can i know what, why and how this Rth is parallel / how its calculated?

answer by jonk: thevenin analysis

i mean my solution would be:

I1 = I2+ I


I1 = (9-V2)/10k
I2 = V2/20k
I = V2-0.7/20k


I1 = I2+ I
(9-V2)/10k = V2/20k + V2-0.7/20k
(9-V2)/10k = (2*V2- 0.7)/20k
2*(9-V2) = (2*V2- 0.7)
edit: V2 = 6.23v // missed up here
edit: V2 is actually 4.675v

V = V2 - 0.7V = 5.53
i = 5.53/20k = 0.276mA

this supposed to be example from a university lecture, what im missing here?

Edit: how this nodal analysis is different than above thevenin analysis

  • \$\begingroup\$ I addressed a similar question here. But it seems as though you may not be familiar with the process of forming a Thevenin equivalent circuit with one voltage source and one series impedance from a pair of voltage sources and a pair of impedances. Do you know how to turn a voltage source and a resistor divider into a thevenin voltage and a series resistance? \$\endgroup\$ – jonk Mar 25 '19 at 3:23
  • \$\begingroup\$ no, i just looked it up, my method above was nodal analysis i still don't see why they are different. i will check Thevenin and study to validate what's wrong \$\endgroup\$ – Hasan alattar Mar 25 '19 at 3:42


Let's start with a simple voltage divider. I'll use the name \$V_\text{CC}\$ for the name of a given voltage source for the divider, for lack of a better name. The point of the exercize is to replace the circuit on the left with an equivalent circuit on the right. This is the process of creating a Thevenin equivalent.


simulate this circuit – Schematic created using CircuitLab

For the above case, the right-hand side Thevenin equivalent is \$V_\text{TH}=V_\text{CC}\cdot\frac{R_2}{R_1+R_2}\$ and \$R_\text{TH}=\frac{R_1\cdot R_2}{R_1+R_2}\$.

So, now your circuit can be transformed using the above Thevenin idea (if you need a proof, there are plenty available on the web.)


simulate this circuit

From the right side, you can compute the current as \$I=\frac{6\:\text{V}-700\:\text{mV}}{6.667\:\text{k}\Omega+20\:\text{k}\Omega}=198.75\:\mu\text{A}\$ (the diode voltage subtracts from the total voltage available, prior to dividing by the total series resistance) and the voltage at \$V\$ as being \$V=I\cdot 20\:\text{k}\Omega=3.975\:\text{V}\$. (\$V_\text{X}\$ would be \$700\:\text{mV}\$ above \$V\$, of course.)

As you can see, this appears to match the answer you presented at the outset.


Using nodal analysis I get:

$$\begin{align*} \frac{V}{R_3}&= I_{D_1}\\\\ \frac{V_\text{X}}{R_1}+\frac{V_\text{X}}{R_2}+I_{D_1}&=\frac{9\:\text{V}}{R_1}+\frac{0\:\text{V}}{R_2}\\\\ V_\text{X}&=V+V_{D_1} \end{align*}$$

This is three equations and three unknowns. Given \$V_\text{CC}=9\:\text{V}\$ and \$V_{D_1}=700\:\text{mV}\$ and your resistor values, this provides:

$$\begin{align*} I_{D_1}&=198.75\:\mu\text{A}\\\\ V&=3.975\:\text{V}\\\\ V_\text{X}&=4.675\:\text{V} \end{align*}$$

Same results as before.

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  • \$\begingroup\$ thank you i found out i ve stupid algebra mistake \$\endgroup\$ – Hasan alattar Mar 25 '19 at 5:19
  • \$\begingroup\$ @Hasanalattar Oh! We all make such mistakes. No problem. \$\endgroup\$ – jonk Mar 25 '19 at 5:20

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