bountyhunter said:School must be back in session.....
me said:That is sort of like cheating.
Nope you did not, that method is usually how I compute the drop.I hope I did not screw this up
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.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?
Analog is not my bag but that sounded right to me. I wrongly said voltage instead of current in the original post.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.
Electronworks said:Put a load on, and your 'bottom resistor' will be the parallel combo of the bottom resistor and the load...
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.
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?
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.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.
Thanks Nigel I stand corrected. Does anyone care to explain why ?Nigel said:Not true, a voltage divider provides 'some' degree of regulation, far more so than a simple series resistor