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Registered Member #4266
Joined: Fri Dec 16 2011, 03:15AM
Location:
Posts: 874
Hi im trying to find out the capacitance size for a triple but there isn,t a resistor to use to workout the time constant and the fake current. Below is the type of system, any help thanks
Edit the 500ohm resistor is replaced with a 95.5mH inductor
Registered Member #4266
Joined: Fri Dec 16 2011, 03:15AM
Location:
Posts: 874
Hi , I would like 1.5 amp at 1kv, planning on getting close to the max kwh of my house wiring say 8 amp at 240 volt input. Cheers
I plan to have 250joules block(cap and jgbt) that gets discharged throught a load in 0.2RC and recharged at 1_5RC, would like each block to fire at one pulse per second, with multiable blocks.
Registered Member #42796
Joined: Mon Jan 13 2014, 06:34PM
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Posts: 195
for a simple full wave rectifier the smoothing cap can be calculated with this formula C = I x t / V where I is the load current V is the voltage ripple we chose and t is the AC half cycle or charge time
so if we need a rectified mains (230Vrms 50Hx) of 320Vdc at 1 A with a maximum voltage ripple of 10% the cap would be C=1*0.01/(0.1*230*1.41)=312uF
in case of a voltage multiplier the situation is worse as the charging time for the final stage is a multiple of the multiplier stages but i'm not sure how to calculate....
for Cockcroft walton Voltage Multiplieer (the circuit from your schematic) i found a calculator here
but instead of a CW multiplier why not use a simple voltage tripler which use half the parts
i've seen this circuit used in many times over the CW and for obvious resons: half the parts and less capacitance due to shorter charge times
for both you still need parts to have a voltage rating of twice the Vin though...
Registered Member #72
Joined: Thu Feb 09 2006, 08:29AM
Location: UK St. Albans
Posts: 1659
Instead of a tripler, a half-wave rectifier opposite a doubler of the opposite polarity will provide the same output voltage with smaller components for the same output current, as long as you don't mind both terminals hot.
Registered Member #4266
Joined: Fri Dec 16 2011, 03:15AM
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Cheers dexter, I keep on change the volts,watts and capances . Can you connect to extension cords to two different wall sockets, and parrellel them without problems.
Dr slack, not sure if the driver circuit will function as expected, but still working on that cheers.
Registered Member #54278
Joined: Sat Jan 17 2015, 04:42AM
Location: Amite, La.
Posts: 367
Using that formula posted by dexter [derived from i(t)=C dV/dt] i.e. C=it/V, we can come up with a useful, somewhat generalized, approximation formula for 'required capacitance for allowed ripple', by setting i=1 amp and V=1 volt. Then in the equation, the i/V component becomes 'one amp per volt' and when a specific 't' is plugged in (yielding a C-value), the equation specifies what capacitance (C) filters the output such that there is one volt of ripple for each amp of load current.
Assuming a line frequency of 60 Hz (T=16.67ms; T=period) and full-wave rectification, we can state that: C = (T/2) amps per volt. So, if we plug in t=(16.67/2)ms = 8.33ms, we get C=0.008333F or C=8333uF. This indicates that for a full-wave rectified 60Hz AC input, we will get one volt of ripple for every amp drawn from this filtered power supply when the filter capacitance is 8333uF. If we double the filter capacitance to 16,667uF we would get only a 1/2 volt drop at one amp, or 1 volt drop at two amps, etc.
The period (T) was divided by two since, for a full-wave output, this is the time between charging peaks of the full-wave rectified AC signal. You would adjust accordingly for 50Hz. Also the capacitance requirement is doubled if the input is only half-wave rectified.
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voltage triple capactance, with no reistor for tc
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