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trying to fix a small board with few parts from an arcade game, and can't

Pommie

Well-Known Member
Most Helpful Member
When the sensor is off (no reflection) the output will be connected to Vcc via the resistor.
When the sensor is on (reflection) the output will be shorted to ground.

Mike.
Edit, note the resistor is 47k so only a tiny current will flow when the sensor is on.
 

rjenkinsgb

Well-Known Member
Most Helpful Member
Where does that board connect to on the main circuit diagram you posted? The large circuit only appears to have a DC supply, so if it comes from that is must presumably be DC, with just the polarity switched around. Swapping the polarity will control which sensor is active.


how does this opb742 work in this schematic.
The resistor defines or limits the current through it, in proportion to the voltage across it.

With the transistor off, there is no significant current flow through anything, no real voltage across the resistor and the input to the 4093 is "high".

With the transistor on, it is shorting the lower end of the resistor to ground - but the current through the resistor is still tiny, eg. if "+VCC" was 15V the resistor current would be roughly 1/3000th of an amp; actually 15 / 47000 amps, the voltage divided by the resistance.

Most conventional transistor amplifier and simple logic switching circuits work like that, using resistors as loads and in combinations to divide voltages etc.

eg.
 

doeby

New Member
the board i'm working on is DC. the main board swaps the polarity to this board to move the motor to the position it wants, only 2 choices.
you two helped me figure out that it looks like the rest of the board is function properly, just not the OP sensors. there are 2 of them, and i figured the odds of them both breaking at the same time was slim. i was trying to go deeper on them, to figure out if 1 is broken, causing the other not to function correctly.

Am i correct in saying the opb742(OP) for the side that is getting negative 15 volts, the LED would not turn on. so no ground out on the other side, and VCC will be going threw resister(it still is reducing, 15/47000?) and onto the 4093. {same as +15 volt side with no reflection,not activated}

then opb742(OP) +15volts and with reflective.{activated} will cause ground out of the resistor, (would get a little hot?)
leaving 15/47000 going to 4093? or nothing going to 4093?

thank you for all the help so far, i'm worse than someone who knows nothing, i think i know a little....
 

rjenkinsgb

Well-Known Member
Most Helpful Member
As Pommie said, the sensor output voltage will be near enough the same as VCC with no illumination and should be near zero when it is illuminated.
The resistor will not get noticeably warm, the power dissipated in it is trivial; roughly 0.005 watts...

It's as you say, whichever side has it's LED unpowered (from negative supply) should have its output permanently high, near VCC.
The other side that has LED power should switch between high with no object and low with something sensed, & the motor run when the output is low (less than 1/3 VCC).
 

gophert

Well-Known Member
Most Helpful Member
The 47k resistor (R3) is "pulling up" the wire at the collector to Vcc voltage with the LED is off (because the phototransistor is not conducting when the LED is off.

Once the LED is on, the transistor conducts and the wire coming off of the collector is near 0V (ground). The 47k resistor is commonly called a "pull-up" resistor.

The node at the phototransistor's collector is only intended to turn on/off logic circuitry. The pull up resistors usually limit to signal switching levels of current and not intended to directly drive motors or lamps or ...
 

doeby

New Member
that makes more sense, thank you, so it is kinda acting like a relay..
got another dumb general question. what makes the vcc follow a certain path, such as why would the vcc (full voltage)not just always go to the4093, whether there is the ground present of not. Is it because of it taking the path of least resistance to ground when one 1 present. Or does that resistor need a ground in line to work. more of an theory question, rather than specific to this board....
 
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Pommie

Well-Known Member
Most Helpful Member
Vcc is the voltage and will cause a current to flow to ground anyway it can - there will be lots of pathways. If the transistor is OFF then it will flow through the resistor to the 4093. If the transistor is on then the 4093 is effectively connected straight to ground.

Mike.
 

gophert

Well-Known Member
Most Helpful Member
that makes more sense, thank you, so it is kinda acting like a relay..
got another dumb general question. what makes the vcc follow a certain path, such as why would the vcc (full voltage)not just always go to the4093, whether there is the ground present of not. Is it because of it taking the path of least resistance to ground when one 1 present. Or does that resistor need a ground in line to work. more of an theory question, rather than specific to this board....
The inputs on modern logic chips (even the 4093) are essentially little capacitors or, mosfet gates. They are not a path to ground and no current flows in/out. Therefore, to apply Vcc or ground to a pin, you need to provide a path to Vcc or ground.
 

gophert

Well-Known Member
Most Helpful Member
Vcc is the voltage and will cause a current to flow to ground anyway it can - there will be lots of pathways. If the transistor is OFF then it will flow through the resistor to the 4093. If the transistor is on then the 4093 is effectively connected straight to ground.

Mike.
I know what you are trying to say but, let's not confuse the OP. Let's make sure when we say it flows TO the 4093, it only goes TO the pin. It doesn't continues flowing through the pin. The pin provides no path back to ground. It is a floating gate and not connected to anything that allows current to flow.
 

Pommie

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Most Helpful Member
I know what you are trying to say but, let's not confuse the OP. Let's make sure when we say it flows TO the 4093, it only goes TO the pin. It doesn't continues flowing through the pin. The pin provides no path back to ground. It is a floating gate and not connected to anything that allows current to flow.
It doesn't have infinite input impedance so a current, albeit tiny, will flow.

Mike.
 

KeepItSimpleStupid

Well-Known Member
Most Helpful Member
probing the CD4093 including Vcc should offer some hints for us. the first gate is the NAND function, the second does an invert and the 3rd basically increases the current drive. datasheet: http://www.ti.com/general/docs/suppproductinfo.tsp?distId=26&gotoUrl=http://www.ti.com/lit/gpn/cd4093b


NAND is NOT AND.

AND IS (Both inuts high, the output is high)
Inputs, AND, NAND
00 0 1
01 0 1
10 0 1
11 1 0

NAND has the output inverted, so 1, 1, 1, 0

Would need the actual voltages at the pins.

When the inputs are the same. you get inversion.

The voltage across the LED portion of the sensor should give an idea about aging. The LED intensity will drop with age. Compare the reflective surfaces with a known good sensor.

If you have the sensor pointed at the surface, you can look at

At Vcc = 5V, you have to be below 0.9V for a valid low.

The board looks easy enough to power by itself and use an isolated DC source for IN What is Vcc?

So, I see aging LEDs which will reduce the light output, Reflective surface issues and not adjusted to maximize signal.

When there is enough light, the voltage between the b-e junction could be as low as 0.4 V. Generally 0.6 to 0.7 V is assumed if you don;t know. In this case, it's important.
 
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doeby

New Member
well back to the drawing board, i replaced both OP sensors with same ones, and get the same result. so far i have check
replaced both OP sensors with opb742.
i switched the 4093 chip with another working boards one, it checked good
i checked all the resistors on the board, they all show the right ohms
i checked the diode 1N4002, current only flows towards silver end, but does read about 560 ohms resistance threw it.
15 V positive and negative input check out, put wires on other working board, functioned properly
i can reflect OP sensor with disc, and it will operate the motor,(so OP1 sensors LED is on working)

did some hands on testing, OP1 sensor LED is on, and operates correctly, but when i make the main board switch polarity,
OP2 doesnt seem to have any LED(Brand new). no matter what i do with the reflective disc, it does not respond.

So since the bridge rectifier is the only part before the LED, (resistor, but it tests good) is that what i should assume is wrong.
or could it also be the ground on the OP2 sensor on the recieving side or "pulling resistor"(tested with right resistance like a regular resistor)


i would think i should force it to turn on LED for OP2, and test between the pins. to see if the LED has power(and is on). then we know which side of the OP sensor is the problem is .
if this is what i should do, i would need someone to let me know exactly how to do it.
 

doeby

New Member
thank you all for all the help... i am victorious..
It needed 2 new OP sensors, neither of them were putting out LED light. original problem
when i soldered the new OP on, 1 of them had a trace on the component side of the board, and it was not making connection
so on OP2, there was no connection between the resistor and LED. just put on a jumper wire, and it works perfect. (self made problem)

So i'm back, i now have a 11 foot putting green, that changes between 18 different contours.. like 18 different mini golf holes. the kids love it

Thanks again for everyone's time.
120550120551
 

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