ts a shame because i know how to get the output voltage from my divider by changing the resistor but i still dont know why it works
can anyone explain post 2 but in laymans terms? the clamping bit i dont understand how it is doing that?
Generally diodes
only conduct current in one direction when a DC voltage greater than the diode turn on voltage is applied.
The diode turn on voltage is around 0.6V to 1V depending upon the diode type
so:
The triangle of the diode symbol is the Anode and the Bar is the Cathode.
If the voltage applied between the Anode [+v] and the Cathode [-] is less than 0.6V, the diode will not conduct current, when its greater than 0.6V the diode will start conducting.
If we now reverse the voltage applied to the diode, so that Anode[-v] and Cathode is [+v] the diode will not conduct.
You will see on your diagram two diodes are in reverse parallel connection.. the anode of one diode is connected to the cathode of the other diode.
So now if a voltage less than 0.6V is applied across the diodes , neither will conduct.
If the applied voltage is say +1V, the forward biassed diode will conduct and clamp at +0.6V,,, if now the applied voltage is reversed -1V the other diode will conduct and clamp the voltage at -0.6V.
The resistors of the first divider will limit the current flowing thru the diodes
This clamp diode circuit will therefore never allow the voltage to exceed +/-0.6V across the diodes.
This clamped voltage is applied to the output resistive divider.
Do you follow OK.?