# Bad Boy power factor and such..



## ga2500ev (Apr 20, 2008)

steelneck said:


> I just made a quick n dirty charging circuit of a large 70mf cap. and a bridge rectifier screwed to a piece of alum. as a heat sink, now trying to wrap my head around this very simple circuit.
> 
> ```
> ~ ----||---------|
> ...


The cap on the AC side functions as a resistor. However, the power factor is off because the current and voltage do not match as they do with an actual resistor.

I really must insist that before you continue on this path, to please take a read of this thread on this type charger and specifically the safety post that Lee Hart posted on the page 16 which the link should take you directly to.

ga2500ev


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## PStechPaul (May 1, 2012)

One real problem with this is that both ends of the battery pack will be floating WRT neutral, and thus will have a potentially lethal voltage to ground. You may be OK if you use a GFCI outlet, but it is still dangerous. You might try a doubler circuit, which will give you about 200 VDC, and maintains a neutral connection to the negative side of the pack:









This circuit has a PF of 551/1047, or 53%, and has an efficiency of 94%. Using 50 uF capacitors and removing the output filter, it draws 1.73A at 119V or 206VA and 55.9W, with output power of 55.7W, so PF=27%, and efficiency is close to 100%. But the output has a lot of AC ripple without the filter.

However, I see that you have 235 VAC, which may or may not have a neutral, depending on your local power system. Here is a simulation of your circuit:










It does not register exactly the same as you read, but close enough. The power factor in this case is 35/1473 or 2.3% or using input power, 70.2/1473 or 4.8%. The actual efficiency seems to be about 50%, but it's not really straightforward when considering current into a battery. There is 2.3 watts in C1 (which has 50mOhm ESR), and 9.4 watts in each diode, so that is almost 40 watts of losses.


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## steelneck (Apr 19, 2013)

ga2500ev said:


> The cap on the AC side functions as a resistor. However, the power factor is off because the current and voltage do not match as they do with an actual resistor.


Yes, that part i understand.



> I really must insist that before you continue on this path, to please take a read of this thread on this type charger and specifically the safety post that Lee Hart posted on the page 16 which the link should take you directly to.


I have read that thread and it was from there i got the idea. I also understand the safety issues being non isolated. I will not use a charger like that for the EV i am about to build, this is only something i threw together as a temp. solution since i do not own a real charger yet. I may use it to scare some life into some old half dead lead acids, maybe to fast charge my lawnmower battery (but not unattended). I started this thread mainly to understand this simple circuit better.

I will experiment with it more when i get my self a decent amp-meter. Today i only got a very cheap and simple volt/ohm meter. But i must say i find it quite amazing that a charger that can be used for just about any battery can be built this simple.

@PStechPaul: I am afraid your post was way over my knowledge.


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## dedlast (Aug 17, 2013)

Steelneck,
These are the important parts that you need to know from Paul's post (at least in my opinion):



PStechPaul said:


> One real problem with this is that both ends of the battery pack will be floating WRT neutral, and thus *will have a potentially lethal voltage to ground.* You may be OK if you use a GFCI outlet, but it is still dangerous...
> 
> Here is a simulation of your circuit:
> 
> The actual *efficiency seems to be about 50%*, but it's not really straightforward when considering current into a battery. There is 2.3 watts in C1 (which has 50mOhm ESR), and 9.4 watts in each diode, so that is *almost 40 watts of losses*.



Bill


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## steelneck (Apr 19, 2013)

dedlast said:


> Steelneck,
> These are the important parts that you need to know from Paul's post (at least in my opinion):


Yes i know it can be lethal, just as my 256V car battery pack once put together, but that was not the question here.

How do i measure PF for a circuit like this?

Without a battery load connected i get 250V DC (probably some kind of average), how big pack is it possible to charge? I guess it would do just fine with a pack of 96 V or so, but how high could it go within reason and how to calculate that?


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## PStechPaul (May 1, 2012)

You should learn to use LTSpice so you can determine how your circuit may work under various conditions. I would rather not give any more specific advice because of the danger involved with such a circuit. At least you can't hurt yourself running simulations (aside from eyestrain and carpal tunnel issues from using the computer). I have attached the ASC file for your FWB simulation, but I had to rename it TXT. The simulator program is free for the downloading at http://www.linear.com/designtools/software/

In general, this circuit will supply maximum current with large values of C1 and low cell voltages, but this will also have the worst power factor and efficiency. As you connect it to higher load voltages, such as a full pack, the current will decrease but more power will be transferred at better power factor and efficiency. The highest power transfer will probably occur when the load is about half the open circuit voltage, so I would guess you might get 2-3 amps on a 125V pack. 

View attachment 240AC-DC_60Hz_FWB_Simple.txt


*WARNING:* Also, remember that AC is much more dangerous than DC, as even 20-50 mA can cause ventricular fibrillation, which can be fatal and requires defibrillation. The same DC current will lock up your muscles and can stop your heart, but you may survive if you can break loose. Please, at least use a GFCI, and test it.


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## steelneck (Apr 19, 2013)

PStechPaul said:


> You should learn to use LTSpice so you can determine how your circuit may work under various conditions.
> 
> (...)
> 
> As you connect it to higher load voltages, such as a full pack, the current will decrease but more power will be transferred at better power factor and efficiency. The highest power transfer will probably occur when the load is about half the open circuit voltage, so I would guess you might get 2-3 amps on a 125V pack.


Thanks, but i rather test in practice than use closed source software.
You where first to actually give me an answer. I was wrong when i thought the PF would be better connected to low voltage batteries, i guess when i understand why, then i will be able to fill in the blanks by my self.


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## kennybobby (Aug 10, 2012)

steelneck said:


> ... i connected it only to one cell, now it went down to 54 W, but still 5 A at the same voltage, i thought the power factor would be better with this simple circuit the lower the battery voltage is, using a larger portion of the sine wave. This indicates something that i do not understand. Can someone enlighten me?
> 
> How long would you estimate the charging time would be for a single 3.2 V nom. 100Ah cell with this setup?


i estimate it would take 20 hours to charge a single empty 100 Ah cell with this circuit. 

i'm curious about why do you even care about power factor, the ratio of real resistance R to the complex impedance Z? But to answer your question the capacitor on the input makes this circuit almost entirely a reactive (complex) load so the power factor will be ~ 1/Xc or less, i.e. about 2.2%. 

How fast and how large a pack could this thing charge? That is determined and limited by the energy stored in the capacitor running off the 50Hz grid...which is less than 2 J.


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## PStechPaul (May 1, 2012)

Actually, I found that the input and output RMS current in this circuit are just about equal at about 5 amps for a 125V battery pack. And your estimate of 20 hours is dead on for a 100 Ah pack. The close equivalence of input and output current also applies for just two cells, and the efficiency remains good even when power factor is poor. So as long as you do not exceed the current rating of the source, you can increase the size of the capacitor to get more current. You may be able to use 200 uF (three 70 uF in parallel) to get about 15 amps and a 7 hour charge time.


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## steelneck (Apr 19, 2013)

PStechPaul said:


> Actually, I found that the input and output RMS current in this circuit are just about equal at about 5 amps for a 125V battery pack. And your estimate of 20 hours is dead on for a 100 Ah pack. The close equivalence of input and output current also applies for just two cells, and the efficiency remains good even when power factor is poor. ..cut..


Can you explain a bit further about the math behind those numbers?


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## PStechPaul (May 1, 2012)

I just changed the parameters of the simulation and read the values. But it is also intuitively obvious that any current going through the capacitor will go through the bridge, and the current out of the bridge can only go through the load (battery pack). This ignores any diode leakage, but that should be minimal. And the peak open circuit output voltage will be 240*sqrt(2), or about 339 volts, so there will be charging current if the battery pack voltage is less. The simulator takes care of the math. 

But you can get an idea of the current by using the formula for the reactance of the 70 uF capacitor at 60 Hz (you may need to use 50 Hz which will result in 15% less current):

X(c) = 1 / (2 * PI * F * C) = 1 / 26389 = 37.9 ohms

I = V / X = 240 / 37.9 = 6.33 amps

At 50 Hz it would be 5.28 amps.

If you measure the current and/or voltage you will need to use a true RMS meter for optimum accuracy.


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## steelneck (Apr 19, 2013)

PStechPaul said:


> ... any current going through the capacitor will go through the bridge, and the current out of the bridge can only go through the load (battery pack). This ignores any diode leakage, but that should be minimal.


Thank you Paul. So, if it draws 5 A from the wall, it will also charge at that amperage (minus small losses), only the voltage will drop to whatever is needed to get that amperage through. 

When i connected it to one single cell at 3.20 V i got 3.24 V and that at close to 5 A per above. That is 16.2 W and that in turn translates to 19.7 hours to charge a single 3.2V 100A (320Wh) cell. Now i understand this better. The little part i missed, making me not understand this simple circuit, was that the drawn current will be close to the same in/out. Thank you Paul!

The part about calculating the capacitor i knew, and it turned out to be very precise once i plugged in the circuit.


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