That makes sense. So when the ratio changes, the accuracy shifts. I don't need too high accuracy, ±0.2% is fine heck ±1% would probably be OK it's only a basic power supply.
I think this one is simply additive, so if your divider has 0.1% accuracy, and your reference is 0.1% accuracy, your final output is 0.2% accuracy.I'm using a precision TI voltage reference ~ 2.048V ±0.1%, so I wonder, if I have a ±0.1% reference, and a ±0.1-0.2% divider accuracy, what is my overall accuracy? Thanks.
If you are asking that from the standpoint of statistical analysis, there are relatively simple formulas for calculating the expected imprecision of a combination, given that the individual imprecisions are defined as mean ± some number of standard deviations (or relative standard deviation = coefficient of variation = %).I'm using a precision TI voltage reference ~ 2.048V ±0.1%, so I wonder, if I have a ±0.1% reference, and a ±0.1-0.2% divider accuracy, what is my overall accuracy? Thanks.
Thanks for confirming that. I actually started my reply before post #12 and had included some stuff about standard errors (S.E.M.), etc. After an hour or so, I deleted it and decided the analysis was really quite simple based on the assumed definition of the resistor's tolerance limit. I didn't see the intervening posts until after I posted.I did statistics for GCSE, got an A in it, but we were mainly talking about probability. We didn't do much on standard deviation and distributions and the like, but I do understand it well enough...
hgmjrIn this derivation, R1 is assumed to be at the high end of its tolerance range and R2 is assumed to be at the low end of its tolerance range. Δ1 represents the tolerance value associated with R1 and Δ2 represents the tolerance value associated with R2.
\(\frac{V_o}{V_i}\ =\ \frac{R_2-\Delta_2 R_2}{R_1+\Delta_1 R_1+R_2-\Delta_2 R_2}\)
\(\frac{V_o}{V_i}\ =\ \frac{(1-\Delta_2) R_2}{(1+\Delta_1) R_1+(1-\Delta_2) R_2}\)
Note that the two terms in the denominator are two of the terms in the polynomial formed by the product of
\(\normalsize [(1+\Delta_1)+(1-\Delta_2)](R_1+R_2)\)
To show this to be true the polynomial is expanded.
\(\normalsize [(1+\Delta_1)+(1-\Delta_2)](R_1+R_2)\ =\ (1+\Delta_1) R_1\ +\ (1-\Delta_2) R_2\ +\ (1+\Delta_1) R_2\ +\ (1-\Delta_2) R_1 \)
By rearranging the above expression and subtracting the two terms in the expanded polynomial that are not present in the denominator from both sides of the expanded polynomial, a handy equivalency is formed that can be used to simplify the expression for Vo/Vi.
\(\normalsize (1+\Delta_1) R_1\ +\ (1-\Delta_2) R_2\ \ =\ [(1+\Delta_1)+(1-\Delta_2)](R_1+R_2)\ -\ (1+\Delta_1) R_2\ -\ (1-\Delta_2) R_1\)
The terms to the left of the equal sign above are the same two terms that are in the denominator of the expression for Vo/Vi. This allows the expression on the right to be substituted for it in the expression for Vo/Vi. Below is the resulting expression.
\(\frac{V_o}{V_i}\ =\ \frac{(1-\Delta_2) R_2}{[(1+\Delta_1)+(1-\Delta_2)](R_1+R_2)\ -\ (1+\Delta_1) R_2\ -\ (1-\Delta_2) R_1}\)
Further simplification of the denominator yields
\(\frac{V_o}{V_i}\ =\ \frac{(1-\Delta_2) R_2}{[(1+\Delta_1)+(1-\Delta_2)](R_1+R_2)\ -\ [(1+\Delta_1) R_2\ +\ (1-\Delta_2) R_1] }\)
Next divide both numerator and denominator by R1 + R2
\(\frac{V_o}{V_i}\ =\ \frac{\frac{(1-\Delta_2) R_2}{R_1+R2}}{\frac{[(1+\Delta_1)+(1-\Delta_2)](R_1+R_2)}{R_1+R_2}\ -\ \frac{[(1+\Delta_1) R_2\ +\ (1-\Delta_2) R_1]}{R_1+R_2} }\)
More simplification:
\(\frac{V_o}{V_i}\ =\ \frac{\frac{(1-\Delta_2) R_2}{R_1+R2}}{[(1+\Delta_1)+(1-\Delta_2)]\ -\ \frac{[(1+\Delta_1) R_2\ +\ (1-\Delta_2) R_1]}{R_1+R_2} }\)
Based on the assumption that both Δ1 = Δ2 then it should be possible to further simplify the expression for Vo/Vi:
\(\Delta_1\ =\ \Delta_2\ =\ \Delta\)
\(\frac{V_o}{V_i}\ =\ \frac{\frac{(1-\Delta) R_2}{R_1+R_2}}{[(1+\Delta)+(1-\Delta)]\ -\ \frac{[(1+\Delta) R_2\ +\ (1-\Delta) R_1]}{R_1+R_2} }\)
Perform a bit of rearrangement and some simplification yields:
\(\frac{V_o}{V_i}\ =\ \frac{R_2}{R_1+R_2}\ {\frac{(1-\Delta) }{2\ -\ \frac{[(1+\Delta) R_2\ +\ (1-\Delta) R_1]}{R_1+R_2} }\)
\(\frac{V_o}{V_i}\ =\ \frac{R_2}{R_1+R_2}\ {\frac{(1-\Delta) }{2\ -\ \frac{[R_2\ +\ \Delta R_2\ +\ R_1\ -\ \Delta R_1]}{R_1+R_2} }\)
\(\frac{V_o}{V_i}\ =\ \frac{R_2}{R_1+R_2}\ {\frac{(1-\Delta) }{2\ -\ \frac{[(R_1\ +\ R_2) +\ (\Delta R_2\ -\ \Delta R_1)]}{R_1+R_2} }\)
\(\frac{V_o}{V_i}\ =\ \frac{R_2}{R_1+R_2}\ {\frac{(1-\Delta) }{2\ -\ \frac{(R_1\ +\ R_2)}{(R_1\ +\ R_2)}\ -\ \frac{(\Delta R_2\ -\ \Delta R_1)}{R_1+R_2} }\)
\(\frac{V_o}{V_i}\ =\ \frac{R_2}{R_1+R_2}\ {\frac{(1-\Delta) }{2\ -\ 1\ -\ \frac{\Delta (R_2\ -\ R_1)}{R_1+R_2} }\)
\(\Large \frac{V_o}{V_i}\ =\ \frac{R_2}{R_1+R_2}\ {\frac{(1-\Delta) }{1\ -\ \frac{\Delta (R_2\ -\ R_1)}{R_1+R_2} }\)
The final result above contains the product of the expression for the resistor divider without any tolerance and the expression for the factor that permits the calculation of the net tolerance. The expression for the net tolerance is
\(Net\ tolerance\ =\ \Large {\frac{(1-\Delta) }{1\ -\ \frac{\Delta (R_2\ -\ R_1)}{R_1+R_2} }\)
Tom: unfortunately, nobody has given you a good answer yet. The reason for this is that your question is, in general, an indeterminate problem. This is because the "0.1%" is a meaningless number -- in other words, as John pointed out, there's no agreement on what it means.Thanks for the explaination. But does anyone know the answer to my question about the precision reference and the divider accuracy?
> 2.048(0.1%) >V
x: 2.0480 (0.1000%)
> 1(0.1%) >R1
x: 1.0000 (0.1000%)
> 2(0.1%) >R2
x: 2.0000 (0.1000%)
> # Now calculate the output voltage Vo
> <R1 <R1 <R2 + / # Calculate R1/(R1 + R2)
x: 0.3333 (0.2000%)
> <V * # Multiply the result by V
x: 0.6827 (0.3000%) # Our answer
> ivc
x: <0.6806, 0.6847>
> iva
x: 0.6827 +- 0.0020
Thread starter | Similar threads | Forum | Replies | Date |
---|---|---|---|---|
A | Digital to analog resistive circuit | Analog & Mixed-Signal Design | 44 | |
M | Resistive loads in DC/DC converters or inverters | Power Electronics | 11 | |
Basic question from one rusty on passives | General Electronics Chat | 6 | ||
H | BJT Resistive Divider Biasing | Homework Help | 13 | |
H | AC Capacitive Voltage Divider Ratio Versus Resistive | Homework Help | 11 |
by Jake Hertz
by Duane Benson
by Aaron Carman