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I am not an electronics professional, and know little about circuitry other than Ohm's law and the difference between series and parallel circuits. I want to build a two-way telegraph for my first project, consisting of two keys made up of an LED and a push button switch, and a center consisting of an OR gate and a power source. My problem is not in designing the circuit, it's finding out what value to make the resistors/power source. My LEDs will have a voltage drop of 2.2 and my regular diodes 1.1. I cannot seem to wrap my head around the concept of voltage drop when multiple components are involved. I know that the voltage goes down by 1.1 when current passes through a diode, but what happens when a circuit branches into two diodes that then recombines?

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  • \$\begingroup\$ Pressing either button in your circuit will turn on both LEDs and send current through the first resistor. (1) What do you want to happen? (2) Are the two buttons at each end of the line? If they are you need a third wire for the battery or you need two batteries. (3) A silicon diode will drop about 0.7 V when current is flowing through it. \$\endgroup\$ – Transistor Sep 14 '16 at 20:47
  • \$\begingroup\$ I want to calculate the value for the resistors/battery and learn the math involved behind it. What you have described is the desired behavior :) The buttons were going to be about a meter apart \$\endgroup\$ – user123732 Sep 14 '16 at 20:49
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    \$\begingroup\$ Well what are the diodes for? Why do you think you need them? \$\endgroup\$ – Transistor Sep 14 '16 at 20:50
  • \$\begingroup\$ Honestly I just looked up "diode OR gate" on Google and also looked at a Stack Exchange question describing how it works. It looks like you could do without the diodes but I wanted to follow a semi-authoritative description of an OR gate \$\endgroup\$ – user123732 Sep 14 '16 at 20:52
  • \$\begingroup\$ "Diodes in parallel with the same polarity each behave no differently than a single diode" - electronics.stackexchange.com/questions/53190/…... So that being known, the gate would exit with 2.3v and both LEDs would require at least 2.2 with a 3v power source, meaning the parallel resistors aren't even necessary \$\endgroup\$ – user123732 Sep 14 '16 at 20:55
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A key thing to understand in electronics is that sometimes we have to do precise calculations. Sometimes crude approximations are fine. Sometimes you start out with crude approximations and then refine things based on measurements or based on iterating from your first crude soloution.

To a first approximation we assume the voltage across a conducting diode is fixed regardless of the current. This is not strictly true but it's close enough for our purposes.

So if you put two diodes in paralell the voltage accross the pair will be much the same as a single diode. How the current will split between the diodes is hard to predict, it will depend heavilly on manufacturing variation in the diodes. This is why when operating LEDs in parallel it is advisable to use a seperate series resistor for each.

Once we introduce resistors the resistors largely set the current in each branch.

To a first approximation we asusme that the battery produces a fixed voltage, this is also not strictly true, indeed it is probablly less true than the above assumptions about the diodes.

The first design descision we need to make is what current we want through the LEDs. A typical indicator LED will light acceptablly at 5 mA and won't be damaged by 20 mA. So if we aim for 10 mA it doesn't matter if our calculations are off.

The next design descision we need to make is what battery voltage to use. Lets assume you are using a string of regular alkaline AA cells at 1.5V each. 3V is clearly not enough. 4.5V is barely enough but leaves very little voltage drop across the resistors making the circuit more sensitive to variations in the battery and diodes. 6V is probablly sensible.

Now we can start doing some calculations. Lets asusme that your values for the volt drop of the diodes and the LEDs are correct (they seem a little on the high side to me).

So the regular diodes drop 1.1 V, the LEDs drop 2.2 V that leaves 2.7V for the resistors in series with the LEDs.

$$R = V/I = 2.7/0.01 = 270 \Omega$$

The third reistor doesn't really serve any useful purpose, it just makes the circuit burn more power.

BTW if you want to know what current is actually flowing in a resistor of known value it's often better to measure the voltage across the resistor and calculate the current. Inserting an ammeter often adds a lot of extra resistance to the circuit changing the thing you are trying to measure.

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(I am attempting to answer my own question. Feel free to comment with mistakes I surely made.)

Foolish assumptions: A regular diode drops 1.1V and a LED drops 2.2V. This information was gotten from product information on Digi-Key(Wasn't planning on buying from there, I don't know if they sell in small quantities like 5-10, I just needed specs :))'s website. I'm sure my assumptions are wrong.

According to Diodes in parallel or series, "Diodes in parallel with the same polarity each behave no differently than a single diode." Meaning that with, for example, a 6V power source and a diode voltage drop of 1.1, the voltage at the exit of the OR gate would be 4.9V.

Knowing that, the needed resistance for the parallel resistors to limit the voltage to 2.2V for the LEDs would be 270 ohms each with a 6V power source.

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  • \$\begingroup\$ Just note that the voltage drop varies depending on the current through the diode. 1.1 sounds a bit high, that sounds like the value for the maximum current the diode is rated for. You will probably see a bit less, maybe 0.7-0.8 volts. A good circuit designer makes sure that it works properly for a large spread of component values! With your LED circuit, it doesn't matter. Just don't be surprised when you measure it. \$\endgroup\$ – pipe Sep 14 '16 at 21:22
  • \$\begingroup\$ Thank you! I was unaware that current could affect the voltage drop of diodes. I have problems working with current values because I cannot seem to calculate them without actually building a circuit and plugging in an ammeter. \$\endgroup\$ – user123732 Sep 14 '16 at 21:29
  • \$\begingroup\$ Note also that different LEDs have different voltage drops depending on the material they are made from, so check the spec for your LED or measure it with a multimeter, with the LED in series with a limiting resistance. \$\endgroup\$ – Ian Bland Sep 14 '16 at 21:44

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