resistor help.

[SeVeNeVeS]

i've got a meter that I want to connect to a 12v wall wort.






Can anyone please tell me what resistor is required to get the dial to around 8ma when power is switched on.

In an ideal world point me in the direction of where to purchase such a thing in the UK ,maybe e-bay, with a link, I dont want to wate money guessing, buying the wrong thing and blowing the meter. Thank you so much for any help.

[J. Wilhelm]

I'm assuming the power supply is a constant 12V DC, direct current, not alternate.

V=IR

Hence

R = V/I = 12v / 0.008 A = 1500 Ohms or 1.5 kΩ

But it needs to be made to dissipate the amount of power otherwise it will overheat

P = I²R = (0.008A)² * 1500Ω  = 0.96 Watts (~ 1 W standard commercial value).

1 Watts doesn't sound like a lot but it's too much to dissipate for most electronics rated resistors and would get very hot and burn if the rating is 1W or less. Typically resistors on electronic boards run in ¼ W and ½ W ratings, and there's some in a few watts, but they'll get hot even at 1 W.  You'd need something like a big ceramic resistor, with 10 times more power rating, more akin to a speaker crossover resistor. Look for speaker parts, ceramic resistors. I'd give you a website (Partsexpress.com) but they're in the United States.

[SeVeNeVeS]

Many thanks for the info above, would one of these do the job?

https://www.ebay.co.uk/itm/303282579756?hash=item469d0ce92c:g:4PQAAOSwdxldeicd&amdata=enc%3AAQAIAAAAwFFCDMh0XnxFvWXbyBQ3zuDX9lIsuzJX8e5piz7i%2B7BZpOcFvloUznW2lpXE3iKE5sYxr8fdgJK42DyN0LVOlvy7AzpOcL0OLleUynDfT06%2FrcME%2BQDhGG8JaV%2FUYCgpoDgFq8YJehlZQqMTmchZXopLgsaYXP8PrIjDc6Ec6GVO0ECL8HHjg5SywLyXB1ncf%2FQ9B92tLSZEDcSvd4J9alzXUdBc2PdwoDpnfbh%2BxtUqoZl4Xot7g3e7HxQjTDeBZg%3D%3D%7Ctkp%3ABk9SR7rb8uOKYg

I am totally out of my depth here so sorryif any question is a stupid one.

[J. Wilhelm]

Many thanks for the info above, would one of these do the job?

https://www.ebay.co.uk/itm/303282579756?hash=item469d0ce92c:g:4PQAAOSwdxldeicd&amdata=enc%3AAQAIAAAAwFFCDMh0XnxFvWXbyBQ3zuDX9lIsuzJX8e5piz7i%2B7BZpOcFvloUznW2lpXE3iKE5sYxr8fdgJK42DyN0LVOlvy7AzpOcL0OLleUynDfT06%2FrcME%2BQDhGG8JaV%2FUYCgpoDgFq8YJehlZQqMTmchZXopLgsaYXP8PrIjDc6Ec6GVO0ECL8HHjg5SywLyXB1ncf%2FQ9B92tLSZEDcSvd4J9alzXUdBc2PdwoDpnfbh%2BxtUqoZl4Xot7g3e7HxQjTDeBZg%3D%3D%7Ctkp%3ABk9SR7rb8uOKYg

I am totally out of my depth here so sorryif any question is a stupid one.

That may do if you test for heating.  But I was thinking one with a larger package. If It gets too hot to touch, you need a higher rating. The issue is heat dissipation.

But I'm thinking that the resistor in series solution is the brute force approach. There's other ways to feed that Ammeter less power. "Splitting the 12 V into a smaller voltage drop can be done in a number of ways. One quick solution if you must use that power adapter, is to go to your nearest convenience store, and grab a 12 V car cigarette lighter to USB charger adapter. The circuit is a voltage regulator, it's perfectly happy with 12 V and will only feed you 5 V, so you can repeat the calculations with 5V instead of 12 V. Or just skip that 12 V power supply and buy a 220 V mains adapter to USB! Why not take advantage and power an LED or light of some source? Lots of gadgets work with USB power nowadays. Just keep in mind most USB chargers can only hand about 1A and 0.8 A is awfully close; a few of them will be rated 2A. You want the one with the higher rating for safety. It will get hot too, but that's okay if it's within the rating.

[RJBowman]

Why not just use a power source with a lower voltage?

[J. Wilhelm]

Eeeek! I just realized I made an error in my second calculation.  The power dissipated with 12 V and a 1.5 kΩ resistor is:

P= I²R = (0.008A)² * 1500Ω  = 0096. Watts or 1/10th of a Watt.

Sorry about that. But still, you'd better use something rated more than your ant sized garden variety ¼ Watt rated resistor. At this point I'd say the 5W rated resistor that SevenEves posted is more than adequate for the situation. I still wouldn't be comfortable with a ¼ Watt resistor because it'd get hot, but one rated around 1W is fine. You're basically asking the resistor's ceramic package to dissipate heat like a radiator, so physical size counts.

Why not just use a power source with a lower voltage?

Exactly. I suggested that above too. Instead of buying a 12V cigarette plug adapter, just use a mains to USB charger:

R = V/I = 5v / 0.008 A = 625 Ohms (get the closest value you can)

P = I²R = (0.008A)² * 625Ω  = 0.04 Watts

Any resistor will do. Most are rated rated for ¼ Watt at least


Ah! And one more thing: Don't forget that the resistance of the Ammeter also adds into the resistance value. I'm assuming that the Ammeter resistance is very low compared to 1500 Ohms, so I didn't add it.  The needle will not be exactly at 8 milliamperes due to that and the 5% (gold band) or 10% (silver band) tolerance value of the resistor

[SeVeNeVeS]

Thank you so much JW for taking the time and effort to help me out here.

Ive ordered https://www.ebay.co.uk/itm/303282579756?var=602284311061
and https://www.ebay.co.uk/itm/194256368451?var=494513484867   https://www.ebay.co.uk/itm/194256368451?var=494360818633

Not sure when this will be started but if I have the supplies it all helps. A project for a later date me thinks.

I'm only using the 12v becuse I already have it.

[SeVeNeVeS]

You beauty JW.

I got the resistor and tested with a 9V battery before doing the 12V.




Just gotto connect to the 12V and test for potential heat problems but I would say so far so good.

Thank you so much for taking the time to help.