Troubleshooting: Investigating why the LED is so dim (5 of 10)

Let's poke around with a multimeter to see if we can figure out why the LED is so dim.

[alexwhb]

I just wanted to say thank you for making these videos. You seriously teach so well. I like that you go from the ground up and pose questions. Most people seem to assume way too much prior knowledge. please make more videos!

[Massenko]

These series of videos are great, I enjoy watching each one of them

[DeterminedHaphazard]

I have been waiting for "current" word since the last video. Thanks for your videos. I knew the most, but I missed important details.

[maxpolaris]

What a wonderful refresher your videos are! Surely even learn some new ways of thinking and looking from another's experiences and perspective! 👍

[DavidRodriguez-errq]

These videos bare awesome. For those of us that don't have breadboards, components, or MM. You analysis and real world examples are better than any class room lecture

[robertmccully]

Good stuff for beginners, the moral is start small, until you understand it from every possible thought reasoning. Wish they had YouTube when i was young. Was founded in 2005 and i was 50 years old.

[embeddedsystemsguy]

this is exactly the channel i need. Learning so much. Thank you!

[BizillionAtoms]

Could anyone help me? I can't understand how one circuit can have a current of 0.3 mA and the other of 23 mA. Since 5V is the voltage and 220Ω resistance for each circuit, according to Ohm's Law, shouldn't we be getting I = V/R = 5/220 ~ 23mA . So why do we end up with 0.3mA in the upper circuit?

[mukuntha.g]

how do one correctly apply ohms law to the top resistor...some comments below gives the logic about it. But what about writing ohms relation for that resistor?? please answer, thanks!

[jozefsoucik]

best basics i have ever seen. thank you.

[khoavo]

Notice that in both videos Ben never measured the signal's voltage when the loop is closed. But he did measure the amperage when when the loop is closed. That's why a lot of people will apply Ohm's law to the wrong set of numbers.
So why is it that the voltage drops when we connect everything together? I think it's due to the law of Power: P = V * I. A normal power supply will always make it so that the terminal voltage difference will be a fixed value because it provides the power lost throughout the circuit. BUT in the case of the signal, there is no such guarantee.

[vojtatesar]

To add, ability of a power supply to "deliver" current differes just because Ri {=inside resistence = you can imagine it like resistor inside an power supply} is DIFFERENT for DIFFERENT power supplies (Different power supplis are e.g. usb charger; battery; switching-mode power supply...). The Ri depends on construction of a power supply and it is a characteristic of a given power supply.

[maxpolaris]

Thevenize or Nortonize to see how STIIF your current and voltage source is from your circuit? College was 35 years ago for me. 🤔

[ancienttech]

I just found your videos! Thank you so much! These are the fundamentals I've been missing!

[srkki]

I think that I didn't understand this completely. I was thinking about this and couldn't get my mind to grasp the fact that 5 Volt isn't equal to 5 Volt.

What I'm guessing that's happening here is this:
Signals voltage is 5 Volts, but it has some inner resistance (like a real generator would behave), and when we complete a circuit through a resistor and diode, the amperage has to be lower since there is some inner resistance in the signal, therefore an equivalent resistance won't be 220 ohms, it would be much greater, probably even greater that 3kilo ohm (I didn't calculate it)
The USB generator behaves like an ideal generator, providing 5 Volts (with no inner resistance) and therefore the circuits equivalent resistance is 220 ohms.

Is there something I'm missing?

[naveensharma]

Thanks for wonderful series on basics..
One quick question: In a series circuit of 5v [made from USB cable {output 5v dc 1.4 amps}], one resistor, one led. Everything is fixed, 5 volt, 1.4 amps, 330 ohms, how do we make sure to reduce only current or only voltage by resistor ? ..

In the circuit multi-meter readings says, The voltage
- around battery is 5.1 V
- around resistor [330 ohms] is 3.15 V
- around led is 1.95 V
Yes, this proves Kirchhoff's voltage law [ 5.1 = 3.15 + 1.95 ]

The reading for currents/amps
- around battery is 1.4 A
- around resistor [330 ohms] is 185 mA
- around LED is 150 mA

This is not clear. What is the ratio and how much voltage/current will be reduced based a resistor. What if I want voltage drop but not amps or vice versa..

Also, we studied in schools that the current in series circuit remains same but changes at every junction in parallel connection [junction rule].. but here the current is also changing in above mentioned scenario [after every component, resistor and led in this case.]

Kindly explain. thanks in advance.

[typedeaf]

The odd part to me is the forward voltage. There is enough voltgage to push the electrons across the diode, but not enough electrons to illuminate the diode properly.

Does this relate to the controversy of, "Its not the voltage that kills you, its the current'? Lightning must have high voltage, to break the resistance of air. I guess the resistance of the human body composition is ultimately 'what kills you', but surely it must also be hgih current. I believe contact (skin) with 100mA - 1A can be lethal, right, but no voltage is generally given...right?

I guess a 12V car battery has much more potential for harm than a any number of dry cell batteries. Can you explain that?

Does this mean that in schematics, stating the volrage of the power source, without some indication of max current, is implying that you could have 'infinite' current?

[lejnyovn]

Thank you so much Mr. Ben Eater for your videos. For me your videos are like a microscope to understand a new world inside computers and electronics. Great work. I request you for a video on intel 64 bit CPUs.

[anilgr]

Please clear the doubts in the comments here. 😭

[benoitdebrueker]

Hi,

for an accurate measure, shouldn't you be measuring the current with the LED in the circuit (between the input wire and the LED lead) because of the voltage drop thru the diode?

Regards