Regulated or Unregulated 12V DC

I have a project that uses two small 12V DC computer fans that combined use under 1000mA. I also have a Heating pad that uses 12V AC and uses around 800mA. Lastly i have 6 small lights that use 12V AC and 65mA each. I am using a motion sensor to turn on the lights and a pressure sensor to turn on the Fans and the Heating pad. My Arduino controlls it all and the Fan, Lights and heating pad ar connected to external power and use a 12VDC 1A relay.

I am using a standard wall wart that is 12V DC and has plenty of current to drive all my devices.

I use a LM7805 to get a regulated 5V DC for the sensors and Arduino.

My question is... Should i use a LM7812 to regulate the power for the Fan, Lights and Heating pad? I'm asking becasue i think that the wall wart will regulate it enough and that using a LM7812 will be overkill...???

I look forward to any replies and thanks...!!!

PS - I have been told that even though the lights and heating pad are rated to use 12V AC that they will work fine on the 12V DC.

Should i use a LM7812 to regulate the power

If your power supply is providing you with 12VDC you can't use a LM7812 to regulate it because that chip and other voltage regulators like it require there to be about 1.5V to 2V more on the input than the voltage you are trying to achieve.

Measure the output of your wall wart with nothing attached. If it is 12V then there is no need to do anything.

The wall wart is coming in at around 14.34V DC... So I guess that means I need to regulate it and that the LM7812 will work because there is +2.34 volts over the 12V...???

BTW - Mike do you ever sleep...??? U are always such a great help and it is noticed and greatly appreciated....

Ok that sounds like an unregulated supply.
In which case you will have to watch it because it will give out 12V at the maximum current output. As you draw more current the output voltage drops. You might be better of using a 10V regulator just to make sure you have the excess all the way down.

Sleep, whats that?

Ahhh okay I think i get it... So when i tested it there was nothing drawing current so it was over 12V and when i have something connected to it that draws current the voltage drops... I never thought about that so muchas gracias for clearing it up...!!!

WHen u say to use 10V do you mean a 12V wall wart or a 12V regulator...??? What will happen to the devices (Fans, lights, heating pad) if everything combined draws the full current and it drops the voltage to 10V. All the above mentioned devices are 12V...???

Those devices are not very critical as to there voltage input. It's not like they are electronic the are just electric. Running them at a lower voltage will make the lights slightly less bright the motor not as fast and he heater not as hot, but they will last longer, running the at above the rated voltage is the opposite faster, brighter, hotter and not as long lived. But for your application it dosn't matter much.

Okay cool and once again you cleared that up... thank you... And that also answers another question i had about fan speed control too i think... I read quite a few posts on here regarding fan speed control and PWM and I'm starting to think that controlling the fan speed via PWM is not recommended. So, could I slow the fans by supplying a constant lower voltage to the fans...??? If so is there a recommended method or voltage...? I have read that they will need more than 5V to work reliably but i'm not sure how to go about getting them at a good level. I'm mostly concerned with the noise...

Yes lowering the voltage will make them run slower. PWM is OK for control but smoothed PWM is better, this is just like varying a DC voltage. You should be able to get a 12V fan running at v5V with no problem, it is about 3.5V when they won't start, however once started you can go down a bit lower.

I got a 9V 800 mA wall wart and tested it when there was no load on it and it said it was putting out 14V... Does this sound right...???

Then i added a 12V light (65mA) and 2 - 12V fans (250mA + 480mA) that together equaled 795 mA and then tested the voltage and it was still around 10 - 10.25...

If I run a 12V componet on a 9V supply will it draw less current or the same current at the lower voltage...?

If a product says it will draw 250mA will it always draw that much...???

I got a 9V 800 mA wall wart and tested it when there was no load on it and it said it was putting out 14V... Does this sound right...?

While not right it is typical of what you find with an unregulated power supply. I think you might be better off getting a regulated one, they are quite cheap these days.

If I run a 12V componet on a 9V supply will it draw less current or the same current at the lower voltage...?

Ohms law - it will draw less current at a lower voltage.

If a product says it will draw 250mA will it always draw that much

The product will (should) say at what voltage it draws that, no lower the voltage and you lower the current.

So with Ohms law and one of the fans at 12V I take 12V x .25 Amps = 3 Watts...?

Then how do i figure what current it will draw at 9V...??? Do i take the 9V x .25 Amps = 2.25 Watts...??? Is that right??? Isn't it telling me that it will still draw .25A at 9V...??? I guess what i don't understand is how to figure what my devices will draw when ran at a different voltage then what it is meant to be ran at.

I think it is all starting to very slowly sink in and thanks for everyone's patience while i try to make sense of all this. Can someone please tell me if this thinking is right...??? So my circuit will need two voltages from one power source. I plan to use a 9VDC 2A power adapter that outputs a regulated 9V. I will have to regulate some power down to 5V for some of the devices (motion detector, pressor sensor, Arduino, etc.) and use the 9V to power the lights, fans and heating pad which are all hooked up to the 9V rail through their own relays (so 3 relays total; one for lights, one for fans, one for heating pad). So my top current draw on the 5V rail may be up to say 500mA and my top current draw on the 9V could be up to 1500 mA... If I need to regulate 5V from 9V then I will have a voltage drop of 4 Volts. If my entire circuit (5V rail and 9V rail) is drawing 2 Amps of current will I have 8 Watts of heat to dissipate...??? Is that right or is it only figured for each rail...??? So if it is for each rail the 5V rail will have 4V x .5 Amps = 2 Watts and the 9V rail will have none...???

Do you think it would be better to get a regulated wall wart or to get an unregulated one? IF i use an unregulated one like i hooked up today adn it outputs 14V with no load won't i have a lot more heat to deal with? If i'm thinking about this right i think it would make it easier to get a regulated one...

Thanks for any help you can provide....

Ok I think you are going to like this:-

You are right the product of volts and current is power, and this is not constant. However the ratio of volts to amps is a constant. So suppose you know the current at 12V then:-
( 12v / current at 12V ) = ( any other voltage / current at any other voltage)
This ratio of volts per amp is very important.
Suppose you have a heater taking 3 Amps at 12 Volts, then its ratio is 12 / 3 = 4 volts per amp
But instead of saying the electrical property is 4 volts per amp we invent a new unit called Ohms and we call this voltage / current ratio resistance.
So that heater has a resistance of 4 Ohms.
Now we can tell what current it will take at other voltages, say 9V:-
Current = 9 / 4 = 2.25 Amps.

Normally we consider the volts per amp (or resistance to be constant) and most of the time it is near enough, we say it is a linear resistance because if you plot the graph of voltage against current you get a straight line and the slope of that line is the resistance in Ohms.

Neat or what?

FANTASTIC... That makes total sense and it is too cool... I would've searched the web forever trying to figure it out... Thank you for the very clear example... So now I can confidently figure out what my power needs are for my project at 9V and here is my math...

  1. Lights – 12V 65mA
    a. 12V/65mA = 0.1846 Ohms
    b. 9V/.1846 = 48.75
    c. So at 9V each light will draw 48.75mA
    d. This is 0.44 Watts…???

  2. Heater – 12V 800mA
    a. 12V/800mA = .015 Ohms
    b. 9V / .015 = 600
    c. So at 9V the heater will draw 600mA
    d. This is 5.4 Watts…???

  3. Fans – 12v 250mA
    a. 12V/250mA = .048 Ohms
    b. 9V / .048 = 187.5
    c. So at 9V each fan will draw 187.5mA
    d. This is 1.6875 Watts…?

I am still fuzzy on the AC/DC converters and how to figure what my heat dissipation will be at various voltages/currents so I have a couple of questions about voltage drops, heat sinks, etc...

So lets say I have an AC/DC converter that outputs a regulated 9V DC 2A. On my circuit I use two voltages; 9V and 5V. The 9V power rail will need up to 1500mA. The 5V power rail will need up to 500mA.

My questions :

  1. When the power adapter says the output is regulated does it mean that it is going to output a steady 9V no matter if my load is 500mA or the full 2Amps or will there always be a voltage drop...???

  2. When I regulate 9V down to 5V with a LM7805 will it always supply a consistent 5V no matter if the current drawn is 100mA or 500mA?

  3. To calculate my voltage drop in a circuit I first take the difference of the input voltage and the regulated voltage- 9V – 5V = 4V. Then I figure the product of my current draw 2 Amps and this voltage; 4V * 2 Amps = 8 Watts. So my heat to dissipate will be 8 Watts…???

  4. When determining my watts of waste/heat to dissipate will the 9V and the 5V rail be determined together or separately…??? When I think about it rationally I think it will be determined separately but I’m not sure. For instance…

a. If I have the consistent regulated 9V input and then I regulate 5V with a LM7805 and my 5V rail uses up to 500mA @ 5V then I will have a voltage drop of 4V and 2 Watts of heat to dissipate if the current drawn on the 5V rail is 500mA…???

b. Then for my 9V rail I will have up to 1500mA @ 9V needed so with a consistent regulated input of 9V there will be no voltage drop and therefore no heat to dissipate…???

c. Am I thinking about this all wrong or does this look right…???

  1. At what wattage of heat to dissipate do I need to use a heat sink…??? How can I determine the size that the heat sink needs to be…???

  2. Where do the heat sinks go...??? I think for the 5V rail it should be on the LM7805 but what about for the 9V rail?

This is all such a huge help adn i feel like i owe something back...!!! Is there anything i can do to show my appreciation...???

  1. Lights – 12V 65mA
    a. 12V/65mA = 0.1846 Ohms 184.6 Ohms
    b. 9V/.1846 = 48.75
    c. So at 9V each light will draw 48.75mA - yes
    d. This is 0.44 Watts…??? - yes

  2. Heater – 12V 800mA
    a. 12V/800mA = .015 Ohms 15 Ohms
    b. 9V / .015 = 600
    c. So at 9V the heater will draw 600mA
    d. This is 5.4 Watts…??? - yes

  3. Fans – 12v 250mA
    a. 12V/250mA = .048 Ohms 48 Ohms
    b. 9V / .048 = 187.5
    c. So at 9V each fan will draw 187.5mA
    d. This is 1.6875 Watts…? - yes

  1. When the power adapter says the output is regulated does it mean that it is going to output a steady 9V no matter if my load is 500mA or the full 2Amps

Yes that is what it means to be regulated.

  1. When I regulate 9V down to 5V with a LM7805 will it always supply a consistent 5V no matter if the current drawn is 100mA or 500mA?

Yes

  1. To calculate my voltage drop in a circuit I first take the difference of the input voltage and the regulated voltage- 9V – 5V = 4V. Then I figure the product of my current draw 2 Amps and this voltage; 4V * 2 Amps = 8 Watts. So my heat to dissipate will be 8 Watts…???

Yes

a. If I have the consistent regulated 9V input and then I regulate 5V with a LM7805 and my 5V rail uses up to 500mA @ 5V then I will have a voltage drop of 4V and 2 Watts of heat to dissipate if the current drawn on the 5V rail is 500mA…???

That is the heat dissipated by the 5V regulator yes.

Then for my 9V rail I will have up to 1500mA @ 9V needed so with a consistent regulated input of 9V there will be no voltage drop and therefore no heat to dissipate…???

Yes - but there is a regulator inside your power supply and that will dissipate something but that is taken care of by the power supply designers.

At what wattage of heat to dissipate do I need to use a heat sink…

When you exceed the free air dissipation of the device as told to you by the data sheet.

How can I determine the size that the heat sink needs to be

This can be quite tricky, you need to know first the thermal resistance of your heat sink, then you can relate that to surface area, it also depends on if you have forced air cooling or natural convection and if so how good the natural convection is.
Generally it is more of a trial and error situation but this link shows you what you have to consider:-
http://www.thebox.myzen.co.uk/Tutorial/Power.html

This is all such a huge help

Good - No problems :slight_smile:

SWEEEEETTTT..!!!! I fell pretty confident regarding how to figure my power needs and a much better understanding of how things work in general...!!! This is great...!!!

Okay now i have a couple questions regarding the design of teh circuit and how some of the componets are connected; specifically the relays. Should i start a new thread or ask the questions in this one...???

Yes start a new thread, see you there.

Will do... thanx...

-H-

One more quick question regarding power...

Does anyone know where I can get an AC/DC regulated 9V 2Amp wall wart... I'm in the USA and ideally it would have a tip that is 5.5mm O.D.; 2.5mm I.D.; and female....

Thanks again...

SummitSeeker:
Okay cool and once again you cleared that up... thank you... And that also answers another question i had about fan speed control too i think... I read quite a few posts on here regarding fan speed control and PWM and I'm starting to think that controlling the fan speed via PWM is not recommended. So, could I slow the fans by supplying a constant lower voltage to the fans...??? If so is there a recommended method or voltage...? I have read that they will need more than 5V to work reliably but i'm not sure how to go about getting them at a good level. I'm mostly concerned with the noise...

You can buy 4-pin computer fans that take a standard 12v supply and have a separate PWM input. The PWM input needs to be driven by an open-collector NPN transistor to ground. The specified PWM frequency is 25KHz but is probably not critical (you can get 32KHz from some Arduino PWM pins if you reprogram the corresponding timer).