WEBENCH® Tools/LDC1000EVM: How LDC1000 measure parallel resistance of LC tank

Hello Viktor,

The LDC measures Rp by alternating two different current drives, which correspond to RPMIN and RPMAX. The LDC tracks the amount of time it drives the RPMIN current and the RPMAX current, and so it can determine the average current. The LDC also controls the sensor amplitude so that it averages out to a fixed level. By knowing the voltage, the current, and Georg Ohm, we can determine the Rp.

Note - you need to make sure that the triangle modulation pattern is occurring when measuring the Rp, as this corresponds to the alternating RPMIN and RPMAX values. For the LDC1000, you need to make sure the CFB cap is properly set, as instructed in the datasheet. I also strongly recommend using the LDC1101, as it is superior in many ways - better accuracy, better resolution, faster, smaller, and cheaper too.

Regards,

ChrisO

Thank you, Сhris!

I ask questions, because I want to understand the principles of the sensor. This is important, because we want to make a metal detection sensor ourselves. I will analyze your answer, then I'll ask you again.
Do not close the topic.

Hello!

I do not understand the principles to the end. Let's look at a concrete example.
Let:
1. I chose the amplitude of the 4 Volt peak-peak. Hence the positive period has an amplitude of 2 V.
2. Has chosen Rpmin = 1,026 kΩ and Rpmax = 7,182 kΩ
3. The current value of Rp should be 2 kΩ

How does it work? I understand this:
1. LDC calculates the current min. Imin = 2V / 7,182 kΩ = 0,28 mA;
2. LDC calculates the maximum current Imax = 2V / 1.026 kΩ = 1.95 mA;
3. The LDC has a linear rise time from Imin (or not?) To Imax and a linear reduction time to Imin.
4. Since during the modulation the loop impedance is 2 kΩ (as we have assumed), the modulated current passes through the well loop, then the voltage on the circuit will be: at a current of Imin (for Rpmax): 0.28mA * 2kΩ = 0.56 V and Imax(for Rpmin): 1.95mA * 2kΩ = 3.84 V.
3.84V can not be because voltage is limited to 2 V !!!
What am I doing wrong? How does the measurement go on?
I understand how to calculate the voltage by time - this is when it increases linearly, so you are talking about a triangle.
But I do not understand how current modulation, voltage increase, and average current calculation are carried out?

Please help me understand my example.

Below I indicated additional questions and explanations for my problem.
I do not understand the principles to the end. Let's look at a concrete example.
Let:
1. I chose the amplitude of the 4 Volt peak-peak. Hence the positive period has an amplitude of 2 V.
2. Has chosen Rpmin = 1,026 kΩ and Rpmax = 7,182 kΩ
3. The current value of Rp should be 2 kΩ

How does it work? I understand this:
1. LDC calculates the current min. Imin = 2V / 7,182 kΩ = 0,28 mA;
2. LDC calculates the maximum current Imax = 2V / 1.026 kΩ = 1.95 mA;
3. The LDC has a linear rise time from Imin (or not?) To Imax and a linear reduction time to Imin.
4. Since during the modulation the loop impedance is 2 kΩ (as we have assumed), the modulated current passes through the well loop, then the voltage on the circuit will be: at a current of Imin (for Rpmax): 0.28mA * 2kΩ = 0.56 V and Imax(for Rpmin): 1.95mA * 2kΩ = 3.84 V.
3.84V can not be because voltage is limited to 2 V !!!
What am I doing wrong? How does the measurement go on?
I understand how to calculate the voltage by time - this is when it increases linearly, so you are talking about a triangle.
But I do not understand how current modulation, voltage increase, and average current calculation are carried out?

Hi Victor,

There is a scaling factor on the voltage due to the AC signal present. This scaling factor reduces the voltage to one that can be supported by the LDC. The actual current drives are listed in the datasheet.

The modulation (driving RPMIN/RPMAX current drives) is done over multiple cycles of the sensor, and so there is a triangle wave modulation applied to the sensor's signal amplitude.

Regards,

ChrisO