# Basic electronic question which is driving me crazy!

Discussion in 'General Electronics Chat' started by noobiEletron, Aug 1, 2011.

1. ### noobiEletron Thread Starter New Member

Aug 1, 2011
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Hi I'm (very) noobie in electronics an stuff. I really like physics and maths, however i can't get into my mind ohm's freaking law! Can you believe it??

I mix up everything...

Ok. My (ashamed) questions are:

1. In a 9v battery, a (whatever resistance) resistor and a LED circuit, how can I calculate in VOLTS the current that goes AFTER the resistor? If it resists the current it must not be the same before and after the resistor, rite?
(I know it seems dumb but i dont get it...)

Forgot to say: I got LTspice just to improve my skills in electronics, although I see a problem... When i get a 9v battery, a led and a resistor the current that goes after the resistor is X if I increase the resistor value, which should now decrease the current after it... well guess what? instead it increases!! am i doing something wrong? is it the program? Plz help me out!

Thank you, if someone ever answer this question...

Last edited: Aug 1, 2011
2. ### SgtWookie Expert

Jul 17, 2007
22,194
1,763
In a series circuit as you have described, the current is the same through all components in the circuit. The voltage drops across the resistor and LED may be quite different though.

3. ### noobiEletron Thread Starter New Member

Aug 1, 2011
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Well, I'm thankful you have answered, so, if the current is the same through all components, why are resistors there?? If they make no difference?

OMG im so dumb...

4. ### SgtWookie Expert

Jul 17, 2007
22,194
1,763
The resistors are there to limit the current.

Let's say you have an LED that is rated for 20mA current, and at that current has a Vf (forward voltage, or voltage drop across the diode when it is forward biased with a current of 20mA flowing through it) of 2v. As the current through a forward biased diode increases, the voltage drop increases, but not at a linear rate. If you put 9v across an LED rated for 2v, you would get very high current flow through the diode, which would ruin it. So, you need to limit the current.

You're using a 9v battery as a voltage source. You subtract the LED's 2v drop from the 9v, leaving 7v to drop over a resistor. You still want 20mA current through the LED. Since current will be the same through the circuit, you need to figure out what resistance you need to limit the current to 20mA.

Rlimit = 7v / 20mA = 350 Ohms.
350 Ohms is not a standard value of resistance, but 360 Ohms is.
There is a decade table of standard resistance values here:
Use the yellow E12 and green E24 columns.

Now if you want to check the current through this new resistance value, calculate I=E/R:
7v / 360 Ohms = 19.444...mA current, which is fine.

PackratKing likes this.
5. ### kubeek AAC Fanatic!

Sep 20, 2005
4,872
863
You need to get the basics right first. Current is not in volts, current is in amps and voltage in volts.

Second, the current remains the same in the whole circuit because it is a series circuit. Imagine it like a pipe with water, the rate of flow -current- has to be the same everywhere, and the pressure in the pipe - voltage - differs.

So, you want to calculate voltage in the said circuit. Lets assume the negative terminal is our common, so the positive terminal is at +9V. Then you have a resistor between the positive terminal and the diode, so the negative of the diode is connected to the negative terminal.

The diode will drop (means have voltage across it) a constant voltage, lets say 2.5V. This means the rest of the voltage has to be dropped across the resistor, i.e, 6.5V.

Ohms law says that resisctance is equal to voltage drop on a resistor divided by current through that resistor. Lets say the resistor is 100ohms, so you get current of 6.5/100 A, that is 65mA.

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6. ### Wendy Moderator

Mar 24, 2008
21,012
2,744
LEDs and diodes stand outside Ohm's Law. Most solid state parts do. The voltage they drop is a function of the quantum mechanics they use to perform their functions. A silicon diode will drop around 0.6 to 0.7V when forward biased, you subtract this voltage from the total power supply voltage.

Resistors are completely dependent on Ohm's Law. If you know the current through a known resistor, you can predict exactly what voltage is across it. If you have a known resistance and know the voltage you can predict the exact current through it. If you know the current and the voltage on a resistor you can calculate its exact resistance.

LEDs, 555s, Flashers, and Light Chasers

Don't be ashamed not knowing electronics, none of us were born knowing it. Ohm's Law is fundamental, but occasionally I have to write it down just to keep it straight.

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7. ### noobiEletron Thread Starter New Member

Aug 1, 2011
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thanks for responding. I'll just quit eletrconics... bah. I can't even understand LED's Vf... nor using ohm's law...

Sorry bother...

Apr 5, 2008
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9. ### noobiEletron Thread Starter New Member

Aug 1, 2011
5
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I really wanna understand electronics, however I just don't know where to start lol
Can anyone tell me?
I appreciate that

10. ### PackratKing Well-Known Member

Jul 13, 2008
849
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I earned my Electricial noob / rube crown the hard way........ If I had a dollar for every fuse I burned, I would take a nice vacation, or buy some more sophisticated test equipment.

I still have a rough time with some of the more abstract math, and thank God daily for my scientific calculator.

Where does your primary interest lie ?? .......AC DC Analog, Digital, computers, or just repairing whatever comes along ??

The E-books on this forum are going to help a lot.....if only I had something like this magnificent resource back then.............

Still far from an "expert " [ those don't really exist ] If I can do it, so can you.

shoot back.

11. ### noobiEletron Thread Starter New Member

Aug 1, 2011
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Hi there. Well your words have inspired me! I'll try it the way it comes. Thanks. it sometimes get so confused that i feel like giving up. that's it.
ohm's law and resistors are killing me... ans so the many ways of expressing values and none single way. this confuses me... ill try it out, though.

12. ### PackratKing Well-Known Member

Jul 13, 2008
849
218
Where there is a will, there is a way..........and I totally understand the killing you part............

An instructor in the Navy E.Tech course I pestered endlessly gave me some worthwhile philosophy. He was a very patient man.

Considering the graph an AC sine wave is drawn on ? consider it drawn showing two or three complete cycles, and don't even try to wrap your head around a megacycle, just take it for granted that things can move that fast.
I built some very "stone age" experiment boards to help learn the basics, which I still use at times. descriptions on request

13. ### #12 Expert

Nov 30, 2010
17,888
9,297
I will try to make a story for you.

If a wire is like a pipe, a resistor is like a valve. The resistance limits how much current can flow. No matter how much pressure is applied, the valve (resistor) always restricts the flow.

Then we can think about a pond with a dam. This is like the diode in that when you have enough water, the excess overflows. All you have to do is have ENOUGH water to get up to the level of the dam. For a power diode, this is about six tenths of a volt to start the flow. For an LED, it requires about 2 to 5 vlts to get the pond full and start the overflow. After the flow has started over the dam, very little increase in voltage causes very much overflow.

Thus the diode family is not like a resistor.

14. ### Kermit2 AAC Fanatic!

Feb 5, 2010
4,165
1,121

Use this. Draw it out yourself.

Finding Voltage I x R

Finding Resistance E / I

Finding Current E / R

All in the picture. The OHMS LAW picture.

15. ### Wendy Moderator

Mar 24, 2008
21,012
2,744
The above is really a good tool. Focus it on one resistor. If you know 2 of the variables, you can figure the 3rd. If you only know 1, you don't have enough information.

It has enormous practical uses. If you measure the voltage across a resistor and know the resistance, you just did a current check.

Start simple, and build up. That is the ticket. Like I said, no one knows any of this in the beginning.