Because the remaining voltage drops across the resistor.I've known the diode drop the voltage by 0.6V.
But why is the dropped voltage different depend on the resistor at the below pictures?
Hi,I've known the diode drop the voltage by 0.6V.
But why is the dropped voltage different depend on the resistor at the below pictures?
Resistor is in series with the diode and resistance limits the current flow. Higher the resistance more voltage is dropped across the resistor and less current flows through it.Hi,
The most basic answer is that two things in series drop voltages, and the one with the higher resistance drops the most voltage. So if you increase one resistance that resistance will drop more voltage and thus the other device in the series string will have less voltage across it now.
The basic rule is:
The sum of the voltage drops around a series circuit is equal to zero.
Resistor is in series with the diode and resistance limits the current flow. Higher the resistance more voltage is dropped across the resistor and less current flows through it.
I = V/R
If R is increased then I through the series circuit reduces.
V = IR
If R is increased then V (drop across resistor) across resistor increases.
V = Vd + Vr
Vr = (V - Vd) / I
Vd = (V - Vr) / I
I = (V - Vr - Vd) / (Rr + Rd)
The two different values of resistor (10Ω and 10kΩ) result in different amounts of current through the diode. Diode forward voltage is not constant, and 0.6V (or 0.65V or 0.7V, or whatever your preference) is merely an approximation. In reality, diode forward voltage varies as a roughly logarithmically function of forward current.I've known the diode drop the voltage by 0.6V.
But why is the dropped voltage different depend on the resistor at the below pictures?
Hi,The two different values of resistor (10Ω and 10kΩ) result in different amounts of current through the diode. Diode forward voltage is not constant, and 0.6V (or 0.65V or 0.7V, or whatever your preference) is merely an approximation. In reality, diode forward voltage varies as a roughly logarithmically function of forward current.
Thank you.hi JD,
The resistance response of the diodes forward voltage versus current is not linear, ref this image.
E
View attachment 133367
Thank you very much.