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# Voltage divider tolerance derivation

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#### Frozenguy

##### Member
I read an article from Texas Instruments on the derivation of the effective tolerance of a voltage divider on the output of a regulator:
https://www.ti.com/lit/an/slva423/slva423.pdf

I need to know if this derivation would apply to my situation.

I have a 5 V reference from a (TI) REF5050. On that node, I want to put a voltage divider to ground. Vout will be 4.8 V.

Can I use the derivation in the TI article to determine the effective tolerance at Vout of my divider or does that derivation only apply to the top side of the divider when used with a regulator?

Cliff notes quote from the article:
The maximum error is actually inversely proportional to the divider ratio and decreases linearly as the power supply's output voltage approaches its internal reference value.

You can use it. Just rearrange it since in your case, their Vo is your Vin and there Vref is your Vout.

I'd buffer it after that if I were you...not a good idea to run a divider directly as a reference.

All that said, those resistors are going to destroy your reference's native accuracy and stability. It'd be cheaper to buy another reference than to pay for quality metal foil resistors. I don't think even metal foil resistors will cut it unless they are packaged together as a divider in the same package specifically for your ratio ($20-$50 a piece the last time I looked).

What is this being used for? Why must it be a very slightly lower 4.8V instead of 5V?

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You can use it. Just rearrange it since in your case, their Vo is your Vin and there Vref is your Vout.
Great, thank you. I have done that and got numbers that I like. Wanted to make sure because I may need to defend these numbers. Thanks again.

I read an article from Texas Instruments on the derivation of the effective tolerance of a voltage divider on the output of a regulator:
https://www.ti.com/lit/an/slva423/slva423.pdf

I need to know if this derivation would apply to my situation.

I have a 5 V reference from a (TI) REF5050. On that node, I want to put a voltage divider to ground. Vout will be 4.8 V.

Can I use the derivation in the TI article to determine the effective tolerance at Vout of my divider or does that derivation only apply to the top side of the divider when used with a regulator?

Cliff notes quote from the article:

Perhaps this article from my "Old Gold" collection will be of assistance. From the date of this article, I would bet the guy who designed this circuit no longer works for a living.

Ratch

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Perhaps this article from my "Old Gold" collection will be of assistance. From the date of this article, I would bet the guy who designed this circuit no longer works for a living.

Ratch
Interesting. Nowadays you could probably replace almost that entire circuit with just an MCU and dial in whatever ratio you want. I wonder what the noise is that on that circuit.

Interesting. Nowadays you could probably replace almost that entire circuit with just an MCU and dial in whatever ratio you want. I wonder what the noise is that on that circuit.

It says in the article that the noise is under 0.3 mV peak to peak.

Ratch

Perhaps this article from my "Old Gold" collection will be of assistance. From the date of this article, I would bet the guy who designed this circuit no longer works for a living.

Ratch

Very interesting. Thank you for sharing that.

I'm looking for (5) reference voltages. Each separated by about 100 mV. So I'll need to increase the output steps.

It says:
Logic to provide any number of output steps can be added easily.

The logic it's referring to is, I imagine, SN74L00 and SN74L73. But I'm rather confused about how this circuit works, to be honest.

A thought of mine is that I don't need any reference voltage below 4.3. Can I make the ground of the circuit 4 V, with a reference of 5.0 V so the output steps are divided across 1 V? I'll still need to increase the number of output steps.

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You just need an oscillator running at a duty cycle equal to the voltage ratio that you need.
If your Vref is 5.oV and you want 4.9V out, then you need a duty cycle of 98%
96% = 4.8V
94% = 4.7V
92% = 4.6V
90% = 4.5V
88% = 4.4V
86% = 4.3V

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