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Voltage Divider +12V, -12V, Ground.

Timothy

Feb 1, 2015
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Hello,
I have been trying to figure out how to work out the voltage highlighted in the picture. I have tried all sorts of of methods to analyse this. Standard voltage divider doesnt work because of the 24v and 12v paths, and I just can't get to 8.83v.I know it probably a very simple solution.

This circuit is part of a 741 op amp temperature measuring circuit which I'm designing. The resistances are just placeholders to help me work things out. The section of the circuit in the picture is intended to provide an offset so that the output of the op amp is 0-10v between the temperatures of 0 and 100 degrees C.

I'm not looking for an answer because I enjoy the satisfaction working these things out for myself, not to mention it's easier to remember that way. A little guidance would be very helpful though.I'm hoping to arrive at a formula so that I can use it in an excel spreadsheet that I'm using to simulate the circuit, so that I can test different values.

Any guidance will be much appreciated,

Thanks.


upload_2015-2-1_15-27-45.png
 

Merlin3189

Aug 4, 2011
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How about trying to put the circuit nodes in height order in the same order as their potential?
-12 is obviously at the bottom and +12 at the top.
Your ground is presently at the same height as -12, but since it is (presumably?) zero, it should be somewhere midway between the two.
If you then get the other nodes in the right order, maybe the voltage divider will be more clearly shown.
 

Merlin3189

Aug 4, 2011
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Having looked in a bit more detail, my suggestion doesn't help a lot!
Perhaps putting 12 V voltage sources explicitly between the +12, 0 and -12 Volt points will give you a complete circuit (ie. no unconnected points) that you know how to analyse?

Once I calculated the current in one branch, everything else was straightforward.
 

davenn

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Sep 5, 2009
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firstly
how about simplifying the circuit
The voltage in the circled voltmeter is going to be a result of the combination of the 3 lower resistors
R2, R3, R4 .... so work out their combined value for a start
NOTE its a series/parallel combination
 

Merlin3189

Aug 4, 2011
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Have you come across the Superposition Theorem before? It helps with the analysis of circuits with more than one voltage source.

And just get rid of the red circled Voltmeter and it's wires. It's a distraction, until you have solved the circuit.
 

davenn

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And just get rid of the red circled Voltmeter and it's wires. It's a distraction, until you have solved the circuit.

further to that ... get rid of ALL voltmeters till the circuit basics are solved :)
 

Ratch

Mar 10, 2013
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Hello,
I have been trying to figure out how to work out the voltage highlighted in the picture. I have tried all sorts of of methods to analyse this. Standard voltage divider doesnt work because of the 24v and 12v paths, and I just can't get to 8.83v.I know it probably a very simple solution.

This circuit is part of a 741 op amp temperature measuring circuit which I'm designing. The resistances are just placeholders to help me work things out. The section of the circuit in the picture is intended to provide an offset so that the output of the op amp is 0-10v between the temperatures of 0 and 100 degrees C.

I'm not looking for an answer because I enjoy the satisfaction working these things out for myself, not to mention it's easier to remember that way. A little guidance would be very helpful though.I'm hoping to arrive at a formula so that I can use it in an excel spreadsheet that I'm using to simulate the circuit, so that I can test different values.

Any guidance will be much appreciated,

Thanks.


View attachment 18429

Call me stupid or call me dense, but I don't understand the problem. I calculated all the voltages using 2 loops and they are all correct. The voltage across R3 and R4 are 8.03 volts like the schemat says they are, so what is the problem?

Ratch
 

Merlin3189

Aug 4, 2011
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I calculated all the voltages using 2 loops and they are all correct. -Ratch
Well, if you did that easily, then it's me who's a bit dense. I couldn't do it without drawing 3 diagrams, over a page of scribble and using a calculator. With those values, even calculating parallel resistances needed a calculator. My method was to redraw the cct to suit me, then split it into two circuits based on the two voltage sources, solve those to calculate the two contributions to the current through the 543 Ohm resistor, calculate the PD across it and finally the PD across his big voltmeter.

Perhaps you mean you just used the voltages shown (as calculated by the simulator, I suppose?) which do give the 8.03 as required? If not, I'd be interested in your simpler method.

Oh, and to OP -
absolutely agree with Davenn to get rid of the voltmeters first, then you might see the wood for the trees, as they say. And what are your thoughts about the superposition method?
 
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Ratch

Mar 10, 2013
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Well, if you did that easily, then it's me who's a bit dense. I couldn't do it without drawing 3 diagrams, over a page of scribble and using a calculator. With those values, even calculating parallel resistances needed a calculator. My method was to redraw the cct to suit me, then split it into two circuits based on the two voltage sources, solve those to calculate the two contributions to the current through the 543 Ohm resistor, calculate the PD across it and finally the PD across his big voltmeter.

Perhaps you mean you just used the voltages shown (as calculated by the simulator, I suppose?) which do give the 8.03 as required? If not, I'd be interested in your simpler method.

Oh, and to OP -
absolutely agree with Davenn to get rid of the voltmeters first, then you might see the wood for the trees, as they say. And what are your thoughts about the superposition method?

No need to redraw the diagram. Just ignore the voltage meters. The circuit topology is two loops or one node. I will calculate the loop currents first.

The first loop is -12, R2,R1,-12 and the second loop is R4,R3,R2,+12 . Each loop needs an equation, and they are (20000+543)*i1-20000*i2=24 for the first loop and (20000+1000+837)*i2-20000*i1=-12 for the second loop. Solving the equations simultaneously gives i1=0.00584573 amps, and i2 =0.00480445. The voltage you want to find is (1000+837)*i2=8.82577.

As Laplace pointed out, you can calculate the voltage using one node. The equation is (v-12)/543+(v+12)/20000+v/(1000+837)=0 . Solving for v gives 8.82577, which agrees with the answer using the loop current existing through R3 and R4. Obviously the node method is easier for this circuit.

Ratch

Edit: PS I can show you how to solve that problem using superposition if you wish.
 
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davenn

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Edit: PS I can show you how to solve that problem using superposition if you wish.

its not something I have made use of or was ever taught ... So yes please :)
 

Ratch

Mar 10, 2013
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its not something I have made use of or was ever taught ... So yes please :)

cct4.GIF

OK, we have two voltages +12, and -12 that are going to supply current to the circuit. Because it is a linear circuit, the total current will be the total contribution of both voltages taken separately. If we find the total current in the R3, R4 branch, then it will be a trivial task to determine the voltage across those resistors. The method is to activate one voltage source at a time while replacing the other voltage by its resistance, which is zero. The function par(x,y) is defined to be the parallel resistance of two resistors whose values are x and y ohms.

Determining the contribution of the top voltage (+12) is as follows: Total resistance (rt1) is 543+par(20000,1000+837) = 2225.47 ohms. Total current (it1) is 12/rt1 = 0.00539213 amps . Using the current divider law, we find the current in the R3,R4 branch to be ib1 = it1*20000/(20000+1000+837) = 0.00493853 amps . This current is assumed to exist from left to right.

Determining the contribution of the bottom voltage (-12) is as follows: Total resistance (rt2) is 20000+par(543,1000+837) = 20419.1 ohms. Total current (it2) is 12/rt2 = 0000587685 amps . Using the current divider law, we find the current in the R3,R4 branch to be ib2 =- it2*543/(543+1000+837) = -0.000134081 amps . This current is assumed to exist from right to left. Adding the branch current from the two voltage sources ib1+ib2 =0.00493853 +(-0.000134081 = 0.00480445 amps, which is what was calculated with the loop method. Then multiplying R3 plus R4 by the current gives the correct answer.

Ratch
 
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davenn

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OK hope you don't mind .... added the image above your text to save the scrolling up and down
also got rid of the volt meters for clarity

I need to work through that to see if I can follow it

having been on this and a couple of other electronics forums for some years, it has shown up great deficiencies in
what I was taught in electronics classes
Never ever touched on Thevenin and Norton Eqv's or superposition
It has caused more than a little frustration in trying to analyse circuits over the years

Dave
 

Arouse1973

Adam
Dec 18, 2013
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Here is my solution to this is anyone is interested.
Thanks
Adam

Resistor Problem.PNG
 

Ratch

Mar 10, 2013
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It was an oversight on my part for not recognizing earlier that this circuit is a easy application of Millman's theorem.

((-12/20000)+(12/543))*(543||20000||1837) = 8.82577 volts

Ratch
 

Merlin3189

Aug 4, 2011
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Ratch - great summary. Tried all three now and Laplace's method (Kirchoff 1 ?) seems neatest to me. Your method (Kirchoff 2?) gives me more or less the same equations as I used, but in two fell swoops (instead of 4 half swoops!)
All I can say is, you guys must be more in practice than me. I still had to redraw the diagram (Davenn has done it right now - brill! Isn't it so much easier like that) and a page of scribble AND use a calculator (which I'm not very good with.) What I couldn't do without a diagram was to know which way the current was flowing. I had to have my little arrows, otherwise I might have added when I should have subtracted.

When I was at school resistors were always 1, 2, 3, 4, 6, 12, 24 Ohm etc if you had a 12V supply. I pity students now that teachers have calculators. Who ever heard of 1837 Ohm or 543 Ohm? It's like that guy who wants us to use an exponential function for emitter base voltage: for me it's 0.5, 0.6 or 0.7 whichever is easiest. And it doesn't change, however much current I want through it! (Until it goes Pop!)

PS Just noticed Ratch's latest. Millman - that's a new one for me.
 
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