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Allowable parasitic voltage drop ?

dogday

Member
Joined
Oct 29, 2022
Messages
51
Location
Peak District
Car
Mercedes E220 cdi
Hello All,

RE: W212 E220 cdi 2012

My car has never had any electrical issues. It has the original Mercedes (Varta) battery and the car starts easily every time. The car is generally used daily for short journeys. As a matter of precaution I have fitted a new battery (Varta).

On testing I get the following readings:

1. Battery unconnected - 12.8V
2. Battery connected - 12.5V
3. Engine running -14.5V

I haven't tested for current but is a 0.3V drop when connected usual ?

Thanks for any advice.
 
You need to test for current draw with everything off...allowing enough time for the ECUs to "sleep". Voltage drop will occur when connected....but its the current that matters. This is a guide only but typically a normal parasitic draw for a car battery is between 50 and 85 milliamps (mA) for newer cars, and less than 50 mA for older cars. Anything more than that needs investigation.

EDIT....beaten to it by a faster (and more knowledgeable) typist!! 😄
 
typically a normal parasitic draw for a car battery is between 50 and 85 milliamps (mA) for newer cars, and less than 50 mA for older cars

Lay a modern car up for a week and that means you are looking at as much as 15Ah loss with the battery's internal self discharge thrown in. It explains why modern car batteries need to be so big.

The parasitic draw on older cars really was usefully lower.

I still have the figures for my old 190e

15mA alarm off

34mA alarm on
 
....and why I usually upgrade to the biggest one that will fit in the space when I replace them. That meant on my old ALFA I was able to go from 65AH to 90AH with it still fitting in the same tray and using the factory hold downs. If i did not use it for a week it would always start........but after a fortnight it was 50/50....with the big battery it would spin it over easily after a month! 159 models are known to have high current draw through the alarm.
 
I have put mine on charge today, not run for two weeks so will be at best very slow to crank, didn't even try to start it. Hospital visit tomorrow which is less than two miles away, will give it a run afterwards to Madonalds.
 
You need to check the amps.

Connect an Ammeter in series. Lock the car and wait 20 mins for the network to shut down. It should be no more than 0.06A (0.07A or 0.08A is acceptable)
Thanks BlackC55 and for the other replies. I'll give that a try. I can't remember my electronics but I assume that the ammeter just goes in series without having to span any load and should be placed on the negative side (for safety reasons) ? I wouldn't have thought that there would be any fault as the car starts easily each morning even after standing for a week ? My last car drained the battery overnight because the boot interior light was staying on and I only realised when I saw a glow below the parcel shelf one night !!
 
You need to test for current draw with everything off...allowing enough time for the ECUs to "sleep". Voltage drop will occur when connected....but its the current that matters. This is a guide only but typically a normal parasitic draw for a car battery is between 50 and 85 milliamps (mA) for newer cars, and less than 50 mA for older cars. Anything more than that needs investigation.

EDIT....beaten to it by a faster (and more knowledgeable) typist!! 😄
Thanks ALFAitalia. You say "Voltage drop will occur when connected.....". Any idea by how much typically ? Thanks
 
Thanks ALFAitalia. You say "Voltage drop will occur when connected.....". Any idea by how much typically ? Thanks

Typically by a few tenths. here's why it happens.

If a battery had zero internal resistance then there wouldn't be a voltage drop under load. But batteries do have an internal resistance even though it's very low being in the order of milliohms.

To keep it simple lets say the internal resistance is 10 milliohms and we draw 10 Amps.

Voltage drop = current x resistance

Voltage drop = 10 Amps x 0.01 Ohms = 0.1 volt

So in this example a 10 amp load would in theory reduce the voltage by 0.1 volts.

It's more complex than this because a batteries internal resistance is not a fixed value, it changes with temperature, state of charge and the battery's age. You can guess that it increases when a battery is cold, when it's partially discharged and when it gets old. It's this increasing internal resistance that reduces a batteries performance under these circumstances and that makes it hard to give an exact value for voltage drop.
 
....and why I usually upgrade to the biggest one that will fit in the space when I replace them.

Many MBs had an 'uprated battery' factory option - this was £65 on our Vito. The used W203 we bought had been specced with that option as well. This gave you a higher capacity battery and (possibly) a slightly larger tray to accommodate it.
 
Typically by a few tenths. here's why it happens.

If a battery had zero internal resistance then there wouldn't be a voltage drop under load. But batteries do have an internal resistance even though it's very low being in the order of milliohms.

To keep it simple lets say the internal resistance is 10 milliohms and we draw 10 Amps.

Voltage drop = current x resistance

Voltage drop = 10 Amps x 0.01 Ohms = 0.1 volt

So in this example a 10 amp load would in theory reduce the voltage by 0.1 volts.

It's more complex than this because a batteries internal resistance is not a fixed value, it changes with temperature, state of charge and the battery's age. You can guess that it increases when a battery is cold, when it's partially discharged and when it gets old. It's this increasing internal resistance that reduces a batteries performance under these circumstances and that makes it hard to give an exact value for voltage drop.

100%

There will also be some voltage drop in the loom when current is flowing, which you need to bear in mind if you're not measuring voltage at the actual battery terminals.
 
Typically by a few tenths. here's why it happens.

If a battery had zero internal resistance then there wouldn't be a voltage drop under load. But batteries do have an internal resistance even though it's very low being in the order of milliohms.

To keep it simple lets say the internal resistance is 10 milliohms and we draw 10 Amps.

Voltage drop = current x resistance

Voltage drop = 10 Amps x 0.01 Ohms = 0.1 volt

So in this example a 10 amp load would in theory reduce the voltage by 0.1 volts.

It's more complex than this because a batteries internal resistance is not a fixed value, it changes with temperature, state of charge and the battery's age. You can guess that it increases when a battery is cold, when it's partially discharged and when it gets old. It's this increasing internal resistance that reduces a batteries performance under these circumstances and that makes it hard to give an exact value for voltage drop.
A quick check on Google says that a typical internal resistance of a car battery in good condition is between 5 and 10 milliohms so your estimate of 10 milliohms is spot on. Therefore, my 0.3 volt drop equates to 30 Amps or 360 Watts (W=VI). That's the same as having dipped beam, main beam and tail lights all on ! If there was some component or short causing that amount of power drain the car would be dead every morning ! Thank goodness it's gone 6pm. The sun is over the yard arm so time for a beer. :)
 
## EXTRA RE: STOP START ##

Since purchasing my car over a year a go it has never had stop start. Just assumed it wasn't an option on the car as it is 12 years old. When I removed the old battery, after it had been stood unused all day, it read 12.6 Volts unconnected. After fitting the new battery I now have stop start !!!! Just shows that just measuring the battery voltage does not prove the condition of the battery !
 
## EXTRA RE: STOP START ##

Since purchasing my car over a year a go it has never had stop start. Just assumed it wasn't an option on the car as it is 12 years old. When I removed the old battery, after it had been stood unused all day, it read 12.6 Volts unconnected. After fitting the new battery I now have stop start !!!! Just shows that just measuring the battery voltage does not prove the condition of the battery !

Start/stop requires the battery to be healthy and fully charged ... it's disabled if the car thinks otherwise. That doesn't mean it's unusable, just that it's potentially not up to starting the engine multiple times per journey.
 
Start/stop requires the battery to be healthy and fully charged ... it's disabled if the car thinks otherwise. That doesn't mean it's unusable, just that it's potentially not up to starting the engine multiple times per journey.
Thanks for the reply. Yes, the battery is still usable as it starts the car without issue. A reading of 12.6V would seem to prove the battery is fully charged and in good condition but it isn't.
 
A reading of 12.6V would seem to prove the battery is fully charged and in good condition but it isn't.

12.6V sounds a little low to me ... 12.7 is often quoted as 100% charge. But it's possible the car also estimates internal resistance based on how the battery responds to charge/discharge, which is a significant part of how healthy it is.
 
12.6V sounds a little low to me ... 12.7 is often quoted as 100% charge. But it's possible the car also estimates internal resistance based on how the battery responds to charge/discharge, which is a significant part of how healthy it is.
When connected I suppose it's also possible that a small parasitic drain on a battery could be enough such that 'not enough' is left to run power hungry stop start features ?
 
When connected I suppose it's also possible that a small parasitic drain on a battery could be enough such that 'not enough' is left to run power hungry stop start features ?

More likely the resulting (small) voltage drop from an abnormal drain could be enough to make the car think the battery wasn't fully charged.

The start/stop functionality is generally extremely conservative ... there's a big list of criteria that must be met for it to become active.
 
More likely the resulting (small) voltage drop from an abnormal drain could be enough to make the car think the battery wasn't fully charged.

The start/stop functionality is generally extremely conservative ... there's a big list of criteria that must be met for it to become active.
New battery (connected) now reading 12.62V after being sat all night so looking good. Will run a current test later to see if it is within the milliamp parameters as people have suggested. Thanks for everybody's input and have a good Christmas and new year.
 

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