"Simple" way to prevent over-voltage cutoff on battery powered inverter - Page 1
Forum section: electronics / general

author: Ziggo
author_rank: Contributor
timestamp: October 31, 2022, 07:56:58 pm
content: I want to use a "60V Max" battery from Greenworks to power a 1500W inverter. The battery uses 15 Li-Ion cells in series so it's voltage under load will be under 60V. However, the no-load, fully charged, voltage will be above 60V. I would like to use an inverter that has a high-voltage cufoff of 60V.Is there some, relatively simple, way to keep the voltage under 60V to the inverter? The inverter is a Mean Well and I would prefer to stick with a quality brand rather than getting a ~72V max Chinese inverter (which are also not that easy to find). Efficiency doesn't really matter, especially if it's only in effect above 60V.I've found a buck converter that can handle 720W, but not any that can go to 1500W. The only thing I've come up with is a simple power resistor that is switched in and out with a comparator circuit so it's only in-line when the power switch to the inverter is on and the battery voltage is above 60v.

author: f4eru
author_rank: Super Contributor
timestamp: October 31, 2022, 08:14:21 pm
content: QuoteCould you put two of those in paralell ?Also, perhaps make sure those buck can handle a >99% duty cycle, to pass through once the battery has dropped a litttle...

author: David Hess
author_rank: Super Contributor
timestamp: November 01, 2022, 03:40:32 am
content: 60 volts at 1500 watts is 25 amps. How much higher will the voltage be?If it is small enough, then some switched power diodes in series can drop the voltage a couple volts. I might be clever and use power MOSFET body diodes, so turning on the MOSFETs bypasses the diodes once the voltage is low enough.If the voltage is a little higher, then it might be worth using a low dropout linear regulator. This solution is essentially the same as the MOSFET body diode solution, but fewer devices are required.

author: Ziggo
author_rank: Contributor
timestamp: November 01, 2022, 04:22:57 am
content: Quote from: f4eru on October 31, 2022, 08:14:21 pmI'm worried about using them in parallel since I don't think they will share the load well.

author: Ziggo
author_rank: Contributor
timestamp: November 01, 2022, 04:23:49 am
content: Quote from: David Hess on November 01, 2022, 03:40:32 amYes, the diodes should be better than a resistor since they will drop a constant voltage. I think the max will be around 63V. Then, I just need to decide whether (and how) to drop them out when the battery voltage goes below 60V.

author: David Hess
author_rank: Super Contributor
timestamp: November 01, 2022, 07:35:24 am
content: Quote from: Ziggo on November 01, 2022, 04:23:49 am3 volts, or a little more, at 25 amps is feasible to handle with a power MOSFET linear LDO, so I think I would do it that way, and it removes the problem of switching. The efficiency loss happens either way. But this is a real hack of a solution versus an inverter that can handle slightly higher input voltage or a lower voltage battery.A buck converter would give slightly higher efficiency but at considerably more complexity. If there is no room to dissipate about 100 watts of heat, then this or an inverter is the only solution. An transformer inverter to step down say 65 volts to 55 volts would also work with high efficiency.

author: Ziggo
author_rank: Contributor
timestamp: November 02, 2022, 10:01:26 pm
content: Quote from: David Hess on November 01, 2022, 07:35:24 amThanks for the suggestion, but I can't find any LDOs that are in that range (nothing about about 10A). The 100W power dissipation can be managed so my current plan is to just use the body diodes from a series of power MOSFETs, swithcing the MOSFETs on when the voltage drops enough.

author: David Hess
author_rank: Super Contributor
timestamp: November 03, 2022, 03:32:44 am
content: Quote from: Ziggo on November 02, 2022, 10:01:26 pmIt would be a discrete LDO made with either a linear regulator controller and external power MOSFETs or an operational amplifier and associated parts.

author: Siwastaja
author_rank: Super Contributor
timestamp: November 12, 2022, 01:42:19 pm
content: Is it possible to just terminate charging early? Just charge to 60V, which is 4.0V/cell, roughly 80% SoC. Good for battery life, too. This would be simplest, by far.

author: Faringdon
author_rank: Super Contributor
timestamp: November 12, 2022, 05:58:32 pm
content: You can buy step down modules from murata, vicorpower etc.....and they have the reference pin to the error amplifier available...so you can put it on a pcb and pull it down with a current clamp, so that when ever its current goes above max nom, it pegs back its power thruput.....with lots of these like that...you can parallel the step down modules to get your <60v spec done.There are other ways of paralleling step down modules.The way told needs you to do a PCB though....with the external current clamp error amplifier on it.Actually come to think of it...vicorpower actually sell step down modules which are "array capable".....so search for these, as you can parallel them easily to get your 1500w....look for "array" versions.Give me a shout of you want more ways of paralleling converters to get higher power...egUse a load share controllerSingle error amp feedback to allTxconductance error amps with output conencted together.Use a vout degradation method...eg reduce vout as current rises, then they naturally share...or just do this with series ouput resistors...but you have to put the resistor downstream of the vout divider..But in fact, loads of offtheshelf smps's have a port so you can pllel them with others the same.

author: Ziggo
author_rank: Contributor
timestamp: November 17, 2022, 01:07:09 pm
content: Quote from: Faringdon on November 12, 2022, 05:58:32 pmThe issues here are:1. Cost - I want something for around $50, $100 max.2. Complexity / Time - I don't have the expertise needed to do this in a reasonable amount of time (a few hours).

author: Ziggo
author_rank: Contributor
timestamp: November 17, 2022, 01:15:03 pm
content: Quote from: Siwastaja on November 12, 2022, 01:42:19 pmI have considered doing this, but the main issue is that I would need to babysit the battery charging and pull it off at the right time. Or, use some other load to discharge the battery down to a usable level if it gets fully charged.I finally received my Mean Well inverter and tested it with my battery (a 60V, 8.0Ah battery from Greenworks - I already have 4 of these). The battery gets to 63V at full charge and the inverter cuts out above 59V.So, my plan at this time is to use thishttps://www.amazon.com/gp/product/B07YTVK1JK/ref=ppx_yo_dt_b_asin_title_o04_s00?ie=UTF8&psc=1low/high voltage disconnect relay driving a 40A 100V solid state relay (that I already had). When the relay is open, there will be a series of power diodes (which I also already have) to do the 4V voltage drop. When the voltage drops enough and the relay is closed, the diode bank will be shorted out.I still need to wait a week or two for the disconnect relay to get here. It's not a very elegant solution, but I think it will work OK. The battery voltage drops pretty quickly under load so the diodes will only be in line when the battery is full and the load is light.

author: M0HZH
author_rank: Regular Contributor
timestamp: November 17, 2022, 01:57:34 pm
content: A workaround which will add another point of failure, complexity and reduce efficiency. You also don't want to always be very close to the absolute maximum voltage rating of the inverter, for long-term reliability (or nuisance overvoltage cutoffs).Sounds like the inverter is most likely designed with 48V operation in mind, isn't it better to find a 13S battery (which are quite common)?

author: 2N2222A
author_rank: Regular Contributor
timestamp: November 26, 2022, 12:06:32 am
content: I liked the MOSFET body diode idea. How about 6 5A diodes in series. When the current exceeds 5A a relay closes or a MOSFET turns on to bypass the diodes. They could be 10A diodes in a TO-220 package along with the MOSFET. All on the same heat sink.

author: Magnethicc
author_rank: Contributor
timestamp: November 29, 2022, 10:41:15 am
content: Quote from: Siwastaja on November 12, 2022, 01:42:19 pmBy far the easiest solution.OP suggestion is to put diodes in series to drop the voltage but will dissipate ~100W which will need to be dissipated with considerable heatsink.It is much easier to measure the voltage of the battery (resistor divider) and disconnect the charging station when the measured voltage gets to a reference value.OP already have a relay so just connect the output of the charging station to the relay. when measured voltage reaches to a set reference value then disconnect the relay.There is no need to babysit the charging station.The input voltage to the control circuit can be taken from a buck converter from the charging station/battery or use high voltage op-amps and take the input straight from the charging station/battery.The current to the relay can be taken from the charging station/battery through a BJT and a current limiting resistor.