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Problem with resistor divider output

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vbhupendra

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resistor divider output

pls see the resistor n/w
5V
------
|
|
R1
|
|
Vi-------R2------------Vout
|
|
R3
|
|
-------
-2.5V
I want

Vout = 0V if Vi=-40V
Vout = 5V if Vi=0V

what resistor values should i use ? i tried many combination after solving the network but maximum Vout i could get is 4.44V only.

If you can solve it for at least 4.9V output i would be delighted.

Thanks
 

paulux

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Re: resistor divider output

The voltage on Vi & Vout should be the same because no current flow across R2??? Or include the termination of node Vout
 

Davood Amerion

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resistor divider output

i agree with paulux.
Your schematic not compeleted. we dont know current through R2!
and ask your question clear.
this post has ambiguity on it!
 

petarpav

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resistor divider output

Hi. I thing your schematic is wrong, you must swap place on Vi and Vout.

Best Regards.
 

huojinsi

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Re: resistor divider output

Pls upload ur complete schematic!
 

vbhupendra

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Re: resistor divider output

max current n/w can consume is 1.5mA
 

Davood Amerion

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resistor divider output

You can use superposition theorem and make 2 equation with
values of {Vout=4.95, Vi=0} and {Vout=0.2, Vi=-40}.
Then solve this 2 equation. result produce ralativity of resistor values.

Vout= 5*R2/(R1||R3) + (-2.5)*R3/(R1||R2) + Vin*R1/(R2||R3)
 

IanP

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Re: resistor divider output

My gut feeling is that there is no solution for the values of Vin, Vout and Vref (+5 and -2.5V) you suggested ..
Drop the output voltage from 0-4.95 to, say, 0-4V, and use a rail-to-rail 5V opamp to extend the output range from 0-4V to 0-5V ..

Regards,
IanP
 

Kral

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Re: resistor divider output

vbhupendra,
As petarpav suggested, Vi and Vout are interchanged. However the required voltage values given in the downloaded schematic are slightly different from the voltage values that you posted. Which ones do you want:0, 5 or 0.02, 4.95?
Regards,
Kral
 

vbhupendra

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Re: resistor divider output

Kral,
I want o/p to be 0.02V to 4.95V.
and Davood i solved it with different methods applying KCL KVL superposition etc. but the max i could get is 4.44V only. If there any solution to it contraint is i can use Vref as 5V and -2.5V only. or insert some coponents at the input side.
 

Eric Best

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Re: resistor divider output

vbhupendra, IanP is right,

if VRef=+5V and Vr=-2.5V, then the required circuit in "resnw.pdf" for the stated i/o conditions a), b), cannot be realized:

a) Vout1 = 4.95V if Vin1 = 0V
b) Vout2 = 0.02V if Vin2 = -40V

because (Kirchhoff's Current Law):

(1) \[\frac{VRef-Vout1}{R2} + \frac{Vin1-Vout1}{R1} + \frac{Vr-Vout1}{R3} = 0\]

(2) \[\frac{VRef-Vout2}{R2} + \frac{Vin2-Vout2}{R1} + \frac{Vr-Vout2}{R3} = 0\]

Solving the above equations the following ratios for R1, R2 and R3 can be derived:

\[\frac{R3}{R1} = \frac{(VRef-Vr)(Vout1-Vout2)}{VRef(Vout2-Vout1+Vin1-Vin2)+Vout1Vin2-Vout2Vin1}\]

\[\frac{R2}{R1} = \frac{(VRef-Vr)(Vout1-Vout2)}{Vr(Vout1-Vout2+Vin2-Vin1)+Vout2Vin1-Vout1Vin2}\]

substituting Vin1, Vin2, Vout1 and Vout2 from a), b) we obtain:

\[\frac{R3}{R1} = \frac{4.93(VRef-Vr)}{35.07VRef-198}\]

\[\frac{R2}{R1} = \frac{4.93(VRef-Vr)}{-35.07Vr+198}\]

both these ratios must be positive! for the circuit to be realizable, so that

if VRef > Vr (numerator > 0), denominators must also be positive:

VRef > \[\frac{198}{35.07}\] = 5.64585...[V]

Vr < \[\frac{198}{35.07}\] = 5.64585...[V]

(or the second solution: VRef and Vr can change places ;-))

Conclusion:
For instance, if VRef=+5V, then Vr can be, say, +6V (min.~5.7V) or conversely (see an example in the enclosure).
These voltages have to be chosen first and then the above ratios can be calculated and finally resistors R1, R2, R3 chosen (so that the mentioned max. consumption is reached).

Best Regards,
Eric
 
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    vbhupendra

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Davood Amerion

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Re: resistor divider output

This is one solution for your specific problem:

VoH = 4.95 = 5-(R2*7.5)/(R2+R3)
VoL = 0.02 = 5-((R2*(2.5+0.02))/R3) - 40/R3

R1 = 39.4K
R2 = 5K
R3 = 745K
 

Eric Best

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Re: resistor divider output

Davood Amerion said:
VoL = 0.02 = 5-((R2*(2.5+0.02))/R3) - 40/R3

R1 = 39.4K
R2 = 5K
R3 = 745K
Davood Amerion,
your circuit works, of course, but I can't find the reason for Vr (=-2.5V) in the end (I'm not reproving you for it ;-)).
R3 could also be grounded (different value then), it serves just to decrease VoH from 5V to 4.95V when Q1 is turned off.

btw. there's something wrong in your expression for VoL, the last section. Haven't you forgotten anything? The "-40/R3" implies to be a current...
In my opinion it should have been:

\[-\frac{R2(40-Vbe)}{R1} \frac{\beta}{\beta +1}\]

am I right?

Regards,
Eric
 
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Davood Amerion

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resistor divider output

yes i mistak
Vol = 0.02 = 5-((R2*(2.5+0.02))/R3) - (R2)*40/R1

as i said it is solution for specific problem of vbhupendra.
 

vbhupendra

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Re: resistor divider output

dawood
i dont ve enough space to put anything.
 

Davood Amerion

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Re: resistor divider output

vbhupendra
this was your sentence:
If there any solution to it contraint is i can use Vref as 5V and -2.5V only. or insert some coponents at the input side.
isnt it? :?:
 

vbhupendra

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Re: resistor divider output

dawood
yes i did said but now i realized hardly i could keep any component. thanks 4 ur solutions. thanks eric.
 

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