Design a resistive voltage attenuator whose input resistance be 1 MΩ (R1,R2) and able to divide by 5 the amplitude of a DC voltage when the output of the attenuator is connected to a circuit with 1 MΩ input resistance (\$R_i\$.)

The result is:

  • R1=800kΩ
  • R2=250kΩ

My question is how do I get those resistor values out a voltage attenuation equation?

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  • \$\begingroup\$ Welcome to EE.SE! Is this homework? \$\endgroup\$
    – winny
    Oct 9, 2020 at 11:30
  • \$\begingroup\$ no, is just an example I have come across, and after doing my maths my results are totally different. I was wondering if I was doing something wrong, that's all... \$\endgroup\$
    – krm76
    Oct 9, 2020 at 12:31
  • \$\begingroup\$ You need to know how to work out the value of parallel resistors. Can you do that ? \$\endgroup\$ Oct 11, 2020 at 0:26

1 Answer 1


So the total resistance of the divider R1 + R2 ||Ri = 1M\$\Omega\$, thus R2 || 1M\$\Omega\$ = 200K\$\Omega\$ for 0.2 division, so

R1 = 1M\$\Omega\$ - 200K\$\Omega\$ = 800K\$\Omega\$.

R2||1M\$\Omega\$ = 200K\$\Omega\$ so

1/R2 = (1/200K\$\Omega\$) - (1/1M\$\Omega\$) => R2 = 250K\$\Omega\$


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