Battery Charger Power Draw Questions

[Saltyhog]

Originally Posted by goboatingnow

I think it might be useful to have a basic primer in regulator , voltage source , power supply battery charger etc, so that people can actually see what the currents voltages , and resistance etc note all resistance computations are output resistance of the "charger " circuit. well ignore the battery for the moment.

Lets hypothesise the following

Sure, I'll play


Billy has just arrived in college studying EE, he's played around at home with resistors and the like. His lecturer sets him a task, Billy, I have this special battery I want to charge , it has a nominal voltage of 5v and absorption voltage of 5,5 and a float voltage of 4.5V"

Futhermore he says "all you have is this 10V power supply ( say like a simple wall wart) and a potentiometer, a diode( more on that later), and a multimeter." oh and by the way , I want the circuit to indicate when its in "float mode,absorption & bulk", heres some indicators ( with a built in battery) and a switch ". " off you go" he says to Billy

So given the few components heres what billy does. ( see diagram) The diode he rekons will give him a bit of reverse polarity protection or something !. ( see later)

So , firstly Billy set the pot half way, reads the output voltage and as expected gets 5V

Q1. Whats the output impedance( resistance) of the source ( i.e. looking into billy circuit from the output side , ignore the wall wart for the moment).

Assuming the pot is set to half scale, the output impedance is 1/4 of the total pot value. Per your drawing Ro = 1/ (1/R1 + 1/R2). Note that the circuit is unregulated.


The lecturer says , OK lets test it with a load. The lecture connects a restive load, and billy notices that the output voltage decreases to 4V. " its not regulating" says the lecturer. Billy realises he has to be the control circuit. so he tweaks the pot to restore the output voltage to 5V, Billy has become the regulator !.

Q2. Has the output resistance of the regulator( i.e. billy circuit ) increased, decreased or stayed the same.

With Bill in the circuit, it is now regulated. Assuming the circuit stays in regulation, ie the load does not exceed the wall-worts' current spec, and Bill is able to react quickly enough to changes in the load (again, keeping the circuit in regulation), then the output resistance of the circuit is now 0. It matters not where Bill sets the pot, as long as the circuit stays regulated, the effective Rout stays at 0. When in regulation, R1/R2 have no bearing on the output impedance of Bill's circuit. R1/R2, along with Bill's feedback, merely provide the means for said regulation. I think this is where we disagree. How you accomplish regulation is immaterial. Whether you adjust R1-R2 pot as you show or by some other more efficient means, as long as it's regulated, the Rout stays at 0. That is theoretical perfect regulation. In the practical world, Rout will be some small value. In fact measuring Rout, via load testing both static and dynamic, is how you measure the quality of your regulation.

The lecturer gives him the battery, he says , "its fairly discharged, around 4V", he says. Billy connects to to his circuit ( after removing the other load and restoring his pot to mid position) . He immediately notices his voltage on the output is 4V. So in the spirt of being a regulator, he turns his pot, Interesting he can turn his pot ALL the way up, and the voltage is still 4V and he can turn it all the way down and the voltage is 4V, hmmmm.

This must be bulk mode, so he turns on that indicator.

Measuring current, he notices, that when his pot is all the way up , he's outputing 2 amps, while all the way down he's outputting nothing. He realises he cannot "regulate the output"

Q3, is the output resistance ( the resistance looking into billy circuit ) the same irrespective of the pot position even though the output voltage remains the same.
Q3.1 why can't he change the output voltage
Bill's circuit is no longer in regulation. The load has exceeded the circuit's ability to stay in regulation (due to exceeding the wall-wort's current supply abilities. The voltage on the battery is now being controlled by the battery, not Bill's circuit. Bill's circuit is now an unregulated current source.

While pondering what to do, Billy notices that with the pot all the way up and current flowing, output voltage of his circuit is recovering, gradually increasing and he notices the current is reducing

Q4: is this a constant current circuit ?
Q5: if not what is happening ( hint batteries don't really have a linear bulk mode)

No, it's an unregulated current source. A regulated current source, such as we'd find in a quality charger, would keep the bulk current constant regardless of the battery voltage increasing as it charges.

With a shock billy realises that the voltage has risen to 5.5V, , oh we're at the "absorption" point , he notices the voltage starts to climb to 5.6V and he can hear sizzling.

Quickly he turns his pot down, he notices the voltage and current falls and his battery voltage stabilised at 5.5V, He switches on the absorption indicator.

He finds he has to constantly adjust the pot "downwards" to keep the voltage from rising and he notices the current continues to reduce

Q6: billy is " regulating the circuit " to keep it in check, is this constant voltage mode
Q7:With such mode, is the output impedance of billys circuit, constant or varying. ? ( see previous posters contention)

Again, as long as Bill is regulating the circuit, the output impedance is 0.

Billy remembering the warning the lecturer mentioned ( " don't leave it in absorption mode forever"). Looks at his watch, notices two hours have passed and the current is down at 100 mA. He decides, "sure thats enough", Ill engage " float mode". he turns of the absorption lamp and turns on the float mode lamp.

Oh, he remembers, float is 4.5V , so he turns his pot down to try and set the voltage to 4.5V. To his surprise the voltage remains at around 5.2V, even turning his pot all the way down , he can't "set the float voltage at 4.5V

Q7: what can't billy set ( or regulate ) the float voltage to 4.5 ( hint the circuit cannot sink current per the diode, fwd its resistance is zero, reverse its resistance is huge)

Perplexed, billy notices that with his with his pot all the way down, the voltage of the battery is slowly dropping, after some time , he notices its at 4.4V, hah hah he say and he turns his pot till it read 4.5V and he notices some current flowing

After a while funnily he notices it reads 4.55V , no current is flowing and , even turning the pot down the voltage cannot be "dragged" down to the float voltage

Then after a short while, it drops just below 4.5v and he finds he can bring his pot up to 4.5V and some current flows again, and the cycle repeats.

Q8: Can Billys "regulator " actually force the battery into "float mode"
Q9: why not
Q10: Why do people say " my charger goes into float mode and keeps the battery voltage at XX"

Replace billy with a control loop and you have a smart three stage charger !!!

Understand this circuit and you understand smart chargers
Dave

WRT the float stuff, of course the charger cannot "force" the battery into float. The battery voltage is what it is. The battery doesn't know anything about bulk, absorption or float. My point, much earlier in this now tiring thread, is that 2 chargers could begin to switch modes in error due to their interaction with each other. It's similar to chargers being fooled by house loads. I don't understand why no charger manufacturer senses battery charge current right at the battery so that house loads don't play into the equation. That has nothing to due with constant voltage sources or current sources or Billy.

[Saltyhog]

Originally Posted by goboatingnow

That subject is both complex and simple, you either understand where the issues or you don't,

there are issues as to how regulation is done, source resistance matching, current steering, thermal runaway issues, switch mode interference, interference with logic decisions based on voltage , etc etc

its simple and complex.

You're using lots 'o fancy words there pardner . The original question was can you or can't you stack shore chargers. You're saying it to complex and you have to have an engineering degree to decide. Many sailors don't have that degree. I'm saying yes it can be complex, but ask the charger manufacturer if they recommend it or not. That's the simple (not complex) answer.

[Dockhead]

Originally Posted by goboatingnow

I think it might be useful to have a basic primer in regulator , voltage source , power supply battery charger etc, so that people can actually see what the currents voltages , and resistance etc note all resistance computations are output resistance of the "charger " circuit. well ignore the battery for the moment.

Lets hypothesise the following

Billy has just arrived in college studying EE, he's played around at home with resistors and the like. His lecturer sets him a task, Billy, I have this special battery I want to charge , it has a nominal voltage of 5v and absorption voltage of 5,5 and a float voltage of 4.5V"

Futhermore he says "all you have is this 10V power supply ( say like a simple wall wart) and a potentiometer, a diode( more on that later), and a multimeter." oh and by the way , I want the circuit to indicate when its in "float mode,absorption & bulk", heres some indicators ( with a built in battery) and a switch ". " off you go" he says to Billy

So given the few components heres what billy does. ( see diagram) The diode he rekons will give him a bit of reverse polarity protection or something !. ( see later)

So , firstly Billy set the pot half way, reads the output voltage and as expected gets 5V

Q1. Whats the output impedance( resistance) of the source ( i.e. looking into billy circuit from the output side , ignore the wall wart for the moment).

The lecturer says , OK lets test it with a load. The lecture connects a restive load, and billy notices that the output voltage decreases to 4V. " its not regulating" says the lecturer. Billy realises he has to be the control circuit. so he tweaks the pot to restore the output voltage to 5V, Billy has become the regulator !.

Q2. Has the output resistance of the regulator( i.e. billy circuit ) increased, decreased or stayed the same.

The lecturer gives him the battery, he says , "its fairly discharged, around 4V", he says. Billy connects to to his circuit ( after removing the other load and restoring his pot to mid position) . He immediately notices his voltage on the output is 4V. So in the spirt of being a regulator, he turns his pot, Interesting he can turn his pot ALL the way up, and the voltage is still 4V and he can turn it all the way down and the voltage is 4V, hmmmm.

This must be bulk mode, so he turns on that indicator.

Measuring current, he notices, that when his pot is all the way up , he's outputing 2 amps, while all the way down he's outputting nothing. He realises he cannot "regulate the output"

Q3, is the output resistance ( the resistance looking into billy circuit ) the same irrespective of the pot position even though the output voltage remains the same.
Q3.1 why can't he change the output voltage


While pondering what to do, Billy notices that with the pot all the way up and current flowing, output voltage of his circuit is recovering, gradually increasing and he notices the current is reducing

Q4: is this a constant current circuit ?
Q5: if not what is happening ( hint batteries don't really have a linear bulk mode)


With a shock billy realises that the voltage has risen to 5.5V, , oh we're at the "absorption" point , he notices the voltage starts to climb to 5.6V and he can hear sizzling.

Quickly he turns his pot down, he notices the voltage and current falls and his battery voltage stabilised at 5.5V, He switches on the absorption indicator.

He finds he has to constantly adjust the pot "downwards" to keep the voltage from rising and he notices the current continues to reduce

Q6: billy is " regulating the circuit " to keep it in check, is this constant voltage mode
Q7:With such mode, is the output impedance of billys circuit, constant or varying. ? ( see previous posters contention)

Billy remembering the warning the lecturer mentioned ( " don't leave it in absorption mode forever"). Looks at his watch, notices two hours have passed and the current is down at 100 mA. He decides, "sure thats enough", Ill engage " float mode". he turns of the absorption lamp and turns on the float mode lamp.

Oh, he remembers, float is 4.5V , so he turns his pot down to try and set the voltage to 4.5V. To his surprise the voltage remains at around 5.2V, even turning his pot all the way down , he can't "set the float voltage at 4.5V

Q7: what can't billy set ( or regulate ) the float voltage to 4.5 ( hint the circuit cannot sink current per the diode, fwd its resistance is zero, reverse its resistance is huge)

Perplexed, billy notices that with his with his pot all the way down, the voltage of the battery is slowly dropping, after some time , he notices its at 4.4V, hah hah he say and he turns his pot till it read 4.5V and he notices some current flowing

After a while funnily he notices it reads 4.55V , no current is flowing and , even turning the pot down the voltage cannot be "dragged" down to the float voltage

Then after a short while, it drops just below 4.5v and he finds he can bring his pot up to 4.5V and some current flows again, and the cycle repeats.

Q8: Can Billys "regulator " actually force the battery into "float mode"
Q9: why not
Q10: Why do people say " my charger goes into float mode and keeps the battery voltage at XX"

Replace billy with a control loop and you have a smart three stage charger !!!

Understand this circuit and you understand smart chargers
Dave

For those of us who really don't understand anything about EE:

Are you saying that the whole process of "smart regulation" is achieved by varying the voltage, based on current and voltage sensing? The output current of the charger takes care of itself, I guess, based on what the batteries will accept at the set voltage.

That's exactly what I had in my electrically simple mind.

It also confirms, I think, despite all the mass confusion in this thread, that there really is such a thing as a "bulk phase", and that this is what happens when charger is putting out full output and voltage is below the absorption set point -- so it really is "constant current".

[goboatingnow]

Originally Posted by Saltyhog

WRT the float stuff, of course the charger cannot "force" the battery into float. The battery voltage is what it is. The battery doesn't know anything about bulk, absorption or float. My point, much earlier in this now tiring thread, is that 2 chargers could begin to switch modes in error due to their interaction with each other. It's similar to chargers being fooled by house loads. I don't understand why no charger manufacturer senses battery charge current right at the battery so that house loads don't play into the equation. That has nothing to due with constant voltage sources or current sources or Billy.

We agree on everything ( well you have made a mistake around contact current and q9 ) or two except the nonsense that regulation has an output resistance of zero. Simply because billy is turning a potentiometer to keep the circuit in regulation does not reduce the output to near zero.

Two switch mode chargers will not interfere with each other in the manner you mention. The fact that the voltage is held up by the other is not the issue.

The issue has more to do with the difficulties of variation in output impedance and hence you get a lazy charger and a overworked one.

There a good paper from TI on the subject I was reading last week, I'll look it up tomorrow.

Personally I think you are confused by switched mode analysis.

Dave

[Saltyhog]

Originally Posted by goboatingnow

We agree on everything except the nonsense that regulation has an output resistance of zero. Simply because billy is turning a potentiometer to keep the circuit in regulation does not reduce the output to near zero.

A regulated voltage supply is attempting to emulate a perfect voltage source which by definition has an output impedance of 0. The fact that the output impedance is 0 ohms is exactly the point of voltage regulation. I'm not gonna draw you a picture or tell you a story but think about how you measure the output impedance of a power supply. Consider a black box DC power supply that you would like to characterize for its regulation. What do you do? First measure the open circuit voltage. Let's say it's 10 VDC. Then connect a load, say 9 ohms. Now you measure the loaded voltage to be 9 VDC. What is the output impedance of this power supply? (Do your own ohms law arithmetic) It's 1 ohm, pretty poor regulation! Now do the same test with a well regulated DC power supply. The unloaded voltage and the loaded voltage will be the same! 10 volts loaded or unloaded. Why? Because the effective output impedance of the regulated power supple is 0 ohms. That's the whole point, to emulate a perfect voltage source which has 0 output impedance. It's not nonsense. It's basic first year DC circuits.

Two switch mode chargers will not interfere with each other in the manner you mention. The fact that the voltage is held up by the other is not the issue.

The issue has more to do with the difficulties of variation in output impedance and hence you get a lazy charger and a overworked one.

There a good paper from TI on the subject I was reading last week, I'll look it up tomorrow.

Personally I think you are confused by switched mode analysis.

Dave

I really don't think this is helping anybody else. I think that maybe we are having a communication's failure here because this is basic stuff and I'd be REALLY surprised that you didn't "get it". Maybe the terminology is different. I'm trying to give you the benefit of the doubt.

[goboatingnow]

Originally Posted by Dockhead

For those of us who really don't understand anything about EE:

Are you saying that the whole process of "smart regulation" is achieved by varying the voltage, based on current and voltage sensing? The output current of the charger takes care of itself, I guess, based on what the batteries will accept at the set voltage.

That's exactly what I had in my electrically simple mind.

It also confirms, I think, despite all the mass confusion in this thread, that there really is such a thing as a "bulk phase", and that this is what happens when charger is putting out full output and voltage is below the absorption set point -- so it really is "constant current".

So , sort of.

Firstly a charger is not really in control of anything, in the bulk area of a charge curve the charger is typically running out regulation ( using a technique of limiting damage to the regulator called voltage foldback ) the charger is not a constant current source, its simply an unregulated current source.

As the current reduces as the battery charges the regulator can regain control upon reaching the set point then the charger modulates its resistance ( in the case of livers regulators ) to ensure that the voltage remains in regulation

In float mode the charger does nothing until the battery ( and only the battery) floats down to the float voltage ( which an LA will do by self discharge ) until that point the charger is effectively high impedance. Ie it's not really connected.

This is because unless specifically designed chargers can only source current not sink it. ( that was the point of the diode In the test circuit )


Hence once the battery floats down to the new set point then the charger can regain regulation.

The actual output circuit design isbt complicated. Unfortunately many simply can analyse basic circuit operation as we( EE ) a profession of building block engineers. ( my dad called then valve jockeys )

Dave

[Rustic Charm]

Hi guys,

Can some of u electrical gurus, tell me what my volts should indicate when my motor is running? I'm heading off into Bass Strait for Easter and today I though I noticed a drop in volts. It was charging at 13 volts. Is that ok? Or shoukd it be more?

[goboatingnow]

If it's charging at 13v the battery is or was quite discharged. As you proceed the voltage should rise, if not battery is suspect

Dave


Sent from my iPad using Tapatalk

[Rustic Charm]

Originally Posted by goboatingnow

If it's charging at 13v the battery is or was quite discharged. As you proceed the voltage should rise, if not battery is suspect

Dave


Sent from my iPad using Tapatalk

Thanks. I think I need to go for a run in her for a couple of hours then. She's got four 200ah hours and a 160 reserve.