Voltage divider - a question

Thread Starter

puzzle

Joined Oct 30, 2016
53
I'v read this article :
http://www.learningaboutelectronics.com/Articles/Voltage-follower

although, I don't really understand why he have wrote In the second example "Voltage Followers Consume Practically All of the Voltage from a Voltage Divider Circuit" , this sentence : We have a voltage divider between a 10KΩ resistor and a 100Ω load, dividing the 5V output from the first voltage divider". But Its wrong because the two resistor connected in parallel ! so 5V will also drop on the load, but the current will be different compared to the 10K resistor

what do you think ?
 

ericgibbs

Joined Jan 29, 2010
18,766
hi p,
He is explaining the circuit operation in a confused way.
I believe what he is trying to say if you had a voltage divider of two equal resistors you would get Vsup/2 at the junction, but if you added another resistor in parallel with the lower resistor , the Vout would no longer be Vsup/2.

By connecting a high input impedance OPA voltage follower to the junction it would not pull the Vsup/2 down, so the Vout of the OPA follower would be Vsup/2

E
 

AlbertHall

Joined Jun 4, 2014
12,344
We have a voltage divider between a 10KΩ resistor and a 100Ω load
The 10k referred to here is the top resistor in the diagram, the 100Ω is the 100Ω load in parallel with the bottom 10k. This parallel combination is actually close to 99Ω but this itself is very close to 100Ω. So the voltage divider is effectively 10k into 100Ω and the voltage across the load will be about 0.1V.
 

Thread Starter

puzzle

Joined Oct 30, 2016
53
OK! now I understand! I didn't notice this!
So, Does he right ? Those are the uses of a Buffer ? In between a power supply and low resistor and in a voltage divider ? what other uses ?
 

Papabravo

Joined Feb 24, 2006
21,158
A voltage follower/buffer can be used in any circuit where the sensor lacks the ability to drive a load with a low (relative) impedance such as an A/D converter. If you need to do scaling or filtering operations these can be combined with the follower/buffer. Because the non-inverting configuration cannot have a gain of less than 1, you can use a pair of inverting amps to accomplish a more flexible array of functions with the same goal of decoupling the impedance of the source from that of the load.
 
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