# Attenuator

An attenuation network is often used in devices like a (digital) voltmeter, in oscilloscopes, etc. It offers the possibility to expand the input range of the measurement tool and thus makes it more versatile. Usually the attenuation network is build with a couple of resistor as shown in fig. 1. If the input resistance of the rest of devices is considered indefinite (as for example with a J-FET) the voltage is Vo=Vi*R1/(R1+R2). This works very well with DC-voltages.

However, with AC-voltages, things are not quit that simple. There are parasitic capacitances that ruins this simple setup. Figure 2 shows what is happening. For a DC-voltage this setup still works. However, with an AC-voltage it now becomes different. The total resistance is now the parallel impedance of the resistor and the capacitance and is thus lower. The meter will thus show a lower reading and the accuracy of the instrument so carefully constructed is down the drain….

However, this can easily solved by adding some more capacitors. Yes, you read it correct even more capacitors!

$R_1C_1 = \tau_1$ $R_2C_2 = \tau_2$ $Z_1 = R_1//C_1$ $Z_1 = R_1//Z_{C1} = { {R_1 {{1}\over {j \omega C_1} }} \over {R_1 + {{1}\over {j \omega C_1} }} } = {{R_1}\over {1+j \omega R_1C_1} }$ $Z_2 = R_2//C_2$ Likewise : $Z_2 = R_2//Z_{C2} = {{R_2}\over {1+j \omega R_2C_2} }$ ${{V_o}\over{V_i}} = {{Z_2}\over{Z_1+Z_2}} = {{ {{R_2}\over {1+j \omega R_2C_2} }}\over {{{R_1}\over {1+j \omega R_1C_1} }+{{R_2}\over {1+j \omega R_2C_2} } }}$ $\tau_1 = \tau_2 = \tau$ ${{V_o}\over{V_i}} = {{ {{R_2}\over {1+j \omega \tau} }}\over {{{R_1}\over {1+j \omega \tau} }+{{R_2}\over {1+j \omega \tau} } }} {{1+j \omega \tau}\over {1+j \omega \tau}}$ ${{V_o}\over{V_i}} = {{R_2} \over {R_1 + R_2}} qed$