Can charging voltage be suppressed at circuit? - Page 1
Forum section: electronics / projects

author: loop123
author_rank: Frequent Contributor
timestamp: January 20, 2024, 03:21:33 am
content: Let's say you only want 5V dc to be used by the circuit and you use charging voltage of 8 volts regulated to the 5V battery. What kind of design where the circuit would always see 5V instead of 8 volts in spite of it being charged all the time by 8V? Is this even possible?

author: inse
author_rank: Super Contributor
timestamp: January 20, 2024, 06:34:59 am
content: A low dropout 5V regulator could do the job.But the battery operational voltage needs to be slightly higher than 5V, or you are able to supply the circuit with lower voltage.

author: IanB
author_rank: Super Contributor
timestamp: January 20, 2024, 06:57:56 am
content: Rechargeable batteries act as voltage regulators. They will sink current (within limits) to pull the voltage down to battery voltage, and they will source current (within limits) to push the voltage up to battery voltage.So as long as the charging circuit does not abuse the battery and behaves in a moderate way, the circuit will always see the battery voltage.

author: loop123
author_rank: Frequent Contributor
timestamp: January 20, 2024, 07:33:32 am
content: Quote from: IanB on January 20, 2024, 06:57:56 amBut the adaptor is a 24V 2.5A AC/DC switching power supply. So it's possible the circuits could have charging circuit for the 14.4v NiMH battery while the rest of circuit got the power from the adaptor? But then when the adaptor is not used, how could it automatically choose battery power?

author: loop123
author_rank: Frequent Contributor
timestamp: January 20, 2024, 08:27:49 am
content: A NiCad constang charger is only very low in current, so it couldn't be enough to power the entire circuit. So let me shift my question. There must be a circuit that can detect whether there is an AC/DC adaptor (24V 2.5A) plugged in. If not, it will switch to battery power. This is real time such that when you remove the adaptor. It will immediately switch battery. Ok. What kind of circuit can do the switch? What are usually used?

author: Andy Chee
author_rank: Super Contributor
timestamp: January 20, 2024, 09:00:50 am
content: There are DC barrel jacks with a mechanical switch. This is often used to isolate the battery when powering the device from adapter. And when the adapter plug is physically removed, the switch automatically connects the battery.Other than that, a simple diode should also be able to isolate the charger.

author: tooki
author_rank: Super Contributor
timestamp: January 20, 2024, 09:13:44 am
content: Quote from: inse on January 20, 2024, 06:34:59 amJust a minor correction: 8V-5V=3V, which is more than the 2V needed by a standard-dropout regulator like the 7805. So you definitely don’t need to recommend alow dropoutregulator because this isn’t a low dropout situation.

author: loop123
author_rank: Frequent Contributor
timestamp: January 20, 2024, 09:25:16 am
content: Quote from: Andy Chee on January 20, 2024, 09:00:50 amCan you please share the diagram for this automatic switching between battery or adaptor (when plugged in)? (no mechanical switch)? Where does one put the diode?

author: Andy Chee
author_rank: Super Contributor
timestamp: January 20, 2024, 09:49:40 am
content: Quote from: loop123 on January 20, 2024, 09:25:16 amIf adapter Vin is greater than battery voltage, then the battery diode is reverse biased, thus isolating the battery.If adapter Vin is 0V, then the battery diode becomes forward biased and supplies power to Vout.

author: loop123
author_rank: Frequent Contributor
timestamp: January 20, 2024, 10:26:42 am
content: Quote from: Andy Chee on January 20, 2024, 09:49:40 amMost chargers have low amperage only. Supposed you used 2 chargers. One a main bigger one that would power the smaller charger circuit and main whole circuit, how do you put the diode then?

author: Andy Chee
author_rank: Super Contributor
timestamp: January 20, 2024, 10:33:57 am
content: Quote from: loop123 on January 20, 2024, 10:26:42 amI would never use 2 adapters. I would use 1 big adapter, then design a new low amperage charging circuit powered from that big adapter. The big adapter powers both the main circuit and the low amperage charger.

author: loop123
author_rank: Frequent Contributor
timestamp: January 20, 2024, 10:40:12 am
content: Quote from: Andy Chee on January 20, 2024, 10:33:57 amYes. That was what I meant. How does the basic design look like? Where do you put the diode?

author: Andy Chee
author_rank: Super Contributor
timestamp: January 20, 2024, 10:52:33 am
content: Quote from: loop123 on January 20, 2024, 10:40:12 amIf you are asking, "where do you put the charger", then in general put it where the resistor is. The diodes remain the same position.

author: loop123
author_rank: Frequent Contributor
timestamp: January 20, 2024, 11:03:45 am
content: What happens if the battery is out of order and shorted. Would it be able to disable or affect the main power? What kind of fault in the damaged batteries when it can affect the voltage or current in main adaptor making the equipment malfunction?

author: MarkT
author_rank: Frequent Contributor
timestamp: January 20, 2024, 10:42:27 pm
content: That's what a fuse can help with.

author: loop123
author_rank: Frequent Contributor
timestamp: January 21, 2024, 09:45:17 am
content: In the circuit the above is used (but 47 Ohm) between the NiMH batteries and ac/dc adaptor. That is. The adaptor dc output is 24Volts 400mA. Then it passes through a diode and in series with the above and directly to the 15V NiMH at 1500 MAh.Can the resistor above lower the ampere maybe 10 times less to become 40mA instead of 400mA?Also would the circuit see 24V or 15V? IanB said: "Rechargeable batteries act as voltage regulators. They will sink current (within limits) to pull the voltage down to battery voltage, and they will source current (within limits) to push the voltage up to battery voltage." So the circuit will really see 15V?

author: loop123
author_rank: Frequent Contributor
timestamp: January 21, 2024, 07:04:57 pm
content: A picture is worth a thousand words.Here is the actual circuit. From the 24v 400ma ac/dc adaptor. It goes directly to the black diode at top left (what is it?) then goes directly to the Xicon and from the Xicon directly to the positive wire of the 15V NiMH 1500MAH batteries. Why does it use a Xicon? It's just a resistor, isn't it? To decrease the adaptor ampere from 400ma to 40ma maybe? How do you calculate it? What kind of charger is that? Andy. It has no diode in the battery part in your sketch. So it's not isolating the batteries and all seem to be in parallel directly powering the main circuit.

author: IanB
author_rank: Super Contributor
timestamp: January 21, 2024, 07:21:45 pm
content: Suppose the NiMH battery has twelve cells in series. Then when it is discharged it will have a voltage about 12 x 1.2 V = 14 V, and when it is fully charged it will have a voltage about 12 x 1.4 V = 17 V.The charging calculation with the resistor is to look at the difference between the supply voltage and the battery voltage.If the supply is 24 V, and the battery is at 17 V (fully charged), then the charging current will be (24 V - 17 V) / 47 Ω = 150 mA.If the battery capacity is 1500 mAh, then the charging current of 150 mA is 1/10 of the capacity, or a charging rate of 0.1C. It is usually considered safe to charge an NiMH battery at a rate of 0.1C for a long time without need to protect against over charging (this is what battery manufacturers would call a "standard charge").However, "a long time" is not the same as "forever". Although the power adapter can be plugged in for a long time (e.g. a day) to charge the battery, it should be unplugged once the battery is fully charged, otherwise premature failure of the battery could result.

author: loop123
author_rank: Frequent Contributor
timestamp: January 21, 2024, 09:41:59 pm
content: Quote from: IanB on January 21, 2024, 07:21:45 pmI thought it was the 400mA that is the charging current. So if you use 3A adaptor, it would still be 150mA?Also I never imagine the the charging calculation with the resistor is to look at the difference between the supply voltage and the battery voltage. Why not just the supply voltage. How is the difference derived?Lastly. Would the circuit see 24V or 17V, and why exactly.Btw.. in cars battery. Is there a diode near the battery that isolates it while charging, or is there none. And would the car sees 12V or alternator charging voltage of 15V (?). And if there is no diode and the battery is shorted while running, the car would malfunction? I assume the diode being there near battery means even if the battery is shorted, no problem. See Andy sketch in the following again.

author: loop123
author_rank: Frequent Contributor
timestamp: January 22, 2024, 12:11:47 am
content: Quote from: IanB on January 20, 2024, 06:57:56 amHow come I can't find any reference about this. This is important because when we are repairing old stuff, how many times we removed the bad batteries and just use the AC power. If the batteries can pull the voltage down to battery voltage. Then removing it will put higher voltage at the circuit, and can damage it. Is it not? In cars, after we turn on the engine, you know we can remove the batteries. But will the car circuits see the alternator 18V instead of just the 13V or so of the battery?

author: IanB
author_rank: Super Contributor
timestamp: January 22, 2024, 12:45:24 am
content: Quote from: loop123 on January 22, 2024, 12:11:47 amCorrect.If the battery is a fixed installation you cannot remove it without opening the case. There is often a warning on the outside that says something like, "There are no user-serviceable parts inside. If repair is needed, take it to a qualified repair center."So if you remove the battery without appropriate knowledge and you damage the equipment as a result, then the blame lies with you.If the battery is removable, then there may be a warning that says, "Do not plug in the power adapter without the battery present." I once had a calculator like this.These considerations typically apply to older equipment that had NiCd or NiMH battery packs. They are not likely to apply to newer equipment with LiIon batteries. For example, you can typically operate a laptop on mains power with the battery removed.However, you cannot operate a car with the battery removed. This is an absolute no.

author: loop123
author_rank: Frequent Contributor
timestamp: January 22, 2024, 12:56:29 am
content: Quote from: IanB on January 22, 2024, 12:45:24 amKindly give references about the circuit only seeing the battery voltage and not the charger voltage. It's so counterintuitive that I want to ask if you are sure of what you said. Where did you heard about it? Again you mentioned "Rechargeable batteries act as voltage regulators. They will sink current (within limits) to pull the voltage down to battery voltage, and they will source current (within limits) to push the voltage up to battery voltage.".I can't find it at google. Its very obvious the circuit can be exposed to the higher voltage of the charger. But somehow the batteries can suppress it totally? Please explain how (elaborate than your 2 sentences explanations above) or give other examples if no other sources of this in the internet. Thanks.

author: IanB
author_rank: Super Contributor
timestamp: January 22, 2024, 02:44:37 am
content: Quote from: loop123 on January 22, 2024, 12:56:29 amI didn't hear about it. I am teaching it. It is a natural outcome ofKirchhoff's laws, which you can read about at the link.In particular, Kirchhoff's voltage law says that the sum of the voltages around a closed loop is zero. Since a battery charging circuit and a battery form a closed loop, it follows that the battery terminal voltage and the battery charging voltage are always equal to each other.The corollary of this, is that if 24 V appears at the terminals of a 15 V battery, then the battery is either already destroyed, or is about to be destroyed.Since the battery is clearly not being destroyed, it means the battery terminal voltage must remain around 15 V, plus or minus some small delta.

author: loop123
author_rank: Frequent Contributor
timestamp: January 22, 2024, 04:43:04 am
content: Ok. Kirchhoff's voltage law says that the sum of the voltages around a closed loop is zero. So how exactly the 24V of the ac/dc adaptor at the right vanish? You mentioned that they will they sink current (within limits) to pull the voltage down to battery voltage. So how exactly does the sinking occur? It's not like water where things could just sink. It is a circuit. Please show using some principles and computation how the sinking occur. Thanks a lot.

author: IanB
author_rank: Super Contributor
timestamp: January 22, 2024, 04:56:22 am
content: You have a picture with all the relevant pieces in it (battery, charger and resistor).You have Kirchhoff's voltage law, and you will also need Ohm's law.This is now an exercise for the student. Have a think about it, work it out for yourself, and when you have an answer, post it back here to see if it is correct.If it is not obvious, you will also need to know that the current everywhere in the circuit is the same.

author: loop123
author_rank: Frequent Contributor
timestamp: January 22, 2024, 05:50:11 am
content: Quote from: IanB on January 22, 2024, 04:56:22 amYou gave the formula (24-15V)/47 Ohm = 0.1915AVoltage drop at Xicon resistor = IR = 0.1915 (47 Ohm) = 9V henceIf you put voltage drop of 9V at resistor and reverse the sign. Your idea may work. But I got the original diagram on the internet. Why is the sign of the resistor aligns to the right and not left originally? Does it assume the left has bigger voltage?Initially I was thinking the charger radiates the 9V as virtual particles in the vacuum but thinking about the diagram, there may be more conventional explanation. Thanks.

author: IanB
author_rank: Super Contributor
timestamp: January 22, 2024, 07:06:12 am
content: Yes. If you start at the bottom of the circuit and go around it counterclockwise (the direction of current flow), then you go (up) 24 V, then (down) 9 V, and finally (down) 15 V, bringing you back to zero. This is Kirchhoff's voltage law.Now the resistor has 9 V across it, and a value of 47 Ω, so by Ohm's law the current through the resistor will be 9 V / 47 Ω = 0.19 A. Since the current everywhere in the circuit is the same, this is also the current delivered by the 24 V source and received by the 15 V battery.Note that the diagram has misleading arrows on it, and also misleading + and − signs. The voltage signs on the resistor are backwards, since the right hand side is clearly more positive than the left hand side. The labels drawn in purple are correct.Arrows on the lines are best drawn to show the direction of current flow, and so these arrows should point the other way (the current is flowing counterclockwise in this circuit, not clockwise).

author: loop123
author_rank: Frequent Contributor
timestamp: January 22, 2024, 08:22:53 am
content: Quote from: IanB on January 22, 2024, 07:06:12 amThe original figure came from the following. I was just looking for a drawing in the net to illustrate it.https://isaacphysics.org/concepts/cp_kirchhoffs_laws?stage=allIn the case of the capacitor charging. It's the same concept due to the resistor able absorb some of the voltage. So all charging involves current, and there is no charger on earth where you directly put the source into the battery without using any resistor?

author: loop123
author_rank: Frequent Contributor
timestamp: January 22, 2024, 09:00:01 am
content: Ian, I'm stuck at something. Going to my altered diagram. If you remove the battery. It dooes't look like the violet terminal at left would take the full 24V, isn't it? What voltage will appear at the violet terminals? How do you compute for it?

author: loop123
author_rank: Frequent Contributor
timestamp: January 22, 2024, 12:37:47 pm
content: Quote from: loop123 on January 22, 2024, 09:00:01 amI found the (partial) answer here.https://www.quora.com/What-is-the-maximum-power-that-a-12v-DC-source-with-internal-resistance-2-ohms-can-supply-for-a-resistive-loadSo if your load is equal to 47 Ohms, then the voltage will be divided from 24V to just 12V. It's even lesser than the battery of 15V. How can it fry your circuit if you only use the charger without any batteries, Ian? But then, for typical equipment. What is total resistance seen by the source usually? Is it 50 Ohms? 200 Ohms? 5k Ohms? 20k Ohms? 200k Ohms? more? I know when it is more, there will be more voltage contribution just like in the concept of input impedance of some amplifier must be in MegaOhms to get maximum voltage/power.

author: IanB
author_rank: Super Contributor
timestamp: January 22, 2024, 05:35:34 pm
content: Quote from: loop123 on January 22, 2024, 09:00:01 amYou can only compute it if you know what is connected between the violet terminals. If you remove the battery and connect nothing, as shown in the picture, then you will indeed have the full 24 V at the violet terminals.If you put some kind of load circuit there, then the voltage at the terminals will depend on the current drawn by the load, and if the load varies then the voltage will vary. If the load is small and the voltage goes up to 24 V it could fry the circuit.Since a varying voltage is not desirable for the circuit being powered, you would not normally remove the battery. The battery acts as a regulating device to keep the voltage stable.There are of course much more complex circuits and charging arrangements that could be designed, but it is not possible to talk about those without introducing a lot more detail.

author: loop123
author_rank: Frequent Contributor
timestamp: January 22, 2024, 10:27:35 pm
content: Quote from: IanB on January 22, 2024, 05:35:34 pmThe above is the computation identical to our circuit. When you said "If the load is small and the voltage goes up to 24 V it could fry the circuit.".. did you mean low wattage or low resistance? Because the lower the resistance. It can only get V=IR= 3A (2Ohm)= 6 Volts or lower so not the full 12V (or 24V in our case) or only getting one half of the voltage of the source. Unless you mean by the load is small is low wattage but high resistance?And have you measured the resistance of the whole circuit or equipment with respect to the battery? What values do you usually get? Because if it's big like 200kOhm. Then by concept of voltage divider, you'd get close to the full voltage with respect to the 47 Ohm in series to to the equipment resistance. What official term is this referred to? I know it is not input impedance because there is no impedance involved in battery source.

author: IanB
author_rank: Super Contributor
timestamp: January 22, 2024, 11:38:40 pm
content: The term "small load" always means low power, or low current draw.You can't really measure the resistance of a piece of powered equipment since it is not a resistor. What you do instead is measure the current draw under normal operation.Suppose in the example case the current draw was 0.01 A or 10 mA. Then the voltage dropped across the 47 Ω resistor would be about 0.5 V, and the load would see a voltage of 24 − 0.5 = 23.5 V.

author: loop123
author_rank: Frequent Contributor
timestamp: January 23, 2024, 12:03:07 am
content: Quote from: IanB on January 22, 2024, 11:38:40 pmIf the current in the circuit is low, and the voltage drop in the 47 Ohm resistor is only 0.5V (0.01A x 47 Ohm), then for the load to be 23.5V with low current of 0.01A, the load resistance has to be high, R=V/I = 23.5/0.01 = 2350 Ohm, correct?

author: IanB
author_rank: Super Contributor
timestamp: January 23, 2024, 12:35:55 am
content: Sort of, but you can't properly call it a "resistance" unless it is actually a resistor. Complex circuits don't necessarily behave like resistors.

author: loop123
author_rank: Frequent Contributor
timestamp: January 28, 2024, 01:49:31 am
content: Quote from: IanB on January 22, 2024, 07:06:12 amTo clarify something. For our 24V ac/dc adaptor with 15V rechargeable battery connected to 47 Ohm resistor producing 0.19A charging current. If the equipment is on and being used while charging, it is getting power from the battery? But the battery is producing current, how can the 0.19A charging current from the ac/dc adaptor get into the battery? Won't the currents collide?Also what components can be damaged if the batteries are removed and the 24V charger creating say 21V on the circuit instead of just half? Because my equipment has partial damage from this incident and I'm trying to analyze what components could be damaged from overvoltage. Thanks so much for the tips.

author: IanB
author_rank: Super Contributor
timestamp: January 28, 2024, 08:26:44 am
content: Quote from: loop123 on January 28, 2024, 01:49:31 amThe charging current would get split. If the equipment draws less than the 0.19 A, then it will take that, and the remainder will charge the battery, but more slowly than before. If the equipment draws more than 0.19 A, then the battery will be drained, and eventually will become empty.QuoteI could not say. Not enough information.

author: loop123
author_rank: Frequent Contributor
timestamp: January 28, 2024, 10:51:13 am
content: Quote from: IanB on January 28, 2024, 08:26:44 amThe manufacturer mentioned that with the 24V charger always plugged in. The device can be used indefinitely. Besides the possibility the device only draws much less than 0.19A with the majority going to the battery. What else is the possibility, like could the device be designed to take in the entire 24V and then split them into partial +Vs and -Vs if the battery is not installed or failed (remember batteries has lifetime so the manufacturer could take that into account instead of planning for the device self-destruction when the batteries fail in future)?Using multimeter. The opamps +Vs and -Vs are not using the battery +7.2V and -7.2V. Somehow it has converted +Vs to 4V and -Vs to 11V. This is why I think it may be damaged although the main functions still work. What kind of component damage could cause this and perhaps the clue to what method is use to turn the battery into separate +Vs and -V-. What circuit or parts or methods are commonly used?

author: loop123
author_rank: Frequent Contributor
timestamp: January 28, 2024, 03:52:59 pm
content: Quote from: IanB on January 28, 2024, 08:26:44 amHi, I have traced the circuit for hours. And here is the weird part. After many detours. I found out that the +7.2V battery is connected directly to the V+ of the AMP01 in the following and the -7.2V battery is connected directly to the V- of the AMP01. Yet somehow the voltage turns to 3.95V for V+ in pin 12 and -11.95V for -V for pin 11 when I measured pin 12 and pin 11. Do you have any idea what happened? And yet the AMP01 or unit is working fine!

author: IanB
author_rank: Super Contributor
timestamp: January 29, 2024, 01:22:16 am
content: How are you measuring negative voltages? Does the battery have three wires? None of the images you have posted have any indication of a ground. Where are you connecting the black probe when you measure voltages?But it's not likely that I or anyone else can help you. Diagnosing possible faults in an unknown piece of equipment requires the equipment to be in the hands of an expert technician with diagnostic equipment, and preferably with a circuit schematic or service manual available.With the limited information you have shared, there is no way to proceed. To have a chance, you need to at least provide the big picture, and not little snippets of information out of context. But even if you did that, I am no expert and you would need other people to chime in.Have you at least removed the battery, tested each cell individually for faults, and if necessary replaced the battery with a good battery? All fault diagnosis starts with the power supply.