Working out how to power a project

Hi All,

I'm very new to electronics and am trying to learn how to create circuits without blowing up anything. The main thing I need to know is how to work out the supply I should be using for my project. I feel like the volts don't matter and its all about the amperage draw on the circuit, however I am still worried that I'm going to burn out my components due to faulty knowledge. Could someone explain to me a simple was of working out a power supply?

Thanks in advance!

Jack

Well, for one thing. You need to think of a power supply like a water bottle and not a fire hose. Your circuit will only "drink" what it needs and as long as the power supply is powerful enough to give your circuit what it wants to consume... you are ok. If you have a LARGE capacity power supply at the correct voltage, you circuit will not care.

If you have too small a supply, it will starve your project and it won't operate correctly. In neither case will you "burn out" your project, presuming it's wired correctly, if your voltage is correct and you have enough current delivery capacity.

Well, it would be helpful if you could mention at a high level what kind of project it is. Is it a simple Arduino with a few blinky lights, or are you going to need to power 6 12-volt motors while doing doing communication back to the pc. Are you going to be able to power your project from a wall socket or do you need to run on batteries? If batteries, how long of run time are you expecting?

For the simplest projects, power it either through USB (which delivers 5 volts to the board) or via a 7-12 volt wall wort that plugs into the 2.1mm power supply. If you have higher voltage/amperage devices, then you may need more complex power setups.

pwillard:
Well, for one thing. You need to think of a power supply like a water bottle and not a fire hose. Your circuit will only "drink" what it needs and as long as the power supply is powerful enough to give your circuit what it wants to consume... you are ok. If you have a LARGE capacity power supply at the correct voltage, you circuit will not care.

If you have too small a supply, it will starve your project and it won't operate correctly. In neither case will you "burn out" your project, presuming it's wired correctly, if your voltage is correct and you have enough current delivery capacity.

Thanks! Thats a really good analogy to work with :smiley:

MichaelMeissner:
Well, it would be helpful if you could mention at a high level what kind of project it is. Is it a simple Arduino with a few blinky lights, or are you going to need to power 6 12-volt motors while doing doing communication back to the pc. Are you going to be able to power your project from a wall socket or do you need to run on batteries? If batteries, how long of run time are you expecting?

For the simplest projects, power it either through USB (which delivers 5 volts to the board) or via a 7-12 volt wall wort that plugs into the 2.1mm power supply. If you have higher voltage/amperage devices, then you may need more complex power setups.

Sorry, I didn't think about that. The project consists on 5v ping sensors and 7v servos which are communicating thorough the PC/Arduino. Naturally, given the output of a usb is 5v, I'll be using a wall socket as I don't want to use a battery supply.

Thanks again,

Jack

I feel like the volts don't matter

They are the thing that matters. Voltage is often explained as being analogous to pressure in a water system: too much pressure and the pipe bursts. Too much voltage and you'll damage the components. Check the datasheet or label of a component and it should tell you what the voltage limits are and what the current draw is likely to be.

Say a motor's datasheet says 4-6V, 200mA (800mA peak).... First you need to look for a supply of between 4 and 6 volts: below 4 and the motor won't spin, over 6 and it'll smoke. Then check that power supply with the right voltage, to see if it can supply 800mA comfortably.. So a 5V, 1A wall wart will be good.

But also think of this.... if you want to hook up two of those motors to the same supply, you don't need to double the voltage supplied but you do need to double the available current... so you would still be ok with a 5V wall wart, but would need to look for a 2A one.

JimboZA:
you don't need to double the voltage supplied but you do need to double the available current... so you would still be ok with a 5V wall wart, but would need to look for a 2A one.

This is the part thats confusing me! Thats why I said volts don't matter, I thought as long as I give a 5v motor 5v it will be fine, however if I push too much current through it (say 2 amps when it should be 1) that could burn out the component. Does this mean that as long as I get the volts right, i can use as much amperage without fear of damaging anything (so 10A for a 1A component say) as it will only draw the amount it needs?

Thanks again,

Jack

dantedraven13:

JimboZA:
you don't need to double the voltage supplied but you do need to double the available current... so you would still be ok with a 5V wall wart, but would need to look for a 2A one.

This is the part thats confusing me! Thats why I said volts don't matter, I thought as long as I give a 5v motor 5v it will be fine, however if I push too much current through it (say 2 amps when it should be 1) that could burn out the component. Does this mean that as long as I get the volts right, i can use as much amperage without fear of damaging anything (so 10A for a 1A component say) as it will only draw the amount it needs?

That is correct. It comes down to first learning basic ohm's law theory, which you should attempt to do as learning basic electronics by memorizing a bunch or rules and always do this and never do that is not a good path to understanding. Ohm's law shows that the current flow is dictated by the resistance of the device being powered by a specific voltage. So as long as the voltage source is capable of supplying that much or more current, then the resistance of the device will 'draw' the current it requires and only that much, even though the voltage source may have a greater current capacity available. I (current flow) = E (voltage) / ohms (resistance).
Lefty

Thanks again,

Jack

dantedraven13:
I feel like the volts don't matter and its all about the amperage draw on the circuit

Volts matter. A lot.

For a given voltage, the amperage used by the circuit will be whatever Ohms law says it is. It cannot be more (or less), it's the law. If this wasn't true then the million amps put out by your local power station would have blown up your computer by now.

Using the law we see that if the resistance of any part of the circuit is too low, the amperage at that point may be too high and burn the circuit. If this happens you need to add an extra resistor there (or increase existing resistors). This actually happens a lot when newbies connect LEDs to Arduinos - they hardly ever put resistors in so the amperage will be far too high for both the LED+Arduino.

My point is... you can't PUSH too much current into a circuit... you CAN push to much voltage into a circuit.

So VOLTS need to be correct.

Does this mean that as long as I get the volts right, i can use as much amperage without fear of damaging anything (so 10A for a 1A component say) as it will only draw the amount it needs?

Yep, you don't push current, just as pwillard said: a component draws the current it needs. If a 5v motor draws 800mA, but your supply is 5v with a capability of 2A, you're good.... you just spent too many $$$.

But if you get a 12V supply for that 5V motor, you'll smoke it.

JimboZA:

Does this mean that as long as I get the volts right, i can use as much amperage without fear of damaging anything (so 10A for a 1A component say) as it will only draw the amount it needs?

Yep, you don't push current, just as pwillard said: a component draws the current it needs. If a 5v motor draws 800mA, but your supply is 5v with a capability of 2A, you're good.... you just spent too many $$$.

But if you get a 12V supply for that 5V motor, you'll smoke it.

Right, that all makes a lot of sense and I've looked at ohm's law bafore, it makes sense but now I think I know how to put it into practice. One last issue though: If I create a circuit with my 7v servos, I would use a 7v supply. However, in my circuit are 5v ping sensors, which this current will burn out. So how would I go about bring the volts down? I'm guessing ohm's law + resistors but I thought I should ask the experts :slight_smile:

Thanks again,

Jack

dantedraven13:

MichaelMeissner:
For the simplest projects, power it either through USB (which delivers 5 volts to the board) or via a 7-12 volt wall wort that plugs into the 2.1mm power supply. If you have higher voltage/amperage devices, then you may need more complex power setups.

Sorry, I didn't think about that. The project consists on 5v ping sensors and 7v servos which are communicating thorough the PC/Arduino. Naturally, given the output of a usb is 5v, I'll be using a wall socket as I don't want to use a battery supply.

You probably should read tutorials on how to power servos/motors separately from the board such as: RCArduino: Servo Problems With Arduino - Part 1

There are shields that have plugs to simplify wiring up higher voltage devices (you plug in the power to the shield, and it has a groups of 3 pins for each Arduino pin that provide power, ground, and signal, and the grounds are connected, the power comes from the external power source, and signal goes into the Arduino pin). You can get cables with 3 wires specifically for servers for this. Note, there are two ways to wire the 3 pins, so make sure your server's pins matches the shield (I have one shield that wires the digital pins different from the analog pins). Or you can do it yourself.

If you need it, there are battery solutions, but you do have to think about how long you need to run before recharging batteries (better to get it working on wall power first and then if needed think about batteries).

dantedraven13:
So how would I go about bring the volts down? I'm guessing ohm's law + resistors but I thought I should ask the experts :slight_smile:

A resistor will limit the amount of amps available. Maybe too much for the low-voltage device to function properly.

There's things called 'voltage regulators' which are designed to drop voltages and not limit the amperage too much.

You would probably want to search for Voltage Regulators on the various vendor sites.

Simplest, but inefficient for reasons we needn't go into here, is a 7805. It converts an input voltage higher than 5, to 5. But it has a so-called drop-out voltage of 2 (which it basically loses) and which coincidentally for you, means the input needs to be 7 or more, to get 5 out.

So, you hook your 7v to the motors; you also hook it to the 7805, then take the 7805's 5v to your sensors. They supply 1A, although I think some makes are up to 1.5

JimboZA:
You would probably want to search for Voltage Regulators on the various vendor sites.

Simplest, but inefficient for reasons we needn't go into here, is a 7805. It converts an input voltage higher than 5, to 5. But it has a so-called drop-out voltage of 2 (which it basically loses) and which coincidentally for you, means the input needs to be 7 or more, to get 5 out.

So, you hook your 7v to the motors; you also hook it to the 7805, then take the 7805's 5v to your sensors. They supply 1A, although I think some makes are up to 1.5

Thats awesome actually! :slight_smile: I have some 5v voltage regulators that I bought for my old 5v servos. I was told though that I need to supply 5v to the left pin, 9v to the right pin and that will give 5v out of the middle pin; which confused me as I'm all ready giving the 5v voltage regulator 5v's to power it, so why not just give that straight to the 5v servo? :frowning: Have I been told wrong?

Thanks again,

Jack

Well if it's a 7805, you put >7v to the left hand pin, and ground to the middle. Then you get 5v out the right hand pin, also using the middle pin as ground.

.....but not all regulators are 78xx so have a look at its markings to see what it is.

JimboZA:
Well if it's a 7805, you put >7v to the left hand pin, and ground to the middle. Then you get 5v out the right hand pin, also using the middle pin as ground.

.....but not all regulators are 78xx so have a look at its markings to see what it is.

I've just had a look and they are 7805's! This would explain why when I used them I got nothing out, I've got my pins muddled :sweat_smile: So if I'm looking at one with the heat sink plat facing away from me: left pin = input current, middle pin = ground, right hand pin = output voltage. Is that right?

Thank again,

Jack

left pin = input current, middle pin = ground, right hand pin = output voltage. Is that right?

Yeah, except it's input voltage....

Remember the input has to be at least 2 more than the output.

Btw you might find this datasheet catalog site useful.... type 7805 in and you'll find a bunch of sheets from various makers. The 78xx comes in various output voltages as you'll see.

JimboZA:

left pin = input current, middle pin = ground, right hand pin = output voltage. Is that right?

Yeah, except it's input voltage....

Remember the input has to be at least 2 more than the output.

Btw you might find this datasheet catalog site useful.... type 7805 in and you'll find a bunch of sheets from various makers. The 78xx comes in various output voltages as you'll see.

Awesome!!! You are probably the most helpful bunch of people I've ever encountered! :stuck_out_tongue_closed_eyes: Thank you SO much for the help!!

All the best,

Jack

You're welcome...

Careful of how hot those 78XX's get.... if it's getting towards too hot to touch, you'll need a heatsink. (Edit... I mean an external one, not the silver back of the 7805)

I tested one this last weekend.... with 12V in and 5V out at 500mA, and measuring it with my Fluke's thermocouple, it got to 110C in seconds before I turned it off, un-heatsunk. With a heatsink it sat at about 40C.

And that's why those things are inefficient. Their design dictates that whatever current you draw on the low side, is the same as on the high side.
Now, power = volts x current.

So Powerin = Voltsin x current and Powerout = Voltsout x current.
Rearrange that and Powerin - Powerout = ( Voltsin - Voltsout ) x Current.
That difference in power is dissipated as heat, and you have to get it away from the device somehow.

That would mean for example, as in my test, if you happen to use 12v to get 5v, you're losing 7v x current in heat, and that's more than the power you're actually using at 5v.