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Please help me, What do ohms do to volts?

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JunaidKhan

New Member
Hi, I am new to this form. I must be the dumbest father on this planet. I have read many, many books and working with electronic breadboards building and learning trying to teach my kid. The thing that is driving me crazy is what do ohms do to volts? See I know about ohms law, and I understand volts is really just pressure, it is current that flows. But what I just don't get is what’s with resistor numbers. I am trying to help my kid and I understand its concept.

Here is an easy example. If I have 9 volts and connect 3300k R to an LED, I see that the voltage dropped. But what I don't understand is what does the resistor mean by 3300k? Does it mean that that’s how many electrons it will let through regardless of volts?

Do you see what I am trying to get at? I don’t know why my mind is not latching on to the number system of resistor. What exactly does the 3300k do to volts to cause the drop. I know what resistors do but but I don't see what is ohms, is it a unit of measure.

Please help
 
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BrAdAsS

New Member
seeing as no one replied as yet,

Ohms is a unit of resistance. The resistors are made from different materials and the properties of each material effects the amount of resistance and tolerance. Each resistor has 4 or 5 different coloured stripes on them which tells you the specific ohm value of that resistor and its tolerance....

You can get a multimeter and set it to 'resistance - 20K scale' (ohm symbol Ω) and measure the resistor. It should read 3.3K. Or if you have it on the 2000Ω scale it will read 3300Ω... catching on yet?
 

Pommie

Well-Known Member
Most Helpful Member
If we keep to the water analogy of pressure and flow being voltage and current then a resistor is analogous to a restriction in the pipe. The more restricted the pipe the higher the resistance and the bigger pressure will be built up around it. A dripping tap is like a very high value resistor and so the pressure before the tap (in the pipe) is high and after the tap is low and so the pressure difference is high. A fully open tap offers little resistance and so the pressure is pretty much the same before and after the tap.

Resistance as you know is measured in ohms but we use the SI units (kilo Meg) in place of the decimal point. So, 2k2 = 2.2k = 2200Ω. 8M9 = 8.9M = 9,800,000Ω. We also use R to indicate no multiplier, 22R = 22Ω and 1R2 = 1.2Ω.

In your example with the LED it has to be calculated backward. The LED will have 2V across it (actually not in this case due to a high resistor value) and so the resistor will have 7 and so the current that will flow is I=V/R = 7/3300000 = 0.000002 Amps. I think you may have the wrong resistor value, a normal value for this example would be more like 330Ω.

Mike.
 

kpatz

New Member
Google "Ohm's Law", you'll find tons of info.

But, in short, if you have 1 volt and your circuit has a resistance of 1 ohm, 1 amp of current will flow through the circuit.

Higher voltage increases the current. Higher resistance reduces the current.

The formulas are very simple:

Amps = Volts / Ohms.
Ohms = Volts / Amps.
Volts = Amps * Ohms.

When you read a schematic, and the resistor has "K" after the value, you would multiply by 1,000. So, a 10k resistor = 10,000 ohms. If it has M after the value, multiply by 1,000,000. So 1M = 1,000,000 ohms. Sometimes a decimal value will use the M or K in place of the decimal point. So, 2.2K = 2K2 = 2,200 ohms.
 

ke5frf

New Member
I will try to explain this in a way that simplifies to the point even a child can understand it.

Think of voltage as pressure being applied to something. Say, for instance, when you are moving a bookcase in your home by pushing it across the floor. When you put your shoulder against the bookcase and push, you are applying pressure against it.

The floor and the weight of the bookcase are opposing forces to your pressure. The floor creates friction which, combined with the weight of the bookcase, opposes your attempt to move it. This is similar to a resistor in an electronic circuit. The resistor opposes the voltage (pressure) that is pushed against it. The weight of the bookcase can be measured, and so can a resistor...thus the DEGREE to which it opposes the voltage is determined by the measurement in OHMS, just as the degree to which the bookcase resists being moved can be measured or determined by its weight and the smoothness of the floor.

Once the resistance of the bookcase is overcome by the pressure exerted upon it by your shoulder, it will begin to slide across the floor. If the floor is smooth and creates little friction, and if the weight of the bookcase isn't particularly heavy, and if you are a fairly strong man, then the bookcase will slide reasonably quickly, with ease. This would be similar to the FLOW or movement of electric current. If the voltage applied is strong and the resistor value is small, the a large current will flow through the circuit. If you are very weak and the weight of the bookcase is very large, then it will be difficult to move it and it will slide slowly. Just as in a circuit, if the voltage applied to a resistor is weak, and the value of the resistor is high, then fewer electrons will be allowed to flow through the circuit.

An increase in resistance will have an inversely proportional effect on current flow given that the voltage remains the same.

I hope my analogy makes sense.
 

JunaidKhan

New Member
In your example with the LED it has to be calculated backward. The LED will have 2V across it (actually not in this case due to a high resistor value) and so the resistor will have 7 and so the current that will flow is I=V/R = 7/3300000 = 0.000002 Amps. I think you may have the wrong resistor value, a normal value for this example would be more like 330Ω.

Mike.

Thank you. Ok I think this is where the problem I am having in ohms law. In my example I used a 3.3k Resistor with a 9v Volt battery.

Let me break down my thinking so you can see where I get lost.

I have a very basic circuit, a 9v connected to a 3.3k resistor connected to an LED. I take my multimeter and take the black wire and rest it on the ‘-’ end of the battery and use my red + wire to check the voltage before the resistor. I know here it will report around 8.70v. My battery is getting a little weak so that’s why it is 8.70. Now after the resistor it shows 3.82v. Ok this is where I get lost. I have not seen any formula in ohms law that shows how the resistor value of 3.3k can be used to see the drop of the volts it will do. Like here when I had 8.70v after 3.3k it dropped to 3.82. How did these numbers happen?

So if I had a 6v and used a 3.3k what would be voltage dropped to after 3.3k resistor?

This is where my minds is not getting it. Ohms law does not show this, it talks about V,I,R,W.

These 2 questions will help me if anyone can anwser these:

"What I want to know is how do the numbers on the resistor work if I want to drop a 6v to 2.5v before it enters into the LED.?"

"If I had a 6v and used a 3.3k what would be voltage dropped to after 3.3k resistor?"
 
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JimB

Super Moderator
Most Helpful Member
What you have here is a very poor example for learning "Ohms Law".
At the risk of the thread turning into a blood bath, I will state that devices like LEDs, diodes and transistors do not obey Ohms Law.
(There has been a lot of heated debate about this in the past!):eek:


So if I had a 6v and used a 3.3k what would be voltage dropped to after 3.3k resistor?
The quick answer is that it would still be 3.82 volts.
LEDs have a nearly constant voltage drop arcoss them which varies by only a small amount when the current through them varies.

If you set your meter to measure current, in your original circuit you should see that the current is (8.7 - 3.82)/3300 = 1.48mA or maybe a bit less because the meter will add resistance to the circuit, making the current a bit smaller.

When you use a 6v battery, the current will be (6 - 3.82)/3300 = 0.66mA. Depending on the LED, it may not be very bright at this low current.

JimB
 

kpatz

New Member
Voltage drop occurs when you have 2 resistances in series in the circuit. This is called a voltage divider.

For example, say you have a 9 volt battery and you connect 2 1K resistors in series across the battery terminals (they don't have to be 1K, but to make things simple, make them equal).

+ terminal : 1k resistor : 1k resistor : - terminal

If you measure the voltage across the battery terminals, you get 9V (or something close to it). If you measure across one of the 1k resistors, you'll see 1/2 the voltage (4.5 volts). Each resistor is dropping half the voltage, since both resistances are equal.

In your example, instead of 2 resistors, you have a LED and the resistor. Diodes (and most other semiconductors) are different from resistors in that they drop a more-or-less constant voltage, drawing as much current as needed to drop that amount of voltage. This is why you need a current-limiting resistor with a LED. So, if your LED has a 2.5V forward voltage, and you're using a 9 volt battery, the LED would drop 2.5V and the resistor would be dropping (9 - 2.5) = 6.5 volts. If you want to limit the current through the LED to 20 mA (0.02 A), use Ohm's Law to calculate the resistor value: R = E/I, Ohms = Volts / Amps, so the resistor would need to be 6.5 / 0.02 = 325 ohms. The closest standard resistor value is 330 ohms, so use that.

So, in short, the formula for calculating a LED's series resistor is:

R = (Vsource - Vled) / Amps

Vsource is your supply voltage (e.g. a 9v battery). Vled is the voltage drop of the LED (get this from the LED's datasheet, will typically be between 2 and 3.5V depending on the LED's color), and Amps is the amount of current you want the LED to draw (keep this below the rated maximum for the LED). Typically this will be in milliamps, so it'll be a number less than 1 (such as 0.02 for 20 mA).
 
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ke5frf

New Member
To understand what you are looking for, in addition to Ohm's law you really should study Kirchoff's laws as well as Thevenin's and Norton's theorems. You should Google those terms and read.

You must understand that the LED in your circuit is a resistance in addition to the value of your resistor. Forget the 3.3k resistor and LED for a moment.

If you have a resistor that is 6 ohms and an LED that is rated at 3 ohms, (numbers used for simplicity) and a 9 volt battery, then you will read 6 volts across the resistor and 3 volts across the LED when the circuit is energized.

The resistor is dropping 66.6 % of the voltage and the LED is dropping the other 33.3 %

This is in accordance with Kirchoff's laws.

Similarly, with your 3.3k resistor, the resistor is dropping 3.82 volts. lets round that to 4 volts for simplicity.

Your supply voltage is 8.7 volts, but lets round that to 9 volts for simplicity.

So 9 volt supply minus (-) 4 volt drop = 5 volts. So the other 5 volts is dropping across the LED.

So, the resistance of the LED must be more than the resistance of the resistor using Kirchoff's Law. It will be a higher resistance in the same proportions as the voltage drop across it.

The resistor is 3.3 Kohms, and with my rounded numbers of 4 and 5 volts for the drops, this means about 44 percent of the voltage is dropping across the resistor. The other 56 % of the voltage is dropping across the LED. This means the resistance of the LED is 12 % higher than the resistance of the resistor.

My math may be wrong here, but I estimate the LED value to be about 4.6 Kohms

(This using my rounded numbers, not your exact values)
 

Pommie

Well-Known Member
Most Helpful Member
The problem here is that ohms law doesn't apply to LEDs. LEDs will have (generally) 2V across them unless the current is very low as in your case.

From your figures the voltage across the resistor is 8.7-3.8 = 4.9V. This is across a 3.3k resistor and so the current is I=V/R = 4.9/3300 = 0.0015A = 1.5mA. I'm guessing that your LED glows very very dimly due to this low current. Change the resistor to a 330 resistor and get a new battery to see a brightly glowing LED. Then measure the voltages and we can do the maths again.

Can you see that the thing that defines the voltage across the resistor is the current through it and vice versa.

Mike.
 

ke5frf

New Member
What you have here is a very poor example for learning "Ohms Law".
At the risk of the thread turning into a blood bath, I will state that devices like LEDs, diodes and transistors do not obey Ohms Law.
(There has been a lot of heated debate about this in the past!):eek:



The quick answer is that it would still be 3.82 volts.
LEDs have a nearly constant voltage drop arcoss them which varies by only a small amount when the current through them varies.

If you set your meter to measure current, in your original circuit you should see that the current is (8.7 - 3.82)/3300 = 1.48mA or maybe a bit less because the meter will add resistance to the circuit, making the current a bit smaller.

When you use a 6v battery, the current will be (6 - 3.82)/3300 = 0.66mA. Depending on the LED, it may not be very bright at this low current.

JimB


I think for simplicity's sake we need to avoid specifics like this while someone is trying to wrap their mind around a basic concept such as Ohm's and Kirchoff's laws. It doesn't sound like the OP is at the stage where he or his son is ready for that.

I do agree, however, that he should be using pure resistors for demonstrations instead of active components or components with unusual characteristics that undermine the underlying principals.
 

kpatz

New Member
The problem here is that ohms law doesn't apply to LEDs. LEDs will have (generally) 2V across them unless the current is very low as in your case.
Well, Ohm's Law does apply but not in the usual manner. The LED's resistance isn't fixed, it changes in order to maintain a fixed voltage. Resistors are fixed, a 1k resistor will always be 1k regardless of the voltage placed across it (tolerances and thermal characteristics excluded). Increase the voltage, and the current increases, as does the voltage across the resistor.

With a LED (or other semiconductor), the device drops a more-or-less fixed voltage, and its internal resistance changes to maintain that fixed voltage. So, instead of R being constant, E is constant. Thus, if an LED is supplied a voltage, it will draw more and more current until the voltage drops to the LED's nominal forward voltage. If the current supply isn't limited, the LED will attempt to draw more current than it can handle and will burn out. The series resistor limits the current, allowing the LED and resistor to act as a voltage divider, allowing the LED to maintain the forward voltage it "likes".

To use the water analogy, voltage = water pressure, current = amount of water, and resistance = the thickness of the pipe carrying the water. A semiconductor like an LED would be like a pressure regulating valve. More or less water (current) will flow through the valve to maintain a set pressure (voltage).
 

Pommie

Well-Known Member
Most Helpful Member
I think for simplicity's sake we need to avoid specifics like this while someone is trying to wrap their mind around a basic concept such as Ohm's and Kirchoff's laws. It doesn't sound like the OP is at the stage where he or his son is ready for that.

I was thinking we should avoid things like Kirchoff, Thevenin and Norton (from your earlier post).

Mike.
 

ke5frf

New Member
I was thinking we should avoid things like Kirchoff, Thevenin and Norton (from your earlier post).

Mike.


Wanted to follow up on this by saying I wasn't trying to be rude when I suggested that discussions about particular voltage/current characteristics of LEDs should be avoided. What I was getting at is that the fellow is obviously trying to understand the fundamentals, and fundamentals are about the "rules and laws". Certain devices "seem" to defy these rules and create exceptions that require detailed explanation. I believe it is helpful to understand the rules first, then tackle all the exceptions. It is, to me, backwards and confusing to have all the exceptions dumped upon you along with the rules and then try to sort it all out at once.

Kirchoff's laws are as fundamental as Ohm's law. Once Ohm's law is understood, more complex series/parallel circuits can be studied using Kirchoff's laws. Then it follows that Thevenin's and Norton's theorems can be studied once Ohm's and Kirchoff's laws are understood. At least this is the way I ALWAYS see it in textbooks, ie the "building block" approach.

Perhaps some of you may learn things by a different method, but that would be the "exception" to the "rule"...(couldn't resist)
 
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mbarazeen

Member
the resistor is basicall to limit the current, in your example you can see it limitting, but you cant achive a certain voltage only to be blocked between a ressitor for different voltages.

the definition for Ohm may give you some idea.
the value of a resistor required to establish 1A (you can say in charge flow Q/S) when connected across a 1V source is 1 Ohm.

resistor is to control the current not the voltage.
 

sram

Member
I think you guys are making it hard on the guy. IMO, he should put the LED aside and understand Ohm's law and how it works in simple circuits containing only resistors. He should also understand how resistors are connected (series and parallel) and how that divides voltage and current.

OP, in your original circuit, if there is no LED, and you only connected the 9v battery to the 3300 ohm resistor, a current of I = V/R = 9/3300 = 0.002727 amps will flow in the circuit(Assuming the battery is exactly 9v). The voltage drop on the resistor will be 9v since it is the only component in the circuit, Or it can also calculated by the same ohm's law. Vr = Ir = .002727*3300 = 9v where I is the total current.

Now, if there is another R connected to the previous 3300 ohm resistor in series like this:

https://mysite.mweb.co.za/residents/vdwestch/r_series.jpg

Then you will have to first calculate the total resistance which is basically the sum of the two resistors(in this case) before you can calculate the total current and assuming the second resistor is 1000 ohms then I = V/R = 9/(3300+1000) = .00209 amps

In this case the voltage will be divided and it will be

Vr1 = I*r1 = .00209 * 3300 = 6.9 v

Vr2 = I*r2 = .00209 * 1000 = 2.9 v

If this same second R is connected like this

https://image.tutorvista.com/content/electricity/parallel-resistors-energised-battery.jpeg

which means it is connected in parallel then the total R is calculated differently and the law for that is this

https://www.radio-electronics.com/i...lel_series_resistor/resistors_in_parallel.gif

which when we apply, it will give a value of 767.44 ohms which will give a current of V/Rt = 9/767.44 = .011727 amps. This is the total current, and in this case the current is the one that is going to be divided between the two resistors. Some of it will go to the first R and some will go to the second R in a way so that both R's have the same voltage.

==> Ir1 = V/r1 = 9 / 3300 = .002727 amps

and Ir2 = V/r2 = 9 / 1000 = 0.009 amps


This is how it works. You should understand it first with resistors, then go to the more complicated stuff.
 
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