Author Topic: Battery drain in a split supply configuration  (Read 774 times)

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Offline DrMagTopic starter

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Battery drain in a split supply configuration
« on: March 06, 2019, 11:43:10 pm »
I'm puzzling through an idea where I'm not sure what actually happens. Consider a bipolar supply using a two cell battery; Cell A positive terminal is V+, Cell B negative terminal is V-, and the junction between the cells is Gnd. The battery is used to power a circuit that has both analog components (that swing to negative voltages) and digital components (that are strictly positive).

In this configuration, since more power is used at positive voltages, does Cell A drain faster than Cell B? If not, how does the digital circuitry get energy from Cell B?

I've gone back and forth on what my gut tells me, so now I genuinely don't know what to think.
 

Offline ataradov

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Re: Battery drain in a split supply configuration
« Reply #1 on: March 06, 2019, 11:54:09 pm »
Yes, of course cell A will drain faster. What would be the argument against that?
Alex
 

Offline digsys

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Re: Battery drain in a split supply configuration
« Reply #2 on: March 06, 2019, 11:54:27 pm »
Quote from: DrMag
... does Cell A drain faster than Cell B? ...
Yes. If there is NO active charge equalization, as sometimes found in lithium packs, then A will drain first. PLUS, depending on how you charge them AFTER
they are "flat", you may not get them back to equal status !!
Quote from: DrMag
...  If not, how does the digital circuitry get energy from Cell B? ...
Doesn't. Only a complicated DC-Dc charge pump can, and that's not worth it.
What is GENERALLY done - You use the cells as a single supply and create a virtual 0V. Depending on how much current you need, it could be a simple op-amp.
Hello <tap> <tap> .. is this thing on?
 

Offline Audioguru

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Re: Battery drain in a split supply configuration
« Reply #3 on: March 07, 2019, 12:17:31 am »
Please post the schematic that uses a bipolar supply and we will show you how easy it is to use one battery.
 


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