Need help reading a schematic.

[Spork Schivago]

Re: Need help reading a schematic.

I understand about the electromagnet thing. I read an article about that. Some state regulation or something. Some states, they're supposed to have odds of one out of twelve, other states, one out of fifteen. I guess they have 4-6 PSI on normal tries, it takes around 10 PSI to grab a toy. On the nth time, the PSI increases. If they don't win on an increased PSI try, the odds get better or something. But in this laundromat, the actual arm had screws. I would go in and win on a regular basis. One day, I went in, and the owner or someone actually loosened the screws so the arms where wobbly. There were cameras and everything. I played once, noticed how loose it was, I shook my head and stopped playing. When ever I won, I'd just leave the toy on the counter for children. A week or so later, I came back, they were tightened again. It could've been that he didn't know how to program the control system or it could also be that the screws just came loose on their own.

[stj]

Re: Need help reading a schematic.

probably he didnt know how to control it so he just screwed it up and then fixed it after it became obvious.

[Spork Schivago]

Re: Need help reading a schematic.

probably he didnt know how to control it so he just screwed it up and then fixed it after it became obvious.

Yeah. I know some of the people that worked there (I believe it was family owned / operated) didn't speak English at all. So it's very possible the family member in charge of the Crane Machine couldn't read the english instructions on how to set the odds. Thanks for pointing out the PSI thing. That was a nice read and something I didn't know!

[stj]

Re: Need help reading a schematic.

some of the best come from here:
http://elaut.com/amusement/Belgium/index.php

[Spork Schivago]

Re: Need help reading a schematic.

I want this one! http://elaut.com/amusement/Belgium/g...cketCircus.php

[stj]

Re: Need help reading a schematic.

you must have a lot of space!!

[stj]

Re: Need help reading a schematic.

manual for your learning.

[Spork Schivago]

Re: Need help reading a schematic.

Oh cool, thanks! We don't have a lot of room. Two bedroom, one was turned into a work room. But I'd like to own a house one of these days. A two floor house would be nice, with an arcade / video game room. I could own a bunch of the old consoles and even some of the new. Hook them all up to a TV using some sort of fancy setup. Mabe even have a pinball machine or two, a soda machine and maybe even one of those coin changer do-dads.

[Spork Schivago]

Re: Need help reading a schematic.

So, I now want to modify the circuit and replace some of the components with SMD versions. For example, R2 is a 16 ohm resistor. I know wattage = voltage * amperage. But how can I tell what's the smallest wattage surface mount resistor that can safely be used as a replacement for the R2 resistor there? I believe the through-hole one is a 1/4 watt resistor but do I need to use a 1/4 watt SMD resistor? Thanks.

[Spork Schivago]

Re: Need help reading a schematic.

Would I use something like P = (v * v) / R? Where V would be the 5V that goes to the resistor, R = 16 ohm, which is the resistance of the resistor.

Something like:
P = (V * V) / R
P = (5 * 5) / 16
P = 25 / 16
P = 1.5625

I read that I should double that value when using SMD resistors, so
P = 3.125. That can't be right at all! What am I doing wrong here?

The current going through that resistor would be 0.3125 amps, right? Using I = V / R, I = 5 / 16, I = 0.3125.

[stj]

Re: Need help reading a schematic.

what's this about?

[Spork Schivago]

Re: Need help reading a schematic.

The original schematic I posted on page 1, the PDF. I'd like to replace the through hole resistors with SMD resistors but I'm not 100% sure how to calculate what the power rating for the SMD resistors need to be. Ie, do they need to be 1/16th of a watt, an 1/8th of a watt, a 1/4 of a watt, etc. The through-hole one, R2, is an 1/4 of a watt through hole. But how do I calculate what's the smallest wattage SMD resistor I can use in replace of the through hole 16 ohm through-hole resistor.

[redwire]

Re: Need help reading a schematic.

You can calculate a resistor's power dissipation two ways- knowing voltage across it, or knowing current flow.

For resistor R2, it does not have 5V across it because the LCD backlight uses some. Check the LCD datasheet, it might be 3.2V say depending on color/size. You can also read the backlight current rating.
So R2 sees (5-3.2V)/16=0.20W (as an example)
Good practice is to use a part at ~2/3 of its rating. Why? Resistors are usually rated at 70°C/170°F which is a bit hot for long life.

To dissipate heat, physical size is all that matters and SMD resistors still need to be fairly big. Get your backlight numbers and try the math. You can also use two (SMD) resistors together and double the power handling.

[stj]

Re: Need help reading a schematic.

R2 would depend on the display, modern displays can use a single high-eficiency led with a good light-pipe,
older ones could use anything from 2 to 10 or more - specially if they are actually "back" lit..

[Spork Schivago]

Re: Need help reading a schematic.

Thank you RedWire! So my math formulas were straight up wrong then? I used the magic circle. Was I calculating how much wattage the device was using instead of how much power it was actually dissipating? I know my numbers were wrong and what you said make sense except for the formula. I thought V / R = I. I thought wattage was found using something like:
V * I = P
V^2 / R = P
I^2 * R = P

The schematics say it's an HD44780 for the LCD display, but for some reason, what I got was a BC1602A ( ) It's supposed to be blue in colour but I haven't had a chance to actually power it on yet to check. Still gotta finish soldering all the UV LEDs.

If I'm reading the datasheet correctly, current for the backlight would be typically 20mA, right? For the backlighting voltage, it's using 3.5V for the blue type screen, right?

So V / R = P (I thought V / R == current, not watts)
5 - 3.5 = 1.5V
1.5V / 16 ohm = 0.09375 Watts

Then I just want to take 2/3s of 0.09375 watts.
0.09375 watts * (2 / 3) = 0.0625 watts

0.0625 as a fraction would be 1/16th, so I could safely use a 1/16th watt 16 ohm resistor or I could safely use two 1/32th watt 8 ohm resistors, in series?

[Spork Schivago]

Re: Need help reading a schematic.

R2 would depend on the display, modern displays can use a single high-eficiency led with a good light-pipe,
older ones could use anything from 2 to 10 or more - specially if they are actually "back" lit..

No, no, not the resistance. I want to know how to calculate the smallest wattage resistor I can safely use. IE, is an 1/8th watt 16 ohm resistor the smallest wattage resistor, or is 1/16th watt 16 ohm or 1/32th watt 16 ohm. I want to use SMD resistors but I see a lot of SMD resistors that are small wattage, like 1/16th seems common, but there's also 1/32th of a watt. Most through holes I see are always 1/8th watt or usually they're a 1/4 of a watt. Never smaller.

[Spork Schivago]

Re: Need help reading a schematic.

Redwire,

Google says this about finding a resistors power dissipation (I didn't know what I was looking for was called power dissipation, thank you!)

First, we use Ohm's law ( V = I x R ), to find the current through the resistor. The voltage across the resistor is V = 9 V. The resistance of the resistor is R = 100 Ω. Then, we can use the power rule ( P = I x V ), to find the power dissipated by the resistor.

The current the back light uses is 20mA though, so can I just use P = 20mA x 1.5V? Or P = 0.02A * 1.5V
P = 0.03 Watts
0.03 * (2 / 3) = 0.02 Watts

0.02 as a fraction would be 1 /50th. That just seems reeeally small! I guess I'm still confused here.

[redwire]

Re: Need help reading a schematic.

The Ohm's Law formula you use depends on what numbers you know. (I might have forgotten the E squared in my post, sorry if that confused).
Two ways to do it, look at voltage across the resistor or current through the resistor.

Applying Ohm's Law: P= I^2*R or P=E^2/R

For your LCD display:
White/Blue backlight likes 20mA and about 3.4V: (probably 1 LED)
R2= E/I = (5-3.4)/20mA = 80 ohms
P= I^2*R = 20mA^2*80 = 32mW
LED power P= E*I = 3.4V*20mA = 68mW

Yellow-Green backlight likes 100mA and about 4.2V: (probably 10 LED's)
R2= E/I = (5-4.2)/100mA = 8 ohms
P= I^2*R = 100mA^2*8 = 80mW
LED power P= E*I = 4.2V*100mA = 420mW

SMD 1206 is rated 250mW or 167mW derated by 2/3
SMD 0805 is rated 125mW or 83mW derated by 2/3
SMD 0603 is rated 100mW or 67mW derated by 2/3
I'd use an 0805. They aren't super small like pepper flakes.
A (more modern) low power backlight also means the voltage regulators will run much cooler.

Maybe voltage drops are confusing you. Start with power 5V and then subtract the two voltage drops (across R2 and then backlight LED) and then you hit zero 0V (GND).

[stj]

Re: Need help reading a schematic.

The schematics say it's an HD44780 for the LCD display, but for some reason, what I got was a BC1602A

HD44780 is the interface, originally it was the part number of the chip but they are cloned to hell now.

BC1602A is the display that is using the chip - part number suggests 16characters by 2 lines, A is probably related to led colour or viewing angle.

[Spork Schivago]

Re: Need help reading a schematic.

The Ohm's Law formula you use depends on what numbers you know. (I might have forgotten the E squared in my post, sorry if that confused).
Two ways to do it, look at voltage across the resistor or current through the resistor.

Applying Ohm's Law: P= I^2*R or P=E^2/R

For your LCD display:
White/Blue backlight likes 20mA and about 3.4V: (probably 1 LED)
R2= E/I = (5-3.4)/20mA = 80 ohms
P= I^2*R = 20mA^2*80 = 32mW
LED power P= E*I = 3.4V*20mA = 68mW

Yellow-Green backlight likes 100mA and about 4.2V: (probably 10 LED's)
R2= E/I = (5-4.2)/100mA = 8 ohms
P= I^2*R = 100mA^2*8 = 80mW
LED power P= E*I = 4.2V*100mA = 420mW

SMD 1206 is rated 250mW or 167mW derated by 2/3
SMD 0805 is rated 125mW or 83mW derated by 2/3
SMD 0603 is rated 100mW or 67mW derated by 2/3
I'd use an 0805. They aren't super small like pepper flakes.
A (more modern) low power backlight also means the voltage regulators will run much cooler.

Maybe voltage drops are confusing you. Start with power 5V and then subtract the two voltage drops (across R2 and then backlight LED) and then you hit zero 0V (GND).

Yeah, the missing squared was throwing me off! I was looking at my circle and just kept on seeing I = V / R. Now that I know you was definitely using the power formula (wattage), it all makes sense! I do struggle a little with the voltage drops some times but I'm getting better. I'm okay when it's just one device and I'm calculating the voltage drop across that device, but I seem to struggle when there's more than one device.

Also, the guy originally used a 16 ohm resistor but if I really understand this, with the white / blue background, I should be using an 80 ohm resistor? Because the Voltage / Current shows us the resistance is 80 ohm. Or am I misunderstanding that too?


Eventually, I want to replace the LCD screen with a touch screen QVGA LCD type screen. Thanks for all the help and showing me how to figure that out!