Why does voltage at the node of a voltage divider change if a diode is connected/disconnected?
Assuming that Vdd is a positive Voltage then the diode is forward biased. The diode is in series with Vdd, R1 and the other resistor (R3). The diode`s internal instantaneous resistance in the forward direction changes with the magnitude of the current through it such that diode current times diode instantaneous resistance is always equal to approximately .7 Volts. In the circuit described the diode and R3 form a second branch for current flow through R1 to Vdd. This additional current flow through R1 causes an additional Voltage drop across R1 leaving less of Vdd present at the junction of the diode`s anode,R1 and R2. If the diode is disconnected the additional current no longer flows and the Voltage drop across R1 decreases to it`s original value which leaves more of Vdd present at the junction of R1 and R2. This should make it clear why the node Voltage changes at the junction of R1 and R2 when the diode is connected in the circuit. Additional information concerning “Or is there more
