Heres an example using the protoboard instead of a breadboard. You solder your traces on the back of the protoboard after you've pushed your "through hole" (TH) components in. You can solder jumpers also. You can use a straight edge and snap the protoboards in half or snap/break them along the perforations/holes to any size you need to start out with. Kinda like skoring sheetrock and snapping at the skor, but with protoboard no skoring is needed.
Here you can see where I scratched away some of the copper on the back of the protoboard just to be safe that there wasn't any conduction between the traces I was a bit uncertain of...
This little board was originally being used to test LED ccts, its a rudimentary capacitive dropper. It was wrapped in electrical tape so to isolate the user from the energized cct. I don't use it anymore, if I were to use it Id want to add a fuse, maybe a MOV, ect. You have to be careful with un-isolated mains power supplies, that's one reason transformers are so widely used, isolation from mains.
Breadboard is great for testing your schematics because you don't need to solder, just push your components into the breadboard and then use jumper wires to connect it all together. Each row is isolated from the next but the holes in each row are all connected together with a metal trace hidden under the holes in the breadboard, they are not very visible from looking down through the holes, but maybe you'll see a bit of a metallic shimmer at the right angle.
Here on the left, 2 little breadboards are linked together to build a bigger cct, the breadboard on the right has an esp32 on it with several LEDs being PWMed by the ESP32 via a seperate power source and switching MOSFETs on/off (PWM) to allow the other power source to power the LEDs. The gate of each MOSFET is linked to a different GPIO...
Looking at someone else's breadboard will be confusing, especially at first. Same with schematics. Most times people design a cct or schematic and then explain how it operates in steps when posting online or even in the data sheets it gets explained thoroughly. I guess what I'm saying is don't feel bad if you don't know what's going on in a cct when you first look at it, break it down into its baser components that you are aware of and read the descriptions on how the cct works to gain an understanding. Pretty much everything is made up of resistors, capacitors, diodes, inductors, and transistors. You arrange these base components in different configurations using different quantities, different sizes, ect. These little groups of components or ccts then get strung together to create more complex ccts. You'll see block diagrams in data sheets, most every block or shape will be representing a small cct, and all these smaller ccts are connected together to provide the end user with the final product. The SSR I saw someone writing about is actually a cct, its not just a simple component but an entire cct with all sorts of components ect, you can build them yourselves if you wanted.
EDIT (more stuff on capacitive droppers):
If your mains is 120VAC and 60Hz, then 0.04524 × μF = Amps of constant current. Otherwise if your mains is not 120VAC, then use the capacitive reactance formula to determine how much current will flow.
Capacitive reactance = 1 ÷ (2·PI·Hz·F)
PI = 3.14...
Hz = mains frequency, ie 50Hz or 60Hz
F = farads, need to convert the μF into F, ie divide μF by 1,000,000.
In the posted cct, there's 8 capacitors in parallel
(typically only 1 or 2 are used but I needed all the caps I had to achieve ~50mA of constant current, I could have just used 1 cap rated for 1.08μF but all the safety capacitors I had were very small so I had to wire them in parallel to make a bigger "X2" capacitor, "X2" is a safety capacitor) which means you'd add all the capacitors capacitance up together to arrive at a total capacitance to determine the current flow. I only had 8 safety capacitors (X2 style) laying around, 5 × 0.15μF, and 3 × 0.11μF. Together they add up to 1.08μF. The 0.11μF caps didn't state 0.11μF, rather they were rated in Joules, so I found their 0.11μF capacitance using algebra.
Capacitive reactance = 1 ÷ (2 × PI × Hz × F)
2 × PI × 60Hz × 0.00000108F
=
0.00040715
1 ÷ 0.00040715
=
2,456.1Ω
Ohms law:
V = I·R
V/R = I
120VAC ÷ 2456.1
=
0.0488A; 48.8mA
The cct will flow 48.8mA regardless the load. The current will begin to drop as the load voltage increases. If you had 1 LED then 48.8mA would flow, if you had 10, then almost 48.8mA would flow, but once you start getting into high voltages needed, like if you had 30 LEDs in a row then the current will drop even more depending on the size of the capacitor. You'll want some way to limit in rush current so that if you coincidentally energized the cct at peak mains voltage your LED or LOAD won't blow, like perhaps some series resistance possibly.
Glad you found that diode.