[QUOTE=DrLuke2;32833958]draw your own potentiometer with a pencil (draw a thick graphite line on paper, then connect one end to + the other end to gnd, and there you have your selfmade potentimeter)[/QUOTE]
The fuck?
[QUOTE=Asgard;32845658]The fuck?[/QUOTE]
It works, graphite conducts ever-so-slightly, even moreso when applied as a layer on paper. You can even make a touchpad with it. [url]http://afrotechmods.com/forums/index.php/topic,8501.0.html[/url]
-never mind that, chryseus is on steam-
[QUOTE=Asgard;32845658]The fuck?[/QUOTE]
Just make sure the layer is thick enough!
[QUOTE=ROBO_DONUT;32836467]Because the site is pretty awful in general and because they're so relentless about monetization when it's their users who write all the content.
It's pretty scummy to exploit a community like that.[/QUOTE]
Experts Exchange? :downs:
Made a TTL to RS232 "board" that can be attached to a microprocessor/PC to communicate serially. The soldering isn't too good though, any tips on soldering? Whenever I try it I always end up soldering together 2 pins that are next to each other by accident, and end up spending time to unsolder/check everything works again.
Try using less solder and only apply it directly to the pin while soldering?
[QUOTE=amazer97;32868533]Made a TTL to RS232 "board" that can be attached to a microprocessor/PC to communicate serially. The soldering isn't too good though, any tips on soldering? Whenever I try it I always end up soldering together 2 pins that are next to each other by accident, and end up spending time to unsolder/check everything works again.[/QUOTE]
Use less solder and apply the solder to the pin, not to the soldering iron or you just burn the flux immediately. Make sure you don't keep heat on too long or you'll burn the flux.
I think Subby has been drinking rosin again, he wants me to ask you what is more efficient:
A) A half-wave rectifier
B) A full-wave rectifier
Let's wish him luck with his 250V (non-isolated) buck regulator.
Hopefully he won't kill himself...
[QUOTE=Chryseus;32879062]I think Subby has been drinking rosin again, he wants me to ask you what is more efficient:
A) A half-wave rectifier
B) A full-wave rectifier
Let's wish him luck with his 250V (non-isolated) buck regulator.
Hopefully he won't kill himself...[/QUOTE]
Keep in mind it will be having less than 40-50W going through it
Oh and, [url]http://en.wikipedia.org/wiki/Electrical_efficiency[/url]
doesn't two diodes have double the drop of one!??!!?
[QUOTE=SubbyV-2;32879151]Keep in mind it will be having less than 40-50W going through it
Oh and, [url]http://en.wikipedia.org/wiki/Electrical_efficiency[/url]
doesn't two diodes have double the drop of one!??!!?[/QUOTE]
The benefits from not throwing away half of the waveform with a half-wave rectifier kind of outweighs the minuscule effect that an extra couple diodes would have for the full-wave rectifier.
[QUOTE=Lapsus;32881220]The benefits from not throwing away half of the waveform with a half-wave rectifier kind of outweighs the minuscule effect that an extra couple diodes would have for the full-wave rectifier.[/QUOTE]
what kind of benefits?
i cant think of any for 40-50W drain?
Going to get an Arduino, but I'm looking for a good Canadian supplier. Currently, Newark looks like the best but I would prefer something with a warehouse in canada to avoid slow/high priced shipping :3
Does anyone know of a decent Canadian supplier?
Edit:
This place is quite amazing:
[url]http://www.robotshop.com[/url]
They have a huge selection :downs:
[QUOTE=SubbyV-2;32885356]what kind of benefits?
i cant think of any for 40-50W drain?[/QUOTE]
If your diodes are dissipating 50W of power, you've got other problems. :v:
[QUOTE=Lapsus;32886667]If your diodes are dissipating 50W of power, you've got other problems. :v:[/QUOTE]
40-50W going through the diodes, i see no benefit except ripple reduction and even then its going to be bucked anyway, pretty sure i never said anything about 40-50W power dissipation?
[QUOTE=SubbyV-2;32899001]40-50W going through the diodes, i see no benefit except ripple reduction and even then its going to be bucked anyway, pretty sure i never said anything about 40-50W power dissipation?[/QUOTE]
Watts are a unit of power.
If you have a 50W load, then it's 50W of power dissipated in that load. You don't have to have diodes that are rated for 50W.
The power dissipation of the diodes is going to be ~0.7V * current (in amps) for silicon diodes.
[editline]22nd October 2011[/editline]
[QUOTE=SubbyV-2;32879151]doesn't two diodes have double the drop of one!??!!?[/QUOTE]
They would [i]if they were in series[/i].
They're in parallel for bridge rectifiers.
[editline]22nd October 2011[/editline]
[QUOTE=Chryseus;32879062]I think Subby has been drinking rosin again, he wants me to ask you what is more efficient:
A) A half-wave rectifier
B) A full-wave rectifier
Let's wish him luck with his 250V (non-isolated) buck regulator.
Hopefully he won't kill himself...[/QUOTE]
You don't really need to be a douche about it.
I have a noob question:
How would I go about lighting an LED on this device?
[IMG]http://img818.imageshack.us/img818/9127/screenshot2011102200333.png[/IMG]
It only has 16 pins
I don't know if this means anything :v:
[IMG]http://img402.imageshack.us/img402/1167/screenshot2011102200341.png[/IMG]
There are 64 LEDs but only 16 pins?
8 rows, 8 columns
8 * 8 = 64.
To illuminate an LED you pull its row to V+ and its column to V- (or vice-versa). MCU pins are three state -- HIGH (V+, VCC, or VDD), LOW (V-, ground, VEE, or VSS), and input/high-z (effectively off)
If you want display an image, you probably need to loop through and display each row individually and make use of persistence of vision.
Not sure if you need external current limiting resistors or if they're built-in, check the datasheet. If in doubt, add them anyway. Better safe than sorry. I wouldn't go lower than 200 ohm.
[QUOTE=ROBO_DONUT;32902067]16 rows, 16 columns
16 * 16 = 64.
To illuminate an LED you pull its row to V+ and its column to V- (or vice-versa).
If you want display an image, you probably need to loop through and display each row individually and make use of persistence of vision.
Not sure if you need external current limiting resistors or if they're built-in, check the datasheet. If in doubt, add them anyway. Better safe than sorry.[/QUOTE]
I have no idea how to pull V+ or V- :v:
This component does not have a datasheet provided on the website I got it from which was partly why I was confused on how the pins worked.
Anyway, I'm completely new to this stuff so I'm going to just start with some buttons and blinking LED's but I bought the 8x8 grid for later when I have slightly more knowledge. I was just curious on how I would end up operating the device.
I'll throw a resistor on just in case then just start switching wires around to see what lights up :rolleyes:
Bah you quoted me before I could fix my terribad arithmetic.
[QUOTE=ROBO_DONUT;32901692]Watts are a unit of power.
If you have a 50W load, then it's 50W of power dissipated in that load. You don't have to have diodes that are rated for 50W.
The power dissipation of the diodes is going to be ~0.7V * current (in amps) for silicon diodes.
[editline]22nd October 2011[/editline]
They would [i]if they were in series[/i].
They're in parallel for bridge rectifiers.
[editline]22nd October 2011[/editline]
You don't really need to be a douche about it.[/QUOTE]
[url]http://en.wikipedia.org/wiki/Diode_bridge[/url], Pretty sure the current passes through two diodes though?
Just used LTspice to simulate half wave and full wave, ripple is pretty much a non issue for my application, the peak current for the half-wave (going through the diode) is about ~7A where as the full-wave gets ~4A through the diodes,
Anyway, this is what i mean by efficiency, full wave suffers the diode drop of two, 1.4*4=5.6W where as half wave only suffers a single diode drop, .7*7=4.9W (also, the full-wave suffers it twice as much)
A half-wave rectifier is the better choice for the application, smaller, easier to source and more power efficient.
Oh looky here what arrived today.
[img]http://i.imgur.com/Gs9UE.png[/img]
What could it be? Anthrax? Let's hope.
[img]http://i.imgur.com/hkpN6.png[/img]
Afdjkmsd fucking packages.
[img]http://i.imgur.com/qLTuU.png[/img]
Ooh, what could that be? Can you guess?
[img]http://i.imgur.com/18iZR.png[/img]
How about now?
[img]http://i.imgur.com/GItzu.png[/img]
I wouldn't suggest taking a picture while sneezing. Anyways, it must be pretty clear what it is now!
[img]http://i.imgur.com/VHXJb.png[/img]
Yes! It's stuff!
[img]http://i.imgur.com/t8hio.png[/img]
Stuff!
[img]http://i.imgur.com/xfgew.png[/img]
Stuff!!!!
[img]http://i.imgur.com/yTVi3.png[/img]
STUFF!!!
[img]http://i.imgur.com/F4xzZ.png[/img]
AAH!! MORE STUFF!!
So much stuff..
Gentlemen, I've got work to do
[editline]22nd October 2011[/editline]
The sweet smell of creating the very first thing ever and it works.
[img]http://i.imgur.com/0zQ9z.jpg[/img]
Nice!
I thought I might as well present an older project.
[img]http://localhostr.com/files/RvkZtk5/diagram.png[/img]
Basically, it's a dice. An automated one.
This schematic is not 100% up to date, I don't have Crocodile Physics on this computer, but the only differences are the [I]data collection start[/I] button is not hooked to 5V, but rather 24V to actually trigger the digital sensor in the PLC we used for this and we scrapped the [I]PLC reset memory[/I] button because my software is [B]FUCKING FLAWLESS[/B]. It was not necessary, either way.
Anyway, the idea is that a dice is a very old piece of technology and we wanted to improve the possibilities with a physical dice (hell, we had to come up with [I]something[/I], it's for some exam project).
We did this by still allowing manual dice "rolls", represented by the diodes to the right in the schematic. This is made possible with the binary counter hooked to the 555-timer circuit, dimensioned to 50 Hz. The button near the 555 chip is the actual manual dice "roll" button.
Then we have the [B]data collection[/B], which is where it gets ~automated~ and cool. Basically, instead of pressing the button, we allowed the user to define a number of desired dice rolls from a PLC GUI (touchscreen and shit), as seen below. Excuse the Danish.
[img]http://localhostr.com/files/yzpDQfK/number.png[/img]
To pull it off we allowed the PLC to override the button by sending out an analogue value of 16384, which corresponds to 5V, in front of the button, essentially acting like a button, only automatic.
So what we decided to do, since unless we found a way to modify the 555-timer circuit on the fly we didn't really have any other option, was to modify the length of the [I]AO_slag[/I] pulse (see the schematic). Now from what we could gather, PLC don't have any way to call a random number, so we had to come up with a way to do it ourselves:
[img]http://localhostr.com/files/DUFTitc/randum.png[/img]
This is our formula, which suffices both for sequence length and dice roll length for our needs.
Here's an example, assuming the start-seed is 560:
seed = ((560+1)*75) MOD 1024-1 = 90
seed = ((90+1)*75) MOD 1024-1 = 680
seed = ((680+1)*75) MOD 1024-1 = 898
The start-seed is 500 by default, but the user is allowed to change it, as seen below:
[url]http://localhostr.com/files/Uw6xMvM/seedz.png[/url]
So we wanted to use the data, but it needed to be in an edible format, which is what our digital-analogue circuit is for. Basically for each dice roll, it's a different voltage getting to the PLC.
1 = 0.0V
2 = 1.25V
3 = 2.5V
4 = 3.75V
5 = 5V
6 = 6.25V
Converting these to the analogue values, we could work with them, using the data for statistics and various other probability-calculations.
The last part of the software is [B]simulation[/B], which essentially is a modified data collection algorithm to see how many tries it needs to go through in order to reach whatever dice roll you desire.
[img]http://localhostr.com/files/Ss41it0/simul.png[/img]
For anyone interested, here's the source. Some of the variables and all the comments are in Danish.
[code]cyclecounter := cyclecounter + 1;
(* Simulering *)
IF sim_init = 1 THEN (* sim_init fungerer som en hovedafbryder *)
IF step = 0 THEN (* Det indebærer at man ikke kan begynde at simulere midt i en data collection, for eksempel *)
(* Ved initialisering af simulering: Clear hukommelse der har med simulering at gøre *)
sim_slag := 0;
sim_totalkast := 0;
step := 0;
step := 4; (* På den måde undgår vi konflikter fra brugerens interferens *)
END_IF
ELSE
IF step = 4 THEN
(* Ved evt. afbrydelse midt i simulering: Clear hukommelse der har med simulering at gøre *)
sim_slag := 0;
sim_totalkast := 0;
step := 0;
END_IF
END_IF
CASE step OF
0:
IF EDGENEG(DI_data_collection) THEN
IF seed > 0 THEN
(* Clear hukommelse *)
totalslag := 0;
total_1 := 0;
total_2 := 0;
total_3 := 0;
total_4 := 0;
total_5 := 0;
total_6 := 0;
procent_1 := 0;
procent_2 := 0;
procent_3 := 0;
procent_4 := 0;
procent_5 := 0;
procent_6 := 0;
step := 1;
END_IF
END_IF
1: (* Data collection *)
seed := ((seed + 1) * 75) MOD (1024 - 1); (* Udregn det nye seed til vores pseudo-tilfældige formular *)
t_kast := cyclecounter + seed; (* Dette beskriver længden af terningeslaget *)
AO_slag := 16384; (* For 5V gælder 32767 / 2 *)
step := 2;
2: (* Data collection *)
IF cyclecounter >= t_kast THEN
AO_slag := 0;
(* Intervalinddeling af analog input fra terning *)
IF (AI_terning > -1) AND (AI_terning < 1000) THEN
(* 1'er *)
total_1 := total_1 + 1;
step := 3;
ELSIF (AI_terning > 3000) AND (AI_terning < 5000) THEN
(* 2'er *)
total_2 := total_2 + 1;
step := 3;
ELSIF (AI_terning > 7000) AND (AI_terning < 9000) THEN
(* 3'er *)
total_3 := total_3 + 1;
step := 3;
ELSIF (AI_terning > 11000) AND (AI_terning < 13000) THEN
(* 4'er *)
total_4 := total_4 + 1;
step := 3;
ELSIF (AI_terning > 15000) AND (AI_terning < 17000) THEN
(* 5'er *)
total_5 := total_5 + 1;
step := 3;
ELSIF (AI_terning > 19000) AND (AI_terning < 21000) THEN
(* 6'er *)
total_6 := total_6 + 1;
step := 3;
END_IF
END_IF
3:
IF (totalslag + 1) = slag_antal THEN (* Når det brugerbestemte antal slag er opnået... *)
procent_1 := (total_1 * 100) / slag_antal; (* Udregner den procentvise del 1'erne betyder *)
procent_2 := (total_2 * 100) / slag_antal; (* Udregner den procentvise del 2'erne betyder *)
procent_3 := (total_3 * 100) / slag_antal; (* Udregner den procentvise del 3'erne betyder *)
procent_4 := (total_4 * 100) / slag_antal; (* Udregner den procentvise del 4'erne betyder *)
procent_5 := (total_5 * 100) / slag_antal; (* Udregner den procentvise del 5'erne betyder *)
procent_6 := (total_6 * 100) / slag_antal; (* Udregner den procentvise del 6'erne betyder *)
step := 0;
ELSE (* Hvis ikke: tæl og gå tilbage til step 1 for at udføre det næste terningeslag *)
totalslag := totalslag + 1;
step := 1;
END_IF
4: (* Simulering *)
(* For simulering gælder et lavere max seed fordi terningeslagene skal udføres hurtigere (FOR lave vil resultere i utroværdigt resultat) *)
seed := ((seed + 1) * 75) MOD (512 - 1); (* Udregn det nye seed til vores pseudo-tilfældige formular *)
t_kast := cyclecounter + seed; (* Dette beskriver længden af terningeslaget *)
AO_slag := 16384; (* For 5V gælder 32767 / 2 ; send 5V til terningen for at påbegynde slaget *)
step := 5;
5: (* Simulering *)
IF cyclecounter >= t_kast THEN
(* Slut for analog output (5V) til terningen fordi slaget er fuldført *)
AO_slag := 0;
(* Intervalinddeling af analog input fra terning *)
IF (AI_terning > -1) AND (AI_terning < 1000) THEN
(* 1'er *)
sim_slag := 1;
sim_totalkast := sim_totalkast + 1;
step := 6;
ELSIF (AI_terning > 3000) AND (AI_terning < 5000) THEN
(* 2'er *)
sim_slag := 2;
sim_totalkast := sim_totalkast + 1;
step := 6;
ELSIF (AI_terning > 7000) AND (AI_terning < 9000) THEN
(* 3'er *)
sim_slag := 3;
sim_totalkast := sim_totalkast + 1;
step := 6;
ELSIF (AI_terning > 11000) AND (AI_terning < 13000) THEN
(* 4'er *)
sim_slag := 4;
sim_totalkast := sim_totalkast + 1;
step := 6;
ELSIF (AI_terning > 15000) AND (AI_terning < 17000) THEN
(* 5'er *)
sim_slag := 5;
sim_totalkast := sim_totalkast + 1;
step := 6;
ELSIF (AI_terning > 19000) AND (AI_terning < 21000) THEN
(* 6'er *)
sim_slag := 6;
sim_totalkast := sim_totalkast + 1;
step := 6;
END_IF
END_IF
6: (* Simulering *)
IF sim_slag = sim_oensket THEN (* Hvis udfaldet af terningen matcher det ønskede udfald er vi færdige *)
sim_init := 0;
step := 0;
ELSE (* Hvis ikke, så prøver vi da bare igen: tilbage til step 4 og gentag (i samme ånd som ved data collection) *)
sim_slag := 0;
step := 4;
END_IF
END_CASE[/code]
[editline]22nd October 2011[/editline]
There's a lot more detail to this, we wrote a 70-page report on it, ask if you want to know something.
[QUOTE=SubbyV-2;32902643][url]http://en.wikipedia.org/wiki/Diode_bridge[/url], Pretty sure the current passes through two diodes though?
Just used LTspice to simulate half wave and full wave, ripple is pretty much a non issue for my application, the peak current for the half-wave (going through the diode) is about ~7A where as the full-wave gets ~4A through the diodes,
Anyway, this is what i mean by efficiency, full wave suffers the diode drop of two, 1.4*4=5.6W where as half wave only suffers a single diode drop, .7*7=4.9W (also, the full-wave suffers it twice as much)
A half-wave rectifier is the better choice for the application, smaller, easier to source and more power efficient.[/QUOTE]
Took me a while, but I see what you're saying.
I still feel like there should be some disadvantage to throwing out half the wave. I mean for an ideal transformer, the current in the primary should be proportional to the current in the secondary, so the primary won't be sourcing any current for that half of the wave. However, I really doubt it's that simple for practical transformers. There will probably be some loss due to currents induced in the core for that half of the waveform. I am unsure of how significant this is.
What is this for, anyway? I can't think of many situations where ripple really doesn't matter that aren't totally trivial.
[QUOTE=ROBO_DONUT;32905785]Took me a while, but I see what you're saying.
I still feel like there should be some disadvantage to throwing out half the wave. I mean for an ideal transformer, the current in the primary should be proportional to the current in the secondary, so the primary won't be sourcing any current for that half of the wave. However, I really doubt it's that simple for practical transformers. There will probably be some loss due to currents induced in the core for that half of the waveform. I am unsure of how significant this is.
What is this for, anyway? I can't think of many situations where ripple really doesn't matter that aren't totally trivial.[/QUOTE]
He is trying to make a switch mode (buck) that connects directly to mains.
[QUOTE=ddrl46;32906076]He is trying to make a switch mode (buck) that connects directly to mains.[/QUOTE]
Without isolation?
Saying this is a bad idea would be an understatement.
[QUOTE=ROBO_DONUT;32906131]Without isolation?
Saying this is a bad idea would be an understatement.[/QUOTE]
Without isolation.
[QUOTE=ddrl46;32906299]Without isolation.[/QUOTE]
I think I understand what Chryseus was saying now. I think the first time I read his post I unconsciously thought something along the lines of "nobody in their right mind would do that" and discarded that part as nonsense.
The reason he wants to use one diode instead of 4 or a all in one rectifier package is since he wants to "save" space.
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