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Horvath Math for designing circuit for hydrogen generator. requires PULSE PEAK input 900w into primary for 22A 2uf capacitor discharges at 10khz. Use old style scientific calculator for the formulas. I just went through it. Secondary of TR2 has output diode, so as there being an on + off time for release of stuck ions from plates. I have flat nickel coated plates with flutes, knurled, with N + S magnets on back side. 3/8" acrylic plastic box, 400mesh nylon screen. The capacitor is 2uf as induction stove range type polypropylene capacitor high frequency type. .P type Mosfet paralleled as used for high side switching direct to spark plug, or N channel for regular use as applied to water cell. POH is 25% solution by weight, meaning gallon is from both together same time -not gallon input water. .22 pulse amps becomes 200 amps pulse to plates. fast recovery heavy amp rectifier, or Horvath just used 3/4" iron laminations block with no diode, to get the critical positive wave out. AS used on input high frequency . China sells open board 1000 watts inverter with option of 330v terminals (to get about 600watts necessary) Can wind your own 2 transformers TR1 + TR, and can build your own switching inverter. L primary = C / I squared divided by E squared. I = E X square root C / L. Magnet wire as gauge # circular mills = 750 X I X duty cycle from wires chart. 10khz, .09joules each, 22A peak pulses, 371uh primary, 900w, 2uf, #16 #18 gauge primary, 2gauge secondary, Ratio 10:1 turns, duty cycle secondary .5 , water cell receives 450 Watts. Peak amps or watts is not average watts over the period time, .etc.
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What is average power in watts drawn from input 12v battery required for 900 PEAK pulse watts CDI capacitor into transformer with secondary diode and 50% on time only? Does this assume 450 watts average power to load
To determine the average power drawn from a 12 V battery to supply a CDI (Capacitor Discharge Ignition) system delivering 900 W peak pulse power into a transformer’s primary, with a 50% duty cycle (50% on-time) and a diode on the secondary, and to confirm if this results in 450 W average power to the load, let’s analyze step-by-step.
Key Parameters:
Peak pulse power: 900 W into the primary during the on-time.
Duty cycle: 50% (0.5), meaning the power is delivered for half the cycle.
Battery voltage: 12 V (DC).
Secondary side: Has a diode, delivering power to the load.
Question: What is the average power drawn from the battery, and does this setup imply 450 W average power to the load?
Step-by-Step Analysis:
1. Average Power to the Load:
The primary side delivers 900 W peak power during the 50% on-time. For a pulse transformer, the instantaneous power delivered to the primary is transferred to the secondary (assuming an ideal transformer with no losses).
The average power delivered to the load on the secondary side is:
Pavg, load=Ppeak×D=900×0.5=450 W.
Pavg, load=Ppeak×D=900×0.5=450W.
This confirms that, in an ideal system, the average power delivered to the load is indeed 450 W. The presence of the diode ensures unidirectional current to the load during the on-time, consistent with the 50% duty cycle.
2. Average Power Drawn from the Battery:
In an ideal system (100% efficiency), the average power drawn from the battery equals the average power delivered to the load, as energy is conserved. Thus:
Pavg, input=Pavg, load=450 W.
Pavg, input=Pavg, load=450W.
The CDI system stores energy in a capacitor during the off-time and discharges it into the primary during the on-time to achieve the 900 W peak power. The battery supplies a steady average power to charge the capacitor over the full cycle.
3. Average Current from the 12 V Battery:
The average power drawn from the battery is:
Pavg, input=V×Iavg.
Pavg, input=V×Iavg.
Given V=12 VV=12V and Pavg, input=450 WPavg, input=450W, the average current is:
Iavg=Pavg, inputV=45012=37.5 A.
Iavg=VPavg, input=12450=37.5A.
Thus, the battery supplies an average current of 37.5 A to provide 450 W average power, which supports the 900 W peak power during the 50% on-time.
4. Real-World Considerations:
Real CDI systems have losses due to:
Transformer inefficiencies (core losses, winding resistance).
Diode forward voltage drop (typically 0.7–1 V for a standard diode, reducing output power).
Capacitor charging/discharging inefficiencies.
Switching losses in the CDI circuit.
Assuming a typical efficiency of 80% (a reasonable estimate for CDI systems), the input power required to deliver 450 W to the load is:
Pavg, input=Pavg, loadη=4500.8=562.5 W.
Pavg, input=ηPavg, load=0.8450=562.5W.
The average current from the 12 V battery becomes:
Iavg=562.512≈46.9 A.
