# How to get 80 mV?

Discussion in 'General Electronics Discussion' started by Electro132, May 12, 2015.

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1. ### Electro132

261
3
Feb 12, 2013
Hi,

I want to know how to get 80 mV. I have applied a 1.5 v batt and placed a 10 ohm, 100 ohm and 220 ohm resistors on their own (in circuit with the battery) but got only 1.4 V using a multimeter on all of them. I usually thought getting a higher resistance would mean a lower voltage would go through it but the result is the same on my multimeter. The current i need is at least 0.08 amps. Can anyone help? Thanks.

2. ### Harald KappModeratorModerator

11,513
2,651
Nov 17, 2011
What are you doing?
Placing a resistor (or a series of resistors) across a battery's leads will drop the voltage only a little, as you have measured, due to the internal resistance of the battery. What you're looking for is a voltage divider.

Where is the connection between this requirement and the heading of your post "How to get 80 mV?" 80mV <> 80mA is comparing apples and oranges. Make up your mind and tell us what you really want to achieve.

3. ### Viktory2k1

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Apr 14, 2015
Get some high wattage resistors and it should bring it down, along with the battery life. I am not to bright here, thats why I joined.
I know you can drop 12vdc to <>3v with 27ohm 5 watt resistors but they will get warm. Not sure of current though. I breadboard to figure out these things.

4. ### Harald KappModeratorModerator

11,513
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Nov 17, 2011
Why high wattage? A voltage divider can be built from high resistance resistors dissipating only milliwatts or microwatts.

How do you do that without building a divider? Of course, when you put a very heavy load on a battery, the voltage will drop, but uncontrollably and at the cost of high losses, as stated by you.

A voltage divider is a basic circuit in electronics and you really should understand the concept.

5. ### Viktory2k1

41
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Apr 14, 2015
I am not an electrical engineer and was just giving to advice that I know works. I do not know what a voltage divider is unless you are speaking of mutliple loads in the same circuit

6. ### Viktory2k1

41
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Apr 14, 2015
High wattage to absorb the extra energy and absorb heat, like a ceramic resistor. How would I know these things? I know some basics and apply them to the breadboard and see what happens, thats how I learn things.

7. ### Harald KappModeratorModerator

11,513
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Nov 17, 2011
See the link in my post #2.

Doing a little bit of theory is not one of the worst ideas.

8. ### Viktory2k1

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Apr 14, 2015
Say converting 12vdc to 3vdc, I would use a 33 ohm 5 watt resistor if you had no other choice. I learn quite a bit with my answers alone.

Yes, post 2 was what I meant, didn't even know it had a tech name. Now I do so I learned something, right?

Don't need all those equations just for something simple, in the case of this thread, I think it might because of the precision of the ma he is trying to do but I use this way to power low cost stuff like a 350Ma led using 12v

I have no theory to go by, that is why I joined this forum. I use the breadboard and test. I don't try to weld off the thing and know common sense and usually things work out unless it's something I am trying to make. Then it doesn't work. I am trying to power a 5M string of RGB LEDs from a Arduino and am unsure of the resistor and, for now, transistor wiring which I think is a KA7905 that I desoldered from a pc psu. Should be the same as a tip120 I think. So far, nothing worked yet but I don't spend much time on it.

Last edited by a moderator: May 12, 2015
9. ### Harald KappModeratorModerator

11,513
2,651
Nov 17, 2011
Show us a circuit diagram how you would achiev this. A single resistor across a 12 V source (battery) will not drop the voltage down to 3 V.

If you plan to move ahead in the area of electronics, you will need equations. No way to bypass these.
The op doesn't tell us anything about the required precision. He's using 80 mV in the heading and 80mA in the text, no tolerances/precision at all.

That is a diffferent issue. In this case the resistor does limit the current through th eLED which is quite a correct way of doing this (albeit at the cost of power dissipated, but that's another story). However, you do not generate a specific voltage (like e.g. 3V) by this. What you see is the voltage drop across the LED as a result of the limited current.

So this is a good point to start with some of the theory.

Of course, trial and error is one method to go along, but a risky one. With low voltages like the 5V or 3.3V of an arduino, currents in the mA range and no theory behind your actions, you risk at worst destroying the arduino. If that doesn't matter to you - fine.
When you move to higher voltages and/or currents you risk lifes - your's or other people's. That is no longer fine but irresponsible.

With a bit of theory you'll have a sense of achievement much sooner.

Last edited by a moderator: May 12, 2015
witsender and Arouse1973 like this.
10. ### Gryd3

4,098
875
Jun 25, 2014
This thread looks like it drifted a little...
A single resistor won't do it for you as mentioned previously. If the battery were perfect you would see 1.5V instead of 1.4V. (think of the battery as having a very small resistor inside. When you connect your circuit, 0.1V drops across the internal resistance in the battery and the 1.4 remaining across the resistor you added. (The voltage drop across the internal battery resistance will vary depending on the resistance you 'add' to it.)

Lets look at a basic voltage divider:
1.5V battery (Perfect, with no internal resistance)
2x 50kΩ resistors.
You connect all 3 items in series:
-Total resistance of the circuit is 100kΩ, with 1.5V this allows 15μA of current through.
-15μA of current through a 50kΩ resistor will create a 0.75V drop across it.

But if you want a smaller voltage, you will need to use different valued resistors because even in this example above, the 'sum' of the voltage across both resistors MUST equal the battery voltage. So using 100000kΩ resistors instead will give you the same result... but mismatched values will give you something different
100K + 5.6K resistors will give you 0.0795V across the 5.6K resistor, and 1.420V across the 100K resistor. You will get the same answer using 1000K + 56K as well because the ratios are the same.
There is a catch though... When you hook something up to it to draw that voltage, you will change the resistance! So the math will be wrong and the voltage you get may be different. (This can be accounted for, so don't worry. Just keep it in mind)
The other problem is using something that is not constant... if a device varies during operation and pulls different amounts of current, the voltage will NEVER be stable.

Now. If it's merely 80mA your after, that's a bit different. You need to know the resistance of the device you will hook up, and you simply add a resistor in-line with it so that the total resistance of the device + the resistor only allows 80mA.

** In theory this is easy, but in practice the internal resistance of the battery must be accounted for. This internal resistance can be tricky to measure and will be different depending on the battery used and temperature. (And will vary as the battery dies) . Good news is this can usually be ignored depending on how accurate you need to be.

Best of luck!

11. ### witsender

22
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Dec 12, 2013
Yes, Harald is doing his usual, wonderful thing here... giving good advice.
I can't improve on what Harald says - he's exactly right. I can put it another way though, and sometimes that's what people need, so here goes...

The general subject is called "Ohms Law", and it's so useful!

You need to understand that the amount of electrons (current) flowing through a resistance is defined by the relationship I=V/R.
• "I" is the number of electrons per second flowing through the resistances. This is called current.
• "V" is the amount of force driving electrons through the resistive circuit. Twice (or 3 times) as much force mean there will be twice (or 3 times) as many electrons flowing.
• R is the "resistance". The best way to understand resistance is to see that resistance limits current in a circuit.
You also need a good. preferably instinctual, understanding of the numbers.
For that, you need to be able to translate from algebra to English. It's easy!
"Divided by" means "per", or "for every". For instance, when someone tells you they drive at 100 km per hour, that means 100km/hr, or 100 km for every hour traveling.

Let's apply the translation to Ohm's Law.
Say you have a 10,000Ω resistor, and you connect it to a 5V supply.
Ohms Law is I=V/R, so current (I) is found by a process of division. Current is proportional to Volts (V) per Ohm (Ω). For more volts you get more current, and for more resistance you get less current.
A 5V supply will drive "5 divided by 10,000" Amps. The arithmetic gives a result of 0.0005 Amps. (This is usually called 0.5mA or 500μA)

Now imagine that your 10,000Ω resistance has been made by connecting 2 resistors in series. Imagine there's a 9000Ω resistor in series with a 1000Ω resistor. We already know that the current is 0.0005 Amps, so this time we "invert" Ohms Law, and use it in another form in order to find out the 2 voltages on the 2 resistors.
V=IR, so we multiply the 0.0005 Amps by each resistance, then check...
Ohms Law says V=IR.
9000*0.0005 = 4.5V
1000*0.0005 = 0.5V
4.5V + 0.5V =5V​
From the calculation we see that 0.5V can be measured across the 1000Ω resistor, and 4.5V can be measured across the 9000Ω resistor. Thus we have made a "resistive potential divider" and we can use it to produce voltages of 4,5V and 0.5V

I know it must make your head hurt a bit (it did mine at first) but it all becomes very easy once you get some practice. Find yourself a book! (Does anybody have a recommendation for Electro, please?)
Or even better find someone with some experience, who can give you some face-to-face help - which makes it all so much easier!

Mark

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12. ### witsender

22
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Dec 12, 2013
Ooops! Here's the drawing I made for you

Gryd3 likes this.
13. ### Viktory2k1

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Apr 14, 2015
Maybe posts 8 and 1.
I was electrocuted once by a welder that shorted out and blew me 10 feet to the floor so my back is gone. I just see what is trying to be done here as a simple choke on voltage, no need to get involved in theroy unless he likes that sort of thing. My point was try this or that and see what happens. No way will a 1.5v battery kill him, even if he ate it which I wouldn't recommend. I just hooked up a MB for the pc that is going to run mu pinball and jumped the PSU and it is liquid cooled and heard a nice sound of coolant flowing. little did I know it was squirting a over the floor so I ran out to get some paper towel only to come back and see my 2 cats drinking it off the floor. I cleaned it up quickly so another theory to test. I told you I am dumb when it comes to this unless it works, then I reverse engineer what I did, I have no schooling for this and am here to learn. Sorry to offend.

14. ### Viktory2k1

41
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Apr 14, 2015
The above drawing is what I was trying to explain, I do not know the tech terms fow what it is but can tell you what it does but I think thats overkill on what this project is trying to accomplish. Tell me how to make these then, notice how everything is turned in such a way that you can't read the components:
http://www.zebsboards.com/index.php...dwiz-control-boards/bare-bones-booster-detail
I know that there are 16 MOSFETS but dont know what the other things are and thats why I break out the breadboard. No way can this handle 50A at any voltage, the tracers would burn up. Sorry, don't want to see some dead cats and now a new MOBO may be in order after spilling coolant all over it, MAN, I just bought new memory for it. Obviosely, the hoses are junk.

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Apr 14, 2015
16. ### Electro132

261
3
Feb 12, 2013

Sorry Harald, i meant 80 mA not 0.08 mA.

Thanks guys for the references and circuit. I'll go and take a look at it, cheers

17. ### Harald KappModeratorModerator

11,513
2,651
Nov 17, 2011
Wait.
it is still not clear what you need: 80mV or 80mA?
To get 80mV from 12V you use a divider.
To get 80mA from 12V you use a series resistor, the value of which depends on the load, specifically the voltage drop across the load at 80mA current flow.

18. ### Electro132

261
3
Feb 12, 2013
Its ok. I will play around with the mA, mV and V. I have decided to just use a simple circuit containing a 1.5V batt connected to a resistor and LED. I will also test it using a divider. I wanted to get 80 mV but figured learning how to lower down voltages from a higher input source would be more of an advantage in the long run.

Thanks Harald and everyone.

19. ### BobK

7,682
1,688
Jan 5, 2010
Electro. Could you please explain what you are trying to do in an unambiguous way?

Saying you want to get 80mV or 80mA does not mean anything.

If it is 80mV, the question is across what?
If it is 80mA, the question is through what?

If it is both, the answer to both of these questions better be a 1 Ohm resistor.

Bob

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20. ### Electro132

261
3
Feb 12, 2013
well i am trying to get 80 mV first to come out of a battery even if i have to place a capacitor into the circuit. I know this is possible since it is only 80 mV so how hard can it be? I'm planning to use either a 1.5V or 9V batt.