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Registered Member #4266
Joined: Fri Dec 16 2011, 03:15AM
Location:
Posts: 874
Hi, I'm trying to take a trickle current and send out large pulses. I've got the schematic below which have been mucking around with options, but it showing 11watts from a total of 225watt. The source is 450volt dc with a 100uF cap and a simulated 555 timer.
Is there anything to increase the output watts closer to the input and have it pulsed.
Registered Member #96
Joined: Thu Feb 09 2006, 05:37PM
Location: CI, Earth
Posts: 4061
What are you trying to charge? Pulse charging works well with NiMH and PbSO4 , not recommended for Li-Ion due to the plating issue. Some folks have made chargers which adapt their pulse widths depending on charge state to extend cell life, as lead acid don't particularly like excess voltage at full charge as it causes + plate corrosion.
Registered Member #72
Joined: Thu Feb 09 2006, 08:29AM
Location: UK St. Albans
Posts: 1659
There are two aspects to this
1) What can you do with the circuit to improve its power factor? 2) In what way is the simulator's power meter giving you wrong answers?
The circuit has so few elements, it's easy to analyse by hand, at least approximately for a limiting case or two.
Assume no power is lost in the transistor. case 1 - Assume the output pulses are short compared to the capacitance of C1, and the duty cycle is low, so that the voltage on C1 stays essentially constant. Now if (for instance) the voltage on C1 is settled at 300v, then there's 150v across R2, 300v across C1, so the input power is split 33% to R2 and 66% to the output. The power factor will be 66%.
case 2 - Assume the output pulses are long, or the duty cycle is high, so that C1 gets fully discharged. While discharged, the input power will split so that 900/950 ends up in R2, and only 50/950 ends up in R1, with a power factor of 0.053.
The actual power factor will be somewhere between these two limiting cases. Your meter shows 0.049, lower even than the lower bound I've just calculated, so something about what the meter is actually calculating is screwy, there's an invalid assumption somewhere. Plot the input and output current and voltage waveforms, and do the power sums by hand.
To improve your power factor to approach 100%, with a low duty cycle (<< 50/900), increase C1 to minimise the mean voltage, and so power lost, in R2.
If you have a duty cycle approaching or higher than 50/900, then this won't work and you will burn a lot of power in R2. Either decrease its value, or replace it with a boost converter, complicated but efficient.
Registered Member #4266
Joined: Fri Dec 16 2011, 03:15AM
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Posts: 874
I trying to combine a bunch pulses to a steady dc value, with high amps high voltage, save building or buying a transformer and series parallel them.
Assume no power is lost in the transistor. case 1 - Assume the output pulses are short compared to the capacitance of C1, and the duty cycle is low, so that the voltage on C1 stays essentially constant. Now if (for instance) the voltage on C1 is settled at 300v, then there's 150v across R2, 300v across C1, so the input power is split 33% to R2 and 66% to the output. The power factor will be 66%.
I've changed the timing and that makes 450ish across c1, so the output power should be close to 100% ? Adding inductors or resistors or changed values still tops out at 61watt out of 225, the 50ohm is the load which I can't change, and the power supply only supplies half a amp, so can't change that resistor, is there anything apart from changing c1 and the timming. Adding a second cap after the transistor makes the power factor go close to 1.0 but drops as it charges.
Registered Member #4266
Joined: Fri Dec 16 2011, 03:15AM
Location:
Posts: 874
No,I had to search to find what power factor means, I adding a second transistor were R1 is makes the power factor go to 1.0. Would R1 and R2 make a voltage divider so the cap only gets charged to that voltage?
Registered Member #72
Joined: Thu Feb 09 2006, 08:29AM
Location: UK St. Albans
Posts: 1659
Adding a second transistor after (instead of) R1 changeing the power factor significantly sounds like it's making some odd assumptions about what power factor means.
Graph the input and output currents and calculate the power manually. That way you will understand what the circuit is doing, where the power is going, without being confused by an opaque 'power factor' number that may, or may not be, relevant or what you think it is.
Registered Member #4266
Joined: Fri Dec 16 2011, 03:15AM
Location:
Posts: 874
Plotted the graph , if I charge for 10ms, and only discharge for 3ms, the cap should slowly reach 450volt ?
I plan to use a pic to do the switching, what would I need for the switch , transistor, mosfet, if it charged to 450volt, with the resistance 1-50 ohm? Was thinking about using motor start caps
Registered Member #72
Joined: Thu Feb 09 2006, 08:29AM
Location: UK St. Albans
Posts: 1659
No. If you charge for 10mS through 900ohms, and discharge for 3mS through 50ohms, I'm sure you can use the hint of conservation of current at the capacitor terminal to calculate the eventual voltage, in the limit of a very large capacitor. The total charge accumulating at that node must be zero, and the only degree of freedom is the voltage at that node to control the currents, and hence integrated charge, flowing through the resistors.
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Trickle charge pulse discharge
