Understanding how to calculate voltage drop?

Thread Starter

SlowCoder

Joined Mar 25, 2012
26
Alrighty, my next question ...

I've looked around the web, and have thus far not found the appropriate answer. Most of the answers to the question seem opposite of my intention.

My question: How to drop voltage? Say I have a 9 volt battery that I want to drop to 5v?. I start out with (9 (starting volts)-5 (desired volts))=I*R, right? I may not have a circuit to be powered by it at the time, I may just want to make the supply for future use in projects I haven't yet built or designed. So I don't know the presumed consumed I (amps) for the calculation.

What's the best method for obtaining the resistance? Just toss ohms at it until the numbers are correct?

I'm assuming some known I values for the heck of it:
4=.5(I)*8(r) would be 8ohms resistance for anything less than 500mA at 5V?
4=1(I)*4(r)=4ohms resistance for anything drawing less than 1A at 5V?

What happens if I plug the 5V 1A supply into a circuit that only draws 250mA? Then wouldn't my 5V actually become 20V, per Ohm's Law? And if that's the case, how would I go about keeping the voltage at 5v so I don't fry the 250mA circuit?

But wait, here's a snag, how would I be providing 20V from a 9V source?
 

MrChips

Joined Oct 2, 2009
21,664
So you want to drop from 9V to 5V, hence you need to drop 4V.
You need a resistor which can be calculated using Ohm's Law R = V/I.
But you need to know the current I and this is unknown and is changing.
So R must change as the current I changes.
The way to do this is with a transistor whose pass resistance changes.
This is where a voltage regulator comes in. The voltage regulator has a pass transistor whose resistance changes in order to keep the output voltage fixed.
The most popular voltage regulator for 5V output is the LM7805.
 
Last edited:

cowades

Joined Feb 6, 2012
11
If you want a general supply to operate future 5V projects off a 9V battery you would need a whole cabinet of resistors. Your resistor value in each case would be (9V-5V)/current. More specifically for a 250ma amp load you would need a 16 Ohm resistor in series with the battery and your circuit.

This is probably not what you want because the next project will be a different current and require a different resistor to drop the 4V.

You really want a LM7805 voltage regulator. These are readily available for less than a couple of bucks at any supplier like Radio Shack or DigiKey etc. Application notes are available on line. They are very easy to use, and way more fun than resistors.
 

Audioguru

Joined Dec 20, 2007
11,249
DO NOT use an ordinary 7805 voltage regulator with a 9V battery because it fails to regulate when the battery voltage discharges a little.
Instead use a low dropout 5V regulator that still regulates when the battery is almost dead.
 

Thread Starter

SlowCoder

Joined Mar 25, 2012
26
If your circuit draws 250 ma at 5V, then the resistance of your circuit is 5V/250mA=20 ohms. R=V/I is Ohm's law.
Am I using Ohm's Law incorrectly in my calculations? I prefer to see Ohm's Law as V=I*R, as it's easier to visualize in my head.

I'm not actually building a supply. But I'm trying to grasp how I would properly make such a supply work for multiple applications. I can take a "wall wart" power supply and plug it into any device, as long as I don't exceed the rated voltage and amps. I'm trying to deduce HOW this is accomplished.
 

mrmount

Joined Dec 5, 2007
59
If the power supply is rated 5V/1A and connected to a circuit that needs only 250 mA, it means the power supply needs to work only 1/4th of its full capacity.
It does not mean that since the circuit needs only 1/4 of the rated current (250 mA) the voltage will increase to 20 to keep V*I constant.
 

Thread Starter

SlowCoder

Joined Mar 25, 2012
26
So you want to drop from 9V to 5V, hence you need to drop 4V.
You need a resistor which can be calculated using Ohm's Law R = V/I.
But you need to know the current I and this is unknown and is changing.
So R must change as the current I changes.
The way to do this is with a transistor whose pass resistance changes.
This is where a voltage regulator comes in. The voltage regulator has a pass transistor whose resistance changes in order to keep the output voltage fixed.
The post popular voltage regulator for 5V output is the LM7805.
If you want a general supply to operate future 5V projects off a 9V battery you would need a whole cabinet of resistors. Your resistor value in each case would be (9V-5V)/current. More specifically for a 250ma amp load you would need a 16 Ohm resistor in series with the battery and your circuit.

This is probably not what you want because the next project will be a different current and require a different resistor to drop the 4V.

You really want a LM7805 voltage regulator. These are readily available for less than a couple of bucks at any supplier like Radio Shack or DigiKey etc. Application notes are available on line. They are very easy to use, and way more fun than resistors.
Ok. These both answer my question, and that's what I thought it would be. That the resistance would have to change for each project, and I HAVE to know what current is for each circuit.

I understand that I *could* use a voltage regulator, but I'm interested in knowing how it all works. That would be somewhat of a shortcut, but not what I'm targeting for learning.
 

shortbus

Joined Sep 30, 2009
7,847
For most circuits a 9 volt battery, the rectangle shaped ones with the little snaps for terminals, is not a real good choice any way.
 

BillB3857

Joined Feb 28, 2009
2,533
If you measure the output voltage of most wall-warts with no load attached, the output will be much higher than what is shown on the label. They are designed to operate a specific device with a known current demand and the current actually causes the output voltage to drop. Higher end wall-warts will have internal voltage regulation either by means of devices such as the LM78XX regulators or, if the input is rated for 120/240VAC, it will in all probability use a switching regulator - an entirely different animal.
 
Top