# Improving Power Factor of switching load

1. Feb 14, 2012

### I_am_learning

I have a heater coil (purely resistive) of resistance say R. If I hook this resistance to the mains, I get power of
V_source_rms ^2 / R.
I don't want that much power. So, I use some power electronics switching device, and hence reduce the effective V_load_rms to a lower value. So, my new output power is
Now, From the point of view of source,
Since, there is no other paths for current, I_load_rms = I_source_rms
So, it appears, If I want to consume reduced power by using the same heater, by method of switching, then my power factor becomes poorer.
Is there any way I can avoid it, or can compensate for it?
thanks.

2. Feb 14, 2012

### Staff: Mentor

Yes. You would use a switching power supply configured as a Power Factor Correcting circuit. The simplest is a Boost-Buck topology, where the Boost stage follows the AC input voltage sine wave, and converts just enough energy to drive the load. The Buck stage is used to reduce the output voltage to whatever you want for your application.

You can read about Power Factor Correcting power supplies at wikipedia or at many manufacturer's websites (the companies that make switching power supply components). ON Semiconductor has a nice application note:

http://www.onsemi.com/pub/Collateral/HBD853-D.PDF

.

3. Feb 16, 2012

### I_am_learning

Hi, berkeman,
I tried reading the document you linked, but I couldn't quite follow it and I couldn't find out some specific circuits for my need.
Anyway,
I dug around and have come-up with a simple solution to my problem.
My gross need was to have a variable heater power with as good a power factor as possible.
So, I came up with this simple boost converter as my solution,

When the switch duty-cyle is around 1%, the effective rms voltage across the Heater is same as the source voltage, i.e. 230V rms. At such condition, the heater power = 230^2/1000 = 52.9 watts.
This is the minimum achieveable power.
Now, as the duty cylce is increased, the effective load RMS voltage is also increased.
I simulated this ckt. using L=0.1 mH, and the pulses to the switch applied at 50Khz,and at 99% duty cycle.
I got,
Effective Vrms across heater = 2624V !
Heater Power = 6.8 Kw
Rms Source Current = 32.81 A
power factor = .91

So, it appears (after some playing around), that I can acheive power control from 52 watts to around 7 Kw with very good power factor.
Only bad thing, I find in this scheme is that, I need heater coil rated for some 3 Kvs and the switching device needs to be rated for same KV as well.

What are other draw-backs, and what could be better options?

4. Feb 17, 2012

### I_am_learning

It seems quite weird to me that, I can't find similar scheme for variable heater power with good power factor, but that works on lower voltage levels.
I came across SMPS with active/passive Power Factor Correction, but all of them seem too complex. Isn't there any simpler scheme, like the one I came up with?

5. Feb 17, 2012

### jim hardy

Since it's a heating element you might try this trick.......

take advantage of its thermal capacity.

You can buy an electronically controlled relay that turns on at precisely the zero crossing.
With resistive load that's what you want because it makes no RF interference and there's no inrush.
(For an inductive load you'd want to switch at AC peak, but that's another story - and you can buy those relays also...be sure you get a zero switcher)

This one's overkill but you get the idea....
http://parts.digikey.com/1/parts/760211-relay-ssr-40a-220vac-zero-cross-g3na-240b-dc5-24.html

Now since it takes a few seconds for your heating coil to cool or heat, you could switch your relay on and off a couple times per second and it should be not apparent that switching is going on.

A simple 555 PWM controller driving an electronic relay should do the job. Power it from a "wall wart" transformer for isolation and safety.

Also sharpen up your pencil and slide rule - since power as you said is V^2, you'll want a non linear relation between duty cycle and dial setting to make linear heat output. Clever arrangement of pulse width adjustment resistors can give you that, if you want to go to that level of detail.

6. Feb 17, 2012

### I_am_learning

hi jim,
So, you are saying about integral cycle switching using relays, like
one cycle on/one cycle off if I want medium power
5 cylce on/ one cycle off if I want high power
etc.
But don't this scheme also have poor power factor, because, all you are doing here is effectively reducing Vrms across the load, which as previously calculated would inherently create lower power factor when load-current = source-current.
Although Vrms is calculated over long duration in this case.

I want the heater to have good power factor because it is used as dummy load for "Electronic Load Controller" used in Micro-hydro power. So, I don't want to load the generator with excessive KVA in addition to the minimal required KVA for given heater KW.
Here is a brief description of what is "Electronic Load Controller"

7. Feb 17, 2012

### jim hardy

That is a very non-standard use of the term "power factor".

And i dont quite understand it...

In the standard terminology "Power Factor" refers to the difference between (Volts X amps) and (Watts) as you suggested,
but the reason they're not equal is not because of unequal volts or amps,
instead they're not equal because volts and amps are not necessarily in phase.
Watts is Volts X Amps X cosine(angle between volts and amps)..

Now in a resistive load volts and amps ARE in phase, so angle between them is zero, and cosine(0) = 1.
Which makes V X A = Watts.
Power factor is cosine(angle betwen V and A) and is 1 for pure resistive load.

What you have defined as "power factor" would be more correctly called "duty cycle", and that fits with your idea of switching load on and off.

One of us is confused. For any pure resistance load like a heater , KVA = KW because there's no phase shift.

And as you pointed out, since I (load) = I(source), KW(load) will = KW(source) because when current is zero , Volts X Amps = 0 in both of them . So even YOUR power factor P(load)/VA(source) will be one.
(AHA - that's it - zero in denominator while switch is open. )

Can you figure out which of us is the confused one and clarify?

thanks, old jim

8. Feb 18, 2012

### Mike_In_Plano

You're a sharp feller, Jim, (hats off). I was shaking my head with all the nonsense about adding power factor pre-regulators and switching supplies. Zero voltage switching is precisely what you use for this app. You simply turn your PWM frequency down.

If you're doing anything more advanced than a bang-bang controller, you'll probably do better with a micro for control. Otherwise, the control system time constants will be difficult to create using RCs.

If you do want a PWM generator, a very versitile one is the TL494, the great-grandchild of the Silicon General SG1524. This great little chip is cheap and versitile hence it's survival since the 70's.

9. Feb 18, 2012

### I_am_learning

Hi jim,
Peoples sometimes really don't give a damn, if they feel the other person is talking nonsense or simply if they couldn't quite get what they want to tell. I really appreciate the attitude you have shown here.

But I am pretty sure, you are at the confusion here.

The thing in Bold in above quote would only apply if the circuit contained only linear elements like resisters, inductors, and capacitors. However, we have here, a Resistance which is placed behind a switch. And that makes this circuit non-linear, and that makes your statement inapplicable.

To demonstrate how switching load gives poor power factor, lets look at a simple half-wave rectifier supplied Resistive load.
_____|\|____/\/\/\/\_____
|........|/|.........................|
|.....................................|
AC..................................|
|_____________________|

Say, AC source voltage = 10 V peak
Diode drop = 0V
Heater resistance = 10 ohms
So, during the positive half cycle, current flows and is in-phase with source voltage and has a peak of 1A.
During the -ve half cylce, current = 0.
Since in a whole cycle the current isn't sinusoidal, we can't simply apply P = V*I*cos(phi) formula, because phi is defined only for sinusoidal current.

So, lets work out the power factor of the load in this configuration.
Whatever be the wave-shape of current, the power dissipated in resistive load is Irms^2*R
So, in our case, I is sinusoidal for +ve half cylce with amplitude 1A and 0 for negative half cycle.
So, we can compute Irms = 0.5A. (college mathematics)
So, P = Irms^2*R = 0.5^2*R = 2.5 watts.
Now, VA input from source is simply, VA = Vsource_rms*I_source_rms
Obviously, I_source_rms = I_load_rms = 0.5A
Vsource_rms = 7.07 Volts (for 10V peak sinusidal wave)
Therefore, VA = 7.07*0.5 = 3.535
Thus we have, pf = 2.5/3.535 = 0.707

Also, VA = V_source_rms*Irms (remember, I_load_rms = I_source_rms = Irms)
Thus, pf = V_load_rms / V_source_rms
Thanks.

Last edited: Feb 18, 2012
10. Feb 18, 2012

### jim hardy

You sure did make sense .. !! wow...

I dont think this is right for rectified current.

Just as RMS is the (square root of) integral of sin^2(wt) over an entire period, divided by the period,

Volt-Amps should be the integral of Volts X Amps over the entire period, divided by the period.

I think you need to integrate the volt-amp product, not take product of separate volt and amp integrals.
If you integrate them separately you dont multiply volts by zero amps when you should.
Vrms integral probably was evaluated from 0 to ∏/4 and quadrupled, that's what we do for a whole cycle
but VA product integral would only have been doubled because second half cycle is zero..
so in line quoted above you're multiplying Irms by a larger number than you should.

Does that make sense?

Check my logic. Your math is sharper than mine, surely, please check integral of product vs product of integrals... forty years ago i could have done it quickly.

my alleged brain is throbbing over this one. I edited it five times to get it short enough to read...

Please advise --- i think you know i'll admit when i'm wrong. The goal is to get our thinking straight.

old jim

Last edited: Feb 18, 2012
11. Feb 18, 2012

### jim hardy

@ Mike -

i LOVE the TL494.

We used it in an analog computer for time division multiplier/divider. Good for calculating fluid flow from dp cell across a venturi . Bernoulli.

I think Yungman woud have been proud of us. We sure could have used his help..

old jim

Last edited: Feb 18, 2012
12. Feb 19, 2012

### I_am_learning

Or more precisely, RMS of a quantity is --Square root of integral of of quantity^2 over its entire period divided by the period squred.
In our case, we need to go by the second definition to find I_rms and V_load_rms, because the wave aren't sine-waves. It resembles a sine wave for +ve half-cycle but in negative cycle, its 0. So, it don't qualify as sine wave.

When we do, what you have suggested, i.e. Integral of Volts*Amps, what we get is the real power or Active Power or Watts.
Remember, P_active_instantaneous = V_instantaneous*I_instantaneous , whatever be the nature of the V and I wave.
Then P_active_avg = Average of P_active_instantaneous over a period = Integral of V*I / period.
That is what Watt-meters do. They have two coils, Voltage Coil and Current Coil and the deflection is proportional to product of current in those coils. i.e.
Deflection_instantaneous = V*I (both instantaneous)
The inertia of the mechanical system automatically does the averaging Job, and we get the reading of Active Power.

We generally, measure VA in lab by simply measuring, Irms using Ammeter and Vrms using voltmeter and multiply them.

Also in the previous example of rectifier, think in this way,
from the perspective of source, the load is consuming 0.5A (Irms) current. If the load were purely resistive (no Diodes present), it would have indeed possible to provide,
V_source_rms*0.5A = 7.07*0.5 = 3.53 Watts of real power.
But the load is consuming only 2.5Watts here.
The load is consuming only 2.5 watts but putting current_demand on the source similar to 3.53 watts load, that's why it has poor power factor.

I really hope, I got it straight now.
Thanks.

13. Feb 19, 2012

### jim hardy

Well, when you threw in the rectifier it became no longer sinewaves.

I have never seen the term "Power Factor" used in non-sinewave before, so if you define it as 'difference between sinewave and non-sinewave apparent power" for your project i sure dont object.

In fact --i can't object because i went looking for a "scholarly source" to refute you and guess what i found -
U of Massachusetts is in complete agreement with you.
www.faculty.umassd.edu/xtras/catls/resources/binarydoc/1851.ppt

see slide twenty............ i think it's your math exactly.

I Stand Corrected, SIR.
If i learn something every day i could someday know something...Thanks for your patience and persistence.
Hmm........ i'll remember this lesson by "beware of geeky waveforms bearing power factors"..

Now - back to original question -
IF you constrained the zero-crosing switch i suggested to passing only whole cycles, ie no rectification just on for n cycles off for m cycles n and m both [strike]even,[/strike] err make that integers
Then what would be power factor?
I say one point zero.
Would that fix your "low power factor" quandry?

old jim

Last edited: Feb 19, 2012
14. Feb 19, 2012

### I_am_learning

hi,
I don't think that will work,
Here is what I think,
Just like a half-wave rectified wave form, that resembles sine-wave for +ve half cycle but 0 for -negative half, we can't call it sine-wave.
Similarly, in this case, a wave that resembles a sine wave for say n cycles but 0 for m cycles isn't a sine-wave on the long run!
So, we need to find the Vrms_load = Square root of [ Integrate V_load^2 for (m+n) cycles / (period of (m+n) cycles).]
We still need to control Vrms_load and as previously discussed (p.f. = V_load_rms/V_source_rms), this inevitably degrades p.f.

The point in trying to improve power factor is reducing I_source_rms so as to minimize heating losses in the source.
If you increase n and m cycles to very long value, such that, the heating done by previous n cycles (in the source) would have already cooled when we come at next n cycles,
then I think this method would work even though the p.f. would be still be poor.
But, due to its nature of application (i.e. it needs to work as dump load for Electronic load Controller), we can't be that slow at switching.

I thought about it again, and I now think, if you increase n and m, you are not only increasing cooling time but also increasing heating time, so I think the peak temperature you would push the source would be higher instead. Have your say.
Think this way,
There is a good old incandescent bulb.
You constantly switch it on/off , 1 sec ON / 3 sec off. (25% duty cycle)
Another scenario, you constantly switch it on/off, 10 sec on / 30 sec off. (again 25% duty cycle)
Its similar to what we are doing to the source by using integral cycle switching as you suggested.
Now, can you think, which scheme the source would be happy with? Think about peak temperatures.

15. Feb 19, 2012

### jim hardy

Well the source has to be sufficently stout for its load.

If that 1.5 KW load is run at 70.7% duty cycle, half power, it still needs a 1.5 KVA source. You can't call it a 750 watt load and wire it to a 0.75kva source.

Are we both saying that?

In my power plant we had a 400KW resistor bank that maintained steam pressure in a tank. It was controlled by a variable duty cycle zero switching controller, 480 volts 480 amps. It did N cycles on, M cycles off to make duty cycle N/(M+N)

By your definition, there is a "power factor" for my controller that's related to its duty cycle?
And it would be different yet if rectification occurs?
I have to pick up this terminology for it's new to me.

Thanks again,

old jim

16. Feb 20, 2012

### jim hardy

""The point in trying to improve power factor is reducing I_source_rms so as to minimize heating losses in the source.""

Well - DC switching regulators work on duty cycle principle, passing >full load current at near zero voltage drop and only part of the time, which minimizes their heating.

All the current that goes through a source heats it.
So minimize both the voltage drop inside it and the time current flows through it.

Have fun - and i learned a new term !

old jim

17. Feb 20, 2012

### I_am_learning

Yes, I am trying to say that, but your maths don't quite agree.
For that 1.5 KW load, if we run it at 70.7% duty-cycle, Then it would run at .707 * 1.5 KW. To, run it at 750W the duty-cycle would need to be 50%. If we make dutycyle be 50% then, V_load_rms = .707 * V_source_rms
(remember you need to calculate V_load_rms by integrating over the whole n+m cycle, if you do integral cycle switching)
then, pf. would be .707, so, you need to provide, 750/.707 = 1.060 KVA.
So, in your words, you can't call that 750 W load, and use 750 VA source, you need to use 1.060 KVA source becasue, your load assembly has quite a poor pf.

For your second case, Yes, I would say it has a power factor less than unity.
But, I now come to realize it that, when the switching times are long, like minutes or hours, a different term is used, called the load factor. I believe you already know about this term.
Its also a measure of how much energy a source would be able to provide at if the various loads consumed more-or-less a constant power over how much energy its providing in the current diversity of loads.
I find it similar to pf, the difference being only the scale of time.

18. Feb 21, 2012

### jim hardy

I think i am beginning to understand your thinking.

and switched ON for 707 cycles, OFF for 293 (that is, on for 70.7% of time)

what would be duty cycle,
and power delivered to load ?

I think we're almost there.
I grew up unaccustomed to "Power Factor" applied to anything but sine wave. To find same word used for non-sine waveform correction was a surprise.

Thanks again,

old jim

19. Feb 21, 2012

### I_am_learning

I would call that 70.7% duty-cycle.
And it would make the ratio V_load_rms / V_source_rms = Sqrt(.707) = .84089
And make the power delivered to load = .707 * 1.5 KW = 1.06 KW

Additionally, the p.f. would be .84089 . So, KVA demand = 1.06 / .8409 = 1.26 KVA.
So, the source still needs to be of 1.26 KVA to supply 1.06 KW.

20. Feb 21, 2012