I am using the below circuit (fig1) to step down the voltage, which I found on the internet. I want this circuit to be operate properly from 1 kHz to 50 MHz. When the input is a square wave. But this circuit work properly in the range of MHz. But when the kHz frequency is given I am getting waveform like this (fig2)enter image description here.

How to calculate the value of R1 and C1?


The ratio of R1 to R2 has to match the impedance ratio of XC1 to XC2. If that condition is met then the network will have a flat frequency response.

So choose R1 to have the right ratio at DC (say 10 x R2) and pick C1 so that its impedance is ten times that of C2. This inevitably means that C2 is ten times C1 in capacitance terms.

  • \$\begingroup\$ If biasing is applied to the output of this circuit ,will it affect the waveform? \$\endgroup\$ – Akshay Jul 31 '18 at 11:00
  • \$\begingroup\$ Any biasing resistors added to the output will be in parallel with R2 and thus the effective value of R2 is lowered and therefore , to keep the frequency response flat, you will need to make C2 higher in capacitance by the same proportion that R2's effective value is reduced. \$\endgroup\$ – Andy aka Jul 31 '18 at 11:20
  • \$\begingroup\$ yeah i agree, the principle behind this circuit is to match R1C1=R2C2.IF the R1C1>R2C2,The output will be as shown. But does anyone know how did we get the equation R1C1=R2C2? \$\endgroup\$ – Akshay Aug 2 '18 at 5:26
  • \$\begingroup\$ To the OP: Basically you have a resistive divider and a capacitive divider. In order for the divider ratio to the same at all frequencies, R1/R2 must equal C2/C1. Hopefully you can see that intuitively. From there, you can re-arrange to get R1C1 = R2C2. Otherwise, the divider ratio will be different for different frequencies. \$\endgroup\$ – mkeith Aug 2 '18 at 5:36
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    \$\begingroup\$ With caps that small you need a low capacitance high impedance probe and not just an ordinary o scope probe. \$\endgroup\$ – Andy aka Aug 2 '18 at 11:34

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