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modified 8 years ago

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01:53:24

I'm relatively new to circuits, I've burnt myself on little bits and pieces that I've accidentally pumped way too much into more times than I'd like to admit. I'll say it: I'm lost. How do I determine what resistance I use for each component? Is there a law or an equation someone can point me to? 

published 8 years ago

8 years ago

You should look into how to analyze a circuit containing diodes. You'll find there are simplified models based on how the device should operate, or at least approximately operate. Google, books and practice are your friend. Start simple with a single diode and voltage source, analyze the circuit by hand using things you've learned, then use EC or something like it to cbeck your work. Then build it IRL.

8 years ago

First off, it's great that you're interested in electronics. For linear components (voltage-current relativity is a line, hence "linear"; the linear components are resistors in DC circuits; resistors, inductors and capacitors in AC circuits), Ohm's law is sufficient: R=U/I. For nonlinear components like diodes, transistors, etc, you would have to find the resistance by knowing how it acts at certain voltages/currents flowing through them. Diodes and LEDs, for example, have a parabole-like characteristic at positive polarity. The opposite polarity is designed to block current, therefore it's useful for preventing certain components from getting damaged, as well.

8 years ago

Did you download the program ElectroDroid, it's associated with EveryCircuit and has all sorts of tables useful for doing all sorts of calculations like this.

8 years ago

Wattage rating of components is important. Components that can handle voltage and amperage and cool themselves are a consideration.

8 years ago

We have no idea how far your knowledge in electronics is. But listen closly to what the guys above said. I also try to see how you are in understanding this problem. Check this http://everycircuit.com/circuit/5577240497881088

8 years ago

To answer your question above, for a simple circuit like this use Ohm’s law V=IxR. In some parts of the world it is U=IxR. If you look at the LED specification you’ll see it is rated at 2V and 20mA. As you have a 9V supply you need to drop 7V across the resistor. Using Ohm’s law 7=20mA x R. Rearranging, R=7/0.02=350 Ohm. Each LED will need a resistor in series. If this is unclear please ask.

8 years ago

Try downloading Circuit Jam, sister app to this one, it teaches you how to use various components and sets challenges to complete. It's free too.

8 years ago

I'll look into that other app. As for ohms law, I knew that was a thing and I knew the equation but I had no idea how to apply it. I think that might solve my problem, maybe. Thank you guys so much!

8 years ago

Source voltage minus 2V divided by used current for the LED 20mA

8 years ago

2 resistors 450ohms

8 years ago

Actually 350 Ohm if you work it out. In reality you’d use a preferred value 360 Ohm.

8 years ago

And in real-reality were people really solder things together and know what technologie can do today, would use a 1kOhm resistor, cuz todays 20mA LEDs are even with 5mA very shiny. But Mr LED resistor is splitting hairs. This over years, only this topic were he tries to collect points by the community. But dont tell Mr hairsplit, that log(e)(x) is a wrong syntax in its formula when he mean ln(x). Cuz log(e) is just a constant equal 0.434 multiply x and not ln(x). Conclusion, put from time to time your mobile anway and do learn some thing new. Isn't it boring to continue as "Mr LED resistor" if it does not fit to reality for most LED applications.

8 years ago

I agree that in reality it would be more likely you’d use a lower LED current. But you don’t just make this valid point, you must show again your hate. You give @D264 a plot to study while I am the only person that actually answered his question! @Techno made a simple miscalculation which I pointed out, not to show superior knowledge but, avoid @D264 getting confused. In EC, the default LED current is 20mA. The simple formula demonstrates how to calculate the component values to achieve it. If the people who support you had the guts to comment on your post honestly then perhaps you’d finally realise what a jerk you are. I dare them!

8 years ago

Let’s see if they’ve got the balls to risk losing your favour.

8 years ago

Your March holiday is cancelled, Jerk.

8 years ago

And, @D264, avoid people on EC whose only purpose is to big themselves up and score points, those who have to post a list of circuits where people have thanked them for helping them. Pathetic!

8 years ago

The principle was alreaady clear, but you still gave a tipp for a practical resistor choice. This absolut useless, cause you do not know when you never did that for yourself. So for years its your service to the community to swagger about LED resistors. The rest of your energy you spend in hate and troll propaganda against me, cuz you trapped years ago by calling other users NOOBS while you actually was the ignorant noob. Go aheead till end of your live.

8 years ago

Thanx robbi, thats a good idea. I will extend the helping hand list a little. Just for you.

8 years ago

If you want to look an even more desperate fool then proceed.

8 years ago

I offered you a chance to end this nonsense but you were too stupid to grasp it.

8 years ago

Who continue to be a foolmouth with an response time of about 20ms?

8 years ago

You?

8 years ago

I guess you’re not going to wish me a happy birthday for tomorrow then?

8 years ago

There is no reason to request cake on ten threads in parallel. Birthday is only ones a year.

8 years ago

“Is only ONCE a year”. I correct only to help. No criticism.

8 years ago

You are also going on this lebitch track... uuuuhhhhh

8 years ago

No, I’m not. Like I said, help not criticism.

8 years ago

Thats what LeButch would also say, right.

8 years ago

You just can’t take help can you?

8 years ago

Log(e)=0.434

8 years ago

Actually it’s 0.4342944819 to 10 decimal places.

8 years ago

Still wrong, test ln(10^0.4342944819) = 1.00000077619 but suppose to be exact 1.0!

8 years ago

Wrong and infantile.

8 years ago

You must be bored. You’re certainly boring.

8 years ago

Mr LED-Resistor, or do you prefer Jerk? No, I’m not. Like I said, help not criticism.

8 years ago

Meh!

8 years ago

You need to learn before you try to teach. Suggest you look and work through Examples section then look up on Internet Log(e).

8 years ago

I wonder that your friend @2ctiby dont stop by and help you to understand whats your missunderstand in math is. Cuz he's much more experienced in math then you, I guess. Lets see if he can be a good teacher for your low level Jerk problems.

8 years ago

Infantile granddad. Get a life.

8 years ago

don’t what’s misunderstanding because than let’s

8 years ago

English as protecting shield for you foulmouth strategie. Robbi be a little more creative. You getting more and more boring/uninteressting/useless/waste of time. What have you learned about electronic the last YEARS? Nothing new, cause you know already everthing? LOL, definitive not, but nobody knows why are you here. Accept being a troll.

8 years ago

Another fool comment from a fool. Only sufficient intelligence to talk about petty things and conduct petty trolling. Reassess yourself and do something useful.

8 years ago

I spend 90% for the community in electronic. 10% I waste for you and other trolls. You spend as Mr-LED-preresistor 1% for the community and 99% you talking to "fools" like me and orhers. What a senseless life you have.

8 years ago

You are very good at borrowing my words to fire back at me. You admit to wasting 10% of your time. Shameful.

8 years ago

And 90% winding people up - “look at my circuit ideas - but if you don’t think they are good you’re an idiot and probably a troll too”.

8 years ago

This thread is ended. You do not warrant any further comment.

8 years ago

This is exactly what you are doing, since you are on EC "active" you created hate, useless wrong information, zero help, bullshit at perfection. Wow, great job.

8 years ago

Meh!

8 years ago

Or you do just "Meh", right.

8 years ago

Meh!

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