# Voltage divider question for an equation

#### fredric58

Joined Nov 28, 2014
252
Greetings everyone. Short and to the point. I understand how voltage dividers work. My project requires 3.5v to 5v. Power will be "D" batteries, advertised as 10,000 mA each. Case 1: Use (3) 1.5v in series = 4.5v and estimated 30,000 mA and we're good to go. Case 2: Use (4) 1.5v in series = 6v w/ (est) 40,000 mA and a voltage divider to bring it down to 4.5 - 5 v. HERE'S THE QUESTION: How do you figure out the LOSS of power when using a voltage divider. I'm sure there is one and that there must be an equation. Thank you in advance for your input. Regards, Fred

#### ElectricSpidey

Joined Dec 2, 2017
2,593
Batteries in series increases the voltage but not the current.

And the formulas for the power loss in the divider is Ohm's law and the power laws.

First determine the current in the divider then times that by the voltage across the divider and that will give you the wattage.

And I don't even know what to say about your 10 amp rating for a D battery, perhaps you mean mAh. (which will also stay at 10,000 regardless of how many you place in series)

Last edited:

#### BobTPH

Joined Jun 5, 2013
7,542
You do not use a voltage divider to supply power because the voltage will vary with the load. You use a voltage regulator to drop voltage and keep the output voltage constant. The efficiency of any linear regulator is simply

Vout / Vin

For more efficiency, you use a switching regulator AKA a DC to DC converter. These are about 80 to 90% efficient.

#### LowQCab

Joined Nov 6, 2012
3,454
You need to provide as much information as possible about your project to
get the most appropriate and useful answers.
.
.
.

#### WBahn

Joined Mar 31, 2012
29,164
Greetings everyone. Short and to the point. I understand how voltage dividers work. My project requires 3.5v to 5v. Power will be "D" batteries, advertised as 10,000 mA each. Case 1: Use (3) 1.5v in series = 4.5v and estimated 30,000 mA and we're good to go. Case 2: Use (4) 1.5v in series = 6v w/ (est) 40,000 mA and a voltage divider to bring it down to 4.5 - 5 v. HERE'S THE QUESTION: How do you figure out the LOSS of power when using a voltage divider. I'm sure there is one and that there must be an equation. Thank you in advance for your input. Regards, Fred
Where are you finding D cell batteries being advertised as 10,000 mA?

I think you will discover that what is being advertises is 10,000 mAh, which is a completely different and unrelated specification.

Putting batteries in series does not increase the amp-hour rating -- the resulting rating is the lowest of any of the batteries. So putting four D cells in series leaves you with a battery that starts at about 6 V and that has a 10 Ah capacity.

Voltage dividers are seldom suitable for providing power to anything. In order to have decent regulation, you need to use such low resistance in the divider that only about 1% to 10% of the current from the battery is actually used to power the circuit.

Use a low-dropout voltage regulator or a switch-mode converter instead.

#### Ian0

Joined Aug 7, 2020
8,390
You might understand voltage dividers, but you haven't grasped the difference between current and battery capacity.

Anyone who is advertising batteries as 10,000mAh instead of 10Ah is trying to cheat you.
Do you think Yuasa would ever advertise a battery as 7000mAh?

#### fredric58

Joined Nov 28, 2014
252
Hello, actually, I know quite a bit about batteries. But after 16 chemo treatments, I have what’s considered chemo brain fog sometimes. I understand lead acid, AGM, deep cell, Lipo, etc. etc. that if you run them and series you increase the voltage but the amp hours remain the same. If you run them in parallel, the voltage stays the same and the amp hours increase. https://uk.rs-online.com/web/content/discovery/ideas-and-advice/d-batteries-guide is where I got the 10,000mAh figure for a traditional D cell size battery. So what I could do is run 3-D cells in series which would give me 4 1/2 V and 10,000 Milla amps and then run another 3D cell batteries in series for the same 4.5 V 10,000 Milla amps and run those two in parallel and come up with 4.5 V and 20,000 Milla amps?

#### fredric58

Joined Nov 28, 2014
252
Where are you finding D cell batteries being advertised as 10,000 mA?

I think you will discover that what is being advertises is 10,000 mAh, which is a completely different and unrelated specification.

Putting batteries in series does not increase the amp-hour rating -- the resulting rating is the lowest of any of the batteries. So putting four D cells in series leaves you with a battery that starts at about 6 V and that has a 10 Ah capacity.

Voltage dividers are seldom suitable for providing power to anything. In order to have decent regulation, you need to use such low resistance in the divider that only about 1% to 10% of the current from the battery is actually used to power the circuit.

Use a low-dropout voltage regulator or a switch-mode converter instead.
thank you for the reply.https://uk.rs-online.com/web/content/discovery/ideas-and-advice/d-batteries-guide is where I found the info. I’ll run 3 in series, another 3 in series then run the both in parallel. 4.5v - 20,000mAh now no need for a reg or divider. More details in my other reply. On a phone right now. Can’t see everything at the same time.

#### fredric58

Joined Nov 28, 2014
252

#### sagor

Joined Mar 10, 2019
861
D cells are indeed rated 10,000mAh, but under certain load conditions. The bigger the load, usually the overall capacity decreases. A very light load, and capacity can be rated bigger in certain cases.
https://en.wikipedia.org/wiki/D_battery
Do not mix up mAh with mA, they are two different terms. If 3 cells will last under load for the duration you need, then just go with that, less parts. Without knowing your load in mA, hard to tell if you should try 4 cells and a regulator of some sort, or just stick with 3 cells directly.
Check the load curve vs duration for Duracell:

https://web.archive.org/web/2012052...uct_Data_Sheet/NA_DATASHEETS/MN1300_US_CT.pdf

Last edited:

#### BobTPH

Joined Jun 5, 2013
7,542
So what I could do is run 3-D cells in series which would give me 4 1/2 V and 10,000 Milla amps and then run another 3D cell batteries in series for the same 4.5 V 10,000 Milla amps and run those two in parallel and come up with 4.5 V and 20,000 Milla amps?
Yes, but the unit is still mAH, not mA. mA is a current, maH is a capacity. Batteries sometimes come with a max current rating, which is usually expressed in in C, which is ratio to capacity. So a battery might be rated 10AH and 3C, which makes the max current 30A.

#### crutschow

Joined Mar 14, 2008
32,903
mAH is actually giving the charge storage capacity of the battery in coulombs (1 ampere for 1 second = 1C), thus 10,000mAH equals 10Ah*3600 sec/h = 36000 coulombs.

#### Audioguru again

Joined Oct 21, 2019
6,158
The datasheet for UK RS-online has no graphs that show the battery voltage running down as it gets used. The voltage starts at 1.5V per cell then quickly drops to 1.3V then slowly drops to 1V. The voltage drops quickly after dropping to 1V per cell. At 0.8V per cell then the battery is almost dead. Will your product still work properly with a voltage of only 0.8V x 3= 2.4V?

#### WBahn

Joined Mar 31, 2012
29,164
You might understand voltage dividers, but you haven't grasped the difference between current and battery capacity.

Anyone who is advertising batteries as 10,000mAh instead of 10Ah is trying to cheat you.
Do you think Yuasa would ever advertise a battery as 7000mAh?
View attachment 292615
They aren't trying to cheat you any more than someone selling you a 10,000 µF capacitor is.

Small batteries, particularly primary batteries, are typically specified in mAh, even if the capacity is large enough to make Ah a more justifiable unit. These range of capacities span more than three orders of magnitude starting at just a dozen or two mAh, so using a fixed unit for capacity ratings is less likely to lead to confusion for consumers.

Frankly, I wish grocery stores would adopt this attitude. While they are required to provide unit pricing so that consumers can more easily do comparison shopping, they make no attempt to use the same units. So if you look at some product, say canned tomatoes, and look at different sized containers from the same brand, you are likely to see one priced per oz, one priced per lb, and one priced per can. Totally useless.

#### WBahn

Joined Mar 31, 2012
29,164
Hello, actually, I know quite a bit about batteries. But after 16 chemo treatments, I have what’s considered chemo brain fog sometimes. I understand lead acid, AGM, deep cell, Lipo, etc. etc. that if you run them and series you increase the voltage but the amp hours remain the same. If you run them in parallel, the voltage stays the same and the amp hours increase. https://uk.rs-online.com/web/content/discovery/ideas-and-advice/d-batteries-guide is where I got the 10,000mAh figure for a traditional D cell size battery. So what I could do is run 3-D cells in series which would give me 4 1/2 V and 10,000 Milla amps and then run another 3D cell batteries in series for the same 4.5 V 10,000 Milla amps and run those two in parallel and come up with 4.5 V and 20,000 Milla amps?
You are again conflating millamps with milliamp-hours. They are NOT the same thing!

#### WBahn

Joined Mar 31, 2012
29,164
thank you for the reply.https://uk.rs-online.com/web/content/discovery/ideas-and-advice/d-batteries-guide is where I found the info. I’ll run 3 in series, another 3 in series then run the both in parallel. 4.5v - 20,000mAh now no need for a reg or divider. More details in my other reply. On a phone right now. Can’t see everything at the same time.
There's no way to know if this will be adequate without more information about your application. What is the current draw that you need? How long do you need the batteries to last?

What is the highest and lowest voltage that your system can run properly on? Is that what the 3.5 V to 5 V that is mentioned in your first post refers to?

The effective capacity of alkaline batteries is pretty sensitive to the actual current draw. In general, the higher the current, the less the available capacity is.

#### Ian0

Joined Aug 7, 2020
8,390
They aren't trying to cheat you any more than someone selling you a 10,000 µF capacitor is.

Small batteries, particularly primary batteries, are typically specified in mAh, even if the capacity is large enough to make Ah a more justifiable unit. These range of capacities span more than three orders of magnitude starting at just a dozen or two mAh, so using a fixed unit for capacity ratings is less likely to lead to confusion for consumers.
Maybe - or maybe not. Capacitors have always been in thousands of microfarads, to avoid confusion with old terminology where MFD was used for microfarad, so mF for millifarad could easily get confused.
Batteries have always been in both mAh and Ah depending on the size. The use of 10000mAh is a new phenomenon which appeared at around the same time as fake lithium batteries with smaller batteries in cardboard tubes etc. I remain deeply suspicious - it looks like deliberate obfuscation to me.
Frankly, I wish grocery stores would adopt this attitude. While they are required to provide unit pricing so that consumers can more easily do comparison shopping, they make no attempt to use the same units. So if you look at some product, say canned tomatoes, and look at different sized containers from the same brand, you are likely to see one priced per oz, one priced per lb, and one priced per can. Totally useless.
I agree with that one! Even here in Britain where they are forced to use metric units, they can still manage to create confusing by using "per kg" for one product and "per 250g serving" for another, and then measure some by mass and others by volume - do you know the density of tinned tomatoes? The least well off are the ones who suffer as they tend to be the ones that didn't do well at school especially in maths.

#### Audioguru again

Joined Oct 21, 2019
6,158
Many food products are priced per kg and the same product from another manufacturer is priced per liter.
They both have extra water added.

#### fredric58

Joined Nov 28, 2014
252
You are again conflating millamps with milliamp-hours. They are NOT the same thing!
Sorry I was using voice dictation. Typing on a phone is quite difficult when you have peripheral neuropathy. So is buttoning your shirt. I understand amp hours. EX: A 10 ah battery will provide a device that draws 1 amp in a perfect world 10 hours of power. however depending on the type of battery you can only use 50% to 80% of the batteries power. lead acid about 50% lithium ion phos around 80% before damaging the batteries. I will use just my computer on future posts. I don't want to annoy anyone.

So as far as milliamps go. For me that is a measurement on my multi meter for low powered projects. is that correct.

#### sagor

Joined Mar 10, 2019
861
You still have not stated what the load current is (or will be). For example, the Duracell charts (in the PDF I posted earlier) show various values for "service life". Assuming you want to keep 3 cells each at 1.2V as a minimum (3.6V total for 3 cells), you get various "life" of the battery (at room temperature).
At 150mA load current (approx. 10 ohm load), you can expect about 76-78 hours before the battery reachs 1.2V from 1.5V
At 750mA load current, you can expect about 3.5 hours before the cell drain down to1.2V
At 1A load current, expect around 1.5 hours to reach 1.2V left in the battery.
The relationship is not linear. You have to determine the load you will be putting on the batteries before anyone can estimate how long they will last.