Say, I have this circuit (please ignore the zeros, and the diode should be some load that draws 10ua):

enter image description here

So, \$V_{in}\$ be 12V. I would say that D1 gets \$\frac{R_2}{(R_2+R_1)}\cdot2V\$ as voltage. Then, as I know that it draws 10ua, the resistance there should be \$\frac{12V}{10 \mu a} = 1,200,000\$. Then, the resistance for \$D_1\$ plus \$R_2\$ should be \$1/(1/1,200,000 + 1/R2)\$. If I add this up with \$R_1\$ since they are in series, I get the total resistance.

So, is the total current in this circuit according to Ohm's law: \$\frac{12V}{ (R_1 + \frac{1}{(\frac{1}{1,200,000} + \frac{1}{R2})})}\$ ? If not, where am I wrong?

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    \$\begingroup\$ It would be a big help if you drew a proper schematic. If you don't know the resistance values then delete the zeros. The 1N4004 will clamp the voltage at about 0.7 V so if it's not a diode then what is it? I suggest that you edit to explain why you are asking the question and you will get much more relevant answers. \$\endgroup\$ – Transistor Apr 28 '19 at 12:33
  • \$\begingroup\$ You can add one in using the CircuitLab button on the editor toolbar. Double-click a component to edit its properties. 'R' = rotate, 'H' = horizontal flip. 'V' = vertical flip. Note that when you use the CircuitLab button on the editor toolbar an editable schematic is saved in your post. That makes it easy for us to copy and edit in our answers. You don't need a CircuitLab account, no screengrabs, no image uploads, no background grid. \$\endgroup\$ – Transistor Apr 28 '19 at 12:34
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    \$\begingroup\$ As you say, your resistors form a divider. The equivalent Thevenin resistance is \$R_\text{TH}=\frac{R_1\cdot R_2}{R_1+R_2}\$ and equivalent Thevenin voltage is \$V_\text{TH}=V_2\cdot\frac{R_2}{R_1+R_2}\$. From this, the current in the diode will be \$I_\text{D}=\frac{V_\text{TH}-V_\text{D}}{R_\text{TH}}=V_2\cdot\frac{1}{R_1}-V_\text{D}\cdot\left[\frac{1}{R_1}+\frac{1}{R_2}\right]\$. Whether or not this will be \$10\:\mu\text{A}\$ is a different question. However, you will not be able to force that condition onto the circuit. It will be a sound result of analysis, not an assumed input to it. \$\endgroup\$ – jonk Apr 28 '19 at 16:48

No, if you put something in parallel with R2 then the voltage divider equation no longer applies. The voltage divider equation assumes that the current through R2 is exactly the same as the current through R1.

If you know that your "diode" draws 10\$\mu\$A then you should model it as an ideal current source of 10\$\mu\$A...don't try to calculate an equivalent resistance.

Having added the ideal current source you can use a variety of techniques to solve the circuit...nodal analysis, mesh analysis, etc. I've given you enough hints to get started.


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