Need help in figuring out why voltage drops under load

Tonyr1084

Joined Sep 24, 2015
7,853
Here's my little bit to this thread: Ohms law says that current times resistance is equal to voltage; voltage divided by current is equal to resistance and voltage divided by resistance is equal to current. You probably already know that much.

So "IN THEORY" if you have a 12 volt battery and a 1 ohm resistor comprising a circuit then you have 12 amps flowing through it. Let's bring this closer to real world: You have a 12 volt car battery capable of 100 cold cranking amps. You put a 1 ohm resistor across the terminals (and let's ignore a couple things - wattage and the absolute real world reactions to these theoretical circuits). Since the capability of the car battery is great, the 12 volts (let's assume 12 volts exactly) will hardly drop when you draw heavy current through it. The 12 volt battery (let's say) drops to 11.9 volts with the 1 ohm resistor in the circuit. The current is now 11.9 amps, not 12 amps. Now lets do the same test again but use a small 12 volt battery, something capable of 50 amps (ok, that's not so small) (stick with me here). When the 1 ohm load is placed in the circuit now the battery voltage drops to 11.8 volts and the current drops to 11.8 amps. Here's the important detail - the resistance doesn't change (in theory). It continues to be 1 ohm. So as the voltage drops so does the current. If you had a 12 volt battery capable of 10 amps and you put the 1 ohm resistor in circuit the battery voltage will drop significantly. Let's assume it drops to 9.8 volts. The current is going to be - yes, you guessed it - 9.8 amps. Real numbers will definitely be lower, but I use these as an example to explain how supply voltage depends on the ability to deliver current at that voltage. Exceed the capacity and the voltage drops significantly - until it reaches some equilibrium. Remember, the voltage changes and so does the current, but the load remains constant. In theory. Resistors DO change value with heat, so the notion that resistance doesn't change is not entirely accurate. But for the sake of understanding, assume a 1 ohm resistor is going to always be 1 ohm. Any voltage applied across it is going to result in the same current as the voltage.

What happens if we introduce ZERO ohms? In theory we have infinite amperage. Which in practical terms doesn't happen. 12 x 0 does not equal infinity. It usually equals a failed circuit. Or a lot of heat.
 

Thread Starter

Aleksey Shurtygin

Joined Dec 21, 2018
75
1. Measure voltage on resistor 82Ω and divide it in 82, then you will have current of one LED.
2. Click on "S" here:
View attachment 166214
and choose symbol:
View attachment 166215
EDIT:
Current of one led at V_battery = 3.5V:
I_LED = (3.5V - V_LED) / 82 = (3.5V - 3.3V) / 82 = 0.2 / 82 = 0.0024A.
Current of 120 LEDs is equal 0.0024 * 120 = 0.288A.
Seems it can lights relatively long time from battery...
They won't last very long. If you are actually delivering 5 W then your total current is about 1.67 A.

https://d2ei442zrkqy2u.cloudfront.net/wp-content/uploads/2016/03/QU1500_US_UL1.pdf

At 1 A the service life to 1 V per cell for fresh Duracell Quantum batteries is only about an hour.

There are a couple of ways. If you click the 'S' icon on the toolbar at the top of the Reply box it will list a bunch of symbols and you can select the one you want. Depending on your browser it either inserts it where the cursor is or it inserts it as the first character in the post and you have to move it.

Another way is to hold down the Alt key and type 234 on the number pad (NOT on the top row of the main key area).

You need to know the effective resistance of the meter.
Thanks for the tips on the Ω sign. Batteries lasted for about half a year. Obviously you do not keep the pantry cabinet open for long at a time. You open and then close. So, its open roughly in average for about a minute or two per day.

I also have some updates. I measured voltage with a 10Ω resistor. Also, I just realized that my other meter (which is digital) can measure up to 10A. Soo...

The AC adapter supplies 5.17V without any load.

With a resistor and LED:
Voltage drops to 4.96V. Voltage on resistor is 1.80V and on LED is 3.08. Current is 0.15 A which is inline with my analog meter measurements. Which makes the total resistance to be around 33Ω. Subtract 10Ω and we get 23Ω. And power now becomes 3.08*0.15=0.46W??

Without resistor:
Voltage drops to 4.48V and current now reads 0.65A and resistance comes down to 6.8Ω and 2.9W.

Do you think that make sense?
 

MrChips

Joined Oct 2, 2009
30,720
I also have some updates. I measured voltage with a 10Ω resistor. Also, I just realized that my other meter (which is digital) can measure up to 10A. Soo...

The AC adapter supplies 5.17V without any load.

With a resistor and LED:
Voltage drops to 4.96V. Voltage on resistor is 1.80V and on LED is 3.08. Current is 0.15 A which is inline with my analog meter measurements. Which makes the total resistance to be around 33Ω. Subtract 10Ω and we get 23Ω. And power now becomes 3.08*0.15=0.46W??
At those current and voltages, the internal resistance of the ammeter will have a significant effect.
The only fact you can determine is the current through the known 10Ω resistor.
Current = Voltage / Resistance = 1.80V / 10Ω = 0.18 A
Without resistor:
Voltage drops to 4.48V and current now reads 0.65A and resistance comes down to 6.8Ω and 2.9W.

Do you think that make sense?
No. For two reasons.
1) You have not taken in account the resistance of the ammeter. (See above).
2) The resistance of an LED is not constant. It changes with current.

What are you trying to do?
 

BobTPH

Joined Jun 5, 2013
8,813
Finally, 24 posts in, someone (MrChips) gives a piece of critical information: An LED is not a resistor! You cannot calculate a resistance via Ohm's law at one voltage and use that resistance to determine the current at a different voltage. For example. A typical white LED might pass 20 mA at 3.3V. So, by Ohm's law it's resistance is 165 Ohms, right? So, if you cut the voltage to 1V, you should get 1 / 165 = 6 mA, right? Wrong. You will get almost no current at 1V. Try it and see, it will not even pass 1uA at 1V.

Bob
 

Tonyr1084

Joined Sep 24, 2015
7,853
That's a nonsensical statement that makes exactly as much sense as saying that someone's height is the same as their weight. Meaningless.
Any voltage across a 1 ohm resistor will have the same current.

100 volts across a 1Ω resistor is going to have 100 amps. Here's the math:
100V ÷ 1Ω = 100A
12V ÷ 1Ω = 12A
3.3V ÷ 1Ω = 3.3A

Using 1 ohm as a load always means the voltage and the current will be the same. If I'm wrong please show me how and why. Otherwise, can I get a little respect? Even if I am wrong, which I've been wrong enough times in the past, this would be a chance to educate me. Us (me and the thread starter).

That makes exactly as much sense as saying someone's height is the same as their weight. {edited}.
In 1G, a person's mass is exactly the same as their weight. On the moon a person's weight is 1/6 their mass. It's all relative. And I've got some pretty heavy relatives.
 

WBahn

Joined Mar 31, 2012
29,979
Finally, 24 posts in, someone (MrChips) gives a piece of critical information: An LED is not a resistor! You cannot calculate a resistance via Ohm's law at one voltage and use that resistance to determine the current at a different voltage. For example. A typical white LED might pass 20 mA at 3.3V. So, by Ohm's law it's resistance is 165 Ohms, right? So, if you cut the voltage to 1V, you should get 1 / 165 = 6 mA, right? Wrong. You will get almost no current at 1V. Try it and see, it will not even pass 1uA at 1V.

Bob
Uh, you might reread Post #2. I seem to recall something like this being mentioned:

With this notion of how a battery works, we can now talk about how an LED works. As you apply a voltage to it, initially very little current flows until you start getting close to the "knee" voltage. At that point, the current increases significantly as you increase the voltage. For rectifier diodes, this is a very sharp, exponential increase. For light-emitting diodes it is much softer and quickly looks like there's a resistor involved. But in either case the voltage vs. current curve is highly nonlinear and so you can't just take the voltage value and figure out how much current is flowing.
 

bertus

Joined Apr 5, 2008
22,270
Hello,

How about this theory:

Although we generally consider a cell or battery in a circuit to be a perfect source of voltage (absolutely constant), the current through it dictated solely by the external resistance of the circuit to which it is attached, this is not entirely true in real life. Since every cell or battery contains some internal resistance, that resistance must affect the current in any given circuit:





The real battery shown above within the dotted lines has an internal resistance of 0.2 Ω, which affects its ability to supply current to the load resistance of 1 Ω. The ideal battery on the left has no internal resistance, and so our Ohm’s Law calculations for current (I=E/R) give us a perfect value of 10 amps for current with the 1 ohm load and 10 volt supply. The real battery, with its built-in resistance further impeding the flow of electrons, can only supply 8.333 amps to the same resistance load.

This comes from this page of the eBook:
https://www.allaboutcircuits.com/textbook/direct-current/chpt-11/battery-construction/

Bertus
 

WBahn

Joined Mar 31, 2012
29,979
Any voltage across a 1 ohm resistor will have the same current.

100 volts across a 1Ω resistor is going to have 100 amps. Here's the math:
100V ÷ 1Ω = 100A
12V ÷ 1Ω = 12A
3.3V ÷ 1Ω = 3.3A
Congratulations, by properly tracking your units you've proven that they are not the same. 100 V is NOT the same as 100 A. 3.3 V is NOT the same as 3.3 A.

How tall is someone that weighs 100 kg if their weight is the same as their height?

Say someone's weight is 83 kg and they are 6 ft tall? Is their weight the same as their height? Is it even close? What does the question even mean? But you are probably saying that they aren't even close since 83 is not close to 6. You might even assert that the person is nearly 14 times heavier than they are tall.

Now consider that 83 kg is 182.984 lb and 6 ft is 182.88 cm. Aren't these very close to being the same? So now, magically, the person's weight is almost exactly the same as their height even though neither the person's weight nor their height has changed. They didn't get 14x lighter or 14x taller.

Two things that have incompatible units are NEVER the "same", regardless of what numerical coefficient multiplies the unit. Talking about the equality of such things is nonsensical.

Going the other way, a table that is 4 ft wide IS exactly the same width as a table that is 48 inches wide and the width of those are both the same as the diameter of pool that is 1.2192 m in diameter. The fact that the numerical part of the measurements are different is completely immaterial.

If two quantities are the same, then when you divide one by the other you must get 1 or something equivalent. Not 1 with some unit attached to it. Just 1.

If you divide 48 inches by 4 feet, you get 1.

If you divide 182.984 lb by 182.88 cm, you get 1.0006 lb/cm, a dimensioned quantity..

If you divide 3.3 V by 3.3 A, you get 1 Ω, a dimensioned quantity.

Keep in mind that, just like any other dimensioned quantity, there is nothing fundamentally different about 1 Ω compared to any other resistance. Just like there is nothing fundamentally different about a distance of 1 mile compared to any other distance. We can change the dimensional units to any other consistent unit and the numerical part changes accordingly. Instead of 1 Ω, which is merely 1 newton-meter per ampere-squared second (Nm)/(A²s). We just choose to attach the name "ohm" to that combination of dimensions. I can play all kinds of games with those units, such as expressing energy in ft·lb instead of N·m or A in electrons/minute and attach a name to the result. I can't then take "one" of those and claim that the current through a resistor of that size is the same as the voltage across it.
 

Danko

Joined Nov 22, 2017
1,829
May be couple of diagrams for 120 LEDs strip and link will help.
Difference between simulation and TS's measurements may occur
from poor wall wart (adapter):
A. 5V power is too low.
B. Insufficient ripple smoothing after 5VAC rectifying, so
voltage and current measurements become inaccurate.

For experiments TS can use good quality 5VDC from computer power supply.
upload_2018-12-23_5-27-31.pngupload_2018-12-23_2-20-36.png
Click on thumbnails
 
Last edited:

Tonyr1084

Joined Sep 24, 2015
7,853
Congratulations, by properly tracking your units you've proven that they are not the same. 100 V is NOT the same as 100 A. 3.3 V is NOT the same as 3.3 A.
No. You're right. Volts is not current. But that's not what I was trying to convey. The voltage will be the same as the amperage. If you have 100 volts - you have 100 amps.
Aassume a 1 ohm resistor is going to always be 1 ohm. Any voltage applied across it is going to result in the same current as the voltage.
Any voltage applied across the resistor will result in the same current as the voltage. OK, maybe that wasn't as clear as it could have been. 100, whether volts or amps, 100 is the same. It was that point I intended to drive home.

To be completely clear on this matter, you (in theory) can have 0.01 volts and 100,000 amps. (IN THEORY). And voltage is not current, and current is not voltage. Voltage is "Electric Pressure". Current is the flow of electricity - at any voltage. Yes, Wbahn, you're correct. Saying someone is as tall as their weight IS ludicrous (my word, not yours). Wasn't saying voltage and current were the same things. I was referring to the math.
100V ÷ 1Ω = 100A
12V ÷ 1Ω = 12A
3.3V ÷ 1Ω = 3.3A
100 = 100. 12 = 12. 3.3 = 3.3. I may not be the sharpest tool in the chest, but I'm also not a rake. For a moderator to speak in such a way, I've felt (maybe wrongfully so - but it's how I feel) like I've been called stupid. Not asking for an apology, just for some understanding as to what I was saying about the numbers.
 
Hi everyone! I am trying to find out more about why a voltage drops when you put a load onto the source. Also when using Ohms Law formula what voltage should I use. The V with or without load.

I have a basic battery powered LED strip that uses 3 AA batteries. So, when using voltmeter it shows 4.5 volts without load. However voltage drops to 3.5 when LED is connected. So assuming I know the current, do I use 4.5 or 3.5 to find the resistance, given I know the current.

I've been trying to research it but nothing useful is coming up. All text books and sites simply seem to refer to rated voltage of the battery and do not mention anything about this "drop". I did find more about voltage drop or voltage sag. These appear to be slightly different..

Could someone please either explain me why this happens or let me know what this behavior is called so that I can maybe try to find something relevant.

Thanks!!
I appreciate it!!
You are talking here about one of the most important things in electrical engineering; Ohm's Law. So how do you apply Ohm's Law to a battery or any other voltage source. An equivalent circuit for a battery is a voltage source V with a resistance R in series with it. When you put a load on a battery, the voltage sags because current is flowing through the resistance R. V is simply the no-load voltage measured with a volt meter. R is a little more tricky. Put a load such as your LED light strip onto your voltage source and measure how much the voltage sags (call the sag voltage S). Now use an ammeter to measure the current I. Now use Ohm's law resistance=voltage/current or R=S/I. Now you have a complete equivalent circuit for your battery. You can now calculate the amount of sag you will get with any given load current.
 

Tonyr1084

Joined Sep 24, 2015
7,853
Battery or power supply, all voltages will drop with load. Depending on how strong the supply is (battery or PS). The stronger the supply the less the "Sag". It's like a train on a track. The heavier the cars the more work the engine has to do. The more fuel us used. If the train is moving and maintaining a certain speed then the flow of the cars is the same all the way through the string of cars. The cars weight does not change (except for maybe wind resistance, but lets stick with hypothetical) and neither does the resistance (in theory). The more pressure the locomotive puts to the tracks the higher the speed (current) of the cars. Still, the weight remains the same.
 

WBahn

Joined Mar 31, 2012
29,979
No. You're right. Volts is not current. But that's not what I was trying to convey. The voltage will be the same as the amperage. If you have 100 volts - you have 100 amps.
Any voltage applied across the resistor will result in the same current as the voltage. OK, maybe that wasn't as clear as it could have been. 100, whether volts or amps, 100 is the same. It was that point I intended to drive home.

To be completely clear on this matter, you (in theory) can have 0.01 volts and 100,000 amps. (IN THEORY). And voltage is not current, and current is not voltage. Voltage is "Electric Pressure". Current is the flow of electricity - at any voltage. Yes, Wbahn, you're correct. Saying someone is as tall as their weight IS ludicrous (my word, not yours). Wasn't saying voltage and current were the same things. I was referring to the math.
100 = 100. 12 = 12. 3.3 = 3.3. I may not be the sharpest tool in the chest, but I'm also not a rake. For a moderator to speak in such a way, I've felt (maybe wrongfully so - but it's how I feel) like I've been called stupid. Not asking for an apology, just for some understanding as to what I was saying about the numbers.
I never said anything about you, let alone about whether you were stupid or not. I made a comment about a statement being nonsensical -- and the comment that the voltage will be the same as the current being nonsensical applies regardless of who made it, whether it was you, me, or Albert Einstein.

If saying that someone is as tall as their weight is ludricrous, then why isn't saying that the voltage is the same as the amperage just as ludricous? What is different?

If a truck gets a fuel economy of 1 km/liter, is the distance it drives equal to the fuel it consumes?
 

Tonyr1084

Joined Sep 24, 2015
7,853
@WBahn Post #22 you quoted me. I felt like your comment was directed at me. But I've had enough of this discussion. I stand by my comment and you stand by yours. Nothing more can be accomplished by bantering back and forth.

@Aleksey Shurtygin Whatever voltage you're using, whatever the load, as you change the voltage you also change the current in like proportion. Conversely, the more current you draw the more the voltage changes. Since the load does not change, if you draw 2 amps your supply must be capable of supplying that much. If it can't then the voltage will drop. Sorry if I've confused anything for you.
 
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