# Analog Voltage Divider (DFR0051)

This is a placeholder topic for “Analog Voltage Divider” comments.

The Voltage Divider can detect the supply voltage up to 25V. The DFRobot Voltage Divider module is based on the resistor divider principle.

Hi - what is the resistance of this module - just trying to figure out the current drain on a 12 v supply?

Thanks

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Hi John,

I grabbed one of these and luckily the resistors are large enough to still have codes on the top:

This puts them at 7.5k, 30k, and 4.7k respectively. Plus the current draw of the LED

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Hi James
As a point of interest where did you get the 7k5 from. I can decipher 30k and 4k7 but the other one seems to have “858” on it. How did you arrive at 7k5 from that??

Unless there is a sneaky resistor on the other side.

Also a bit misleading to make the blanket statement that it is a 5:1 division ratio. This is all good as long as the load resistance is very high. But what is some unsuspecting maker connects something like a 10k load?? this 10k has to be considered in parallel with the 7k5 (I assume the lower leg of the divider). This reduces the lower leg of the divider to approximately 4k3 which changes the division ratio to about 8:1.
So our unsuspecting maker applies 25V to the input and gets about 3.1V at the output instead of the expected 5V.

As I think not all makers are familiar with the real world mechanics of a resistive voltage divider it would not do any harm if maybe DfRobot or someone made mention of these little points. Even specifying a minimum load resistance for a reasonably accurate output might help.
Cheers Bob

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As shown by @James and by examination of the tracks in the pic.
The input voltage is across the 858 and 303 resistors and a 472 resistor and LED.

The 858 is actually 85B which is 7.5k as @James said.
https://kiloohm.info/eia96-resistor/85B

The Wiki says it divides 25v to 5v. The current through the LED could be 1 to 5mA.

If we use the figures @James stated.
30k plus 7.5k the current drain at 12V would give a current drain of 320uA.
Therefore the LED current is the greater load.
For a 12V input you could say the current is about 2mA.

Cheers
Jim

PS My estimation of the circuit diagram.
Easy to see the LED provides the greater load to limit the effect of source impedance on the voltage divider.

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Hi James46717
You have the 30k and 7k5 reversed. As you have shown it the division ratio is 1.25:1.

EDIT 3: James46717 has modified his sketch above. It is now correct.

Still does not answer my question, how do you get 7k5 out of 858. Unless 752 is on the other side of the resistor. 30k and 7k5 are the correct values for a ratio of 5:1.
Cheers Bob
PS The load presented to a circuit by this device will change a bit depending on the applied voltage as the LED current will change but at 25V it will be a bit less than 4k7. Assuming a 2V drop across the LED i actually make the combined resistance to look like 4k496 or close enough to 4k5.

EDIT: Sorry James46717 you did answer my question. I did not read your reply properly. I was not aware of that coding system. I do note however that 4k7 does not get a mention in that chart so a mixture of codes would be required anyway. All very confusing.

EDIT 2: Inserted the word “combined” here “make the combined resistance”.

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Hi James46717
A little question

Where did @James say that ?? Nothing like that has come up in my version of the Forum. Unless you have a different version of the Forum somehow. Your post is the first mention I have seen.
Cheers Bob

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FYI:

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Thanks for picking that up, edited and changed the pic.

Comes from too much haste and not enough checking.
Regards
Jim

PS @john236194 was interested in what the load on 12V would be.
In my estimation the LED provides the most load and is a couple of mA.

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Hi James46717
At 12V the combined resistance is 4k9. That would be 2.45mA. Slightly modified if a too low resistance used on the divider output as I described above but would be OK if driving Arduino or similar input which is very high resistance.

If this is a problem for John it would be a simple matter to remove the LED or 4k7 resistor then the load would be 320µA or 0.3mA.
Cheers Bob

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Hi Gramo
Yes even I could come up with that by a simple process of elimination.
I was aware of what the “303” and “472” meant but I was asking what the relationship between “858” or as it turned out “85B” was. James did not elaborate on that and I was unaware of that coding system.

Unfortunately since I had to give up full time work I have lost contact with some of these coding systems and “standards”. Sometimes I think each manufacturer thinks up their own.
Thanks for the info. Cheers Bob

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Hi John,

Welcome to the forum!

Dont forget about the trusty perfboard as well!
That might be the best option if you need a custom ratio or need to increase the resistance of the divider.

Liam

Hi Liam
The same thing applies to a resistive voltage divider no matter where you put it. With the one in question though I fail to see any valid reason for a LED indicator except to chew up a few more mA.
Cheers Bob

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Hey Bob,

Seems a lot of boards these days have indicator LEDs which are a definite problem for portable projects when trying to figure out power consumption. Seems to be an aesthetic or user feedback trend at this point. Some give the ability to disable them, but its a definite issue that isn’t heavily looked at by those that design these boards.

Cheers,
Blayden

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Hi All,

I think that the indicator LED is doing a bit more than just showing that there is power, its also showing that the module is stepping DOWN the voltage, and won’t fry and other parts if connected incorrectly.

i.e.
If the terminal blocks of the divider module have 12V and GND connected backwards, then the project is powered from a step-down module there will be a 12V short through this module.

Just my 2c.
Liam

Hi Liam
I don’t see how this would be any guarantee. If for instance you wanted to step from 5V to 1V (which the divider would happily do) I doubt there would be enough current through the LED to provide much of an indication. In saying that, this might be the sort of LED that will glow on the slightest sniff of current. The characteristics are unknown.

Irrespective what you are doing reversing the connections is usually a recipe for disaster and the fitting of this LED will not prevent that. When a person is fiddling with this sort of thing I think a certain amount of knowledge and care could be assumed. You could be fairly sure that it would probably only happen once. This sort of thing is usually a lesson well learned.

In some cases reverse connection protection is effective but I don’t think this is one of them.
Cheers Bob

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Never a good idea to generate a voltage using a simple resistive voltage divider, unless it’s at the input of a high-impedance MOS device. Far better to use a cheap voltage regulator, or simply a few diodes to obtain a voltage drop.

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Totally agree with that but a divider does not “generate” a voltage but merely translates an existing voltage down to a more manageable or suitable level.

That is the main criteria
A lower resistance load must be considered as I explained above but that can happen as long as the load remains static. The real problem is if the load resistance is unknown or varies as this will vary the division ratio.

If there is no other alternative is available a divider can be tolerated if a high resistance input interface is provided such as an OP amp configured as a unity gain voltage follower. This will provide the bit of current required by the load and will have a low source impedance (~75Ω) as seen by this load.

Cheers Bob

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