How to determine divider resistor as inout to ADC

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shteo83
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How to determine divider resistor as inout to ADC

Post by shteo83 » Wed Mar 01, 2006 1:58 pm

I going to do a digital oscilloscope project. But my ADC is only -0.6v to +2.6V analog input range. I need a voltage id from negative voltage to positive voltage. In my case suppose to get
-2.6v to +2.6. How to the divider resistor? what should be the resistor value? My ref is 2.6v.

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MrAl
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Re: How to determine divider resistor as inout to ADC

Post by MrAl » Wed Mar 01, 2006 4:06 pm

Hi there,

One way would be to connect two resistors
(say 100k each) in series, then connect a
2.6v reference voltage in series with those
two resistors.
In this way, the center tap of the two resistors
goes to the ADC, and the top resistor open lead
becomes the signal input node. The input voltage
is thus applied between ground and the top
resistor lead.
Voltages applied to the top resistor then
get divided by two and offset by 2.6 volts
which gives us the following table:

input....output
-2.6v 0.0v
+0.0v 1.3v
+2.6v 2.6v

Thus, for inputs from -2.6v to +2.6v you get
outputs (from the center tap of the two resistors)
of 0v to 2.6v which is just right for the ADC.

Of course in the logic after the ADC you convert
0v to 2.6v back to -2.6 to +2.6 volts by
subtracting 1.3v and then multiplying by 2.


Take care,
Al

<small>[ March 02, 2006, 11:08 AM: Message edited by: MrAl ]</small>
LEDs vs Bulbs, LEDs are winning.

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philba
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Re: How to determine divider resistor as inout to ADC

Post by philba » Wed Mar 01, 2006 8:22 pm

clever solution.

of course, one could always use an opamp to do this, too.

shteo83
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Re: How to determine divider resistor as inout to ADC

Post by shteo83 » Wed Mar 01, 2006 9:25 pm

That means the Vref 2.6v is fixed? Let said in the series of resistor, one side fixed with Vref 2.6v and another side use to input signals? Must i put the resistor value to be 100k? If not, how exactly the value of resistor to be determine? My ADC is 8bit, single channel, Vref=2.6v, and operation supply voltage=5v.

<small>[ March 01, 2006, 09:35 PM: Message edited by: teo ]</small>

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Chris Smith
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Re: How to determine divider resistor as inout to ADC

Post by Chris Smith » Wed Mar 01, 2006 10:01 pm

The total R value of the divider is based on the current draw over all, that you can provide safely.

If you use a 10 ohm 10 ohm, or a 1000 ohm 1000 ohm, or a 10,000 ohm 10,000 ohm divider the voltage values will always be the same.

The current draw though the two resistors will be the only thing that changes, and by a factor of 10, 100, 1000 proportionally.

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Re: How to determine divider resistor as inout to ADC

Post by dyarker » Wed Mar 01, 2006 10:33 pm

The 2.6V ref stays fixed. Picking a resistor value depends on the input impedance of the ADC and source impedance of the signal to be measured.
Dale Y

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MrAl
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Re: How to determine divider resistor as inout to ADC

Post by MrAl » Thu Mar 02, 2006 11:06 am

Hi again,

Yes, the 2.6v ref stays fixed and the value of
the resistors depends on the input Z of the ADC
and what you can have your input Z be.
The 2.6v ref can be made from a reference diode.

I made a slight error when posting however,
when i said:
"The 2.6v ref becomes signal ground"
which isnt really true, because it's the bottom
of the ref (which is already ground anyway)
that becomes (or rather 'stays') signal ground.
The input signal to be measured is thus applied
between ground and the top resistor lead.
I'll correct that post now.


Take care,
Al
LEDs vs Bulbs, LEDs are winning.

shteo83
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Re: How to determine divider resistor as inout to ADC

Post by shteo83 » Sun Mar 05, 2006 9:35 pm

Hi... MrAl

The circuit that you mentioned above is it suitable fot the input of the ADC? because my lecturer said if the circuit will produce current, then cannot use? Can u give some comment, may be you more pro. then my lecturer ..;-)

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Re: How to determine divider resistor as inout to ADC

Post by dyarker » Mon Mar 06, 2006 12:31 am

What is part number of ADC, and were is -2.6V to +2.6V input coming from?
Dale Y

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Re: How to determine divider resistor as inout to ADC

Post by shteo83 » Mon Mar 06, 2006 1:50 am

TLC5540INS (8bit 40MSPS parallel) is the number of my ic. I used this ic for my digital oscilloscope project, so the input would be use a funtion generator to privided ac signal. Another thing is i was tested it by just simply supply a dc voltage, my maximum input range is 2.6, and supposed to obtain 11111111. My output is using LED to display but why the output keep alternating?

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Re: How to determine divider resistor as inout to ADC

Post by dyarker » Mon Mar 06, 2006 4:06 am

? 8 LEDs for display?

It is not unusual for least significant bit to change on and off.

If changing between 11111111 and 00000000, then set DC input to 2.59V and see. The least significant bit changing at max input can over-run the byte.
Dale Y

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Re: How to determine divider resistor as inout to ADC

Post by rshayes » Mon Mar 06, 2006 4:14 am

The odd voltage range of the A/D converter uses internal resistors to set up a reference voltage divider that uses the supply voltage as a reference. Any load on that divider will cause errors in either the offset voltages or the scale factor. The A/D converter is also a fairly fast part (40 Mhz). A simple voltage divider will need low resistance values to operate in this frequency range.

The 5 volt supply is not a very good reference. It won't be accurate and it will be noisy. A shunt regulator such as a TL431 can give you a clean 2.5 volt reference and can also supply some current for adding an offset to the signal. The bottom end of the A/D reference divider can be connected to ground. This would make your A/D converter range 0 to 2.5 volts.

You will probably need a unity gain buffer on the input side of the divider and possibly a second one between the divider and the A/D converter.

There is a general equation for the output of resistors connected from several voltage sources to one node. For 3 voltages (Vin, Voff, and Ground) the expression is:

Vout=((Vin/R1)+(Voff/R2)+(0/R3))/(1/R1+1/R2+1/R3)

Voff is constant at 2.5 volts.

By setting Vin to 2.6 volts, Voff to 2.5 volts, and Vout to 2.5 volts you can write one equation.

By setting Vin to -2.6 volts, Voff to 2.5 volts, and Vout to 0, you can write a second equation.

This gives you two simultaneous equations with three unknowns. By choosing a value for one of the three resistors (probably R1), you will have two equations in two unknowns, which can be solved using normal methods.

<small>[ March 06, 2006, 04:24 AM: Message edited by: stephen ]</small>

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MrAl
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Re: How to determine divider resistor as inout to ADC

Post by MrAl » Mon Mar 06, 2006 9:19 am

Hi again,

stephen:
What's the 0/R3 for? Typo?

teo:
No that is correct. A simple resistor divider
will draw current but you said that's what you
wanted. If you want high input Z too then you
need an input buffer (op amp) ahead of the
divider, or design the op amp circuit to provide
the input/output you need.
Sound good?

Take care,
Al
LEDs vs Bulbs, LEDs are winning.

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Re: How to determine divider resistor as inout to ADC

Post by rshayes » Mon Mar 06, 2006 1:56 pm

Hello MrAl:

0/R3 isn't a typo. It is simply a term for a resistor to ground (0 volts).

I think that this is sometimes called "Millman's Rule". It describes the voltage at a single floating node with resistors connected in a star configuration to voltage sources on the other end. Ground is a voltage source with zero voltage.

The numerator of the fraction contains one term for each resistor, consisting of the conductance of the resistor times the voltage it is connected to. The denominator contains one term for each resistor, consisting of the conductance of each resistor.

Vout=((V1/R1)+(V2/R2))/((1/R1)+(1/R2))

Vout=((V1/R1)+(V2/R2)+(V3/R3))/((1/R1)+(1/R2)+(1/R3))

Vout=((V1/R1)+(V2/R2)+(V3/R3)+(V4/R4))/((1/R1)+(1/R2)+(1/R3)+(1/R4))

and so forth.

The resistance of the node is the reciprocal of the sum of the conductances of the resistors.

Rout=1/((1/R1)+(1/R2))

Rout=1/((1/R1)+(1/R2)+(1/R3))

Rout=1/((1/R1)+(1/R2)+(1/R3)+(1/R4))

and so forth.

It looks complicated, but it works out quite simply using a calculator with Reverse Polish Notation (Mostly HP).

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MrAl
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Re: How to determine divider resistor as inout to ADC

Post by MrAl » Tue Mar 07, 2006 3:00 pm

Hi again,

stephen:
Oh ok i see what you are saying here now...
With V3=0 you get the equation:
Vout=((Vin/R1)+(Voff/R2)+(0/R3))/(1/R1+1/R2+1/R3)

I guess i would have added:

"which equates to:
Vout=((Vin/R1)+(Voff/R2)"

or something like that.


Take care,
Al
LEDs vs Bulbs, LEDs are winning.

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