# Solving Open Circuit Voltage and Short Circuit Current

In the two images presented below, I need to solve for the 'Open Circuit Voltage' and 'Short Circuit Current' in loop 1 at the first LED String (Consisting of 10 LEDs in series) circled in red.

I have a base understanding of how to go about solving this, but it hasn't provided a solution thus far. I was hoping someone could explain the steps on how to solve these two problems. My ultimate goal is to the create a load line so that I can compare it to another LED light and find an operating point. If someone could help that would be much appreciated.

Open Circuit Voltage Short Circuit Current • I'm surprised you can't work out the short circuit current. What have you tried? (It's very easy.) I can see that each $3.42\:\Omega$ resistor is almost 1% of the $350\:\Omega$ resistor value. So two of them represents almost 2%. This means that pairing is 50 times more conductive. So the current through the short will be approximately $\frac{50}{51}\cdot 480\:\textrm{mA}\approx 470.6\:\textrm{mA}$. That isn't exact, of course. But it is very, very close. Why can't you get that far or show your work here, if you can? – jonk Sep 10 '17 at 7:06
• Hi, sorry I should have specified that I have tried and compared results with a simulator, but wasn't confident in the answers. I used a mesh technique that gave me: $$3.42i_2 + 3.42i_2 + 350(i_2-i_1) = 0$$ solving for which gave me 470.7 mA. This allows me to confirm those results which is good. The open circuit, however, is more challenging and I believe I should use a Thevenin method of approach but can't wrap my head around it. – N.Jon Sep 10 '17 at 7:49

Okay. With the inclusion of your own results that closely mirrors my "mental hand-waving" result, I'll give this a shot. First, though, by getting a precise theoretical result for the short-circuit current, based upon the comment I made.

First, we need a proper schematic. I wish you'd learn to use the schematic editor that is included with this site. We've all had to spend the time learning it, so we know exactly how little excuse there can be for not using it. Your hand-drawn images do not add any value I can think of, that isn't already available in the editor. So no excuses (for you or me.) simulate this circuit – Schematic created using CircuitLab

Now everything is numbered (as it should be) to save wasted breath and writing. I've used $V_n$ to number each node and I've added a ground symbol so that we know what we are talking about when we specify a voltage, without also saying what it is relative to. These things are important and you should get in the practice when asking questions like this. It's your question, after all, and you should add value to it yourself if you want others to take a moment because it saves them time.

Sorry about the lecture. But it has to be said.

Also note that, as already discussed in comments you received in a different question you asked, it is entirely possible to mentally combine $R_5$ and $R_6$ as a new resistor value because they are both in a series circuit and therefore the current must be the same in both of them (everything to the right of them is mentally a single "circuit element" that is in series with those two resistors.)

But I intend on leaving things exactly as you have them.

With things detailed out, now we can talk again about the short-circuit current, in the case where node $V_2$ is shorted to node $V_7$. Here, the right-hand side of the circuit becomes meaningless and we only need to focus on $R_4\mid\mid \left(R_5+R_6\right)$. The current through the $R_5+R_6$ leg will be:

$$I_{R_5}=I_{R_6}=\frac{480 \:\textrm{mA}}{1+\frac{2\cdot 3.42\:\Omega}{350\:\Omega}} = 470.799\:\textrm{mA}$$

This is formed quite simply by mentally treating resistors as conductances and using the ratio of their conductances, plus 1, to determine how the current splits between them. Just as a note, the same idea also computes the current in $R_4$ in a very symmetrical way:

$$I_{R_4}=\frac{480 \:\textrm{mA}}{1+\frac{350\:\Omega}{2\cdot 3.42\:\Omega}} = 9.201\:\textrm{mA}$$

Just by way of i-dotting and t-crossing, here's the schematic we've been talking about: simulate this circuit

So that's done.

Now, disconnecting the circuit leg between $V_2$ and $V_7$ leaves me unable to isolate the rest of the circuit. However, given your assumption that each LED string (of 10) will have the exact same voltage across them (this is an assumption and in actual practice with real LEDs [or even just realistic models of them] the assumption would be at least slightly wrong), we can simplify the rest of the circuit. simulate this circuit

There are lots of different techniques to use in solving something like this. The following approach is not the one I'd use, since I'm comfortable with nodal analysis and would greatly prefer it here. But I'm just going to follow a more basic approach and hope it communicates the way you want. (You haven't specified a specific approach you want followed, so I'm free to assume anything I want to.)

Here, we've disconnected the circuit leg between $V_2$ and $V_7$. Now we can see that $R_5$ and $R_6$ can be combined into one resistor, $R_7$ and $R_8$ can be combined into one resistor, and also that $R_9$, $R_{10}$ and $R_3$ can also be combined into one (as they could have been earlier, too.) Let's simplify (and also replace your current source and $R_4$ with its Thevenin equivalent): simulate this circuit

(In the above, the new voltage nodes are no longer the exact same ones as in the original schematic. So I will say $V_2^{'}$ when referring to this new node. Just to make sure you understand the distinction.)

But since the voltages across the LED strings are assumed to be exactly the same, this circuit can be further simplified: simulate this circuit

Here you can now easily see that this is nothing more than a series chain of resistors proceeding from one voltage supply to another. The current will be:

$$I_{TOTAL}=\frac{168\:\textrm{V}-27.4\:\textrm{V}}{R_4+R_5+R_6+R_7+R_8+R_2\mid\mid\left(R_3+R_9+R_{10}\right)}\approx 383.906\:\textrm{mA}$$

From this, we can now easily compute:

$$V_2-V_7=V_2^{'}=168\:\textrm{V}-\left(R_4+R_5+R_6\right)\cdot I_{TOTAL}=31.007\:\textrm{V}$$