Help understanding attenuator circuits with diodes

Hello, for a lab we recently constructed a voltage divider that used diodes. We had to obtain a 10:1 ratio using the circuit featured below.

In order to find I_Diode, we had to simulate the circuit below. We had to match the slope up to the resistor value we calculated for the voltage output ratio. We then took the current value at the slope.


Anyway, I'm just wondering how the first circuit works, the voltage attenuator circuit. How does that circuit act as a voltage divider?

Thank you for any possible help. I greatly appreciate it.
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It looks to me,as if the forward biased diode & the 900Ω internal resistance of the AC source form a voltage divider.
The reactances of the capacitors are too low,& the resistance of Rload is too high, to have much effect on the resulting attenuation.

 

The DC current source biases the diode. The dynamic resistance of the diode will typically be r=26/Idc[mA]. So if Idc=2.6mA the dynamic (AC) resistance would be around 10 ohms. The capacitors block the diode DC bias voltage from appearing at either the source or load terminals.

The diode dynamic resistance in conjunction with the source resistance and other elements forms the voltage divider. If as the previous poster suggests the capacitor impedances are low at the source frequency, then the divider is essentially made up of the source resistance and the diode AC resistance.

If r=10 ohms then the AC divider multiplier factor would be of the order of 10/910. To obtain a 10:1 ratio of input to output would obviously require a higher diode AC resistance which would in turn indicate a lower DC bias current for the diode.

The range of input voltage would be quite small if one wished to keep the output linearity to a reasonable facsimile of the input signal.