# Wattage in voltage devider

#### Chillum

Joined Nov 13, 2014
546
Hi, this is my first post. I read all voltage divider threads.

I'm thinking of getting 5v from a 12v DC source. Now I've read the first 2 books of All-in-one Electronics for Dummies, and a bit of reading on the internet, so I don't know much yet. The question:

There's voltage regulators that are made to step down a 12V source to 5V, but they lack severely in mA. So I research voltage dividers, which I've met in my readings, but I'm not sure how the amperage would be calculated, the usable amperage. I think a lower value would mean more amps for the load, but with this: http://www.sengpielaudio.com/calculator-ohm.htm (and http://www.calculatoredge.com/electronics/voltage divide.htm) lower values require higher wattage, thats where the trouble starts, what is amperage dispensed by the resistor and what is amperage left over for all the juicy bits that will become the load? (This entire question is at date hypothetical, but soon I'll use it) Is it more efficient to use a 7805 regulator potentially delivering more than 1A, or a 470R (which I can get a high wattage package of) and then a trimmer 0.1W 470R potentiometer (set it to 335.71, measure with multimeter until its 5V) but if I go below 250R (?) I'll burn my trimmer potentiometer. Any lower values of resistors require more than 0.1W on the potentiometer. And I'm guessing go as low as you can go for highest usable load current, but then again I'm thinking, well its like a water pipe, and if one part is thin, thin is as high as you can go? Regulator 78L05 delivers 100mA.

So maybe I should add: I wan't to experiment with digital electronics, what type of load do I need to be able to carry, and if there is transistors for the load, can I use a second 12DC power source to drive only the load?

OK, thats more than one question...

PS I bought 8 12VDC psu for basically nothing, thats why the "obsession" with 12VDC

#### Chillum

Joined Nov 13, 2014
546
Hi, this is my first post. I read all voltage divider threads.

I'm thinking of getting 5v from a 12v DC source. Now I've read the first 2 books of All-in-one Electronics for Dummies, and a bit of reading on the internet, so I don't know much yet. The question:

There's voltage regulators that are made to step down a 12V source to 5V, but they lack severely in mA. So I research voltage dividers, which I've met in my readings, but I'm not sure how the amperage would be calculated, the usable amperage. I think a lower value would mean more amps for the load, but with this: http://www.sengpielaudio.com/calculator-ohm.htm (and http://www.calculatoredge.com/electronics/voltage divide.htm) lower values require higher wattage, thats where the trouble starts, what is amperage dispensed by the resistor and what is amperage left over for all the juicy bits that will become the load? (This entire question is at date hypothetical, but soon I'll use it) Is it more efficient to use a 7805 regulator potentially delivering more than 1A, or a 470R (which I can get a high wattage package of) and then a trimmer 0.1W 470R potentiometer (set it to 335.71, measure with multimeter until its 5V) but if I go below 250R (?) I'll burn my trimmer potentiometer. Any lower values of resistors require more than 0.1W on the potentiometer. And I'm guessing go as low as you can go for highest usable load current, but then again I'm thinking, well its like a water pipe, and if one part is thin, thin is as high as you can go? Regulator 78L05 delivers 100mA.

So maybe I should add: I wan't to experiment with digital electronics, what type of load do I need to be able to carry, and if there is transistors for the load, can I use a second 12DC power source to drive only the load?

OK, thats more than one question...

PS I bought 8 12VDC psu for basically nothing, thats why the "obsession" with 12VDC
Maybe I should use the trimmer on R1 (the 7V part), and use a high wattage resistor for the load part R2 (5V) That would satisfy my pipe theory, but please explain still.

#### ISB123

Joined May 21, 2014
1,236
Voltage regulator(7805) will output 1A max. which isn't really small current when working with microcontrollers and should be enough.You can use bypass transistor to get more current but you will need a heatsink.
Voltage divider shouldn't be used to bring 12VDC to 5VDC because it will be very inefficient and you will need High wattage resistors which cost as much as voltage regulators.Voltage dividers are used as reference points,to deliver very small amounts of power,bias the base of transistor,etc.....

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#### studiot

Joined Nov 9, 2007
4,998
You do not need simple or fancy voltage dividers at this stage.

If and when you come to trying out some digital circuits there are plenty of devices that can work directly from a 12volt supply.

Your first goal must be to properly understand the relationship between Voltage, Current and Resistance that is Ohm's law.

Your second goal should be to be able to uderstand the difference between series and parallel circuits and tell them apart.

Your third goal should be to understand power and energy in electric circuits and the relationship between power, energy, voltage and current.
You will then be able to understand that the power in a load is determined by the load not by the power supply.
You will also be able to understand and make sensible decisions about the answers to your initial questions.

#### crutschow

Joined Mar 14, 2008
25,416
A resistive divider is not used to deliver power, it's used to generate a voltage for a high impedance load (load impedance much higher than the equivalent resistive divider impedance). Any load on a divider will change its voltage since the load acts as a resistor in parallel with the bottom divider resistor to ground. And the divider will dissipate more power than a linear regulator would for the same output current and voltage.
The easiest way to get a lower voltage from a higher voltage is to use linear regulator. Common ones such as the LM317 (adjustable) or the LM7805 (fixed 5V) can deliver more than a amp with proper heat sinking.
For 12V in and 5V out, the heat power dissipated in the regulator will be [(12V-5V) * I] = 7*I where I is the load current, so you need to size the heat sink accordingly.

#### Chillum

Joined Nov 13, 2014
546
You do not need simple or fancy voltage dividers at this stage.
ok thx

If and when you come to trying out some digital circuits there are plenty of devices that can work directly from a 12volt supply.
ok that's cool

Your first goal must be to properly understand the relationship between Voltage, Current and Resistance that is Ohm's law.
I THINK i do.
Your second goal should be to be able to uderstand the difference between series and parallel circuits and tell them apart.
I THINK I do
Your third goal should be to understand power and energy in electric circuits and the relationship between power, energy, voltage and current.
P=VI
You will then be able to understand that the power in a load is determined by the load not by the power supply.
The pipe theory (smallest pipe determines current) and load may use any amount of the pipe
You will also be able to understand and make sensible decisions about the answers to your initial questions.
Thats where I get confused, what amps is voltage divider and what amps is maximum pipe, ie max load

#### ISB123

Joined May 21, 2014
1,236
How much current does the load need?

#### Chillum

Joined Nov 13, 2014
546
How much current does the load need?
this is all hypothetical, but lets say for arguments sake 2A, and then because we're discussing it 1A and 500mA

#### Chillum

Joined Nov 13, 2014
546
A resistive divider is not used to deliver power, it's used to generate a voltage for a high impedance load (load impedance much higher than the equivalent resistive divider impedance). Any load on a divider will change its voltage since the load acts as a resistor in parallel with the bottom divider resistor to ground. And the divider will dissipate more power than a linear regulator would for the same output current and voltage.
The easiest way to get a lower voltage from a higher voltage is to use linear regulator. Common ones such as the LM317 (adjustable) or the LM7805 (fixed 5V) can deliver more than a amp with proper heat sinking.
For 12V in and 5V out, the heat power dissipated in the regulator will be [(12V-5V) * I] = 7*I where I is the load current, so you need to size the heat sink accordingly.
can you give me some more about the size of the heatsink? 7*I= what unit 'F / 'C and when I have my value = 7, how do I convert 7 into cubic cm?

#### MrChips

Joined Oct 2, 2009
21,855
When doing engineering design it cannot be hypothetical. Come up with expected real values.
What are you powering?
What voltage?
What curent?

#### Chillum

Joined Nov 13, 2014
546
An 7805 vs trimmer for 7V, and 5W (or 3W) resistor for load side; the 7805 is cheaper by almost 35%

#### ISB123

Joined May 21, 2014
1,236
Use voltage regulator,voltage divider was never invented for use as power supply.

#### Chillum

Joined Nov 13, 2014
546
When doing engineering design it cannot be hypothetical. Come up with expected real values.
What are you powering?
What voltage?
What curent?
I'm a student of the WEB, I don't have an application yet, just I know you need 5v for digital electronics, so I thought I'll just start and make sure I'll have power, when I get to a project... so, please, be a teacher I'm hungry ;-)

#### ISB123

Joined May 21, 2014
1,236
A nice start could be to make yourself a small buck converter.

#### Chillum

Joined Nov 13, 2014
546
another question, this time about the 7805, can't find a datasheet I understand: what input current does the 7805 require to deliver 1A?

#### ISB123

Joined May 21, 2014
1,236
7805 Can output/input 1A anything bigger than that requires bigger heatsink or bypass transistor.
It needs minimum of 1.5V to start regulating.
The bigger the difference between input and output voltage the more heat gets dissipated.

Example:

(Vin - Vout)xIo
(12V-5V)x0.5A=3.5W dissipated

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#### Chillum

Joined Nov 13, 2014
546
BOTH output and input! Thats cool, no loss of input current to deliver output current, yeah! Elaborate, how would I calculate heatsink size, and how would I go about a bypass transistor. Got the circuit diagram for standard 5V operation page 7: https://www.sparkfun.com/datasheets/Components/LM7805.pdf

#### ISB123

Joined May 21, 2014
1,236
Using the above formula you can get what heatsink size to use.You would need to use at least 10W heatsink on the regulator since it can output 1A.In case bypass transistor is used it would require heatsink too.

Transistor heatsink:
5Vx3A=15W
10Vx3A=30W
15Vx3A=45W

Regulator with bypass transistor:

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#### Chillum

Joined Nov 13, 2014
546
Ok, thx, didn't know heatsinks were measured in watts... looks simple enough, but don't think I'll be needing/ using it thanks anyways. Now the simple 5v only mode at https://www.sparkfun.com/datasheets/Components/LM7805.pdf page 7 requires a .33uF capacitor between V+ and GND, and 0.1uF between Vo and GND. Can the 0.33uF be electrolytic? And should the 0.1uF be ceramic? *sigh* noob - I know ;-)

#### BobTPH

Joined Jun 5, 2013
2,538
Hopefully it has sunk in by now that you do not use a voltage divider as a power supply.

But what is a regulator really? A regulator is a voltage divider with the regulator replacing the top resistor and the load replacing the bottom resistor. But the magic of it is that the resistance of the regulator changes to automatically keep the same voltage on the load.

Bob