0-24 V Analog voltage conversion to 0-5V digital

[energy_tr]

Hi everybody,

as my first question, I need to convert 0-24 V analog voltage to 0-5V digital... (3.3, 2.5 V is also fine)

Are there any COTS modules for multi i/o conversion, or an A/D ic that has multi input capability?

What I need is to get in sone 0V or 24V data into a pic, and also I have some resistor values to read.

About the resistor values, I thought to use the same IC (if possible) and read the voltage and the voltage of the shunt resistor.

But I don't have much space for this, just a small box about 2 feet x 2 feet for 20 I/O's...

Thanks a lot for any help...

Best Regards...

 

[ericgibbs]

Hi everybody,

as my first question, I need to convert 0-24 V analog voltage to 0-5V digital... (3.3, 2.5 V is also fine)

Are there any COTS modules for multi i/o conversion, or an A/D ic that has multi input capability?

What I need is to get in sone 0V or 24V data into a pic, and also I have some resistor values to read.

About the resistor values, I thought to use the same IC (if possible) and read the voltage and the voltage of the shunt resistor.

But I don't have much space for this, just a small box about 2 feet x 2 feet for 20 I/O's...

Thanks a lot for any help...

Best Regards...

hi,
The simple way to reduce from +24Vdc down to +5Vdc is by using two resistors in a potential divider chain.

The junction of the divider goes to the ADC input, just ensure that the ADC 'sees' a resistance of 10K or less.

Most PICs have a number of +5Vdc ADC's.

 

[makorihi]

Basically something like this.

https://web.mit.edu/rec/www/workshop/voltage-divider.gif

Just calculate what you need R1 and R2 to be for Vout to be +5V when Vin is +24V. I usually make the resistor to ground about 3K because many opamps seem to like that area.

 

[energy_tr]

thanks for the replies.. here is a nice forum definitely..
actually I can'T make a voltage division, because I cannot change the original circuit. The circuit should not be modified much, when I add my box to the car, I need to have the analog speed meter still working..

What comes into my mind, is if I can use a shunt resistor in series, and if the analog voltage is guaranteed to be in range of 0-5V on that resistor, it might work. but the main problem is that, the circuit is kinda an analog automobile speed display, which also has a few (12) variable resistances that I need to measure.

I need to read the resistance values between 0 ohm and 330 ohm. and 0 ohm to 180 ohm. the resistance changes , when I speed my car, and it drops when I push the brakes.

Thanks for your help, I loved this forum, and nice people..

 

[Nigel Goodwin]

Sorry, I've no idea what you mean?.

The previous answers require no changes to your original circuit, just a potential divider between the original circuit and your new meter.

 

[yngndrw]

I think what he means is that the resistance (R1 + R2) from the output to ground will affect (Effect ? I always get the two confused.) other instrumentation.

You could make a voltage follower with an op-amp. This can then feed into a potential divider. (Or you could make an non-inverting amplifier with an op-amp which will scale and buffer the signal in one.)

 

[ericgibbs]

hi energy,
It would be better to tell us what you want to do rather than how you want to do it..

Draw and post a simple sketch of the project with a short description.

 

[Speakerguy]

ADC's like to see a low source impedance, on some PIC uC it is 10K on others it is 2.5k IIRC. If you use two resistors to divide the signal down, parallel their values to figure out what the source impedance is (ie two 10k resistors in series would divide the signal in half and have a source impedance of 5k).

A better way to do it is with an op-amp buffer that will provide very low source impedance to the ADC and you can even incorporate your anti-aliasing filters into the op amp ckt very easily. But this is more costly and complex than two resistors and is not needed if you're only measuring a DC level and you have a signal that will be unaffected by the loading the resistors puts on it in the simple divider method.

 

[richacm]

Just been following this thread as I have something similar I am doing...

As a newbee are there any benefits of using a voltage divider over a zener diode? I use a 5V zener diode to lower the voltage down to 5V, should I consider using a voltage divider instead?

 

[ericgibbs]

Just been following this thread as I have something similar I am doing...

As a newbee are there any benefits of using a voltage divider over a zener diode? I use a 5V zener diode to lower the voltage down to 5V, should I consider using a voltage divider instead?

hi,
If you are talking about Analog division.
If you use a zener diode, you will not get a 'true' reduction over the total range of the source voltage, as the source voltage drops below 5V the zener will stop conducting. So the zener/resistor will work for a source voltage of +10v down to +5V.

For digital reduction from +10V to +5V using a 5Vz and a series resistor to 0V, with the zener/resistor connected to the digital input, thats OK..

Do you follow.?

 

[energy_tr]

The problem is in the attached jpeg picture...

Hi,

I need to read 3 varying resistance values of the car (as speed of the car increases, the resistance that is "seen" changes from the driver's display side of the circuit.

Also the rpm and gas displays are similar.

Out of gas and over temperature warnings are 24 V and 5 V on/off lights in the analog driver's display...

I need to get them into my pic, and add the digital data into the computer, so that I can create a digital display...

I cannot "load" the observed resistances, that's the reason I wonder is I can use a voltmeter/ammeter type of readout. That I need ADCs, but they are mostly operating in 0-5V analog input ranges, and I have 0-24V analog inputs.

Please help me.. thanx anyway...

 

[Nigel Goodwin]

I cannot "load" the observed resistances, that's the reason I wonder is I can use a voltmeter/ammeter type of readout. That I need ADCs, but they are mostly operating in 0-5V analog input ranges, and I have 0-24V analog inputs.

As we've told you all along, a simple potential divider is all you need, possibly with a buffer amp following it so you can keep the resistor values high.

But as we've no idea what the source impedance is, we can't really suggest values.

 

[energy_tr]

what I still can't figure out is, if I use a voltage divider for the variable resistor between 2-90 ohm, than how can I divide 2 ohm, without loading the circuit?

if I use 0,05 ohm, is it appropriate, as I don't want the car's speedometer to show wrong results in the driver's analog display... (showing 40 mph while going with 70 mph)

If that's possible, than I will use 0,05 ohm as a shunt resistor and measure the voltage across it for current, and another voltage measurement is needed for the 2-90 ohm varying resistor.

Than, the only question that I still can't resolve is:

how can I measure the voltage on 2-90 ohm resistor, if the voltage span is above 5 V...
if I need to divide the voltage, how should I make the division, so that the original car speedometer is not effected?

thanks for your patience..

 

[energy_tr]

I might need to add that, I only have the two wire ends of the varying resistor.

I can't touch the varying resistor, as the gasoline meter resistance is somewhere else in the car...

the two wires come to the driver's analog display, and I only have the two wire ends...

and I (probably) can't alter the engine of the car.. I prefer only to touch the driver's analog display...

 

[ericgibbs]

what I still can't figure out is, if I use a voltage divider for the variable resistor between 2-90 ohm, than how can I divide 2 ohm, without loading the circuit?
The change in resistor value must be detected in the 'main' unit by a change in the current flowing in the resistor, [usually done by measuring the change in voltage across the resistor], OR the current flow thru a resistor in the main unit.

If you add a resistor in series with 'sensor' resistor you will most likely affect the calibration of the main unit.

if I use 0,05 ohm, is it appropriate, as I don't want the car's speedometer to show wrong results in the driver's analog display... (showing 40 mph while going with 70 mph)

If that's possible, than I will use 0,05 ohm as a shunt resistor and measure the voltage across it for current, and another voltage measurement is needed for the 2-90 ohm varying resistor.

Than, the only question that I still can't resolve is:

how can I measure the voltage on 2-90 ohm resistor, if the voltage span is above 5 V...
if I need to divide the voltage, how should I make the division, so that the original car speedometer is not effected?

thanks for your patience..

Hi,
I would suugest you do the following tests.
Use a DVM and measure the actual voltage across the individual 'sensor' resistors.
At maximum and minimum limits of the sensor resistors.

Tell us what you measure... OK

 

[energy_tr]

I will measure the voltages asap, and place the readout values...

it might take a day or two... but I'll let you know...