[Newbie]: I have some difficulties constructing this breadboard power module.

Thread Starter

babaliaris

Joined Nov 19, 2019
55
Hello!

I figured out how to wire the components but I have some questions.

1) These nail pins won't fit in the board holes. I even tried to pierce the board with the pins using a hammer.
2) The ground, ends up on pin number 2?
3) Pins 3 and 4 are the output, right?
4) Do these grey markings on the boards means that its a short-circuited? So these two capacitors (orange) with the two diodes are connected
to each other?

The kit has an ac adapter with 4 different diameter barrel plugins but no power jack to connect to the board. I bought this one (I hope its
the right one).

Also, I don't have that soldering tool, so I just thought to wire the components together with the help of pliers and insulating tape :p

Also if you watch carefully, next to c4 (the black capacitor) and below it there are some leftover holes. I don't know why.

IMG_20191119_181118.jpg
IMG_20191119_181152.jpg
IMG_20191119_181233.jpg
 
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ericgibbs

Joined Jan 29, 2010
10,835
hi b,
Welcome to AAC.
That looks a poorly designed PCB, do you have a photo to post of the other side/face of the PCB??
Did the 'nails' come with the kit.?
E
 

KeithWalker

Joined Jul 10, 2017
1,224
Hello!

I figured out how to wire the components but I have some questions.

1) These nail pins won't fit in the board holes. I even tried to pierce the board with the pins using a hammer.
2) The ground, ends up on pin number 2?
3) Pins 3 and 4 are the output, right?
4) Do these grey markings on the boards means that its a short-circuited? So these two capacitors (orange) with the two diodes are connected
to each other?

The kit has an ac adapter with 4 different diameter barrel plugins but no power jack to connect to the board. I bought this one (I hope its
the right one).

Also, I don't have that soldering tool, so I just thought to wire the components together with the help of pliers and insulating tape :p

Also if you watch carefully, next to c4 (the black capacitor) and below it there are some leftover holes. I don't know why.

View attachment 191934
View attachment 191935
View attachment 191936
If you are going to experiment with electronic circuits, it is essential that you obtain a soldering iron. A cheap 30 watt iron will get you started. Twisting the leads together will not give you very reliable connections and they will probably become intermittent over time.
The board you have in the kit is a printed circuit. The grey markings indicate where there is copper on the board. The correct way to assemble it is by pushing the component wires through the holes and soldering them to the bare copper lands on the other side. The wires are then clipped off close to the solder joint. The assembly will be much more robust if you push the component wires through the holes far enough to make the component lie close to the board.
I think you have assembled the kit with the components on the wrong side of the board but I can't tell for sure without seeing the other side. The way to tell is to see which side has bare copper around the holes. The component wires are inserted from the other side,
The circuit is correct. It does not matter which way round you connect the AC adapter to pins 1 and 2.
Pins 3 and 4 are the regulated 5 volts DC output.
Welcome to electronics and good luck.
Keith
 
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dendad

Joined Feb 20, 2016
3,602
C4 needs to be placed in the circle, the same as C3. As you have it, C4 is shorted out.
C3 and C4 both have the board designed to allow radial or axial capacitors to be used.
"Homework" for you is to discover what that means ;)
A small drill can be used to make the pin holes larger. A firm fit is best.
And the parts do need to sit flush to the board. That is mechanically a lot stronger.
And I echo @KeithWalker in that you will need a soldering iron. Just placing the parts in the board is no good, and will result in tears and frustration.
A good habit to get into when soldering is to wear safety glasses and always wash your hands after soldering. Safety First.
Electronics is a great hobby and I hope you really enjoy your journey in it.
 

dl324

Joined Mar 30, 2015
11,565
Welcome to AAC!

Cropped photo of the board:
croppedPsBoard.jpg
1) These nail pins won't fit in the board holes. I even tried to pierce the board with the pins using a hammer.
I suspect they're to be used for input and output to the board. If the pins don't fit, you can try using a drill bit to enlarge the holes. A tapered reamer can also be used. Another option is to use pins that fit. If the spacing is 0.1", you can use 0.025" square header pins.
2) The ground, ends up on pin number 2?
Pin2 is not ground. Pin 4 is ground.
3) Pins 3 and 4 are the output, right?
Pin 3 is 5V and Pin 4 is ground.
4) Do these grey markings on the boards means that its a short-circuited? So these two capacitors (orange) with the two diodes are connected to each other?
What grey markings? Connected is a more accurate description than "short-circuited".

The kit has an ac adapter with 4 different diameter barrel plugins but no power jack to connect to the board. I bought this one (I hope its the right one).
An obvious mistake on their part. What is the voltage of the supplied adapter? It looks like you noted that the transformer secondary is 5VAC. Is that what the kit called for? That's uncomfortably close to the minimum input voltage for 7805. The dropout voltage is given as 2V typical.
Also, I don't have that soldering tool, so I just thought to wire the components together with the help of pliers and insulating tape :p
Get a soldering iron. The connections will be more secure and you won't have unnecessarily long leads on the solder side of the board.

You should also learn how to form the leads using needlenose pliers or a lead former. Your projects are a reflection of your workmanship.
Also if you watch carefully, next to c4 (the black capacitor) and below it there are some leftover holes. I don't know why.
It looks like the kit has a provision for using an axial capacitor for C3. You can determine this by following the trace from the anode of the large capacitor.

It looks like C4 isn't installed correctly.

This is a power supply of my own design (homemade PCB):
wilpwr.jpg
It provides adjustable VPP, VCC, and switched VCC for an EPROM programmer.
 
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Thread Starter

babaliaris

Joined Nov 19, 2019
55
Thanks for the help!
But I still have some questions at the end of this post.

This is the other side of the board. This is the correct face to place the components?
IMG_20191119_230645.jpg

Also, I managed to use the adaptor to check the led. The following image shows the adaptor and its info:
IMG_20191119_230743.jpg

1) I'm still confused with the ground. In theory, the ground is the cable that returns to the negative (-) side of the voltage source, so if pin 1 is + and pin 2 is the - of the source then why ground is pin 4?

2) Why do I need this power supply circuit? Isn't the adapter doing all the work I need (It gives me 5vdc output)? As you can see the led works with the adapter as the source. So why do I need to combine the adapter and that circuit to power a breadboard?
 

BobaMosfet

Joined Jul 1, 2009
1,209
That adapter does indeed provide 5VDC output.

Top side of the PCB is:

1574198839880.png

component markings (silk screen) mean this is the top. You can solder on the bottom side.

Not sure why you have a kit when the wall adapter does what you need to get 5VDC. The 'kit' has a problem in that it's schematic shows a transformer to drop input to 5VAC, and a rectifer to convert that to DC- but the LM7805 needs at least 7VDC to convert to 5VDC....
 

dl324

Joined Mar 30, 2015
11,565
2) Why do I need this power supply circuit? Isn't the adapter doing all the work I need (It gives me 5vdc output)? As you can see the led works with the adapter as the source. So why do I need to combine the adapter and that circuit to power a breadboard?
You're right. In fact, the board won't work with a 5VDC input.

I bought this for $0.71 shipped from Aliexpress:
https://www.aliexpress.com/item/33044576815.html
It accepts power from USB or a 7-15VDC input and has jumperable 3.3V or 5V output for each breadboard power rail. For the latter, the USB provides 5V output.
 
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Thread Starter

babaliaris

Joined Nov 19, 2019
55
LOL. Even that I'm noob I knew something was odd. Well, I'm not surprised because the shop where I order this stuff, they did not even send me all the parts. For example, the breadboard was missing! I called them but they never send it (and I paid with a credit card...).

So, in other words, the adapter does the work I need right? It provides me with 5vdc which I need to work with basic circuits on a breadboard.

The instructions say that "When you finish with the constructions, in pins 1 and 2 connect the secondary wrapping of a transformer 220/5V/0.5A and if everything were placed correctly the led will light up".

Also, I have another question now :p

A transformer takes as input a voltage and gives as output another value of voltage without changing the waveform of the signal? For example
if it is ac you still get ac.

While an adapter does the same thing but also changes the waveform? For example, if you have ac it makes it dc.
 

dendad

Joined Feb 20, 2016
3,602
Your board is designed to run AC into it.
Pin 1 is not the +ve source, and pin 2 is not "gnd". They are AC input pins.
DC input on pins 1 and 2 can be used too, but then the -Ve or "gnd" pin of your plug pack is not to be tied to the power supply "gnd" output pin. It is treated as an AC input, just always of the same polarity.
Do not connect them together!

You could connect a DC source straight across C3 and not feed in via the diodes. The DC input voltage must be at least 2V higher than the output voltage so the regulator has something to work with.

The adapter will give you the 5V you need. But the power supply may be a cleaner 5V supply. Switch mode supplies are more efficient but noisier than analog ones.
If you run the power supply board, you will have to find another plug pack of higher volts. A 9V one would be fine. And a heatsink will be needed on the regulator.
 

dl324

Joined Mar 30, 2015
11,565
A transformer takes as input a voltage and gives as output another value of voltage without changing the waveform of the signal? For example if it is ac you still get ac.
The board you bought will work with AC or DC input. The output is 5VDC; providing the input voltage is high enough. With DC input, the input should be about 9VDC.
While an adapter does the same thing but also changes the waveform? For example, if you have ac it makes it dc.
The adapter that came with the kit, it converts 220VAC to 5VDC and makes the board unnecessary.
 

Thread Starter

babaliaris

Joined Nov 19, 2019
55
Did they suggest the hammer? :eek:
No that was my idea :p

This is what I have in mind when I see the circuit in the paper:
IMG_20191119_235858.jpg

This is what we do in theory. The current must always return back to the source. So I'm still confused about the ground. If my drawing is wrong, then pin 4 which is the ground where does it end up? Where does the current is returning to, through pin 4?
Probably there is some knowledge which I don't know and I'm hoping you will figure out what it is.
 

dl324

Joined Mar 30, 2015
11,565
This is what we do in theory. The current must always return back to the source. So I'm still confused about the ground. If my drawing is wrong, then pin 4 which is the ground where does it end up? Where does the current is returning to, through pin 4?
Probably there is some knowledge which I don't know and I'm hoping you will figure out what it is.
The connection between the junction of D2 and D4 to ground doesn't exist in the kit schematic.
 

Thread Starter

babaliaris

Joined Nov 19, 2019
55
The connection between the junction of D2 and D4 to ground doesn't exist in the kit schematic.
This is the problem. I'm trying to figure out what is the difference between pin2 and ground. If the ground is not the wire that returns back to the ac source, then where does it end up to? Where does the current go?
 

dendad

Joined Feb 20, 2016
3,602
No, your circuit is wrong. Look at the supplied one. No side of the Ac is at ground. Pin 4 does not go to the D2/D4 junction, but the D1/D2 junction.Try again :)
 

Thread Starter

babaliaris

Joined Nov 19, 2019
55
The ground DOES NOT return to the source!
Why do you think it needs to?
This helped me a little bit.

In-circuit analysis what I have done so far was analyzing circuits that connect to a voltage source ac or dc. In these circuits, the ground was always returning to - of the source. Something like this:



But this power supply looks different. It seems like there are two circuits at the same time. It's like the source of the 5vdc is that bridge
of diodes and that source is connected to another ac source.
 

dl324

Joined Mar 30, 2015
11,565
This is the problem. I'm trying to figure out what is the difference between pin2 and ground. If the ground is not the wire that returns back to the ac source, then where does it end up to? Where does the current go?
Pins 2 and 4 don't need to be connected.
 

Thread Starter

babaliaris

Joined Nov 19, 2019
55
Anyway, this is probably because I must do more reading. Can someone tell me if these are incandescent bulbs? I tried to plug them directly to the 5 volts without a resistor and I tried polarizing them in both ways and they work no matter what.

IMG_20191120_004159.jpg
 
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