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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?

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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.

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  • \$\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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