I'm learning electronics because I want to make a robot (just for fun). Before diving into robot building and programming I just want to grasp the basics of electronics (and the tools). I just created a circuit (led, resistor), in series, and performed some calculations and after that picked my multimeter to check for the correctness of my calculations. I'm using a pack of batteries of 6.12V. I placed a resistor (325 Ohm) and a led on a breadboard and according to Ohms law the current should be 18,83 mA, however checking with my multimeter it is reading a 11,20 mA. Can anybody explain to me where the 7,63 mA has gone?
I also created a voltage divider circuit, but I don't know how I can measure the output voltage using a multimeter. I want to measure the output voltage to check if my voltage divider circuit is correct.
The 18.83mA you are looking for is the current through a 325 Ohm resistor with 6.12V
18.83mA = 6.12V
325Ohm
As you have an LED in series with the resistor you have to take into account the voltage drop of the LED.
For example, if the LED is a standard red one the voltage drop of the LED is approximately 1.7V
So:
Current = 6.12-1.7 = 13.6mA
325
Depending on the type of LED you are using the current will be different, hence the 11.2mA you are reading.
Also to measure the voltage of your potential divider with the multimeter,
Connect the black COM of the multimeter to ground, and connect the red V lead to the junction between the two resistors.
There are three laws that apply when looking at DC circuits:
Kirchoffs First and Second (see wikipedia)
Ohms Law
The statement above that "The 7.63mA is being used by the LED" is not right. Don't even think it.
The voltage round the circuit must sum to zero. That means 6.12 into the LED and resistor are balanced by the voltages across the resistor and LED.
The LED is a non linear device - it does not conduct until the voltage across it rises to about 1.7V. Thereafter the voltage across it stays nearly constant. So 6.12 = voltage across LED + voltage across resistor. Ohms Law gives the values for the resistor.
The current round the circuit is constant. The same current flows through the battery, the resistor and the LED. The current flowing into any point of your circuit is the same as that flowing out of it. E.g. the current into the LED is the same as the current out of it.
The other things to remember are that components are approximate. Meters are approximate. Also, especially when looking at a voltage divider, the resistance of your multimeter might be significant. If you had a voltage divider of 2 10MOhm resistors, and your multimeter had a resistance of 10 MOhm (which one of mine has), you change the circuit from 10MOhm + 10 MOhm to 10MOhm + 5 MOhm when you put the meter on. By using Kirchoffs Laws and Ohms law you could work out the effect of that.
Luckily, for logic circuits, you can often approximate them as DC circuits in one state or another.
I bought the book and I also have the first edition back in the 1990's.
So let see... your are learning electronics ? That is great. Build in steps, Learn in steps.
What about your programming skills ? You need that for the programming. Beging with small things first.
And what about your mechanical / construction skills ? You need that for the building and figure out the mechanical aspect of it.
All thanks for you responses. Including the voltage drop in the calculation makes sense. Made my first circuit and I understand it! Now continuing with a more difficult circuit: both series and parallel...
One more thing about calculating the voltage of the voltage divider... Dean said to connect the multimeter at the junction and the ground, but I don't think I get the proper readings. I'm using two equal resistors and I expect to get a voltage reading of 3V, but the multimeter is reading 0v!? Maybe because there is no load connected? So I've put a led in the circuit but then the multimeter is reading a 1.80v? This is the same as the voltage drop of the led, so I don't think the multimeter is reading the correct value, or more likely its me using the meter wrongly!
@Techone: I'm a programmer, so programming the Arduino will not be a problem...
Hm, not sure about your voltage divider. Can you give the circuit, including the values you are using. Then try it again with two different resistors! Is your connection Battery+ (A) Resistor (B) Resistor (C) Battery-, and you are measuring voltage between B and C?
It turned out that my circuit was wrong... Looked at the site robtillaart suggested, and now (altougth very slowy) things are making sense... Still have a lot to learn.
I guess it's a long way before I can finally build my robot... But I won't give up!
It turned out that my circuit was wrong... Looked at the site robtillaart suggested, and now (altougth very slowy) things are making sense... Still have a lot to learn.
I guess it's a long way before I can finally build my robot... But I won't give up!
Thanks to you all!
That is awsome makes me smile that a person is excited about learning something new, and that the're looking forward to a challenge.
Keep at it , and keep asking and learning . the small and large rewards will be worth it ( it great to have those A HA !!! moments).
If I may venture to make a suggestion, you should try the MITx online course for circuit design. It's an introductory course with graded homeworks, labs and exams, plus it's free. It's already three weeks in, but the lowest 2 labs and homeworks are dropped (if you even care about that) so it's not too terribly late to join in. I'm taking it now, so I just thought I would suggest it.
