Battery Protection Circuit - is my voltage reading correct?  Is this working? - Page 1
Forum section: electronics / projects

author: pistolplc
author_rank: Newbie
timestamp: October 23, 2023, 01:33:07 pm
content: I have created a circuit that can switch between series and parallel configurations for a pair of 9v batteries. To protect the batteries, I've added Schottky diodes on the + of each battery.Now I want to test this circuit's function, and I'm confused by the result I'm getting. Specifically, in the "9v" configuration, when I remove one of the batteries and test the voltage across the battery leads, I'm reading 9v-ish. I would have expected that to be 0v because of the diode on the + leg. Where have I gone wrong in my understanding of this circuit? Also, what should I expect to read across that battery terminal in the 18v config?I have verified that there is a relatively small current running through the open leads by placing an LED/resistor across the leads - I get a very dim illumination (likely corresponding to the 2mA leakage current of my Schottky diodes). So it seems the circuit is probably working as intended, but still confusing me as to why I read 9v across that open battery terminal.Finally, if I'm reading 9v and getting 2mA of current through that battery, am I effectively protecting it from the possible damage/risk of reverse voltage? Or is that amount of voltage/current going to possibly damage the battery here if there are voltage imbalances?

author: Kim Christensen
author_rank: Super Contributor
timestamp: October 23, 2023, 09:37:47 pm
content: Quote from: pistolplc on October 23, 2023, 01:33:07 pmWhy are you confused? You already know about the reverse leakage current of the diode. You meter has a 1 to 10 Meg-ohm input impedance, so it makes sense you'd see almost 9V there with no additional loading. If it lights a LED, you're going to see voltage on your meter.QuoteWhat type of 9V batteries are these? If they are just alkaline, I wouldn't worry about it. If they are rechargeable it'll bit of a problem with a -2mA "charge" current if one battery goes completely dead and the other is still very much alive.

author: pistolplc
author_rank: Newbie
timestamp: October 23, 2023, 10:50:50 pm
content: Quote from: Kim Christensen on October 23, 2023, 09:37:47 pmI think I'm confused because my very elementary understanding of the diode is that it has extremely high resistance to reverse voltage. E.g., shouldn't the voltage on the anode be very low? (9V through some very high resistance should be lower than 9v.) But I'm a mechanical engineer, so this is not really my bailiwick.Stated another way, I thought this battery protection circuit would "prevent" the batteries from being subjected to a voltage (which is maybe the wrong concept) from the other battery.QuoteI could use alkaline or Lithium Ion recharegable 9v. Is the alkaline safer just because if there is some sort of problem with the circuit or large charge imbalance, the resulting "battery damage" is minimal (compared to say a flaming li-ion battery)? Or is it because the 2mA just won't have much effect on the alkalines?Thanks so much for your response and help.

author: Kim Christensen
author_rank: Super Contributor
timestamp: October 24, 2023, 02:29:13 am
content: Quote from: pistolplc on October 23, 2023, 10:50:50 pmThe reverse diode leakage current is generally quite low, but it's not zero. So if the only load is your voltage meter, and it has an input impedance of 10M, then only 0.9uA of leakage current would cause 9V across it.ie: A diode has extremely high resistance to reverse voltage but this resistance is not infinite and varies with the voltage across it.QuoteMore that the alkalines are disposable so "killing" one with 0% charge left in it doesn't really matter since it's already dead. A rechargeable would get damaged by being reverse charged.You would be better off using a diode with a much lower leakage current. What diode have you chosen that has 2mA of leakage with only 9V across it? I'm sure you can find something with better specs or change your design to protect the batteries in a better way.Maybe post the entire circuit and explain why you'd need to switch between 9V and 18V supply voltages.

author: pistolplc
author_rank: Newbie
timestamp: October 26, 2023, 02:41:13 pm
content: First, thank you again for your responses!As to the question about why to switch voltages - I make guitar amplifiers using an LM386 chip, and I can get different sounds when supplied with 9v vs. 18v, so I have a little circuit board that switches between series and parallel battery connections. The diodes are to protect the batteries from becoming reverse biased in the parallel configuration.The diodes I've picked are 1N5820 Schottky diodes. The 2ma leakage is at the rated blocking voltage of 20v, so I'm probably not really getting 2ma - that's just worst case scenario. Maybe these diodes are overkill @3 amps - the LM386 power output seems to be around 1W for its most "powerful" version. At 18v supply, the max current from the batteries will be about .1A. The 1N5819 diodes appear to have a leakage current of .5mA at 20v. Do you think that would be a better option in this scenario?

author: mikerj
author_rank: Super Contributor
timestamp: October 26, 2023, 07:53:38 pm
content: Quote from: pistolplc on October 23, 2023, 10:50:50 pmSchottky diodes tend to be quite leaky, and the leakage can rise significantly with increasing temperature. In general the lower the Vf the higher the reverse leakage current.

author: Kim Christensen
author_rank: Super Contributor
timestamp: October 27, 2023, 02:57:06 am
content: Quote from: pistolplc on October 26, 2023, 02:41:13 pmI wouldn't worry so much about it in the parallel situation. 2mA leakage would mean that one battery might try to charge the other one at 2mA. In reality, both batteries will be of similar states of charge if they are the same capacity and age and were both fully charged when installed. In other words the voltage differential between the batteries will be so low that you'll have next to zero leakage in that scenario. (Because the voltage difference is less than a diode drop)Diodes in series with the batteries won't protect them when the batteries are in series configuration. You'd need two more diodes for that.QuoteWith a 16V supply, the LM386N-4 can output 1W into a 32load with 10% distortion. Peak current from the supply rail will be less than 0.5A so either diodes will work fine. RMS current is going to be much less, so the 1n5819 would be a better fit.