4
\$\begingroup\$

Enter image description here

I know this is a basic question, but I'm just not good at solving diodes in parallel circuits.

I tried applying the current divider rule to get the current flowing through the diode, such that:

Enter image description here

\$\endgroup\$
  • 2
    \$\begingroup\$ Where did the number 70 come from ? \$\endgroup\$ – AJN Nov 29 '20 at 17:10
  • \$\begingroup\$ The current is 33.548 mA for an ideal diode. What did your number come to and ditto the @AJN comment. \$\endgroup\$ – Andy aka Nov 29 '20 at 17:11
  • \$\begingroup\$ This would be an approximate answer since the application of the current divider formula neglects the presence of the diode. The approximation will get worse and worse as the current gets smaller and smaller. After solving, compare the voltage drop across the diode and the voltage drop across 200||90. If the voltage across 200||90 is not significantly larger than 0.7 V, the approximation may not be valid. \$\endgroup\$ – AJN Nov 29 '20 at 17:13
  • \$\begingroup\$ *sorry it should be 10 instead of 70 @AJN \$\endgroup\$ – ks0937 Nov 29 '20 at 17:21
  • 2
    \$\begingroup\$ You shouldn't leave out the units. Even better is give the resistances and voltage source names, R1, R2, R3, V1, etc. and only insert actual values until you have solved the equation (at the very end). That applies even if it is not specified in the homework or exercise. \$\endgroup\$ – Peter Mortensen Nov 30 '20 at 10:23
12
\$\begingroup\$

but I'm just not good at solving diodes in parallel circuits

So, in general for problems like this (which you do hit in real life if you're doing circuit design, so it's not just academic), you start by replacing the diode by an open circuit:

schematic

simulate this circuit – Schematic created using CircuitLab

Solve the circuit using whatever means you want. If \$V_d\$ is negative, or less r than the voltage drop you're using for your diode (0V, 0.6V, etc. -- or whatever the prof says), then you're done -- the diode isn't conducting, and life is wonderful. If \$V_d\$ is greater than the designated diode drop, then replace it with a voltage source:

schematic

simulate this circuit

And again, solve the circuit by whatever means you want. Me, I'd do it by a series of simplifications, replacing supplies with their Thevenin equivalents and parallel resistors by their parallel equivalents, to the extent that I could. Unless I have a headache and can't think, and the simplifications just aren't working -- then I'd use node analysis.

In the event that you have multiple diodes, life gets more complicated -- you probably want to do as above, replacing the opens with 0.7V sources where it's clear to do so then repeat -- you may find that as you add diodes other diodes get "turned on" (the obvious case here is a bunch of diodes in series).

As you do a lot of these, you'll either develop an intuition for it or not. I certainly have worked with accomplished analog circuit designers who cannot just glance at a schematic and get it right, so if you can't do it off the bat that doesn't mean you're in the wrong field.

\$\endgroup\$
8
\$\begingroup\$

You should really simplify stuff before launching into the math. For instance: -

enter image description here

Can you see how much easier this is to simplify following the standard source transformation I did above? Do you know what the next step is?

\$\endgroup\$
  • \$\begingroup\$ What is the name for the voltage source transformation? \$\endgroup\$ – Peter Mortensen Nov 30 '20 at 10:29
  • \$\begingroup\$ @PeterMortensen you can use thevenin or millman's theorem for this. \$\endgroup\$ – Andy aka Nov 30 '20 at 11:16
5
\$\begingroup\$

You can solve such circuits by iteratively simplifying them.

Whenever you see two resistors in parallel, such as the 200\$\Omega\$ and 90\$\Omega\$ resistors, combine them. (as you have done.)

Same thing if you see to resistors in series.

Whenever you see a voltage supply, followed by a voltage divider, you may replace that with it's Thevanin equivalent, a reduced voltage with a single resistor.

If you do those operations, your circuit will now look like this (if I have done my math right)

schematic

simulate this circuit – Schematic created using CircuitLab

\$\endgroup\$
  • \$\begingroup\$ Since it's now series, and we're finding the mesh current, I'd go one step further and swap D1 with R2 then combine R1+R2 so that D1 is now driven by a Thevenin equivalent source of 2.25V through (62+5)=67ohm... next steps would depend on whether the diode is modeled as a simple ideal diode or a complicated exponential-law diode. \$\endgroup\$ – MarkU Nov 30 '20 at 0:42
  • \$\begingroup\$ @MarkU I wanted to leave some work for the OP. :-) \$\endgroup\$ – Math Keeps Me Busy Nov 30 '20 at 1:14

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.