Battery Drain - Relay x 2 - SOLVED.

So I have a project which uses a 9v battery to power my arduino that controls an LCD and a TPIC595, which turns two relays on or off (no duty cycle playing here).

The relays are OJ-SH-105NM:
5V
55.6 Ohms Coil
91mA coil current

When the relays are on, the voltage from the battery gets lower and lower. This is immediately obvious even without a DMM from the LCD gradually getting more and more dim.

I have measured the current in line with the battery:
Relays are off = ~100mA
Relays are on = ~160mA

Do I need to put a current limiting resistor on my relays to try to decrease the operating current when on?

Or is a 9v battery simply not enough juice for two 5v relays like the ones I have, plus the bright LCD display.

Please forgive the crude diagram:

If you mean what I know of as a PP3 then they are good for smoke alarms and not much else. You need a bigger battery or mains power supply.

Is that because of there small Ah? ~1.2amp Hours.

Also, thank you for noting PP3, I didnt know they were reffered to as this.

If you have 1.2Ah ones they must be lithium. The more usual capacity is around 500mAh.

If they really are 1.2Ah then I'd expect them to perform a bit better than you describe. If they are the standard ones then their performance is not good for the kind of things you are using them for. Get some higher capacity batteries or a mains power supply.

I just took the max from a quick Google search, the PP3 I have its just the bog standard one.

6x AAs might do the job, just need to see if I have room.

If you are using the built in linear regulator you are wasting about half the battery anyway as the excess voltage is disapated as heat. A buck converter wastes a lot less of the battery.

What are you switching with the relays? There are other ways to switch things that use less current.

Can I just replace the stock one for a LTC3525 on the nano PCB itself?

Idea from: » Arduino misconceptions 6: a 9V battery is a good power source.

Never used one but I did a quick search and I'm not sure what to say. If you use one of those it's a step up converter so the input would be from 2 cells @3V I guess. Unless someone can comment I suggest you look into it more and maybe try one. Are you short of space?

Not massively short on space. 4xAA would be ideal, I can stretch to 6xAA.

Do you have an alternate to the linear regulator you use?

There are plenty of ready built buck converters available on your favourite retail website. I suggest you use 6 cells and convert to 5V. 6V is too close to 5V. I use LM2574 or similar, but that's just the chip, they need other components around them. I believe Pololu have some pretty good buck converters, but I have never tried one. Pololu have a good reputation for quality.

Okay I will take a look tomorrow and get back with an idea.

With the external buck, do you just remove the linear regulator on the board and solder the buck output to where the regulator output would have been?

With the external buck, do you just remove the linear regulator on the board and solder the buck output to where the regulator output would have been?

No, leave the existing regulator where it is, feed 5V into the 5V pin on whatever board you have.

Say if I were to use a MP1584EN, which has an input voltage minimum of 4.5V, would I not be loosing half the juice of all the batteries?

Or do batteries work a little different to this, in the sense that by the time that the series voltage is 4.5V, the total current supply will be very low?

Maybe a MT3608 buck boost would be better with inputs as low as 2V?

Still, does battery current drain quicker than the voltage level they will be at?

I don't understand what you are asking. I looked that up and it's a buck converter so needs an input voltage higher than the required 5V output. I didn't check the specification so I don't know how much higher.

Please ask your question in a different way.

Check the efficiency of the buck boost converter, I think (not sure) they are less efficient than buck converters.

Sorry for my bad questions.

With a buck converter, they take a higher voltage and produce 5V correct? - so when the voltage of the batteries drops to <5V, the buck will not supply the required 5V? Does this not mean I would loose a lot of the voltage left in the batteries?

Hence the possible idea of the buck boost, to use every last bit of the batteries - giving up some efficiency like you mention.

With a buck converter, they take a higher voltage and produce 5V correct? - so when the voltage of the batteries drops to <5V, the buck will not supply the required 5V?

Correct. Higher than 5V because it won’t convert 5V to 5V.

Does this not mean I would loose a lot of the voltage left in the batteries?

The question suggests that you don’t understand batteries. If a 9V (for example) battery has discharged to 6V that doesn’t mean 1/3 is gone with 2/3 left. It means there is pretty much nothing left.

PerryBebbington:
Correct. Higher than 5V because it won't convert 5V to 5V.

Awesome thanks.

PerryBebbington:
The question suggests that you don't understand batteries. If a 9V (for example) battery has discharged to 6V that doesn't mean 1/3 is gone with 2/3 left. It means there is pretty much nothing left.

Is this not the case for 6 series 1.5V batteries though?

If 6 x 1.5V batteries, 9V, discharge to 6V, does that not mean that each battery has only discharged by 0.5V. So each battery has 1V left?

Sorry if I don't understand, you will be teaching me here so thank you a lot!

No, batteries don't discharge like that. An Idea battery would keep exactly the same voltage until it was discharged then drop quickly to zero. Real batteries don't quite do that but they come close. Once it drops by 1/3 a battery is pretty much gone.