diodesvoltage
Can someone show me how to calculate voltage drops and current on this circuit?
simulate this circuit ā Schematic created using CircuitLab
I have 9 V source and 100 Ohm resistor.
So for the resistor on its own, the current would be 9 / 100 = 0.09 A across the whole circuit.
But I don't know what to look up to determine the voltage on the diode, or what the current will be on the circuit, now that it is added.
I think that when \$V_i < V_d\$, the diode does not conduct, so the circuit behaves live a potential divider, with \$V_o=V_i\dfrac{R2}{R1 + R2}\$. Otherwise, the diode behaves like a short circuit, so \$V_o=V_iāV_d\$.
One problem (maybe the only one) with your analysis, is that the diode behavior doesn't depend on \$V_i\$ but on the voltage across the diode itself, which is not the same thing in this circuit.
If you consider that the diode is "on" when \$V_i\dfrac{R_1}{R_1+R_2} > 0.7 \mathrm{V}\$ instead of just \$V_i > 0.7\mathrm{V}\$, the behavior in the "on" and "off" states will match up better at the transition.
Also, for this kind of simple modeling, when the diode is "on" we don't consider it as a short circuit, but as a fixed voltage of about 0.7 V.
They are simulating a (more or less) real 9V battery. They've modeled the battery as an ideal voltage source of 9V with a series resistance of ~1.5 ohms.
simulate this circuit ā Schematic created using CircuitLab
If you work this out, the current is 9V/(11.5 ohms) = 0.783A, so the voltage across the 10 ohm resistor must be 7.83V.
Best Answer
Try to find a datasheet on the internet. Find about which voltage would be across the LED when the current you calculated without the LED (0.09A) runs through the LED now tou added the LED. Next apply Kirchoffs Voltage Law and use this LED voltage, recalculate the current through R1 and repeat the steps above.
You see that after 1 or 2 iterations the LED voltage hardly changed; you're done