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Marko Buršič
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schematic

simulate this circuit – Schematic created using CircuitLab

It might be simpler to make a high precision current source, rather than measure the current. You could be using the same voltage reference for DAC,ADC and current reference. The instrumentation amplifier shall have all resistors with ultra low TCR.

EDIT:

Version 2. More complex, but AC:

schematic

simulate this circuit

You do use a transformer and triac phase control. At very low conducting angle we cold say:

$$sin \ \omega t\approx tan\ \omega t \approx \omega t$$

Therefore

$$V_{SMA}sin \ \omega t\approx V_{SMA} \omega t$$ $$R_{SMA}=\dfrac{\Delta V_{SMA}}{\Delta I_{SMA}} = \dfrac{V_{SMA} \omega T_1}{I_{SMA} \omega T_2}$$

Instead of using ADC, you could be using a capture input to measure pulse time coming from window comparator. The window shall be adjusted so that sine wave is at very end of cycle, let say 0.5ms before zero cross. You do control the triac and you fire at each second a short pulse, to take the reference reading. Then you measure times each cycle, you increase conduction angle and you compute the \$\Delta R/R\$.

enter image description here

$$V_{comp_V}\approx V_{ref}\omega T_{Vref}$$ $$V_{comp_V}\approx V\omega T_{V}$$ $$\dfrac{V}{V_{ref}}=\dfrac{V_{comp_V}\omega T_{Vref}}{V_{comp_V}\omega T_{V}} = \dfrac{ T_{Vref}}{T_{V}}$$ $$\dfrac{I}{I_{ref}}=\dfrac{V_{comp_I}\omega T_{Iref}}{V_{comp_I}\omega T_{I}} = \dfrac{ T_{Iref}}{T_{I}}$$

$$\dfrac{R}{R_{ref}}=\dfrac{V\cdot I_{ref}}{I\cdot V_{ref}} = \dfrac{ T_{Vref}}{T_{V}}\cdot \dfrac{ T_{I}}{T_{Iref}}$$

schematic

simulate this circuit – Schematic created using CircuitLab

It might be simpler to make a high precision current source, rather than measure the current. You could be using the same voltage reference for DAC,ADC and current reference. The instrumentation amplifier shall have all resistors with ultra low TCR.

EDIT:

Version 2. More complex, but AC:

schematic

simulate this circuit

You do use a transformer and triac phase control. At very low conducting angle we cold say:

$$sin \ \omega t\approx tan\ \omega t \approx \omega t$$

Therefore

$$V_{SMA}sin \ \omega t\approx V_{SMA} \omega t$$ $$R_{SMA}=\dfrac{\Delta V_{SMA}}{\Delta I_{SMA}} = \dfrac{V_{SMA} \omega T_1}{I_{SMA} \omega T_2}$$

Instead of using ADC, you could be using a capture input to measure pulse time coming from window comparator. The window shall be adjusted so that sine wave is at very end of cycle, let say 0.5ms before zero cross. You do control the triac and you fire at each second a short pulse, to take the reference reading. Then you measure times each cycle, you increase conduction angle and you compute the \$\Delta R/R\$.

$$V_{comp_V}\approx V_{ref}\omega T_{Vref}$$ $$V_{comp_V}\approx V\omega T_{V}$$ $$\dfrac{V}{V_{ref}}=\dfrac{V_{comp_V}\omega T_{Vref}}{V_{comp_V}\omega T_{V}} = \dfrac{ T_{Vref}}{T_{V}}$$ $$\dfrac{I}{I_{ref}}=\dfrac{V_{comp_I}\omega T_{Iref}}{V_{comp_I}\omega T_{I}} = \dfrac{ T_{Iref}}{T_{I}}$$

$$\dfrac{R}{R_{ref}}=\dfrac{V\cdot I_{ref}}{I\cdot V_{ref}} = \dfrac{ T_{Vref}}{T_{V}}\cdot \dfrac{ T_{I}}{T_{Iref}}$$

schematic

simulate this circuit – Schematic created using CircuitLab

It might be simpler to make a high precision current source, rather than measure the current. You could be using the same voltage reference for DAC,ADC and current reference. The instrumentation amplifier shall have all resistors with ultra low TCR.

EDIT:

Version 2. More complex, but AC:

schematic

simulate this circuit

You do use a transformer and triac phase control. At very low conducting angle we cold say:

$$sin \ \omega t\approx tan\ \omega t \approx \omega t$$

Therefore

$$V_{SMA}sin \ \omega t\approx V_{SMA} \omega t$$ $$R_{SMA}=\dfrac{\Delta V_{SMA}}{\Delta I_{SMA}} = \dfrac{V_{SMA} \omega T_1}{I_{SMA} \omega T_2}$$

Instead of using ADC, you could be using a capture input to measure pulse time coming from window comparator. The window shall be adjusted so that sine wave is at very end of cycle, let say 0.5ms before zero cross. You do control the triac and you fire at each second a short pulse, to take the reference reading. Then you measure times each cycle, you increase conduction angle and you compute the \$\Delta R/R\$.

enter image description here

$$V_{comp_V}\approx V_{ref}\omega T_{Vref}$$ $$V_{comp_V}\approx V\omega T_{V}$$ $$\dfrac{V}{V_{ref}}=\dfrac{V_{comp_V}\omega T_{Vref}}{V_{comp_V}\omega T_{V}} = \dfrac{ T_{Vref}}{T_{V}}$$ $$\dfrac{I}{I_{ref}}=\dfrac{V_{comp_I}\omega T_{Iref}}{V_{comp_I}\omega T_{I}} = \dfrac{ T_{Iref}}{T_{I}}$$

$$\dfrac{R}{R_{ref}}=\dfrac{V\cdot I_{ref}}{I\cdot V_{ref}} = \dfrac{ T_{Vref}}{T_{V}}\cdot \dfrac{ T_{I}}{T_{Iref}}$$

edited body
Source Link
Marko Buršič
  • 24.5k
  • 2
  • 21
  • 34

schematic

simulate this circuit – Schematic created using CircuitLab

It might be simpler to make a high precision current source, rather than measure the current. You could be using the same voltage reference for DAC,ADC and current reference. The instrumentation amplifier shall have all resistors with ultra low TCR.

EDIT:

Version 2. More complex, but AC:

schematicschematic

simulate this circuitsimulate this circuit

You do use a transformer and triac phase control. At very low conducting angle we cold say:

$$sin \ \omega t\approx tan\ \omega t \approx \omega t$$

Therefore

$$V_{SMA}sin \ \omega t\approx V_{SMA} \omega t$$ $$R_{SMA}=\dfrac{\Delta V_{SMA}}{\Delta I_{SMA}} = \dfrac{V_{SMA} \omega T_1}{I_{SMA} \omega T_2}$$

Instead of using ADC, you could be using a capture input to measure pulse time coming from window comparator. The window shall be adjusted so that sine wave is at very end of cycle, let say 0.5ms before zero cross. You do control the triac and you fire at each second a short pulse, to take the reference reading. Then you measure times each cycle, you increase conduction angle and you compute the \$\Delta R/R\$.

$$V_{comp_V}\approx V_{ref}\omega T_{Vref}$$ $$V_{comp_V}\approx V\omega T_{V}$$ $$\dfrac{V}{V_{ref}}=\dfrac{V_{comp_V}\omega T_{Vref}}{V_{comp_V}\omega T_{V}} = \dfrac{ T_{Vref}}{T_{V}}$$ $$\dfrac{I}{I_{ref}}=\dfrac{V_{comp_I}\omega T_{Iref}}{V_{comp_I}\omega T_{I}} = \dfrac{ T_{Iref}}{T_{I}}$$

$$\dfrac{R}{R_{ref}}=\dfrac{V\cdot I_{ref}}{I\cdot V_{ref}} = \dfrac{ T_{Vref}}{T_{V}}\cdot \dfrac{ T_{I}}{T_{Iref}}$$

schematic

simulate this circuit – Schematic created using CircuitLab

It might be simpler to make a high precision current source, rather than measure the current. You could be using the same voltage reference for DAC,ADC and current reference. The instrumentation amplifier shall have all resistors with ultra low TCR.

EDIT:

Version 2. More complex, but AC:

schematic

simulate this circuit

You do use a transformer and triac phase control. At very low conducting angle we cold say:

$$sin \ \omega t\approx tan\ \omega t \approx \omega t$$

Therefore

$$V_{SMA}sin \ \omega t\approx V_{SMA} \omega t$$ $$R_{SMA}=\dfrac{\Delta V_{SMA}}{\Delta I_{SMA}} = \dfrac{V_{SMA} \omega T_1}{I_{SMA} \omega T_2}$$

Instead of using ADC, you could be using a capture input to measure pulse time coming from window comparator. The window shall be adjusted so that sine wave is at very end of cycle, let say 0.5ms before zero cross. You do control the triac and you fire at each second a short pulse, to take the reference reading. Then you measure times each cycle, you increase conduction angle and you compute the \$\Delta R/R\$.

$$V_{comp_V}\approx V_{ref}\omega T_{Vref}$$ $$V_{comp_V}\approx V\omega T_{V}$$ $$\dfrac{V}{V_{ref}}=\dfrac{V_{comp_V}\omega T_{Vref}}{V_{comp_V}\omega T_{V}} = \dfrac{ T_{Vref}}{T_{V}}$$ $$\dfrac{I}{I_{ref}}=\dfrac{V_{comp_I}\omega T_{Iref}}{V_{comp_I}\omega T_{I}} = \dfrac{ T_{Iref}}{T_{I}}$$

$$\dfrac{R}{R_{ref}}=\dfrac{V\cdot I_{ref}}{I\cdot V_{ref}} = \dfrac{ T_{Vref}}{T_{V}}\cdot \dfrac{ T_{I}}{T_{Iref}}$$

schematic

simulate this circuit – Schematic created using CircuitLab

It might be simpler to make a high precision current source, rather than measure the current. You could be using the same voltage reference for DAC,ADC and current reference. The instrumentation amplifier shall have all resistors with ultra low TCR.

EDIT:

Version 2. More complex, but AC:

schematic

simulate this circuit

You do use a transformer and triac phase control. At very low conducting angle we cold say:

$$sin \ \omega t\approx tan\ \omega t \approx \omega t$$

Therefore

$$V_{SMA}sin \ \omega t\approx V_{SMA} \omega t$$ $$R_{SMA}=\dfrac{\Delta V_{SMA}}{\Delta I_{SMA}} = \dfrac{V_{SMA} \omega T_1}{I_{SMA} \omega T_2}$$

Instead of using ADC, you could be using a capture input to measure pulse time coming from window comparator. The window shall be adjusted so that sine wave is at very end of cycle, let say 0.5ms before zero cross. You do control the triac and you fire at each second a short pulse, to take the reference reading. Then you measure times each cycle, you increase conduction angle and you compute the \$\Delta R/R\$.

$$V_{comp_V}\approx V_{ref}\omega T_{Vref}$$ $$V_{comp_V}\approx V\omega T_{V}$$ $$\dfrac{V}{V_{ref}}=\dfrac{V_{comp_V}\omega T_{Vref}}{V_{comp_V}\omega T_{V}} = \dfrac{ T_{Vref}}{T_{V}}$$ $$\dfrac{I}{I_{ref}}=\dfrac{V_{comp_I}\omega T_{Iref}}{V_{comp_I}\omega T_{I}} = \dfrac{ T_{Iref}}{T_{I}}$$

$$\dfrac{R}{R_{ref}}=\dfrac{V\cdot I_{ref}}{I\cdot V_{ref}} = \dfrac{ T_{Vref}}{T_{V}}\cdot \dfrac{ T_{I}}{T_{Iref}}$$

added 194 characters in body
Source Link
Marko Buršič
  • 24.5k
  • 2
  • 21
  • 34

schematic

simulate this circuit – Schematic created using CircuitLab

It might be simpler to make a high precision current source, rather than measure the current. You could be using the same voltage reference for DAC,ADC and current reference. The instrumentation amplifier shall have all resistors with ultra low TCR.

EDIT:

Version 2. More complex, but AC:

schematic

simulate this circuit

You do use a transformer and triac phase control. At very low conducting angle we cold say:

$$sin \ \omega t\approx tan\ \omega t \approx \omega t$$

Therefore

$$V_{SMA}sin \ \omega t\approx V_{SMA} \omega t$$ $$R_{SMA}=\dfrac{\Delta V_{SMA}}{\Delta I_{SMA}} = \dfrac{V_{SMA} \omega T_1}{I_{SMA} \omega T_2}$$

Instead of using ADC, you could be using a capture input to measure pulse time coming from window comparator. The window shall be adjusted so that sine wave is at very end of cycle, let say 0.5ms before zero cross. You do control the triac and you fire at each second a short pulse, to take the reference reading. Then you measure times each cycle, you increase conduction angle and you compute the \$\Delta R/R\$.

$$V_{comp_V}\approx V_{ref}\omega T_{Vref}$$ $$V_{comp_V}\approx V\omega T_{V}$$ $$\dfrac{V}{V_{ref}}=\dfrac{V_{comp_V}\omega T_{Vref}}{V_{comp_V}\omega T_{V}} = \dfrac{ T_{Vref}}{T_{V}}$$ $$\dfrac{I}{I_{ref}}=\dfrac{V_{comp_I}\omega T_{Iref}}{V_{comp_I}\omega T_{I}} = \dfrac{ T_{Iref}}{T_{I}}$$

$$\dfrac{R}{R_{ref}}=\dfrac{V\cdot I_{ref}}{I\cdot V_{ref}} = \dfrac{ T_{Vref}}{T_{V}}\cdot \dfrac{ T_{I}}{T_{Iref}}$$

schematic

simulate this circuit – Schematic created using CircuitLab

It might be simpler to make a high precision current source, rather than measure the current. You could be using the same voltage reference for DAC,ADC and current reference. The instrumentation amplifier shall have all resistors with ultra low TCR.

EDIT:

Version 2. More complex, but AC:

schematic

simulate this circuit

You do use a transformer and triac phase control. At very low conducting angle we cold say:

$$sin \ \omega t\approx tan\ \omega t \approx \omega t$$

Therefore

$$V_{SMA}sin \ \omega t\approx V_{SMA} \omega t$$ $$R_{SMA}=\dfrac{\Delta V_{SMA}}{\Delta I_{SMA}} = \dfrac{V_{SMA} \omega T_1}{I_{SMA} \omega T_2}$$

Instead of using ADC, you could be using a capture input to measure pulse time coming from window comparator. The window shall be adjusted so that sine wave is at very end of cycle, let say 0.5ms before zero cross. You do control the triac and you fire at each second a short pulse, to take the reference reading. Then you measure times each cycle, you increase conduction angle and you compute the \$\Delta R/R\$.

$$V_{comp_V}\approx V_{ref}\omega T_{Vref}$$ $$V_{comp_V}\approx V\omega T_{V}$$ $$\dfrac{V}{V_{ref}}=\dfrac{V_{comp_V}\omega T_{Vref}}{V_{comp_V}\omega T_{V}} = \dfrac{ T_{Vref}}{T_{V}}$$ $$\dfrac{I}{I_{ref}}=\dfrac{V_{comp_I}\omega T_{Iref}}{V_{comp_I}\omega T_{I}} = \dfrac{ T_{Iref}}{T_{I}}$$

schematic

simulate this circuit – Schematic created using CircuitLab

It might be simpler to make a high precision current source, rather than measure the current. You could be using the same voltage reference for DAC,ADC and current reference. The instrumentation amplifier shall have all resistors with ultra low TCR.

EDIT:

Version 2. More complex, but AC:

schematic

simulate this circuit

You do use a transformer and triac phase control. At very low conducting angle we cold say:

$$sin \ \omega t\approx tan\ \omega t \approx \omega t$$

Therefore

$$V_{SMA}sin \ \omega t\approx V_{SMA} \omega t$$ $$R_{SMA}=\dfrac{\Delta V_{SMA}}{\Delta I_{SMA}} = \dfrac{V_{SMA} \omega T_1}{I_{SMA} \omega T_2}$$

Instead of using ADC, you could be using a capture input to measure pulse time coming from window comparator. The window shall be adjusted so that sine wave is at very end of cycle, let say 0.5ms before zero cross. You do control the triac and you fire at each second a short pulse, to take the reference reading. Then you measure times each cycle, you increase conduction angle and you compute the \$\Delta R/R\$.

$$V_{comp_V}\approx V_{ref}\omega T_{Vref}$$ $$V_{comp_V}\approx V\omega T_{V}$$ $$\dfrac{V}{V_{ref}}=\dfrac{V_{comp_V}\omega T_{Vref}}{V_{comp_V}\omega T_{V}} = \dfrac{ T_{Vref}}{T_{V}}$$ $$\dfrac{I}{I_{ref}}=\dfrac{V_{comp_I}\omega T_{Iref}}{V_{comp_I}\omega T_{I}} = \dfrac{ T_{Iref}}{T_{I}}$$

$$\dfrac{R}{R_{ref}}=\dfrac{V\cdot I_{ref}}{I\cdot V_{ref}} = \dfrac{ T_{Vref}}{T_{V}}\cdot \dfrac{ T_{I}}{T_{Iref}}$$

added 194 characters in body
Source Link
Marko Buršič
  • 24.5k
  • 2
  • 21
  • 34
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edited body
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Marko Buršič
  • 24.5k
  • 2
  • 21
  • 34
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edited body
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Marko Buršič
  • 24.5k
  • 2
  • 21
  • 34
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edited body
Source Link
Marko Buršič
  • 24.5k
  • 2
  • 21
  • 34
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edited body
Source Link
Marko Buršič
  • 24.5k
  • 2
  • 21
  • 34
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Source Link
Marko Buršič
  • 24.5k
  • 2
  • 21
  • 34
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