Multiplex switches - Page 2 - Electronic Circuits Projects Diagrams Free

**3v0** · 4th March 2008, 03:27 AM

Hey we went from 3 to 4 switches

You have the basic idea.

**3v0** · 4th March 2008, 03:29 AM

Quote:

Originally Posted by I²R

Sort of. A pullup resistor is used to maintain a small, stable voltage on an input pin. If you left it off, your circuit MIGHT work correctly, but the pin's voltage may fluctuate, causing unwanted false inputs.

I am not an EE. But I would have said steady voltage and small current?

AtomSoft · 4th March 2008, 03:32 AM

ok cool. So there is no definite way to calculate the divided voltages right? Like i would have to place resistors and manually check each one to determine drop ?
All i know in ohms law is how to get 1 value from 2 others. I don't know how to subtract like 100k ohms from 5v to determine the drop on the resistor.

Quote:

Hey we went from 3 to 4 switches

You have the basic idea.

yeah the concept is in the head but the head doesnt know how to draw lol

EDIT: Found this

I²R · 4th March 2008, 03:41 AM

It's simple. Consider a 20 kOhm resistor in series with a 5 kOhm resistor, with 10 V applied across them. You want to determine the voltage at the point between them. First thing you do is calculate the current through the resistors:

I = V / R

I = 10 / (15 k + 5 k)

I = .5 mA

Now that you know that, you can calculate the voltage drop across each resistor:

V = I x R

V1 = .5 mA x 15 k

V1 = 7.5 V

V2 = .5 mA x 5 k

V2 = 2.5 V

Reality check: the two voltages should sum to the total supply voltage. 7.5 + 2.5 = 10. If you have more than two resistors, you can just extend this process. Now, work out your voltage drops.

I²R · 4th March 2008, 03:44 AM

Quote:

Originally Posted by 3v0

I am not an EE. But I would have said steady voltage and small current?

A negligible current. The inputs are high impedance and measure voltage, not current.

AtomSoft · 4th March 2008, 03:44 AM

Quote:

We know:
# The total voltage
# The individual resistor values

We can calculate the current flow and then the voltage drop across the individual resistors. From the Ohm's law page, we will use the formula:
I = V/R
I = 12/3000 ohms
I = 0.004 amps or 4 milliamps

The current flow through the resistors is 4 milliamps. Since the resistors are in series, we know that the current flow through each resistor is the same.

Then, to find the voltage drop across the 1000 ohm resistor, we can use the formula:
V = I*R
V = .004*1000
V = 4 volts across the 1000 ohm resistor

And to find the voltage drop across the 2000 ohm resistor, we can use the formula:
V = I*R
V = .004*2000
V = 8 volts across the 2000 ohm resistor

Quoted from : http://www.bcae1.com/resistrs.htm
Around the middle of page

so i got to :
I = V/R
then
V = I*R

This would sum out to be 5volts / 4 resistors regardless since all resistor values are equal.

I = 5 / 4000 = 0.00125
then
V1 = 0.00125*1000 = 1.25V
V2 = 0.00125*1000 = 1.25V
V3 = 0.00125*1000 = 1.25V
V4 = 0.00125*1000 = 1.25V

Now if i had
R1 = 1k
R2 = 1.5k
R3 = 2k
R4 = 2.5k

1 + 1.5 + 2 + 2.5 = 7k

I = 5 / 7000 = 5.5555555555555555555555555555556e-4
then
V1 = 5.5555555555555555555555555555556e-4*1000 = 0.55555555555555555555555555555556v
V2 = 5.5555555555555555555555555555556e-4*1500 = 0.83333333333333333333333333333325v
V3 = 5.5555555555555555555555555555556e-4*2000 = 1.111111111111111111111111111111v
V4 = 5.5555555555555555555555555555556e-4*2500 = 1.3888888888888888888888888888888
would this mean i need more power?

how the hell would this work ?

Tried in a ohms law calc and got

0.00071 A over 7K

so
V1 = 0.00071 * 1000 = 0.71
V2 = 0.00071 * 1500 = 1.065
V3 = 0.00071 * 2000 = 1.42
V4 = 0.00071 * 2500 = 1.775
=4.97V

why not 5V ? Will it still work due to the +/-5% thing on resistors?

I²R · 4th March 2008, 03:50 AM

No they can be the same value. In fact, this simplifies the equation since once you know the voltage drop across one, you know the drop across all of them. To determine the voltage applied by each keypress, you simply add up the voltage drops connected to that key.

**3v0** · 4th March 2008, 03:53 AM

Quote:

Originally Posted by AtomSoft

ok cool. So there is no definite way to calculate the divided voltages right? Like i would have to place resistors and manually check each one to determine drop ?
All i know in ohms law is how to get 1 value from 2 others. I don't know how to subtract like 100k ohms from 5v to determine the drop on the resistor.

yeah the concept is in the head but the head doesn't know how to draw lol

EDIT: Found this

Little current flow into the analog port pin. It contains a capacitor that charges to whatever voltage is on the pin. After that no current flows.

So in the all open case the analog port see 5V.

When you close S1 the 100K is in parallel with the 1st 1K so it becomes 990.1 Ohm
(look up parallel resistance calculator)

Which is close enough for government work to still call it 1K

The point is that by making the 100K resistor a large value it has little effect on the voltage divider.

AtomSoft · 4th March 2008, 04:13 AM

ok cool:

So it
V1:V4 = 1.25

then:
button 1 = 1.25v
button 2 = 2.50v
button 3 = 3.75v
button 4 = 5.00v

right?
wait wait:
3960.4 is the total ohms then?
you guys just confused me lol let me figure it out 1 min

I²R · 4th March 2008, 04:21 AM

Close. Take a close look at your schematic and think about what the voltage with respect to ground is going to be for each key.

I'll be back tomorrow. Goodnight.

**3v0** · 4th March 2008, 04:25 AM

With VCC=VDD=5V

All open will be 5V
S1 closed will be 3.75V
S2 closed will be 2.5V
S3 closed will be 1.25V
S4 closed will be 0V

AtomSoft · 4th March 2008, 04:41 AM

Dam i c it now i think i need a vacation lol. I forgot the resistors add up hence smaller voltage towards end and 4 = 0v because it closes gnd and connects straight right?

AtomSoft · 4th March 2008, 05:00 AM

So now for the software part:

In the data sheet :

Code:

To do an A/D Conversion:
1. Configure the A/D module:
• Configure analog pins, voltage reference and
digital I/O (ADCON1)
• Select A/D input channel (ADCON0)
• Select A/D acquisition time (ADCON2)
• Select A/D conversion clock (ADCON2)
• Turn on A/D module (ADCON0)
2. Configure A/D interrupt (if desired):
• Clear ADIF bit
• Set ADIE bit
• Set GIE bit
3. Wait the required acquisition time (if required).
4. Start conversion:
• Set GO/DONE bit (ADCON0 register)
5. Wait for A/D conversion to complete, by either:
• Polling for the GO/DONE bit to be cleared
OR
• Waiting for the A/D interrupt
6. Read A/D Result registers (ADRESH:ADRESL);
clear bit, ADIF, if required.
7. For the next conversion, go to step 1 or step 2,
as required. The A/D conversion time per bit is
defined as TAD. A minimum wait of 2 TAD is
required before the next acquisition starts.

Its like 12am here so im going to relax now and try this tommorow. I should have my junebug by like thursday and hope to have something to make by then and thanks to you all so far so good. So i want to thank you all and wish you a good night (morning for some)

blueroomelectronics · 4th March 2008, 05:12 AM

Hope you get it soon, much more fun than a simulator.

AtomSoft · 4th March 2008, 05:16 AM

i cant wait. Oh yeah i made my first 555 timer(blinker) today. would be my 3rd project so far. Im going to post it and put it in my sig. (its a video (youtube))

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4th March 2008, 03:27 AM	(permalink)
3v0	Hey we went from 3 to 4 switches You have the basic idea. __________________ search engine for electronic parts Junebug USB PIC programmer kit., USB Bit Wacker, Homepage The 15 Minute Printed Circuit Board! (+drill time)

4th March 2008, 03:41 AM	(permalink)
I²R	It's simple. Consider a 20 kOhm resistor in series with a 5 kOhm resistor, with 10 V applied across them. You want to determine the voltage at the point between them. First thing you do is calculate the current through the resistors: I = V / R I = 10 / (15 k + 5 k) I = .5 mA Now that you know that, you can calculate the voltage drop across each resistor: V = I x R V1 = .5 mA x 15 k V1 = 7.5 V V2 = .5 mA x 5 k V2 = 2.5 V Reality check: the two voltages should sum to the total supply voltage. 7.5 + 2.5 = 10. If you have more than two resistors, you can just extend this process. Now, work out your voltage drops.

4th March 2008, 03:50 AM	(permalink)
I²R	No they can be the same value. In fact, this simplifies the equation since once you know the voltage drop across one, you know the drop across all of them. To determine the voltage applied by each keypress, you simply add up the voltage drops connected to that key.

4th March 2008, 04:13 AM	(permalink)
AtomSoft	ok cool: So it V1:V4 = 1.25 then: button 1 = 1.25v button 2 = 2.50v button 3 = 3.75v button 4 = 5.00v right? wait wait: 3960.4 is the total ohms then? you guys just confused me lol let me figure it out 1 min Last edited by AtomSoft; 4th March 2008 at 04:26 AM.

4th March 2008, 04:21 AM	(permalink)
I²R	Close. Take a close look at your schematic and think about what the voltage with respect to ground is going to be for each key. I'll be back tomorrow. Goodnight.

4th March 2008, 04:25 AM	(permalink)
3v0	With VCC=VDD=5V All open will be 5V S1 closed will be 3.75V S2 closed will be 2.5V S3 closed will be 1.25V S4 closed will be 0V __________________ search engine for electronic parts Junebug USB PIC programmer kit., USB Bit Wacker, Homepage The 15 Minute Printed Circuit Board! (+drill time)

4th March 2008, 04:41 AM	(permalink)
AtomSoft	Dam i c it now i think i need a vacation lol. I forgot the resistors add up hence smaller voltage towards end and 4 = 0v because it closes gnd and connects straight right?

4th March 2008, 05:12 AM	(permalink)
blueroomelectronics	Hope you get it soon, much more fun than a simulator. __________________ Bill Smart Kits build Smart People http://www.blueroomelectronics.com

4th March 2008, 05:16 AM	(permalink)
AtomSoft	i cant wait. Oh yeah i made my first 555 timer(blinker) today. would be my 3rd project so far. Im going to post it and put it in my sig. (its a video (youtube)) Last edited by AtomSoft; 4th March 2008 at 05:49 AM.