Charging 3.7v li-ion battery with 4 volts - Page 1
Forum section: electronics / beginners

author: Muku
author_rank: Contributor
timestamp: August 26, 2020, 06:53:23 pm
content: Hello everyone,I have a board with different components that work on 2 different voltages., 3.3 and 4 volts. The main power supply is a 12v supply. So I have a TPS54240 switching IC converting the 12 volts to 4v that goes to one component and then a simple LDO that converts 4v to 3.3v for another component. Now the problem is I also need to implement a li-ion charger to charge a typical 3.7v, 350mAh battery that fully charges to 4.2volts. To implement the charger I am looking at 1 ic MCP73832. It is cheap an readily available. But the problem is I only have 4 volts available from the switching regulator which means the battery will never be able to charge upto 4.2 volts. What could be the solution to this problem? The maximum rating of the 4v component is 4.2 volts, so I was thinking may be I could kick up the switching regulator output a little and run the 4v component at max rating. Would that be a good idea? Or should I just leave the battery charge cuff off voltage at 4v? Or is there any other option?

author: bob91343
author_rank: Super Contributor
timestamp: August 26, 2020, 06:56:12 pm
content: Add a voltage booster and you can get any source you like to charge the battery. Little switching regulators are cheap and could run from the 4V or whatever you have.

author: Muku
author_rank: Contributor
timestamp: August 26, 2020, 07:17:12 pm
content: I already have 2 inductors on the board, I don't really want to add another one. I am going to power the whole thing with 4.3 volts and hope for the best

author: MLXXXp
author_rank: Frequent Contributor
timestamp: August 26, 2020, 07:36:24 pm
content: Quote from: Muku on August 26, 2020, 07:17:12 pmIf you're going to change the 4V regulator to something higher, set it for 4.7V and feed that to the charger circuit. Put a standard silicon diode in series between the 4.7V supply and the 4V circuit. The diode will drop 0.7V to give you your 4V supply (which can still feed the 3.3V LDO).This is assuming the current required by the 4V circuit is low enough that a series diode is practical. You may have to tweak the 4.7V somewhat depending on the voltage drop of the diode you choose, over the current draw range of the 4V circuit.Of course, you could also use a 4V LDO (using an adjustable one) to drop the 4.7V or some other higher voltage, but a diode is much simpler and if your 4V circuit can tolerate 4.2V, it sounds like the poorer regulation provided by a diode won't be an issue.

author: Muku
author_rank: Contributor
timestamp: August 26, 2020, 08:19:37 pm
content: Quote from: MLXXXp on August 26, 2020, 07:36:24 pmHello,Thanks for the suggestion. The datasheet say "The power supply range is from 3.4V to 4.4V but recommended is 4v ". In the absolute maximum rating section it says 4.5v. The diode idea is great. What if a SCHOTTKY diode like bd160 or something similar, it should be able to carry the current and also drop the voltage, according to datasheet around 2amps of current forward voltages goes to 0.8. These are test boards but having a diode instead of LDO would save me a lot of money. Would you dare to use the diode solution on a production board if it saves money? What would be the worst case scenario that can happen from this diode solution besides the obvious failure of the diode?

author: magic
author_rank: Super Contributor
timestamp: August 26, 2020, 08:42:49 pm
content: 4.2V is not the normal state of a Li-ion cell, it's only forced briefly during charging to quickly shove more charge into it. The cell drops down afterwards. I don't know what percentage of rated charge is lost by charging to 4V and this could be tried, but I would be concerned trying it with a dedicated charger IC, because it could do something stupid like fail to terminate the charge and keep the cell floating on 4V as long as power is connected or just outright malfunction. BTW, see what's the minimum rated operating voltage of the charger you want to use.The worst that can happen with the diode solution is that current draw of your circuit becomes low, voltage drop across the diode decreases and the 4V IC is exposed to conditions outside its guaranteed operational ratings. Also, load regulation is somewhat poor, which can be a problem if the 4V circuit is sensitive to supply voltage. Finally, diode voltage drop decreases with temperature.

author: MLXXXp
author_rank: Frequent Contributor
timestamp: August 26, 2020, 08:51:57 pm
content: A schottky diode may work but keep in mind the minimum voltage required for you charger circuit. You will need a certain amount higher input voltage due to the minimum resistance of the main pass transistor in the MCP73832. (This is similar to the dropout voltage of a linear regulator.)The MCP73832 datasheet says the typical Rdson of the pass transistor is 350m ohm. It doesn't say what the maximum is but you'll have to account for some tolerance. If you want to charge a 350mAh battery at 1C you'll need 350mA charge current. 350mA across the transistor resistance of 350m ohm would be 0.123V. Your supply would have to be 4.2V + 0.123V = 4.323V. But like I said, 350 ohm is typical so you probably want to add some leeway. You also have to account for the tolerance of your regulator. So, to be safe your charger supply voltage should be about 4.5V.4.5V is too high for the typical 0.3V drop of a schottky diode, so you may as well set the supply to 4.7V and use a regular diode.Would I use a diode in production? It depends on the difference between the lowest and highest current drawn by the 4V circuit and how tolerant that circuit is of power supply voltage variations. As @magic posted as I was typing this, the load regulation of a diode isn't going to be as good as a true regulator.

author: MLXXXp
author_rank: Frequent Contributor
timestamp: August 26, 2020, 09:12:02 pm
content: Quote from: Muku on August 26, 2020, 08:19:37 pmWhat you could do is set your supply to about 4.6V. This would be high enough to charge your 350mAh battery at 1C. A regular silicon diode probably won't pass any significant current without at least a 0.4V drop (maybe higher). With a 4.6V supply and a 0.4V diode drop, your 4V circuit would receive 4.2V, which is safe.At higher currents, the diode might drop 0.8V, so your 4.6V supply would provide 3.8V to the 4V circuit, which is still well within range.Again, your circuit has to be able to tolerate the poor load regulation, which could appear as noise at whatever frequency the current draw changes. You could probably use high(ish) value bypass capacitors to minimise, at least, high frequency noise.

author: Peabody
author_rank: Super Contributor
timestamp: August 26, 2020, 09:29:27 pm
content: If this is for production, then be sure to consider the safety aspects of lithium battery charging. If you will want to power your device while you are charging the battery, you need a load sharing circuit, and might want to consider the MCP73871 which has load sharing built in.

author: amyk
author_rank: Super Contributor
timestamp: August 27, 2020, 02:43:19 am
content: Quote from: magic on August 26, 2020, 08:42:49 pmGood cells don't drop noticeably from 4.2 after they're charged. Charging to 4V (saturated) is around 80%, I believe, and increases cycle life by over 5x.

author: Muku
author_rank: Contributor
timestamp: August 27, 2020, 05:18:39 pm
content: Thanks for so many reply guys.I think the better option for me would be use a battery charger with integrated buck converter. I was looking around and found MP2617B which has buck converter with a battery charger can deliver 3A of current. But the problem is it's max input voltage is 10V so can't use it in my project. Since my board is going to be inside my motorbike and will be powered from the 12V battery inside. There is another version MP2617H which has max input voltage of 14 volts. Great right?? Not really, I went ahead and measured my bike battery voltage when the engine is running, it was 13.37V well under 14v but as soon as I revved it the voltage jumped over 14 i.e. 14.38v exactly.The datasheet of MP2617H says absolute max voltage rating of 20V so, it would not kill the device but I don't know if it will be a problem in long term. What do you suggest? Would it be a good idea to add a shottky in between 12v battery and the MP2617H so that the circuit has reverse voltage protection and drops some voltage while doing so. If 2.5A of current is drawn then a 0.8v shottky would dissipate 2 watts. That seems like a lot of power.

author: amyk
author_rank: Super Contributor
timestamp: August 28, 2020, 12:22:28 am
content: Quote from: Muku on August 27, 2020, 05:18:39 pm...and it probably jumps higher than that due to ignition spikes and such --- if you had mentioned this in the OP you would've gotten some more appropriate advice. Search "automotive" in this forum for a good start.

author: IanB
author_rank: Super Contributor
timestamp: August 28, 2020, 01:18:52 am
content: Quote from: Muku on August 27, 2020, 05:18:39 pmThis is important information. Have you got properly robust input filtering, stabilization and surge protection on the input to your circuit? Because I have a feeling that if you just connect bare electronics to the 12 V rail of an automobile the electronics won't last very long.Yes, I know you can get USB chargers that plug into a 12 V lighter socket, and how much protection can those have, right? But I bet you they have at least some input protection.

author: Muku
author_rank: Contributor
timestamp: August 28, 2020, 11:48:03 am
content: Went around internet looking for factors that can affect electronics connected to a vehicle and found quiet a lot of material. There was a article by Ti. I think most important thing was what is called a load dump. "The windings of an alternator have a large inductance. When the vehicle battery is being charged, the alternator supplies it with a large current, the magnitude of which is controlled by the current in the field winding. If the battery becomes disconnected while it is being charged the alternator load suddenly decreases. However the alternator's regulator cannot quickly cause the field current to decrease sufficiently, so the alternator continues to generate a large current. This large current causes the voltage on the vehicle bus to increase significantly -- well above the normal and regulated level. The peak voltage of this surge may be as high as 120 V and the surge may take up to 400 ms to decay. It is typically clamped to 40 V in 12 V vehicles and about 60 V in 24 V systems."Since, I am working with 12volt system I should prepare my circuit for 40v at most. Either I could use a charging IC with higher input voltage around 40v. But charging IC with both higher input voltage switching converter and li_ion charger is hard to find . So I am sticking with 14 volts IC.There is a great doc by TI which has circuits for automotive application. And I am looking at this circuit.https://www.ti.com/lit/an/snva717/snva717.pdf?ts=1598604209548&ref_url=https%253A%252F%252Fwww.google.com%252FNo I want to use a 15v zener instead of 18v just to be safe. MP2617H already has soft start feature so I am going to skip the soft start part. I am keeping reverse voltage protection diode. MP2617H also allows for input current limit feature so I guess inrush control is also redundant. Also I am thinking of adding a 4 amp PTC fuse on the input. I am thinking of using just the Over voltage Disconnect part to save cost and space on my PCB. I need some help deciding the current and power rating of the components in OVP circuit. How do I calculate proper ratings for components if my total system current requirement is 3A? Could anyone clear me why is the another 18v clamp after OVP is needed?Also I have another problem, I am using a 15v zener because I need device to function when the Lead Acid is charging or when engine is running. Lead acid is charging around 14.4 when revved so if I use 14v diode the OVP may kick in when engine is running and turn the whole thing off. Using a 15v diode means running the MP2617H slightly over its recommended voltage. Absolute max rating is 20v so it should be fine. But just something to think about. Is there any way to solve this? Or should I just count on the Reverse Protection diode to drop the voltage just enough.Thanks