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How do I set the adjustable reference voltage for TSM102A Dual Op-amp - Dual Comp & Voltage Ref IC.

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Gazza_AU

Member
The TMS102A combines building blocks of a dual operational amplifier, a dual comparator, and
a precision voltage reference.

Recommended Operating Conditions:
VCC+ – VCC– Supply voltage = 30v.
VKA Cathode-to-anode voltage VREF to 36V.
http://www.ti.com/lit/ds/symlink/tsm102.pdf

Were do you add resistors for adjusting voltage reference?

Is there a laymans way of calculating values for a given reference voltage: 23.076923076923076923076923076923v

Notice the voltage has a repeating pattern after the decimal, left there to test the decimal accuracy of TMS102A VREF.
 
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MikeMl

Well-Known Member
Most Helpful Member
You are doing really, really good to make the absolute accuracy of any analog circuit any better than 0.1% (~10 bits). Resolutions of 18 bits are not unheard of...
 

Mikebits

Well-Known Member
Why would you need such a reference?
 

Gazza_AU

Member
Were do you add resistors for adjusting voltage reference?
Is there a laymans way of calculating values for a given reference voltage: 23.076923076923076923076923076923v
I am open in saying I cannot work this out by looking at the datasheet.
If you understand the datasheet and can simplify the sum for layman, please do.
 

ChrisP58

Well-Known Member
Figure 2 on Page 6 of the datasheet shows the equation for the two resistors, and how to connect them.

Vka = Voltage from cathode to anode
Vref and Iref are parameters given in the Electrical Characteristics table on page 5
 
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Gazza_AU

Member
Figure 2 on Page 6 of the datasheet shows the equation for the two resistors, and how to connect them.
I have no problem finding the information.

I used the Chip pin layout pic in post for a reference for those wanting to offer advice on resistor connections.
If you understand the datasheet and can simplify the sum for layman, please do.

 
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alec_t

Well-Known Member
Most Helpful Member
To achieve the accuracy implied by all those decimal places is unrealistic. You would have to compensate for many variants (temperature, atmospheric pressure, phase of the moon, passing traffic, ...... :)).
What is your budget for achieving a reference voltage accurate to within even 0.01% or better? Do you have a temperature-controlled environment for the circuit?
 

Gazza_AU

Member
To achieve the accuracy implied by all those decimal places is unrealistic. You would have to compensate for many variants (temperature, atmospheric pressure, phase of the moon, passing traffic, ...... :)).
What is your budget for achieving a reference voltage accurate to within even 0.01% or better? Do you have a temperature-controlled environment for the circuit?
I have not set out to imply any degree of accuracy.
The number came from calculator and repeats.

The aim is to find a way to understand the math for choosing Reference Components.
 

JimB

Super Moderator
Most Helpful Member
One of the complicating factors is the (Iref x R1) term which allows for the low value of reference pin current flowing in resistor R1.

If we ignore this term we can simplify and kick the expression around and get:

R1 = R2 ((V/Vref) - 1)

Say we want 23 volts, and let us make R2 = 1k Ohm

R1 = 1 x (23/2.5 - 1)

R1 = 8.2 k Ohm

Does that make things more understandable?

JimB
 

JimB

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Most Helpful Member
If we consider the effect of (Iref x R1), the output voltage will be changed by:

8.2x10^3 x 1.5x10^-6 = 12.3mV

JimB
 

AnalogKid

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Posting a desired value that has 30 decimal places most certainly is stating a degree of accuracy. The best you will be able to do with this part and 1% fixed resistors is *approximately* 23.8 V.

The math is exactly the same as the standard gain equation for a non-inverting opamp with a 2.5 V input. plus a small error term for input current.

ak
 
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alec_t

Well-Known Member
Most Helpful Member
For the maths, use the 'typical' values in the spec.
Let's assume R1 = 10kΩ, Vref = 2.5V, Iref = 1.5uA.
You want Vka = 23V.
So,
23V= 2.5V*(1 + 10kΩ/R2) + 1.5uA*10kΩ ≈ 2.5V*(1 + 10kΩ/R2),
23V/2.5V = 1+10kΩ/R2,
9.2-1 = 10kΩ/R2,
R2= 10kΩ/8.2 = 1.22kΩ
 

Gazza_AU

Member
Vka = Vref (1 + (R1/R2)) + (Iref x R1)

I am still trying to work out the basic math.

Thanks Jim, will try this:
R1 = R2 ((V/Vref) - 1)
 

Gazza_AU

Member
For the maths, use the 'typical' values in the spec.
Let's assume R1 = 10kΩ, Vref = 2.5V, Iref = 1.5uA.
You want Vka = 23V.
So,
23V= 2.5V*(1 + 10kΩ/R2) + 1.5uA*10kΩ ≈ 2.5V*(1 + 10kΩ/R2),
23V/2.5V = 1+10kΩ/R2,
9.2-1 = 10kΩ/R2,
R2= 10kΩ/8.2 = 1.22kΩ
The values have helped me try and get my head around, the current in equation threw me off.
I misunderstood the datasheet and was using 10ma in the sum :facepalm:

Is 1.5uA fixed across the Vka range, otherwise it is harder to calculate?

The TMS102A combines building blocks of a dual operational amplifier, a dual comparator, and a precision voltage reference.

It is a very handy IC for adjusting voltage variations to fit the full input range of Arduino.
Seems to be good value for the functions.

What is your budget for achieving a reference voltage accurate to within even 0.01% or better? Do you have a temperature-controlled environment for the circuit?
A bit like the length of string question.
I looked for single or dual op-amp with high adjustable reference,
TMS102A is the only one I could find with high reference,
the others were set at 2.5, 2.6 or 5v.

If you stumble across another op-amp with reference IC of similar specs, let me know!
Always room for increased accuracy for the right price.
I like to look for bargains under $2 AUD per IC.

Cheers
 
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Mikebits

Well-Known Member
Here's a thought. Let's just say you manage to build this ultra accurate reference. How to you intend to verify its accuracy?
 

Gazza_AU

Member
One of the complicating factors is the (Iref x R1) term which allows for the low value of reference pin current flowing in resistor R1.

If we ignore this term we can simplify and kick the expression around and get:

R1 = R2 ((V/Vref) - 1)

Say we want 23 volts, and let us make R2 = 1k Ohm

R1 = 1 x (23/2.5 - 1)

R1 = 8.2 k Ohm

Does that make things more understandable?

JimB
Yes, I was not using the correct Iref.
Your explanation of the math has put me on the right track.
Now the sum is adding up.

If we consider the effect of (Iref x R1), the output voltage will be changed by:

8.2x10^3 x 1.5x10^-6 = 12.3mV

JimB
Before asking on the forum, I was thinking Iref was calculated by the R1/R2 voltage division.
I still do not understand if the 1.5uA is fixed for R1 = 10kΩ across a variable R2 and Vka.


Thanks
 
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ChrisP58

Well-Known Member
Yes, I was not using the correct Iref.
Your explanation of the math has put me on the right track.
Now the sum is adding up.



Before asking on the forum, I was thinking Iref was calculated by the R1/R2 voltage division.
I still do not understand if the 1.5uA is fixed for R1 = 10kΩ across a variable R2 and Vka.


Thanks
Iref is the current flowing into the reference pin of the IC. It flows through R1, but not through R2.
 

Gazza_AU

Member
The TMS102A: http://www.ti.com/lit/ds/symlink/tsm102.pdf

Something in the datasheets is not adding up for me.
What is the Vcc overhead required for Vka 36v Reference source.
How can 36v Vka come from 36v Vcc, or am I missing something?
If it is the case that Vcc = Vka, what is the cost in efficiency or accuracy of IC.

I am sure it is in the datasheet somewhere?
info I have found does not seem to explain it fully.




 
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JimB

Super Moderator
Most Helpful Member
Having had a good look at the datasheet, it seems to be a case of ambiguous/careless technical writing.

For the whole integrated circuit:
The Absolute Maximum supply voltage is specified as 36v.
The Recommended maximum supply voltage is 30v.

For the reference section:
The "output" is specified as Vref to 36v.
Here the author has picked the AbsMax voltage, because the integrated circuit will work up to that voltage.

Be aware that the cathode of the voltage reference is supplied through a resistor which may be connected to a supply voltage which is different from the supply voltage which is used for the rest if the integrated circuit.

JimB
 

Gazza_AU

Member
Be aware that the cathode of the voltage reference is supplied through a resistor which may be connected to a supply voltage which is different from the supply voltage which is used for the rest if the integrated circuit.
JimB
Yes, I would ideally like to operate the IC at the highest Vka available through Vcc.
Leaving good head room for IC tolerance.

If you start with a high voltage needing interface with Arduino.
Say 100v Max.
Divide to the highest reliable Vka, to amplify accuracy at 5v level.
.4% @ 20v = .04% accuracy after voltage division to 2v.
This uses 2v example just to highlight the improved accuracy.

The final voltage on input of Arduino would still be of Max 5v.

I could be wrong with this basic logic, let me know.

Thanks
 
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