How do NiCad and NiMh chargers detect -deltaV to stop charging? (SOLVED)

Voidugu:
As you know, one way to detect whether a NiCad or a NiMh battery is fully charged is when the battery voltage starts dropping (-deltaV). My question is: How do chargers detect such a drop in voltage while the battery is charging? The only way to measure the battery voltage is to stop the charging process and measure the voltage right? How would a microcontroller be programmed to detect such a voltage drop? (ie charge for 2 mins, stop charge, measure voltage, store value, charge for 2 mins, stop charge, measure voltage, store value, compare the present value to the value obtained 2 mins ago)

Can someone please educate me on this matter?

Thanks in advance

That is called a 'peak' charger and it is described in section 5 of the listed link. You just contiguously monitor the battery terminal voltage while charging in a 'rapid' mode and there will be a steady increase in the battery terminal voltage until it reaches a 'peak' value and quickly falls back about 10mv. It's up to your monitoring software to keep track of the raising voltage to determine when 'peak' value has been reached and starting to fall, as it's not a specific voltage value but rather the change in direction of the voltage during the charging process.

Lefty

Something that may be of interest (very loosely) is that I was able to revive some Li-on cells that had dropped down to 0.2 volts and wouldn't charge.
PSP 110 Battery Revive. | Ajb2k3's Weblog Not very informative but it did work.

What i am asking is how do the chargers to this. ie how can it be done with a microcontroller. I would like some technical details. I mean, the "loop" that the microcontroller would have to repeat in order to figure out when this -deltaV has been reached and some circuitry that would be required.

@Mr.Retroplayer:
You kind of confused me there. If i understood correctly, you are basically saying that by knowing the voltage you provide to the battery through the series resistor and the voltage drop across the resistor in series with the battery you can calculate the EMF of the battery ?? (ie EMF of the battery = voltage supplied to the resistor and battery - Voltage across the series resistor)

Wouldn't this equation give me the voltage that is used to charge the battery?
Would it give me the actual EMF of the battery in order for me to log the data down and be abple to determine the -deltaV?
Wouldn't i have to stop the charging process, measure the open circuit voltage of the battery and the continue with the charging process in order to measure the EMF of the battery?
By the way, i know that for lipos, its mostly the cell voltage that changes (drops) while the battery is drained (increase in internal resistance has a smaller effect). Does this not apply for NiCads and NiMhs? Does internal resistance vary greatly while the EMF of the battery remains more or less constant?

@Mr.Lefty:
How do you continuously monitor the battery terminal voltage while charging? What to you mean the change in direction of the voltage?

Again thanks for the help people.

@Mr.Lefty:
How do you continuously monitor the battery terminal voltage while charging?
Continuously monitor = continuously measure = put your DVM across battery while charging and write down voltage every min, see that terminal voltage gradually increases over time as it accepts charge current, reaching a peak value and then falling back slightly, that is end of charge indication.

What to you mean the change in direction of the voltage?

1,2,3,4,5,6,7,6. The transition from 7 back to 6 is a change of direction relative to prior sequence.

Lefty

Voidugu:
@Mr.Retroplayer:
You kind of confused me there. If i understood correctly, you are basically saying that by knowing the voltage you provide to the battery through the series resistor and the voltage drop across the resistor in series with the battery you can calculate the EMF of the battery ?? (ie EMF of the battery = voltage supplied to the resistor and battery - Voltage across the series resistor)

Wouldn't this equation give me the voltage that is used to charge the battery?
Would it give me the actual EMF of the battery in order for me to log the data down and be abple to determine the -deltaV?
Wouldn't i have to stop the charging process, measure the open circuit voltage of the battery and the continue with the charging process in order to measure the EMF of the battery?
By the way, i know that for lipos, its mostly the cell voltage that changes (drops) while the battery is drained (increase in internal resistance has a smaller effect). Does this not apply for NiCads and NiMhs? Does internal resistance vary greatly while the EMF of the battery remains more or less constant?

No. Well, yes.... lol.

I am talking about understanding the relationship between voltage, current, and resistance in a circuit series and parallel circuit properties, and Thevenin's thereom. But you don't need to have an engineering understanding of all that. Just know that your charger is providing a voltage. And in a series circuit, all components will drop a portion of that voltage and that will always add up to the total voltage.

So let's say you have 2 resistors and a 9V battery. Measuring the voltage across one resistor reads 3V. Wtihout even measuring the voltage across the other resistor, we already know it is 6V because all the drops must equal the total voltage.

So we put a small resistor in series with the battery and measure the voltage drop across the resistor. We subtract that from the charger voltage and we know the voltage across the battery. We incidently also know the current through the circuit doing this. Ohm's law.

I think you are overcomplicating it and getting hung up on dVt.

To summarize, you are measuring the volage across the added, known-value, resistor. You subtract that voltage from the known charging voltage.

A great resource is The Battery Handbook

Retroplayer:

Voidugu:
@Mr.Retroplayer:
You kind of confused me there. If i understood correctly, you are basically saying that by knowing the voltage you provide to the battery through the series resistor and the voltage drop across the resistor in series with the battery you can calculate the EMF of the battery ?? (ie EMF of the battery = voltage supplied to the resistor and battery - Voltage across the series resistor)

Wouldn't this equation give me the voltage that is used to charge the battery?
Would it give me the actual EMF of the battery in order for me to log the data down and be abple to determine the -deltaV?
Wouldn't i have to stop the charging process, measure the open circuit voltage of the battery and the continue with the charging process in order to measure the EMF of the battery?
By the way, i know that for lipos, its mostly the cell voltage that changes (drops) while the battery is drained (increase in internal resistance has a smaller effect). Does this not apply for NiCads and NiMhs? Does internal resistance vary greatly while the EMF of the battery remains more or less constant?

No. Well, yes.... lol.

I am talking about understanding the relationship between voltage, current, and resistance in a circuit series and parallel circuit properties, and Thevenin's thereom. But you don't need to have an engineering understanding of all that. Just know that your charger is providing a voltage. And in a series circuit, all components will drop a portion of that voltage and that will always add up to the total voltage.

So let's say you have 2 resistors and a 9V battery. Measuring the voltage across one resistor reads 3V. Wtihout even measuring the voltage across the other resistor, we already know it is 6V because all the drops must equal the total voltage.

So we put a small resistor in series with the battery and measure the voltage drop across the resistor. We subtract that from the charger voltage and we know the voltage across the battery. We incidently also know the current through the circuit doing this. Ohm's law.

I think you are overcomplicating it and getting hung up on dVt.

To summarize, you are measuring the volage across the added, known-value, resistor. You subtract that voltage from the known charging voltage.

A great resource is The Battery Handbook

Well to be fair you also might be making it more complicated then it really is. In context with a nimh peak charger there is no reason for a series resistor to be used to detect when 'peak' voltage has been reached, in fact I'm not sure you could even detect it that way, just measurement directly across the battery terminals will work. A series resistor's voltage drop tells you about the value of current flowing into the battery rather then it's gradually increasing terminal voltage as it accepts a constant current charge from the charger. Now there might have to be a voltage divider wired across the battery terminals to keep the measurement value within say an arduino analogRead() range and scaled back up in software to track when peak voltage has been reached, but that is not effected by the current flow through the battery.

Lefty

retrolefty:
Well to be fair you also might be making it more complicated then it really is. In context with a nimh peak charger there is no reason for a series resistor to be used to detect when 'peak' voltage has been reached, in fact I'm not sure you could even detect it that way, just measurement directly across the battery terminals will work. A series resistor's voltage drop tells you about the value of current flowing into the battery rather then it's gradually increasing terminal voltage as it accepts a constant current charge from the charger. Now there might have to be a voltage divider wired across the battery terminals to keep the measurement value within say an arduino analogRead() range and scaled back up in software to track when peak voltage has been reached, but that is not effected by the current flow through the battery.

Lefty

Perhaps you could explain how you are supplying 7.2V to charge a 7.2V battery and with only the battery in parallel with the 7.2V supply you would expect to read anything different than 7.2V? That makes no sense whatsoever.

Yes, you are measuring current. With Ohm's law, you have three components: voltage, current, and resistance. You need to know two of them to solve for the last unknown. So you use a known resistance, measure the voltage drop (which give you current) and calculate that the voltage across the battery by subtracting that voltage drop from the 7.2V using Thevenin's thereom. This is considered a half bridge circuit, where a wheatstone bridge would be a full bridge.

THAT makes it sound complicated. The OP doesn't really have to understand how all that works to apply it.

But to think that you would expect to ever measure anything different from a 7.2V supply in parallel with one component (the battery) makes no sense at all. And THAT was what the OP was (rightly) confused about.

BTW, I am an electrical engineer in the aerospace and defense business. I would hope that I could understand something as simple as a battery charger. Just saying...

But to think that you would expect to ever measure anything different from a 7.2V supply in parallel with one component (the battery) makes no sense at all. And THAT was what the OP was (rightly) confused about.

BTW, I am an electrical engineer in the aerospace and defense business. I would hope that I could understand something as simple as a battery charger. Just saying...

Again a nimh 'peak' charger is operated in a 'rapid' constant current charge (typically at a 1C rate), not with a constant applied voltage. As such, it's the detection of the peak of the terminal voltage (and then a slight regress) while a constant current is being applied that is 'signal' being used to terminal the constant current charge, as cell temperature will start to rise rapidly if the constant current is allowed to continue past that cell's 'peak signature signal'. This signal is only detectable by reading the terminal voltage as a constant current does not contain the 'signal signature' that peak detection requires because, well because it's a regulated constant current. :wink:

Lefty

and... just how exactly do you think they maintain a constant current? Again by doing exactly what I just said for the feedback (which is exactly what the OP was asking about.) :wink: Anyway, wasn't looking for an argument, just looking to help the OP with his question. Do with it what you will, but all I can say is that the NiHM chargers that I have built work exactly as they should.

@ Mr.Lefty
I know for a fact that in order to charge a battery that has an open circuit voltage of lets say 6 volts, at any current, the voltage you supply to the battery (aka power supply voltage) has to be higher than the 6 volts of the battery. If you connect a voltmeter across a charging battery then you would essentially measure the voltage supplied to the battery by the charger in order for it to charge and not the actual EMF of the battery.

@ Mr.Retroplayer
Say i have a 9 volts supplied to the circuit you are suggesting. The voltage drop across the resistor is 1 volt and the current through it (and the battery etc) is 1 ampere. Then as you suggest the voltage across the battery would be 8 volts, and as Mr.Lefty said earlier, this could be simply done using a voltmeter across the battery, since i am not interested in the current going to the battery (there is no need for the shunt resistor). The voltage across the battery is the voltage provided by the charger to the battery and not the EMF of the battery. I want a way to measure the EMF of the battery, if there is one, other than disconnecting the battery and measuring the open circuit voltage.

Voidugu:
@ Mr.Lefty
I know for a fact that in order to charge a battery that has an open circuit voltage of lets say 6 volts, at any current, the voltage you supply to the battery (aka power supply voltage) has to be higher than the 6 volts of the battery. If you connect a voltmeter across a charging battery then you would essentially measure the voltage supplied to the battery by the charger in order for it to charge and not the actual EMF of the battery.

@ Mr.Retroplayer
Say i have a 9 volts supplied to the circuit you are suggesting. The voltage drop across the resistor is 1 volt and the current through it (and the battery etc) is 1 ampere. Then as you suggest the voltage across the battery would be 8 volts, and as Mr.Lefty said earlier, this could be simply done using a voltmeter across the battery, since i am not interested in the current going to the battery (there is no need for the shunt resistor). The voltage across the battery is the voltage provided by the charger to the battery and not the EMF of the battery. I want a way to measure the EMF of the battery, if there is one, other than disconnecting the battery and measuring the open circuit voltage.

I think you are not understanding the principles used in constant current charging. I never stated that a specific voltage value reached is the signature of reaching 'peak' charge, just that the terminal voltage at the battery terminals will be a seen as a gradually increasing one as it accepts the constant current charge, but there will be a point where the voltage generated by the charger will reach a peak value and then fall back a little all the while maintaining a constant current charge value, and this is the signature that the battery has reached full charge. You seem stuck on charging with a fixed voltage which is just not what happens when using 'rapid' charging methods that utilize constant current charging method and using peak voltage detection to end the charging process. The document I've linked twice is a good source of information about the subject.

Lefty

Ok... done arguing. Good luck.

With that last clarification i think i understand what you are saying mr. Lefty. Please correct me if i am wrong:

You are basically saying that as the battery charges up, its voltage increases. A constant current charger will try to maintain a constant charging current and therefore will also increase its charging voltage more or less in the same manner as the battery itself increases its own voltage (since it charges up). When the battery is at full charge, its voltage (aka EMF) will start dropping. The charger will keep the charging current constant and will therefore have to decrease its battery charging voltage in a similar manner, following the voltage drop (-deltaV) of the battery. By measuring the drop in voltage of the constant current charger we can identify a drop in the voltage and therefore stop the charging process since the battery is full.

Am i correct?

Am i correct?

Yes. And that charging voltage and it's little peak shift have nothing to do with the actual terminal voltage of the battery after you disconnect it. It's a small dynamic voltage change during the charging process to signal that it's time to terminate the charge process. Other means of stopping a constant current charge cycle is to have the temperature sensor inside the battery pack as the temp will start a rapid rise upon full charge being reached and that is how a lot of the early 'rapid' chargers worked. But of course that worked with only higher priced battery packs that had the proprietary temp sensor (compatible with the charger) in them.

Lefty

Retro,

Would you be so kind as to explain for us how that litte magic black box that you are calling a "constant current charger" works internally? Based on the OP's question, I believe THAT was what he was asking. I would assume that if he already HAD a constant current charger he wouldn't need to know what he is asking.

Silly me assumed that he wanted to know how that part actually worked because he wanted to build that little magic box himself.

Yes, if you are using a constant current charger (little magic black box) you just measure across the battery and the voltage will change. But the entire reason that works is because of what is happening inside that little magic black box and that is exactly as I explained.

I am immensely curious how a constant current is maintained without feedback.

I am immensely curious how a constant current is maintained without feedback.

No magic required.
Any constant current regulator does have and uses current feedback internal in it's own circuitry, so it does indeed know what the current is at any specific time and if the load resistance changes causing a change of current the regulator will raise or lower the loop voltage to keep the current at the desired constant current value desired. That part is standard for any charger that uses a constant current charging method. What the OP was asking is how to detect when to stop charging using the -deltaV detection method, which I think I answered satisfactory according to the OP's last posting?

Lefty

I do indeed have a battery charger (Turnigy Accucel 6). I was asking this question because i did not know how it would charge NiCads or NiMhs using the -deltaV method and i was planning on using such technology to charge NiCads or NiMhs with a microcontroller. Thank you very much Retro for explaining what's basically inside the charger and thank you Lefty for answering the question.

Again, thanks both of you. This case is closed :slight_smile:
Have a nice day. Thanx.

AJB2K3:
Something that may be of interest (very loosely) is that I was able to revive some Li-on cells that had dropped down to 0.2 volts and wouldn't charge.
PSP 110 Battery Revive. | Ajb2k3's Weblog Not very informative but it did work.

I think Boeing tried that already :wink:

MarkT:

AJB2K3:
Something that may be of interest (very loosely) is that I was able to revive some Li-on cells that had dropped down to 0.2 volts and wouldn't charge.
PSP 110 Battery Revive. | Ajb2k3's Weblog Not very informative but it did work.

I think Boeing tried that already :wink:

LOL But I don't think Boeing is laughing at the moment, it's costing them a fortune in downtime and whatever rework falls out is bound to be pretty costly as well. Li-po cells are great from a energy density and current capacity point of view, but they do bring along a lot of care and feeding rules to be as safe as past battery types.

Lefty