Current is a gateway drug to magic smoke.

A list of basic stuff to buy when starting out


Do not leave the iron on when you are not using it.

Clean the tip by applying solder, wiping it clean on a sponge or steel wool and then coat it with a thin layer of solder to prevent oxidation before you turn it off and leave it.

If you use a sponge, make sure it's damp but not soaked.


LED strips from “worst” to “best”. Price is according to that.

WS2812B > APA102C > APA102

If you overvolt a capacitor it will fail in moments, if you go above 50-70% of the voltage limit it will fail in months, or years. All electrolytic caps will fail eventually though, even if idle

Alkaline typically doesn't have a mAh rating because it's so current dependant

2 batteries in series will have the same capacity as 1 battery, it's double the voltage. 
In parallel you get the same voltage, but double the capacity.
A 1Ah battery at 1.5V can deliver 1.5Wh in an ideal world. Two batteries can deliver 3Wh, so if you put them in series they'll be at 3V, thus the capacity must be 1Ah still to get to 3Wh
Likewise, if you put them in parallel, the voltage is still 1.5V, in which case you'll get 2Ah from them and end up with the 3Wh


To measure voltage just put multimeter to voltage mode and connect leads to positive and negative of the batteries. If the battery is non-rechargeable and way below it's supposed voltage (think <1V for 1.5V battery) it's pretty much dead.

Laws and equations

The law stating that the direct current flowing in a conductor is directly proportional to the potential difference between its ends. It is usually formulated as V = IR, where V is the potential difference, or voltage, I is the current, and R is the resistance of the conductor.

To explain that in a more useful way - You can calculate for a third variable if you know the other two.

Voltage(V) = Resistance*Current
Current(A or I) = Voltage/Resistance
Resistance(R or Ω) = Voltage/Current

That means if you wanted to know what resistor to use if you had an LED that drops 3.3V, works at 20mA and your power source was 5A - you'd calculate 1.7/0.02 and would end up knowing you need to use a resistor that is 85Ω or close to that value.

If you're wondering where 1.7V came from in the above example, it's the source voltage - 5V minus the voltage drop of the LED.
  • electronics_at_component_level.txt
  • Last modified: 2017/11/10 19:14
  • by c0rn3j