# Missing resistor in a voltage divider network

Discussion in 'Electronics Homework Help' started by Bish70, Feb 25, 2020.

1. ### Bish70

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Feb 25, 2020
I hope someone can help. If I have a series of 6 resistors in a voltage divider network and I'm given the voltage supply and five of the resistor values, how do I find the first resistor value in the chain without being given the current?

2. ### Harald KappModeratorModerator

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Nov 17, 2011
An image of the task description would help a lot.
Without knowing either the current or the voltage(s) at the tap(s) you are lost.
With the 5 known resistances and at least the voltaeg at one tap of the divider you should be able to calculate the remaining voltages, the current and the missing resistor.
However, for more detailed help show us the complete task description.

Note that we will not solve your homework, but you can expect good support that enables you to find the solution by yourself.

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3. ### Bish70

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Feb 25, 2020
Thank you, I was going to post the image but I'm conscious of plagiarism and just wanted to see if there was any guidance. I've been told I can work it out using the Op Amp golden rules, Ohms and Kirchhoff's Laws but as you stated without the Voltage input or current I can't see how! Must be something simple I'm missing in the question

Thanks
Neil

4. ### Harald KappModeratorModerator

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Nov 17, 2011
Ohm's law and Kirchhoff's laws I understand, but where does the opamp come into play?
Without a picture we're lost.
post a photograph of a hand drawn sketch. Only make sure it is crisp and well readable. Note that we have a size limit of ~ 100 kB.

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Feb 25, 2020

6. ### Bluejets

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Oct 5, 2014
Photo too small...rotate right way up while you are at it.

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Feb 25, 2020

8. ### duke37

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Jan 9, 2011
You have five 1k resistors in series with 1V across each so you can work out the current.
The voltage across Rx will be the difference between 12V and the resistor chain voltage.
All this assums that the op-amps take no current.

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9. ### Bish70

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Feb 25, 2020
Thank you, I did look at this but I couldn't document my reasoning for the 1v across the 1Kohm resistor without calculating the initial voltage drop across the initial resistor and for this I needed the value to find the current

10. ### duke37

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Jan 9, 2011
Each op-amp switches at 1V difference from the one next door. You are told this in the question. Why were you told this if it should not to be used?
You do not need the actual current, You can do this without calculating it, You do need to know that voltage is current times resistance. Here the current is constant.

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11. ### Bish70

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Feb 25, 2020
Thank you for your help, it was much appreciated. I guess it's kind of obvious when you point it out, the old adage of 'read the question' never loses its importance.

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Jan 9, 2011

13. ### Bish70

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Feb 25, 2020
I worked out the answer to be 7K ohms

11,655
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Nov 17, 2011
Reasonable.

15. ### hevans1944Hop - AC8NS

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Jun 21, 2012
Now, where is the rest of the circuit that accepts the op-amp outputs and somehow causes the 7-segment display to show the digits 0, 1, 2, 3, 4, or 5 depending on how many op-amp outputs are "high"?

16. ### duke37

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Jan 9, 2011
Not quite 7K ohms. It is 7kΩ, temperature does not come into it.

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17. ### Bish70

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Feb 25, 2020
I would take it from this that you know this and I’m potentially in the mire

18. ### hevans1944Hop - AC8NS

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Jun 21, 2012
I do know this, having played with flash A-to-D converters in the 1970s. I doubt you are "potentially in the mire" because you have the ability to discover how it was done, lo! those many moons ago. And that is why no one that I know actually builds flash converters from discrete components anymore... except as an educational exercise. They already exist as integrated circuits with binary coded outputs.

At one time there was a bar-code display that was used as a VU (Volume Unit) meter in some hi-fi audio rigs. This display device consisted of ten or twelve (I forget how many) LED bars stacked one above the other that would successively illuminate and stay on as the sound level increased. The number of bars illuminated was an indication of the VU level, important for magnetic tape recording. As the sound level decreased, the number of bars illuminated also decreased. A special integrated circuit (LM3914, see post #19 below by @Harald Kapp) accepted audio input and performed a flash A-to-D conversion whose output was very much the same as the circuit you posted. The major difference was the conversion was deliberately non-linear because that's the way VU meters roll. Later, a similar IC (I think this was the LM3814 that Harald linked to in his post below) provided linear steps to light up successive LED bars. I think that one may still be in production somewhere, but I haven't bothered to research that.

There are integrated circuits that will accept binary coded inputs and produce segment-driver outputs for 7-segment displays. It is a "good thing" that you are learning about this stuff, especially if your goal is to eventually become an accomplished integrated circuit designer.

BTW, op-amps are not ideal comparators for flash converters because op-amps are too slow. The only reason that I know of to actually use a flash converter is speed of conversion. With fast comparators, they will out perform any other analog-to-digital conversion circuit I can think of. Software defined radios (SDRs) may use flash converters, as do some military sensor applications, but successive approximation conversion is cheaper and less complicated and usually "gud enuf". You should be learning about these as well as integrating-type analog-to-digital conversion.

EDIT: Thanks @Harald Kapp for finding the modern LM3914!

Last edited: Feb 27, 2020
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19. ### Harald KappModeratorModerator

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Nov 17, 2011
LM3914.

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20. ### bertusModerator

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Nov 8, 2019
Hello,

The LM3914 is not a VU meter, it is linear.
The LM3915 is a 3dB step meter.
The LM3916 has a VU meter arrangement.

Bertus

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