Ten Essential Skills for Electrical Engineers - Stuck on Thévenin Equivalents - Page 1
Forum section: electronics / beginners

author: bbjk7
author_rank: Contributor
timestamp: September 16, 2014, 06:42:10 am
content: Hi all,I understand the basics of analyzing circuits using the Thvenin method, or so I thought. This particular question has me stumped:1.2 A sensor produces a voltage between –5 and +3 V. Theoutput resistance of the sensor is 470 ?. The sensor must beinterfaced to an ADC with an input range of 0 to +3.3 V andan input resistance of 3300 ?. Determine the values of R1 andR2 for the interface network shown in Figure 1.11:Do I treat the 3k3 resistor as the load? I always struggle in tackling these problems

author: f5r5e5d
author_rank: Frequent Contributor
timestamp: September 16, 2014, 07:55:52 am
content: it would help to label the bottom trace ground/0 V just to rule out real possibilities like floating sensors, diff inputADC...but yes an assumption that makes the problem doable is that the input terminals of the ADC are the ends of its (parallel equivalent) input Rthen I think you should appeal to Superposition, although I suppose that can be framed as a way of using Thevenin Equivalents - it may be clearer to state how Superposition applieshttp://www.allaboutcircuits.com/vol_1/chpt_10/7.htmlmay be a start

author: w2aew
author_rank: Super Contributor
timestamp: September 16, 2014, 04:45:19 pm
content: While not answering your question directly, this video that I did regarding R2R ladder DACs gives a pretty good tutorial on the application ofThevenin'sandSuperpositionto calculate a circuit's output:

author: zapta
author_rank: Super Contributor
timestamp: September 16, 2014, 05:02:36 pm
content: Yes, 3.3k is the load. Write the equations for the voltage on the load, one for the -5V input and one for the +3V input (use Thévenin to find the equivalent V,R of everything but the load). These equations are equal to your target load voltages 0V, 3.3V respectively. Now you have two equations with two variables, R1, R2 which you can solve.

author: Kyburo
author_rank: Newbie
timestamp: September 16, 2014, 10:16:39 pm
content: Woah, great video. A little beyond me as I've not until recently looked back to electronics since college 7 years ago but was brilliantly explained and I could follow the circuit diagram and math explanation. Yet another great youtube channel to barrel through!

author: free_electron
author_rank: Super Contributor
timestamp: September 16, 2014, 10:21:20 pm
content: bad example. adc's with such input impedance need an input amplifier... no need for thevenin, you got system problems to tackle

author: suicidaleggroll
author_rank: Super Contributor
timestamp: September 17, 2014, 12:35:00 am
content: Quote from: zapta on September 16, 2014, 05:02:36 pm^ thisJust remove the voltage source from the circuit and instead label the bottom left corner as 0V, the top left corner as -5V, and the top right corner as 0V. Then it's a simple voltage divider and algebra. Reduce it down to a simple function of R1 and R2...(hint: since the top and bottom right corners are 0V, you can knock the 3.3k resistor out and you just have a straight line...5V -> R2 -> 0V -> R1+470 -> -5V).Then change the top left corner to 3V, the top right corner to 3.3V, and again reduce it down to a simple function of R1 and R2.Finally, solve one of the two equations for either R1 or R2, plug it into the other one, solve for the variable, plug it back into either equation, and solve for the other variable.

author: bbjk7
author_rank: Contributor
timestamp: September 18, 2014, 01:25:24 am
content: Thanks all, I was quickly able to find the 1st equation: (R2= R1+470) for when the sensor is at -5V and the ADC is at 0V.It's a bit trickier for when the sensor is at 3V and the ADC needs to be at 3.3V. I can get the equation down to: R2= 1.7/(0.3/(R1+470)+3.3/3300), but after that my algebra is a bit rusty :/How can I get it simpler enough to start substituting the two into each other and solving? No never mind, I shouldn't be asking people on a forum to do my maths for mei'm going to have to sharpen my algebra skills. Can't believe a simple divider network can get this frustrating/complicated

author: suicidaleggroll
author_rank: Super Contributor
timestamp: September 18, 2014, 01:47:39 am
content: Quote from: bbjk7 on September 18, 2014, 01:25:24 amWhy do you need to reduce it further? You already have R2 as a function of R1, so just swap it into the other equation and solve for R1.

author: bbjk7
author_rank: Contributor
timestamp: September 18, 2014, 01:51:38 am
content: I end up with two R1's in the same equation that cancel eachother out? Just bad algebra?

author: IanB
author_rank: Super Contributor
timestamp: September 18, 2014, 02:11:32 am
content: Quote from: bbjk7 on September 18, 2014, 01:51:38 amI don't understand. You first of all wrote:R2 = R1+470Then you had:R2 = 1.7/(0.3/(R1+470) + 3.3/3300)Now, if R1+470 = R2, doesn't an obvious next step present itself?

author: bbjk7
author_rank: Contributor
timestamp: September 18, 2014, 03:37:31 am
content: Quote from: IanB on September 18, 2014, 02:11:32 amyes, i know, i've been down that path.. R1+470 = 1.7/(0.3/(R1+470) + 3.3/3300), but then i end up with the same variable twice on either side... simplifying down to something like A = A or 1 = 1silly me, forgot you could subtract rather than divide variables with something like 2xR1=R1+470

author: IanB
author_rank: Super Contributor
timestamp: September 18, 2014, 07:13:01 am
content: Quote from: bbjk7 on September 18, 2014, 03:37:31 amIsn't it more obvious to write this instead?R2 = 1.7/(0.3/R2 + 3.3/3300)That's what I was getting at. It's simpler than having all those "+ 470" terms lying around...That would rearrange to give:R2 = (3300/3.3)(1.7-0.3) = 1400Then R1 = R2 - 470 = 930