# Thevenin Equivalent Circuit Question

I'm reading from Hayt's Engineering Circuit Analysis and, in it, it says that

Two other methods have a certain appeal because they can be used for any of the three types of networks considered. In the first, simply replace network $$\B\$$ with a voltage source $$\v_s\$$, define the current leaving its positive terminal as $$\i\$$, analyze network $$\A\$$ to obtain $$\i\$$, and put the equation in the form $$\v_s =ai+b\$$. Then, $$\a=R_{th}\$$ and $$\b=v_{oc}\$$.

I'm curious as to the logic here. Ultimately, I understand that this arises because any linear circuit (circuit with only linear circuit elements) must have a linear IV characteristic. I'm wondering if someone can step me through this specific statement though, and why it works.

I get that I can mechanically say that any linear circuit has, at its output nodes (if I can call it that) an IV characteristic of $$\V=V_{oc}-R_{th}*I\$$ and so this equation in Hayt is simply a restatement. I'm just wondering if someone can step me through the why of this fact. Apologies if this isn't clear, it's not even that clear in my head!

• Are you asking why all linear circuits have a linear curve on their I-V graph? Or why (almost) any linear curve on the I-V graph has an equation of the form $V=V_{oc}-R_{th}I$? – The Photon Dec 8 '19 at 19:40
• I suppose an answer to both of those questions would be quite useful! But in particular, why the latter statement is true. – 1729_SR Dec 8 '19 at 20:09
• The latter is just KVL and Ohm's Law – Tony Stewart Sunnyskyguy EE75 Dec 8 '19 at 20:16
• Did you learn in math that the equation for a line is $y=mx+b$? – The Photon Dec 8 '19 at 20:18
• Indeed, funnily enough they taught that to me a long while ago. What I'm looking for is why the response of every linear circuit conceivable can be written in terms of the parameters 𝑉𝑜𝑐 and 𝑅𝑡ℎ for instance. – 1729_SR Dec 8 '19 at 20:27

Imagine one of many loads can be applied. What is the resulting voltage on the battery? simulate this circuit – Schematic created using CircuitLab <=== ignore

ref ESR

I have rounded up the current. Notice one is 50A. Why?

All batteries and power supplies have a series resistance, where rising current , results in a voltage regulation error or drop in voltage.

Batteries are pretty linear and a short circuit would not be safe after a certain time interval due to temperature rise. Op Amps and Power Supplies will be linear until active current limiting, by design.

An example of this is a car battery where a fully charged battery is 12.5V and the amount of current at 0'C that can be sustained for 30 seconds at 7.5V is called the Cold Cranking Amps. This is because batteries reduce in Ah and rise in ESR near and below freezing. thus $$Vbat=Voc - I*ESR$$

## Test

So for a 5V drop what is the series resistance , Rs or ESR if the CCA=750 A? ( I don't use Rth)

## Example 2

A solar PV has Voc= 12V and a short circuit current of 1.2A what is the maximum power operating point and max power? Assume the PV is a nonlinear current source but ideal between short and open circuit. Hint what is the PV Impedance at full solar input? What is the max load power resistance? (if you know MPT theorem)

• Example 2: Voc/Isc = Rth = 10. MPP is when load = Rth, with power Voc^2/(4R) ? – 1729_SR Dec 8 '19 at 21:40
• Note the PV supples 0 current source at Voc and Imax at short cct. So you never get 12V*10A but 12^2/40 is too low. .This is a challenge and not the same as a voltage source, nor is it a constant current source. – Tony Stewart Sunnyskyguy EE75 Dec 8 '19 at 21:50
• I'm not sure I know the answer then - if you can enlighten me or provide a source to read more I would be really appreciative! – 1729_SR Dec 8 '19 at 22:31
• Pmpt = Voc*Isc/√2 with Vmpt~ 82%, Impt~87% at max solar input only then declines with input .. trick question... – Tony Stewart Sunnyskyguy EE75 Dec 8 '19 at 22:36
• Can you explain how you got those values? – 1729_SR Dec 8 '19 at 22:54