Hi all,

I have split 12V down to 9V using a 10K and 30K resistor but the current is too high - I need to get it down to half an Amp.

Is it possible using passive components, is there a pre-made module that I can use?

Any recommendations, as usual, would be gratefully received.

Kevin

 

a 7809 regulator ?
Or a 12 Volt to 9 Volt Buck Inverter 3 terminal IC ?

 

To keep the voltage constant you really need a voltage regulator.
7809: https://www.mouser.co.uk/Search/DownloadDatasheet?url=https://www.mouser.co.uk/datasheet/2/389/l78-974043.pdf
LM317: https://www.google.com/url?sa=t&rct=j&q=&esrc=s&source=web&cd=1&cad=rja&uact=8&ved=2ahUKEwjRioe2rNXjAhUdTxUIHYV0Cb8QFjAAegQIABAC&url=http://www.ti.com/slvs044-aaj&usg=AOvVaw2jDzU5MVeJtrM4bxWkBbBF
You will need a heatsink for 0.5A

Alternatively you can get a ready made module:
https://www.ebay.co.uk/itm/1PCS-LM7...940749?hash=item2ca7fd0fcd:g:KlgAAOSwFMZWsBS5
https://www.ebay.co.uk/itm/LM317-DC...739760&hash=item36501b2bbe:g:u0EAAOSwZvldFSKD

 

?
12V 6A to 9V 0.5A
?

Please describe your output circuitry
? do you need constant current (500mA) or constant voltage (9V) output ?

 

You cannot get 0.5A through a 10K and 30K resistor with 12V input.

Are you asking how to get 9V to power a 0.5A load from a 12V 6A supply? If so, the answer is a voltage regulator. Voltage dividers are not generally appropriate for powering something, they are used to provide a reference or very low current signal, not power.

Bob

 

Is it possible using passive components, is there a pre-made module that I can use?

Not possible with just passive components.

Any recommendations, as usual, would be gratefully received.

Instead of giving us your solution to your problem, describe the problem you're trying to address.

Devices will only draw the current they need. A fuse is generally used for overcurrent protection, but that's mainly to prevent fires, not protect the circuit being powered from damage.

 

In most devices current is a result of the voltage. Current is not an independent variable.
So when you say you have 12V, 6A, it means you have a 12V supply that can provide up to 6A of current, but that doesn't mean it will always supply that current.
The current is determined by how much the circuit needs/takes.

And if you have a 12V battery rated a 6Ah, that is the energy capacity of the battery, not its current rating.
For short periods, it can deliver much more than 6 amperes (if you ever place a short across one, you will immediately realize that ).

 

@dl324 I have a project box with 240v supply. Inside the box I have a 2 gang Extension lead with a 12V 6A PSU for LEDs and a 9V 0.5A PSU for another circuit.

I decided to do away with a lot of the project box innards and use just the 12V PSU and tap 9V from it (10K and 30K resistor). This worked but blew the 9V circuit.

I presumed that the 10A (or whatever it ended up as after the voltage split) was too much for the 9V circuit hence my question, how to reduce the Current.

Kevin

 

@dl324 I have a project box with 240v supply. Inside the box I have a 2 gang Extension lead with a 12V 6A PSU for LEDs and a 9V 0.5A PSU for another circuit.

I decided to do away with a lot of the project box innards and use just the 12V PSU and tap 9V from it (10K and 30K resistor). This worked but blew the 9V circuit.
Kevin

Huh? It had it, you took it out and want to put it back again?
Are you using 30K/10K as a voltage divider? If things "blew" then obviously it did not work.
I guarantee if you give these guys information on what it is your attempting to do you will get plenty of useful information. I do not think its really clear on what it is your doing or have done so not much can be said to help out other then what @AlbertHall already said in post #3.

 

As mentioned before, you cannot control voltage and current at the same time. Voltage V, current I, and resistance R are interrelated according to Ohm's Law.

I = V / R

You can set the voltage. The current will be whatever the load demands.

For your application, use a 7809 three-terminal linear voltage regulator to set the voltage to 9V. You don't have to worry about the current. The current will take care of itself.

 

A 10KΩ + 30KΩ voltage divider cannot "blow" under the conditions you have stated. There must be something else that is "wrong" in your analysis of the situation.

Here is the proper analysis.

With no load attached, the current through the resistors is

I = V / R = 12V / 40KΩ = 300μA = 0.0003 A

Total power = 12V x 12V / 40KΩ = = 3.6 mW = 0.0036 W

Even if your load attached is 0Ω, current is
I = V / R = 12V / 10KΩ = 1.2 mA = 0.0012 A
Total power = 12V x 12V / 10KΩ =14.4mW = 0.0144 W

There is not enough power to kill either resistor.

Your LEDs simply will not be lit under those conditions.

 

Show us exactly how you connected the 10kΩ and 30kΩ resistors to the 12V and to the LEDs.

 

If you need to drop 3V @ 0.5A then you need a 6Ω 3W resistor, not a 10KΩ resistor.

R = V / I = 3V / 0.5A = 6Ω

Power = V x I = 3V x 0.5A = 1.5 W

 

Thanks for you help guys. Being an electronics noob obviously doesn't help - I'll go away and ponder the answers.

Kevin

 

What is your 9V load? What are you trying to power? That is a starting point.
As you have a good 12V supply to start with, it will be quite easy to supply the 9V from it.
The current capacity of the 12V supply only needs to be equal or greater that the 9V load's current.
Just think of a car battery. It is capable of supplying quite a high current, for example, starting the car, but that does not mean it forces that current into a small load. The loads only draw the current they require.
So, if you need 9V at 500mA, just get a regulator, and there are many options.
I quite like these...

Some are adjustable and some fixed. Get an adjustable one, set it to 9V BEFORE connecting to the load!
As long as the reg can supply the current required, that is ok. These boards can supply 3Amps, but as mentioned before, your load will only the the current it requires, so it will only draw 500ma at the 9V output if that is all it needs.
You could make your own regulator if you want, but a plain resistor divider is a no-no.
This forum site has some good tutorials under the "EDUCATION" tab, and I like this one...
http://www.learnabout-electronics.org/
Have a read up on things.

 

Have a look at this one...
https://www.ebay.com.au/itm/5V-3A-D...ge-Regulator-Integrated-Circuits/223211785373
It has solder tabs for voltage selection.

 

http://tinyurl.com/yytwxw4n

• in above simulation there is 100:300Ω voltage divider
• 200Ω is a randomly set Load resistance for assumed 9V circuitry
• 800Ω is a randomly set Load resistance for assumed 12V circuitry
• (there will be a voltage drop and at 12V primary voltage source but it is negligible in common cases ... so we ignored it here)
• there are 2 switches below each "9V" and 12V node to toggle the 200 & 800 Ohm loads ON or OFF

! there is still a chance , if you used 10kΩ:30kΩ divider , that your 9V load did not got any significant current to be turned on ... and it's actually not blown ?

 

! there is still a chance , if you used 10kΩ:30kΩ divider , that your 9V load did not got any significant current to be turned on ... and it's actually not blown ?

@ci139 I plugged the previous supply back in and although the Power LED came on, there is no life from the circuit.
I have ordered a Buck converter as suggested previously and I'll see what happens.

This has been a confusing thread - I am primarily building a Mechanical project and using all of my electronic nous (and a proportion of other peoples). A number of people in this thread say it's not possible and a number of people suggest ways that it is possible.

Kevin

 

my grandpa was electrician , my uncle was freezers service engineer/electrician (i spend most of my childhood summers with) -- is where my interest to the field ...
? call your friends find someone to assist you , the internet is actually the slowest way to get things done (i often forget i have loads of science books with answers lying around allover the appartment -- so insted of 20s lookup i find myself googling some trivia for 20minutes )

 

@ci139 A number of people in this thread say it's not possible and a number of people suggest ways that it is possible.
Kevin

Making a voltage divider is possible yes and it has many uses. Its not much of usable power supply for any type of load tho.
I suggest you do some research on what they are and what limits they have. The thread might not seem so confusing then.