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What's the formula for this?

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Marks256

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8654-screenshotjoc.png


What is the value of Output? I know if the resistors are the same value, then output is half the voltage of Vcc, but what is the formula for that?
 

gaspode42

Member
This **broken link removed** may also help
 

Marks256

New Member
bountyhunter said:
School must be back in session.....

Aaah i knew this would happen eventually :)

Thanks all for the help, but I'm honestly not going to take any of the "oh he's just cheating on his ECE classes" BS. If it's any body's business, which it's not, I was "lucky" enough to not get a single ECE class this semester. That's right. Not a single class that will teach me even the smallest bit of electronics. Actually. Come to think of it, i've never had an electronics class in my life. So. If the average person were to look at this situation, he could easily deduce that i've been left to learn on my own. Weird huh. Never had a single class, or formal education to steer me in the right direction to complete my electronics knowledge. Now. Most of you have seen me here on the forum and know that i've been learning electronics for the past few years, but have been learning it in bits. So, i don't appreciate being accuse of "cheating" when all i'm trying to do is honestly learn on my own in my spare time. I'm terribly sorry that i did not know that this is called a voltage divider. please accept my deepest apologies for ANY inconvenience it may have caused :rolleyes:
 

Karkas

Member
I think you don't have to worrie about it, las semester i should had been taught the voltage divider, but the proffesor missed it(extremely wrong), i think it only happens in my country, but that's not the subject. Now in this semester i had to do exactly what you're doing, but in person with oter students and some proffesors, do you already know about the current divider?
 

Marks256

New Member
I know only what i've stumbled across on the net, or learned from asking help on a forum or IRC.
 

3v0

Coop Build Coordinator
Forum Supporter
me said:
That is sort of like cheating.

I was trying to get you to figure it out.

You already guessed that it was proportional when you said it was 1/2 Vcc when both were equal.

With R1 between gnd and the output and R2 from output to Vcc the voltage is Vcc times the fraction of the resistance that is R1.

Vout = Vcc(R1/(R1+R2))

-------------------------------
another approach
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we know this from ohms law.

V=IR

The current is the same in R1 and R2 if the current drawn by the output is insignificant. In most cases we use a divider to create a reference voltage so that is the case.

We can find the current from gnd to Vcc because we know the total resistance Rt = R1 + R2.

V=IR or I = Vcc/Rt

now that you know the current you can solve for the voltage across R1.

V = I * R1

at this point just plug in the numbers for I and R1.

I hope I did not screw this up ;)

3v0
 
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Mikebits

Well-Known Member
I hope I did not screw this up ;)
Nope you did not, that method is usually how I compute the drop.

Add total resistance then with supply volt compute current through the resistor leg using above mentioned formula, from there a simple ohms law calculation is needed.
E=IR

Well explained 3VO
 
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UTMonkey

New Member
Correct me if I am wrong but that formula is only useful if working from the basis of "no load" on the output?
 

Electronworks

New Member
Hi UTMonkey

Yes you are correct and indeed followed by train of thought. If you make the resistors too high in value and then measure the divider network, you will get a voltage lower than expected as the meter will load the bottom resistor.
Put a load on, and your 'bottom resistor' will be the parallel combo of the bottom resistor and the load...

:eek:
 

Mikebits

Well-Known Member
In this case, I think we are assuming High impedance loads such as a CMOS micro input. That is marks modus operandi...
 
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3v0

Coop Build Coordinator
Forum Supporter
My goal here was to help Mark understand how it worked and how to calculate it. Mark is a long time member and not some kid asking us to do his homework.

UTMonkey said:
Correct me if I am wrong but that formula is only useful if working from the basis of "no load" on the output?
You are only wrong in that I specified that it was for a no load condition. Note that the calculator provided by another posted also assumed no load.

The current is the same in R1 and R2 if the current drawn by the output is insignificant. In most cases we use a divider to create a reference voltage so that is the case.
Analog is not my bag but that sounded right to me. I wrongly said voltage instead of current in the original post.

Electronworks said:
Put a load on, and your 'bottom resistor' will be the parallel combo of the bottom resistor and the load...

Sure but why would you use a voltage divider to drive a load. I would use a simple resistor to drop the voltage. Using a voltage divider in place of a regulator is a trap newbie's fall into. If the draw is steady enough to work without a regulator a simple resistor will work. If not the divider will not work and you need regulation.

3v0
 

Nigel Goodwin

Super Moderator
Most Helpful Member
Using a voltage divider in place of a regulator is a trap newbie's fall into. If the draw is steady enough to work without a regulator a simple resistor will work. If not the divider will not work and you need regulation.

Not true, a voltage divider provides 'some' degree of regulation, far more so than a simple series resistor.

The trick is to ensure that the current through the divider chain is considerably more than the load requires - a minimum of five times more for most purposes.

It's a standard technique used in almost everything - for example base biasing on a transistor.
 

Bob Scott

New Member
8662-screenshotjoc.png


What is the value of Output? I know if the resistors are the same value, then output is half the voltage of Vcc, but what is the formula for that?

I can't wade through all the junk replies. Junk replies made by flippant vendechos are getting to be A VERY BIG PROBLEM lately. The s/n ratio is getting really small.

The output voltage of this attenuator is [R2 / (R1 + R2)] * Vcc.
 
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Bob Scott

New Member
Not true, a voltage divider provides 'some' degree of regulation, far more so than a simple series resistor.

The trick is to ensure that the current through the divider chain is considerably more than the load requires - a minimum of five times more for most purposes.

It's a standard technique used in almost everything - for example base biasing on a transistor.
That's right, because the effective output impedance of the divider is lower than a single series resistor. For output impedance purposes, R1 and R2 are effectively in parallel. The voltage rails are considered to have zero Z between them.

1/Zout = 1/R1 + 1/R2

Bob
 
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3v0

Coop Build Coordinator
Forum Supporter
One sure way to get peoples attention regarding the simple stuff is to be wrong.

Nigel said:
Not true, a voltage divider provides 'some' degree of regulation, far more so than a simple series resistor
Thanks Nigel I stand corrected. Does anyone care to explain why ?

I speculate that having the fixes resistor in parallel with the load is a most stable resistance than the load by itself.

3v0
 
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