Protecting Circuit from placing battery backwards

xsr_Gx

Hello,

My question is how to protect my circuit when batteries are placed the wrong way. I have attached a high level block diagram of our set up as well as the parts we would be using. As previously mentioned, this node requires an average of about 300mA-hrs per day, the max current load is ~ 300mA and all devices run at 3.3V The system works as follows:

When we have sun, the DS2715 turns ON charging the battery bank. Since the Si2351 is an enhanced PMOS, the switch is OFF preventing the batteries to provide current to the power rail. MAX1651 and MAX1793 regulates the voltage to provide a clean and stable 3.3V. When there is no sun, the PMOS closes and the Battery bank can now provide current to the power rail.

Now, in the case that the batteries are put the wrong way, we decided to use a 2A fuse and a schottky diode, however i am not sure about how to properly size the fuse and the diode? Is this the right method to prevent damaging my circuitry? ANy help will be appreciated.

SR

BB.

Re: Protecting Circuit from placing battery backwards

Choose a high current (can handle the current of fast blow fuse) using a Schottky diode. Also check the leakage current--many Schottky diode types can have fairly high leakage current and may discharge your cells (don't know for sure--check spec. sheets and measure as needed).

Question--what kind of NiMH cell will you be using--A standard "D Cell" with the raised button?

If so, you can possibly buy or build battery a battery carrier that will not allow a reversed cell to make contact with the + power contact (shield with a couple plastic bars that would keep the flat battery base away from the power contact).

When you pick your NiMH cell--check everything out (short circuit current through the fuse, etc.) and the polarity of the "button" end of the cell--I believe some cells (maybe Mil-Spec?) may have the "bump" on the negative end of the cell (if you are dealing with "non-consumer" type cells).

-Bill

xsr_Gx

Re: Protecting Circuit from placing battery backwards

Thanks Bill,

yes, i am using D type batteries and has the button on teh + side. Question: is my set up good? Are the diode and fuse where they are suppose to be? Do i have to select a diode that can handle currents higher than the fuse as well? What if i want to add an LED which can turns on when this happens?

niel

Re: Protecting Circuit from placing battery backwards

many protection circuits use an inline fuse that is fed from a diode wired so that if the battery is backwards that it sends power through the diode and blows the fuse on its way back to the battery.
i'll try to illustrate.

---battery+++fuse+++
| diode>

---

|

the lines | indicate a wire connection to that above.

BB.

Re: Protecting Circuit from placing battery backwards

Fuses tend not to "limit current" -- they act like little heaters that "pop" when they get hot enough. A fast blow fuse and a diode that can handle the current...

• V=I*R
• R= V\I = 0.1 volts \ 0.3 amps = 0.33 ohm fuse (limiting resistor to protect circuits)
• I= V\R = 2 volts \ 0.33 ohms = 6 amps (will limit "shorts" to ~6 amps or so)

I am thinking that since you have very low current needs for your electronics and you have D Cell NiMH which have very high surge current capabilities probably in the range of 100 amps or more--so a simple resistor to limit surge/short currents would be a good safety item to add.

Also, the resistor will limit the current to the fuse/diode to less than ~10 amps--a very manageable number.

By the way, how long are you thinking between recharging? 30% self discharge per month at room temperature is not an unreasonable expectation. (I missed solar panel first time--not sure is large enough panel to keep load + battery happy).

-Bill

PS: You could add a reversed biased LED across the fuse + blocking diode + resistor... Or you could also do a "self test"... Push a button and turn on a green LED if voltage is present and over X.X Volts.

niel

Re: Protecting Circuit from placing battery backwards

i don't think the resistor is needed as it's absence will only serve to pop the fuse faster. if the wiring is very small then i can see a resistor, but it must be good enough to handle currents high enough to pop the fuse.

mike95490

Re: Protecting Circuit from placing battery backwards

I think a small resistor IS needed, to prevent a high amp surge that would fry the diode. Otherwise, you are looking at a larger, over spec diode to handle a raw battery short.

niel

Re: Protecting Circuit from placing battery backwards

mike,
the resistor still may not be needed as the fuse should pop first. it depends on the max he wishes to pass to the loads for the fuse size and that high current may still be needed to pop the fuse. we are still siting d size batteries here and they won't be called upon to have fuses in the 100a area. diodes generally have a high surge rating even if only 1 amp diodes and do not need to be special such as a schottky. it is a good cheap method of protection as a battery reversed will surge through to the fuse and pop it. this method has been used on many electronic items including cb radios so i'm sure it'll be fine for him without the resistor.

lorelec

Re: Protecting Circuit from placing battery backwards

How about another P-channel FET is series with the battery? Gate goes to ground, source to battery +, and drain to your circuit. If the battery is connected backwards, the gate will be reverse biased and the FET stays off. I've never tried it, but it looks good on paper.

If you want to use the diode/fuse arrangement, you can always use something other that a Schottky type if you're worried about the leakage current. The only real advantage that Schottky has is it's low Vf, and that's not a particular concern if you're using it for a crowbar. Most diodes can sustain a pretty hefty (very momentary) surge in current, but I would probably oversize the diode anyhow so that that it can take the short circuit current of the battery while the fuse blows. Diodes are cheap.

Also, in looking at your MAX1651 and 1793 parts, the dropout between then is ~500mV. I'm not sure you're going to be able to do a 3.6 > 3.3 conversion with them on battery power, especially considering that you'll also have some drop across your FET.

Marc

xsr_Gx

Re: Protecting Circuit from placing battery backwards

Thanks guys,

Bill: The solar panel is a 12 W, 17.2 max power voltage, .7A max power current, 1.13 short circuit current, 21.6 open circuit voltage and .8 short circuit current.

We will be using ~300mA-hrs per day continuously and will take place mostly at night. This means that during the day, the solar panel would have to provide ~ 50mA-hrs out of this ~300mAhrs. So if there is enough sun, it will charge the battery bank and will provide the ~50mA, but if there is not sun at all ( due to cloudy or night time) the battery will provide the remaining ~250mA.

Lorelec: I understand that i need a diode and a fuse, but i guess i m confused on whether to use a schottky diode or not. Also, how do i know what characteristics to use other than the current it can handle? WHen i look in the datasheets i see tons of specs taht i m not sure if i should take them into consideration.

I like your idea of using a PMOS, how would i choose it based on my needs? like the Vth, Rds(on), and Vds?

lorelec

Re: Protecting Circuit from placing battery backwards

Nevermind on the PMOS. As I have it wired, the FET's body diode will conduct with a reverse battery, completely bypassing the FET. Swapping the source and drain will correct that problem, then then the FET will turn on with charging current, reverse battery or not. It could be made to work with additional components and complexity, but probably not worth the trouble in your application.

As for diodes, you'll want to choose one (in my opinion), that will be able to handle a reasonable surge beyond its rated current -- the "Ifsm" is its surge capacity, specified in amps for so many milliseconds (and usually based on a single or partial cycle of 60Hz current). This will depend on how much short-circuit current your battery is capable of. Also, one rated for your battery voltage (Vrrm) -- nearly any diode will work in your case. Specs like reverse recovery (Tr) and forward voltage drop (Vf) aren't much of a concern for you. You just want something that will switch fast enough to short the battery and blow the fuse. Before the fuse blows, though, the diode will need to absorb all of the energy coming from your battery, so that's your main concern.

Marc

BB.

Re: Protecting Circuit from placing battery backwards

I went back and did a quick review of your MAX 5035 (switch mode down converter) and your DS 2715 (NiMH battery charger).

I have concerns that neither will operate correctly. Both assume a low impedance voltage source (from what I can see)--The solar panel is a very high impedance Current Source -- And since the feed back circuitry for the MAX part is all based on output voltage and load--I don't see how the Current Source will provide "stable power" (most likely, as soon as the MAX output current/voltage loads exceed the current of the solar panel to MAX input--the solar panel voltage will collapse.

And, varying/collapsing input voltage to the DS charge controller will mess with its varying timing and history/state machine based charging algorithms.

Perhaps Marc, Solar Guppy, or others here with more recent experiences/knowledge than mine can review and see if I am right or wrong (hey--you get what you pay for on the Internet ).

Personally, I would scrap the whole down converter / NiMH charge controller and use a Vmp>6 volt solar panel with the appropriate current rating to supply your loads. I am assuming your load will be 300 mA over a 24 hour per day period.

If you will be on the California coast, and we assume 3 hours of sun (winter, foggy summers) then the current rating of the solar panel should be at least:

• 0.3 amps * 24 hours * 1/3 hours of Sun * 1/0.66 cell eff = 3.64 amps

For a "6 volt panel" (assume Vmp=8.5 volts); for 12 volt (assume Vmp=17 volts):

• 6V Panel Watts = 8.5 volts * 3.64 amps = 31 Watts Min
• 12 Panel Watts = 17 volts * 3.64 amps = 62 Watts Min

Of course--I could have misunderstood your power needs--So, you may have to plug in your own numbers...

-Bill

PS: Your 10AH battery pack should probably be 3x 10 AH to operate without a NiMH charge controller using the above setup...

lorelec

Re: Protecting Circuit from placing battery backwards

In briefly looking over the DS2715, it appears that it can be configured as a current source, so you'd want to make sure that the charge rate of your batteries was compatible with your panel. I've never used the DS2715, so I can't really offer any real-world advice. I've used the MAX503x series with solar panels a lot and they work well, provided that you observe the usual caveats about panel loading. Bill's advice about scrapping the whole idea for a simpler design might be worth considering. My main concern is what I stated before, about the dropout between your 3.6v battery and your desired 3.3v regulated output. I think you'll run into problems there. If you can use 2.7v or even 3.0v, you might have a chance.

But if it was me, I'd at least give the experiment a try first. Half of the fun is in the trying!

Marc

xsr_Gx

Re: Protecting Circuit from placing battery backwards

Hello,

Do you think if i add an extra battery to have a nominal votage of 4.8V, would this be a better solution to provide a 3.3V output? However the solar panel would need more voltage to recharge the battery bank.