Voltage Divider
Written by Bryce Ringwood   

(Under construction - for comment)

This program provides a means of determining the Voltage, at the junction of two resistors in an electrial circuit. There is assumed to be a load of some sort on the voltage divider

Uses

Used to calculate the correct voltage at a resistor junction when measuring with a multimeter. For example a 20 000 Ohms/Volt meter on the 10 volt range will look like a load resistance of 20 000 X 10 = 200 K.

Voltage Diider Example pentode circuit used in text
Voltage Divider Example used in text

 

Data Entry Section

Enter the values for supply voltage, R1, R2 and RL. This corresponds to data entry boxes 1,2,3 an4. Leave 5and 6 as zero. (The load resistance is often the meter resistance. This is a constant value for digital voltmeters and VTVMs. For analog meters, multiply the Ohms/Per volt by the measuring range.) If you know the value of the resistors, the voltage wil be calculated - enter V as zero.

If you know the supply Voltage, the voltage V and the load current through RL - enter these via boxes 1,5 and 6, leaving the other boxes as zero. The resistor values will be calculated, including the load resistor. (Entering a voltage here will over ride all other values.)

On input, box 5 is the current through RL. On output, Box 5 is the current through R1.

1. Enter supply Voltage V+ Volts         

2. Enter / Calculate Resistor R1 in Ohms

3. Enter / Calculate Resistor R2 in Ohms

 4. Enter Load Resistor RL in Ohms         

 5. Enter Load / Calculate I - Amps        

  6. Enter / Calculate  V:                    volts

Additional Results

Power dissipated in Watts in   Combination    

 

 

 

Example - Applying and Measuring Screen Voltage

A pentode valve requres a screen voltage of 100 Volts. The HT+ voltage is 250 Volts, and the screen grid current at that voltage is 1.0 mA - What are  values for  R1 and R2 ?

What will the voltage recorded on a 1000 Ohms/Volt meter on its 250 volt range.

Answer1: A "rule of thumb" states that the voltage divider should carry 10 times the current drawn from it.  This is what the program uses. Enter the supply voltage as 250 V. Enter the load current I as 1m (1mA) and enter the voltage V as 100. (Boxes 1,5 and 6-put zeroes in the restR1 will be 15k and R2 will be 11.1k. Note that the equivalent load resistance will be 100k.

Answer 2 The Meter will have a resistance of 1000 Ohms per volt * 250 = 250k.  The total load resistance RL will be the combination of 100k and 250k. This can be worked out from "Resistors in Parallel" = 100*250/(100+250) = 71.25k.

Enter V as 0, re-enter RL as 71.25k, Keep the values for Supply voltage, R1 and R2.(Boxes 1,2,3 and 4 and put zeroes in the rest) See the voltage that will be displayed on the meter 97.63 V. Note the total current through R1 is now 10.16m.

Many old radio manuals specified a meter with a resistance of 1000 Ohms/Volt, although the more modern ones usually specify a 20000 Ohms/Volt meter. From the above calculation, you can see that a simple inexpensive meter is often "good enough".

Theory

R2 and the load resistance are resistors in parallel. Since the same current is flowing through R1 and the R2||RL combination, V is alculated from simple proportion as a ratio.
 

Formulae

 

\textstyle V=V_{supply} \times {\displaystyle{R2 \over {R1+R2||R_L}}}

Where:

{\displaystyle R2||R_L} ={\displaystyle {R2 \times R_L} \over {R2 + R_L}} 

 

 


References

Horowitz P, Hill W "The Art of Electronics"  - Cambridge University Press, 1988

 
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