Make Electronics-Exp 3

The idea is good, but it introduces parallel resistors when more than one switch is closed. There are simpler ways of introducing parallel resistors. It may cause confusion here, and perhaps unwittingly supply too much current to the LED....eg with 25mA supplied to the 20mA LED in this setup when all the bottom three switches are closed.

You are correct. The idea is to play around with the switches and see what happens whether you have 1 or more than 1 resistor at the same time in the circuit and whether you want them connected to the + or the - side of the battery as the voltage in certain parts of the circuit may go from positive to negative. You can also blow-up the LED by removing all the resistors from the circuit. I have to admit I am new to using EveryCircuit and my skills in using it are limited, there may be better ways to do it for sure. I was hoping for a larger canvas to work with but this is apparently not possible therefore I have some crooked connections as part of my schematic... So, not the nicest looking thing but it lets you play around with it.

The LED specifications imply a 2V drop and what should have been as a result 3.18mA in the circuit based on a 2.2K resistor but the EveryCircuit simulation shows 1.81V drop and a resulting 3.27mA current. The difference is not huge but I am not sure where the it is coming from. If anybody knows, please let me know. Thanks.

An LED is not a fixed resistor, but it does exhibit some resistance. The amount of resistance exhibited will depend on the rest of the circuit such as the battery v and any other resistors. This extra LED resistance in line with another resistor will thus alter the overall current and volt drops as seen in your schematic result. The values in the LED settings do not force the rest of the circuit to comply fully to its own values.

Thanks for your input. I had realized that it was coming from the LED but as of now I cannot quantify it via an equation based on the circuit specifications which is really where my itch is coming from. I can get to (9-2)/2200=3.18mA but I would like to know how to get to the 3.27mA & 1.81V drop calculated via the simulation. Maybe it is part of the internal EveryCircuit model and not meant to be calculated but I would like to be able to replicate those numbers if it is feasible.

Класссссс

I have tried modelling the LED with the Schokley equation and the results are not so good when comparing to the numbers generated by EveryCircuit. So either I am doing something wrong or they are using something else. Oh well, at least I tried.

I don't know what equation EC uses either. It would be interesting if someone here has the time to play around and find out... it would give them a good shot of credibility .... set to it guys!

Just to sum things up and to answer my own question here's a little summary. The default setting for the LED was 2V & 20mA with E=9V and R=2.2K Ohms. That implies (9-2)/2200=3.18mA going into the LED. Since this is too far from the (default) 20mA that the LED is expecting to drop 2V, then the properties of the LED used by EC changes the voltage drop at the LED to 1.81V. We could try to model this divergence via a Schokley equation or looking at the specs of the LED being used, but this is not specified by EC, so we just have to live with what the internal EC model churns out. This in turn implies a current of (9-1.81)/2200=3.27mA. If you want a 2V voltage drop at the LED then you have to click on the LED, then the eye and change the current from the default 20mA to 3.18mA (calculated above) and you will get the expected 2V drop at the LED. You will also see that now the LED lights up in the circuit whereas before it was not as the current was too far below the default 20mA setting of the LED. The moral of the story: make sure the LED current is not too far from the circuit's current if you want to have the expected LED's voltage drop.