Finding Thevenin Equivalent in a Bridge Circuit - Page 1
Forum section: electronics / beginners

author: Legion
author_rank: Frequent Contributor
timestamp: February 27, 2014, 11:35:36 pm
content: I'm not sure how to approach finding the thevenin voltage and resistance with a bridge circuit. I'm trying to solve the circuit below:The 2k2 is the load. I know the voltage at point B is 9V, but I can't figure out how to find the voltage at point A. Because the bridge is unbalanced I can't use resistor ratios. I also can't use series-parallel rules to collapse the resistances. I tried using voltage loops, ie. the rule that the total voltage in any given circuit loop should be zero, but that just gives me different voltage drops across the resistors depending on the loop I take.How do I use thevenin with an unbalanced bridge circuit?

author: T3sl4co1l
author_rank: Super Contributor
timestamp: February 27, 2014, 11:38:44 pm
content: What do you mean "the voltage at B is 9V"?With no ground indicated, that voltage isundefined...Tim

author: The Electrician
author_rank: Frequent Contributor
timestamp: February 27, 2014, 11:40:12 pm
content: There's a trick you can use. It's explained here:http://kauko.hallikainen.org/rw/theory/theory6.html

author: c4757p
author_rank: Super Contributor
timestamp: February 27, 2014, 11:40:42 pm
content: Quote from: Legion on February 27, 2014, 11:35:36 pmNo?Yes you can, you just won't get all the answers at once. Try it, you'll get some information, then you can move on with that.

author: Legion
author_rank: Frequent Contributor
timestamp: February 27, 2014, 11:47:39 pm
content: Quote from: T3sl4co1l on February 27, 2014, 11:38:44 pmGround is the negative terminal of the battery. Since point B is common to the anode of the battery, it is 9V above ground.

author: IanB
author_rank: Super Contributor
timestamp: February 27, 2014, 11:56:31 pm
content: Quote from: c4757p on February 27, 2014, 11:40:42 pmSurely you can only use series-parallel formulas where two resistors in a circuit can be replaced with a single resistor connecting between the same end points or nodes? In an unbalanced H-bridge I don't see any two resistors that can be replaced with one one while leaving the circuit unchanged?

author: Legion
author_rank: Frequent Contributor
timestamp: February 27, 2014, 11:59:36 pm
content: Quote from: IanB on February 27, 2014, 11:56:31 pmI can't tell if you're being rhetorical or not. But your statement above is my understanding of why I can't use series-parallel formulas.

author: c4757p
author_rank: Super Contributor
timestamp: February 28, 2014, 12:01:48 am
content: Quote from: IanB on February 27, 2014, 11:56:31 pmMy mistake, I was looking at the one on the right with the missing resistor. 1k+2k7 in series, that in parallel with 3k3, then all in series with 1k5.

author: Andy Watson
author_rank: Super Contributor
timestamp: February 28, 2014, 12:18:43 am
content: Smells like a homework question, but ...There is no formula (until you've completed the problem) for determining the thevenin equivalent. You will have to make several intermediate steps. The first step I would consider is adding the 1k and 2k7 together.

author: Legion
author_rank: Frequent Contributor
timestamp: February 28, 2014, 12:29:03 am
content: Quote from: Andy Watson on February 28, 2014, 12:18:43 amNot a homework question. It's from allaboutcircuits.com. I'm not in school, I'm trying to teach myself electronics.

author: IanB
author_rank: Super Contributor
timestamp: February 28, 2014, 12:37:39 am
content: Quote from: c4757p on February 28, 2014, 12:01:48 amUnfortunately I couldn't see the original image uploaded by legion before (site blocked). So I was just picturing a regular H-bridge configuration as that is what was described. But now I do see the image after all, and I agree the voltage at point A can easily be found by combining series and parallel resistors.However, the circuit on the left will have to be brute-forced by writing current or voltage equations and solving them (or by finding the Thevenin equivalent).

author: Legion
author_rank: Frequent Contributor
timestamp: February 28, 2014, 01:13:37 am
content: I solved it. No real secret to it. I just couldn't visualize it properly. I kept redrawing the circuit until one layout kinda "revealed" the solution.

author: Andy Watson
author_rank: Super Contributor
timestamp: February 28, 2014, 01:24:20 am
content: The easy (easier) way to solve either left or right circuit is to simplify the configuration with star-delta transforms - but that's not really the point of this exercise. Also, star-delta is not usually covered until after thevenin and norton equivalents have been introduced -so I'm guessing that you've not encountered those transforms yet.

author: Legion
author_rank: Frequent Contributor
timestamp: February 28, 2014, 01:49:18 am
content: I figured I'd throw up the solution in case it helps someone.To find the thevenin voltage I redrew the circuit like this:Use series-parallel rules and voltage division to find the voltage at A. Vth = VB - VA.To find the thevenin resistance I redrew the circuit like this:Take point A as the anode and B as ground. Use series-parallel rules to get Rth.Vth = 7.6922VRth = 788.64 Ohms

author: Excavatoree
author_rank: Frequent Contributor
timestamp: February 28, 2014, 01:59:29 am
content: Interesting - two posts in one day where I've realized I've forgotten the delta-wye transformation formulas.

author: IanB
author_rank: Super Contributor
timestamp: February 28, 2014, 02:16:37 am
content: Quote from: Excavatoree on February 28, 2014, 01:59:29 amI never particularly remembered them, but I can always work them out if I need to know them. Just recalling that "there is such a transformation" is as much knowledge as you ever really need.

author: The Electrician
author_rank: Frequent Contributor
timestamp: February 28, 2014, 01:14:29 pm
content: Quote from: Legion on February 28, 2014, 01:49:18 amWhat software did you to make these images, please?

author: Legion
author_rank: Frequent Contributor
timestamp: February 28, 2014, 01:42:51 pm
content: Quote from: The Electrician on February 28, 2014, 01:14:29 pmPhotoshop. But I wouldn't recommend it. It's really slow. I should probably get some schematic drawing software.