I want to build an soldering iron

Everything related to BASIC version 1.x (Oric 1 and Atmos) or HYPERBASIC (Telestrat).
Don't hesitate to give your small program samples, technical insights, or questions...
Post Reply
Brana
Flying Officer
Posts: 158
Joined: Fri Nov 30, 2007 8:30 pm
Contact:

I want to build an soldering iron

Post by Brana » Sat Nov 24, 2018 5:53 am

Yes, I want to build an soldering iron... :)
...And the first step is to - turn On the Euphoric :D

So I created (first) the following program:
It is on Serbian (sorry about that!) but it is really so simple stuff.

- Enter ANY desired two values on the left
(Use Up/Down arrow keys to move, Enter to - enter the value, R to restart).
- As soon as at least two values are entered, program will then calculate all other needed values on the right side.

My soldering Iron should work on 12 Volts
And I would like to have 30 Watts of power.

So (on the left side) I enter 12 Volts And 30 Watts.

And I get 4,8 Ohms impedance for the soldering iron needed,
With the power adapter of 2,5 Ampers strength

CAN ANYONE CONFIRM THIS IS CORRECT?
Really? :)

Source for building the program was Wikipedia and
https://www.automatika.rs/baza-znanja/e ... zakon.html
Attachments
unimer.tap
Program
(8.1 KiB) Downloaded 3 times
unimer.tap
(8.1 KiB) Downloaded 2 times
SCREEN1 copy.jpg
screenshot
SCREEN1 copy.jpg (25.37 KiB) Viewed 164 times
Last edited by Brana on Mon Nov 26, 2018 1:41 am, edited 7 times in total.

Brana
Flying Officer
Posts: 158
Joined: Fri Nov 30, 2007 8:30 pm
Contact:

Re: I want to build an soldering iron

Post by Brana » Sat Nov 24, 2018 3:21 pm

It IS correct! :) :) :)
According to HIS calculator at 02:00

:) :) :)

User avatar
laurentd75
Officer Cadet
Posts: 63
Joined: Sun Nov 22, 2009 4:33 pm
Location: Paris, France

Re: I want to build an soldering iron

Post by laurentd75 » Mon Nov 26, 2018 1:46 am

Well Brana,
I'm not very skilled in electricity or in electronics, but according to the basic Ohm law, yes, your calculations are correct for your example:

Known values:
P = 30 W
U = 12 V

Using the P = U x I equation we can then compute I = P / U = 30 / 12 = 2.5 A

Then using the U = R x I equation we can finally compute R = U / I = 12 / 2.5 = 4.8 Ohm

All good ! :D

Post Reply