0
\$\begingroup\$

schematic

simulate this circuit – Schematic created using CircuitLab

How do I apply it in the wire containing D4 and D3 if it doesn't have a resistance? The way I learned is: \$\frac{V-V_{D4}-V_{D3}}{R}\$ but in this case there is no resistance across \$V\$ and ground. Should I put \$R=0\$?

\$\endgroup\$
5
  • \$\begingroup\$ IN4733 is a 5.1V zener, D3 is a silicon diode so V = 5.75V. (assuming 0.65V forward drop for the 1N4148) \$\endgroup\$ Oct 16, 2016 at 18:12
  • \$\begingroup\$ Thanks, but it is just an example I'm really interesting in knowing if I can apply nodal analysis in this case! \$\endgroup\$ Oct 16, 2016 at 18:13
  • \$\begingroup\$ Can you show your attempt at nodal analysis, so far? You are looking for currents in and out of the node \$V\$. How would you represent the currents (in and/or out) for the branch with \$D_3\$ and \$D_4\$? Any thoughts? \$\endgroup\$
    – jonk
    Oct 16, 2016 at 18:18
  • 1
    \$\begingroup\$ In real life there is always resistance. You can't blindly follow a rule. Whatever the circuit it must follow Kirchoff's laws of voltage and current (and hence Ohm's law). In this case you could treat D4 and D3 as an open circuit because R1,D1 and R2,D2 would produce V = 5V. This is below the 5.75V needed for the zener and silicon diode to conduct. Hence they are effectively open circuit. \$\endgroup\$ Oct 16, 2016 at 18:19
  • \$\begingroup\$ Understood, thanks. I think it is not worth using this method in this kind of circuit, because as The photon showed I need to apply that exponential equation, I think i'll just do KVL when there are only diodes in a wire. \$\endgroup\$ Oct 16, 2016 at 18:41

1 Answer 1

1
\$\begingroup\$

Yes, you can. That is essentially how SPICE works.

If you're doing back of the envelope calculations, you'll likely be modelling D4 and D3 as voltage sources when they are in zener operation and forward bias operation respectively. In that case you cannot apply nodal analysis (nodal analysis doesn't apply to circuits with voltage sources) but you can apply modified nodal analysis.

The way I learned is: \$\frac{V-V_{D4}-V_{D3}}{R}\$ but in this case there is no resistance across \$V\$ and ground. Should I put \$R=0\$?

What you should have learned is to write the equations for KCL at each node:

$$i_1 + i_2 + i_3 = 0$$

assuming \$i_1\$, \$i_2\$, and \$i_3\$ are the current of the three branches connected to node v, defined so they all flow in to the node. Then you need to figure out how to express these three currents in terms of the voltage at node v and the other nodes of the circuits. If the circuit elements were all resistors, you might get an equation like the one you used. But if the elements are diodes (neglecting zener or avalanche operation), you'd need to use the Shockley equation:

$$ I_D = I_s \exp\left(\frac{qV_D}{nRT}-1\right)$$

where \$I_D\$ is the current through a diode and \$V_D\$ is the voltage across the diode (something you'd be able to write as \$v_x - v_y\$ where \$x\$ and \$y\$ are two nodes in your circuit), and \$I_s\$ and \$n\$ are characteristics of the diode. To model the zener diode you'll need an even more detailed model that includes zener behavior.

\$\endgroup\$
11
  • \$\begingroup\$ Perhaps I misunderstand. I use nodal analysis where circuits include voltage sources. Spice does it, too. (I could quote from a book written by a spice software developer about how Spice sets up each step's linear matrices, using nodal analysis.) But perhaps that's what you are calling modified nodal analysis? I guess I didn't know there were two kinds. (I can use L, C, diodes, and in fact ANY non-linear or differential or integral equation in nodal analysis as well, either to set up closed equations for symbolic solution or else to linearize prior to a time step.) It all just works. \$\endgroup\$
    – jonk
    Oct 16, 2016 at 18:28
  • 1
    \$\begingroup\$ I didn't learn nodal analysis from an electronics book or a class (never took classes on electronics.) I learned by examining how Spice does it. Concrete, complete, and works. When my son took his first electronics class (he's had more training than me), I looked at how the book taught mesh, nodal, and superposition approaches and decided that mesh and superposition were insanely tough to apply when things got tough and that their approach to nodal was Rube Goldberg. Not as simply as Spice applies it, IMO, which is actually easier for me to conceptualize. \$\endgroup\$
    – jonk
    Oct 16, 2016 at 18:42
  • 2
    \$\begingroup\$ @JoãoPedro Yes, that's the key. Every component has a behavior model. You need to memorize just a few to get by on most things. You should memorize the one for capacitors (\$I=C\frac{\textrm{d}V}{\textrm{d}t}\$) and for inductors (\$V=L\frac{\textrm{d}I}{\textrm{d}t}\$) and be able to translate those to integral form, too (for example, \$V=\frac{1}{C}\int I_t \:\textrm{d}t\$.) \$\endgroup\$
    – jonk
    Oct 16, 2016 at 18:48
  • 1
    \$\begingroup\$ When I studied calculus, it was the standard limit theorem approach developed by Dedekind and Weierstrass in the mid to late 1800's to "set right" what physicists had been using for solutions for 100 years without mathematical proofs -- fluxions or infinitesimal variables -- which mathematicians refused to accept and went "around the barn" to avoid. I was just sitting, one day, and it suddenly dawned on me that calculus is nothing but the addition of infinitesimal variables to alegbra. It wasn't until Abraham Robinson, 1964, that these were put on solid math grounding. But that isn't taught. \$\endgroup\$
    – jonk
    Oct 16, 2016 at 18:53
  • 1
    \$\begingroup\$ Mesh would be tricky for me, too. Nodal just flows out almost as fast as I can write it and I generally make far, far fewer mistakes, too. It's just natural. The way spice sets things up lays out equations as "spilling outward" on the right side and "spilling inward" on the left side. It doesn't take voltage differences in the numerators, as often shown in books. When working the right side, the node voltage at the node is treated as an absolute value without ANY regard to the nearby voltages. The left side picks those up as a separate process. Superposition. I like that approach. \$\endgroup\$
    – jonk
    Oct 16, 2016 at 19:00

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.