AMC1301: does not getting Example calculation answer, not able to understand how this CHIP calculates temperature for thermocouple, Need more assistance.

Harit Shukla

Part Number: AMC1301

HI,

I am developing "Isolated Thermocouple Temperature measurement " using the given part , My application is to measure (-40C to +1200C) temperature. i was just going through the App note of TI "Thermocouple Temperature Measurements Using Isolated Amplifiers". and i am interested to develop the similar logic circuit as "Figure 5-Gain Stage with AMC1301" on "page-2" , i was just reading and calculating the given examples in the left side paragraph about how this IC will convert temperature in "mV" and after that I also understand how ICs Gain adjusted. My questions are,

1) How to calculate Final output voltage of "AMC1301" at (TH = 1250°C, TCJ = –40°C), i need full explanation if possible, as i am calculating the given example on Right side Paragraph 2.05 V (TH = 1250°C, TCJ = –40°C).

2) i am not getting the answer for the right next example of it which is "–449.6 mV (TH = –40°C, TCJ = 125°C)" , i have tried same method as above example but not getting -449.6mv".

3) so if possible please give me full explanation how you calculate the Final output voltage , (Any other app note or document will also good).

4) i dont understand what is "common-mode output voltage of 1.44 V" in the same example.

explained answers will be really appreciable.

Thanks

Harit S.

over 5 years ago

0 Alexander Smith over 5 years ago

TI__Guru 55342 points

Hi Harit,

The AMC1301 itself will not convert from mV to temperature by itself as it is simply an isolation amplifier. An ADC and microcontroller will be necessary to further process the data.

1.3. 4. Figure 24. Output Voltage vs Input Voltage in the AMC1301 shows the relationship between input and output voltage for the AMC1301. The input range is +/-250mV, with a fixed internal gain of 8.2. This yields a differential output voltage of 2.05V for a 250mV input. The output common-mode voltage is the center-point of the differential output of the AMC1301. If 0V is applied to the input, both outputs will be at 1.44V as shown in Figure 24. Output Voltage vs Input Voltage.

2.Can you share what you are calculating so I can help debug where your issue is?

0 Harit Shukla over 5 years ago in reply to Alexander Smith

Prodigy 70 points

Hi Alex,

I need to measure temperature range for –40°C to 1250°C from K type Thermocople for above given circuit.

I need all calculation form thermocouple sensor input to final output.

1) What will be final output voltage of AMC1301 when thermocouple measures –40°C temperature?

2)What will be final output voltage of AMC1301 when thermocouple measures 1200°C temperature?

3)What will be final output voltage of AMC1301 when thermocouple measures 0°C temperature?

Please provide all calculation so I'm able to calculate final output voltage of AMC1301 at different temperature.

4) Please explain me how this is calculated as shown in below image, "–449.6 mV (TH = –40°C, TCJ = 125°C)" . How in this -449.6mV is calculated?

At page No. 2 of below link: https://www.ti.com/lit/an/sbaa238a/sbaa238a.pdf?ts=1598341720316&ref_url=https%253A%252F%252Fwww.google.com%252F

Thanks & regards

Harit S.

0 Alexander Smith over 5 years ago in reply to Harit Shukla

TI__Guru 55342 points

Hi Harit,

Apologies for the delay. I'm having trouble as well - I'm checking with the Author of the document to verify if there's a typo somewhere.

0 Alexander Smith over 5 years ago in reply to Alexander Smith

TI__Guru 55342 points

Hi Harit,

I have verified with the author that the 449.6 mV is a typo. The correct calculation yields -260mV for the stated conditions.

The signal chain can be summarize as: Vtc -> gain stage (4.8) -> AMC1301 (8.2) -> Output

Where Vtc = S * (TH – TCJ); S = 0.00004.

Vtc = S * (TH – TCJ); XmV = 0.00004 * (1250 - -40); XmV = 52mV as expected.

52mV * 4.8 * 8.2 = 2.05V as shown in technote example.

Vtc = S * (TH – TCJ); XmV = 0.00004 * (-40-125); XmV = -6.6mV

-6.6mV * 4.8 * 8.2 = -260mV, not -449.6mV as shown in the tech note.

-449.6mV AMC1301 would require -11.4mV for Vtc. Assuming TH = -40C, TCJ would have to be 245C, not 125C.

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AMC1301: does not getting Example calculation answer, not able to understand how this CHIP calculates temperature for thermocouple, Need more assistance.