMATTR Value .5 TEXT -366 520 Left 0 !.tran 0 1 0 .1m John KD5YI |
#12
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Mike Silva wrote:
Yes, both C1 and C2 are charged with 1/2 wave power (one pulse every full cycle). However, it appears to me that the difference is that they only discharge during one half-cycle of every full cycle. On the charging half-cycle they are not also discharging, but they only discharge on the opposite half-cycle. This is different from the 2x2 doubler, where both caps are charged on alternate half-cycles, but both caps discharge during the entire full cycle. If I've analysed this correctly, it seems that there should be some improvement in voltage regulation over the 2x2. But I wanted others to look at the circuit and offer their ideas on it as well. 73, Mike, KK6GM looks like a cross between a full wave bridge and a pair of voltage doublers in parallel. |
#13
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On 1 Feb 2005 08:06:26 -0800, "Mike Silva"
wrote: Yes, both C1 and C2 are charged with 1/2 wave power (one pulse every full cycle). However, it appears to me that the difference is that they only discharge during one half-cycle of every full cycle. On the charging half-cycle they are not also discharging, but they only discharge on the opposite half-cycle. This is different from the 2x2 doubler, where both caps are charged on alternate half-cycles, but both caps discharge during the entire full cycle. If I've analysed this correctly, it seems that there should be some improvement in voltage regulation over the 2x2. But I wanted others to look at the circuit and offer their ideas on it as well. 73, Mike, KK6GM It is true C1 and C2 are not discharging while charging. But when they do discharge on the other half cycle they are being discharged more than they would be in a regular doubler circuit. They still each have to supply 1/2 of the output load at that time to recharge the output capacitor. The output capacitor is getting recharged through C1 and C2 only and not directly from the diodes and transformer. So in effect you have 3 capacitors in series for the load rather than 2 in a regular doubler. Although at first glance it would seem that the output capacitor is getting part of its charge through some of the diodes, it only does so on the first cycle at startup . After that the output capacitor is charged to a higher voltage than the transformer supplies directly. That keeps the diodes directly from the transformer to the output capacitor reverse biased so no current flows directly from the transformer to the output capacitor. So the output capacitor ends up only being charged by C1 and C2 on alternate cycles. So it would seem that performance would be worse than a standard doubler circuit. Unless much larger capacitors were used. 73 Gary K4FMX |
#14
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Gary Schafer wrote:
On 1 Feb 2005 08:06:26 -0800, "Mike Silva" wrote: Yes, both C1 and C2 are charged with 1/2 wave power (one pulse every full cycle). However, it appears to me that the difference is that they only discharge during one half-cycle of every full cycle. On the charging half-cycle they are not also discharging, but they only discharge on the opposite half-cycle. This is different from the 2x2 doubler, where both caps are charged on alternate half-cycles, but both caps discharge during the entire full cycle. If I've analysed this correctly, it seems that there should be some improvement in voltage regulation over the 2x2. But I wanted others to look at the circuit and offer their ideas on it as well. 73, Mike, KK6GM It is true C1 and C2 are not discharging while charging. But when they do discharge on the other half cycle they are being discharged more than they would be in a regular doubler circuit. They still each have to supply 1/2 of the output load at that time to recharge the output capacitor. The output capacitor is getting recharged through C1 and C2 only and not directly from the diodes and transformer. So in effect you have 3 capacitors in series for the load rather than 2 in a regular doubler. Although at first glance it would seem that the output capacitor is getting part of its charge through some of the diodes, it only does so on the first cycle at startup . After that the output capacitor is charged to a higher voltage than the transformer supplies directly. That keeps the diodes directly from the transformer to the output capacitor reverse biased so no current flows directly from the transformer to the output capacitor. So the output capacitor ends up only being charged by C1 and C2 on alternate cycles. So it would seem that performance would be worse than a standard doubler circuit. Unless much larger capacitors were used. 73 Gary K4FMX I ran simulations for both circuits. With the basic assumption that the output voltage being developed is large compared to the diode drops, I can't find any advantage to the more complex circuit. In both circuits the transformer supplies energy on each half cycle. The ripple at the top of the single capacitor in the complex circuit is the same as the ripple at the top of the 2 capacitor string in the simple circuit. The ripple at the output in both cases is 120 Hz for a 60 Hz supply with identical magnitude. The ripple at the junction of the two capacitors in the simple circuit happens to be 60 Hz. I varied the load and the regulation appears identical for both circuits and is mostly a function of the source resistance of the transformer (not terribly surprising). The series resistance added by the diodes would generally be an order of magnitude smaller for high voltage supplies and not a big factor in either case assuming these are high current diodes. The simple circuit has the advantage of being able to easily supply both a double voltage 120 Hz ripple output and a standard voltage 60 Hz ripple output. In a power supply I built for a Heathkit SB-101 a long time ago I took advantage of the two outputs from a simple doubler. The 850 V output went to the finals. Then from the 425 V output I added a divider and additional filter capacitor to provide the 300 V supply. By providing a load resistance across the upper cap similar to the load resistance of the divider on the lower cap the equally shared voltge was maintained at the two filter capacitors. The divider on the lower cap allowed me to tap off at a 300 V point to which I added a little additional filtering (remember that point has 60 Hz ripple). By doing this I didn't need a separate winding or transformer for the 300 V supply. I used an old 300 VRMS TV transformer for both my 850 V and 300 V supplies. That supply is still in use with original components 35 years later. For the simulation of the complex circuit I used a 50 uF output filter and for the simple circuit I used two series 100 uF capacitors to provide the equivalent filtering (i.e., 50 uF). So, all I can tell is that the complex circuit takes one extra 50 uF capacitor, and 4 extra diodes to do the same job as the simple circuit with no difference in efficiency that I can find. Curtis Eickerman |
#15
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So, all I can tell is that the complex circuit takes one extra 50 uF
capacitor, and 4 extra diodes to do the same job as the simple circuit with no difference in efficiency that I can find. Curtis Eickerman ============================= A simpler logical analysis - The more complex circuit will have lower efficiency because it does exactly the same job with 5 additional components to lose power in. Back to the drawing board ! |
#16
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Many thanks for running that simulation. Sometimes more is better, and
other times more is just more! 73, Mike, KK6GM |
#17
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Mike Silva wrote:
Many thanks for running that simulation. Sometimes more is better, and other times more is just more! 73, Mike, KK6GM I think a way of looking at the difference is that one is two half wave doublers in series and one is two half wave doublers in parallel. You would have a hard time using the parallel version if you wanted an 800 volt output and intended to use electrolytic capacitors, because each of the 3 would have 800 volts across them but it may have some utility for low voltage outputs. There may be a better way to connect the diodes, but I haven't got that far, yet. -- John Popelish |
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