Home |
Search |
Today's Posts |
#13
![]() |
|||
|
|||
![]()
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? Be sure to load each side of the AC source that represents the transformer winding with a small capacitor to ground (something like ..01 uF) or the simulation will run slow. The source is constrained only by the diode reverse bias capacitance during most of the swing, so it takes a while for the simulation to scale up and down between what is happening when the diodes are on and when they are off. -- John Popelish |
Thread Tools | Search this Thread |
Display Modes | |
|
|
![]() |
||||
Thread | Forum | |||
FCC: Broadband Power Line Systems | Policy | |||
SWR meter vs TLI | Antenna | |||
Battery voltage indicator | Homebrew | |||
Building a psu for linear amplifier | Homebrew | |||
Checking leaky caps | Boatanchors |