Try 20 to 200 uF as a likely range for pfc capacitors. Notice the C value in my simulation. (24.7 uF)
The website below examines an inductive load of several A, and calculates a value of 80 uF for pfc.
https://www.allaboutcircuits.com/te...nt/chpt-11/practical-power-factor-correction/
In simulation a small change can move the Ampere waveform several degrees. Suppose I start with a new load. The Ampere zero crossing wants to remain at a 'pinned' condition of 90 deg lagging or 90 deg leading. I usually need to try many C values, before I can find a range where I can make the Ampere waveform shift left and right, in relationship to the Voltage waveform.
By the way, I spoke the wrong way around in post #336, regarding current waveform advancing with inductive load, retreating with pfc capacitor.
Hi,
The thread is so long, so I don´t remember...
If you don´t mind, please give un update:
* is it a commercial project or a hobby project?
* What exactely do/did you want to achieve?
* Is it research or do you want to build something like an automatic compensation device?
* Do you still need a solution ... or how do you want to go on with the project?
Klaus
Brad,
If 6uF over-compensated the motor, and made it draw more current, how is it supposed to act when having a shunt of 20uF and more?!
Hi,
I´m very sorry for this ... the project result .. and your course delay.
***
Do you see any chance to come to a solution to prevent from loosing two semesters? Maybe with a completely new approach?
Klaus
So if all values above 1.5uF over-compensate the fan, and all tried series-added small caps of 50nF to 0.1uF did not affect at all the current drawing, I believe that some writers about fraud of shunt caps for domestic usage have right.
Maybe only inductive loads, powered at several horse power can be followed by application of shunt capacitance.
The best solution is to withdraw the whole course, and re-do it next year. I will be late for another two semesters. I've arrived to a dead end, and did not see the result of any compensation or power factor improvement.
Using same project title? Power factor measurement/correction using PIC?
Hi,
I didn't relate it to the forum nor the thread /title.
My concern is your course.
A new start - for me - means to start with new specifications, check several possible solutions and hopefully select the best one.
Klaus
You have a valid point.
Very small motors, e.g., fans with 60-100W ratings, have windings made of thin wires and they act more as a resistive loads. Most of the common fans are induction motors (you can make out because they do not have brushes and commutators or slip rings) and they already have a capacitor built in as an aid to start.
When a motor draws several Amperes, then it calculates to a small Henry value, say 10mH to 100mH. The C value must resonate with this L value, at 50 Hz.
If the C value is too small or too large, then excess Amperes are drawn. This means there is a narrow band where smallest Amperes are drawn. It is not easy to find by experiment in a simulation. The C value is easier to find by calculating it. However (as you state) it is not easy to know the L value either.
If I buy another bigger motor of 350W would that be sufficient to correct its power factor? What is the minimum nominal power for a motor to be able to apply a shunt capacitance across it?
Suppose you were to operate at a lesser AC voltage (say 16V)? Hook up a 50mH coil? That would create 16 ohms of load, or 1A. (These are approximate figures.)
Simulation with lesser AC voltage. Power levels are in the tens of W, rather than hundreds of W.
The correct size capacitor improves efficiency of driving the inductive load.
These simulations are not difficult to set up. They are an enormous aid to understanding. By playing with capacitor values, you will see the effect of low or high values.
Simulation with lesser AC voltage. Power levels are in the tens of W, rather than hundreds of W.
The correct size capacitor improves efficiency of driving the inductive load.
These simulations are not difficult to set up. They are an enormous aid to understanding. By playing with capacitor values, you will see the effect of low or high values.
Brad,
The 10 milli-ohms resistor at beginning of the circuit can't be neglected? It is hard to find such a small resistor
It represents some unknown theoretical resistance in the supply. Its value is not easy to determine. It is probably greater than 1/100 ohm. In house wiring it could be a few tenths of an ohm.
It also keeps the simulator happy. Otherwise it sends an error message because the capacitor is connected directly to the power source. Therefore it helps to install some theoretical inline resistance, so the simulator can calculate an RC time constant.
So this has nothing to do with real prototype circuit?
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