TIDA-01513 Isolation monitoring [SOLVED]
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TIDA-01513 Isolation monitoring
I cant understand how those equations were derrived in this page.This is from the tida-01513 reference design datasheet.Im stuck at page 8,9 and 10 and I cant understand whats happening and how the equations were derrived in here.Here is a picture of page 9.Can anyone explain how those equations were derrived and what isi the closed loop current path of the circuit.Does the current flow through the positive op amp input terminals and comes bach through negative side op amp input terminals?and how the ISO_POS and ISO_NEG equations were derrived?
- tom91
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Re: TIDA-01513 Isolation monitoring
Please link the relevant document and not just the page.
What is the reason you are trying to under stand this? Are you looking to build this circuit yourself?
I have looked at this exact whitepaper a few years back but never got around to taking it further then redrawing the schematic and doing more research.
What is the reason you are trying to under stand this? Are you looking to build this circuit yourself?
I have looked at this exact whitepaper a few years back but never got around to taking it further then redrawing the schematic and doing more research.
- uhi22
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Re: TIDA-01513 Isolation monitoring
- You start with the formula of an inverting op-amp. The gain is defined by the ratio of the feedback resistor and the serial resistor.
- For simplification, let's assume the circuit is symmetrical, means the measurement circuits of the HV+ and HV- have the same resistor values. This gives the possibility, to calculate only the half of the circuit. Imagine, to connect the middle point of the battery to the Vref.
- Now you have a simple inverting amplifier, that has a input voltage of U_HV/2.
- The formulas in the application note are missing the factor 2 on the first look. But they are right, because they just use the double resistance in the lower part of the equation, so they come to the same result.
- For simplification, let's assume the circuit is symmetrical, means the measurement circuits of the HV+ and HV- have the same resistor values. This gives the possibility, to calculate only the half of the circuit. Imagine, to connect the middle point of the battery to the Vref.
- Now you have a simple inverting amplifier, that has a input voltage of U_HV/2.
- The formulas in the application note are missing the factor 2 on the first look. But they are right, because they just use the double resistance in the lower part of the equation, so they come to the same result.
Github: http://github.com/uhi22 --- Patreon: https://www.patreon.com/uhi22
Re: TIDA-01513 Isolation monitoring
There is an addendum that maybe more helpful if you intend to implement the isolation monitor.
'TIDA035 Addendum to TIDA-01513 For Two-point Insulation
Resistance Measurements'
I have implemented this isolation monitor including code, happy to share the design.
'TIDA035 Addendum to TIDA-01513 For Two-point Insulation
Resistance Measurements'
I have implemented this isolation monitor including code, happy to share the design.
- uhi22
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Re: TIDA-01513 Isolation monitoring [SOLVED]
Or, an different approach of thinking, which comes to the same conclusion:
- The minus inputs of both op-amps are exactly tied to the vref, because of the op-amp behavior. (It controls the output always in a way, that the voltage difference between + and - is zero.)
- This means, at the input side, we have a simple circuit consisting of a voltage supply (U_HV) and a resistance which consists of all series resistances.
- We can calculate the current in this input circuit: I = U / R = U_HV / (sum of all series resistors).
- Using this current, we can calculate the voltage drop on the feedback resistors: U_feedbackResistor = I * R_feedbackresistor.
- And add the reference voltage, to get the output voltage.
- The minus inputs of both op-amps are exactly tied to the vref, because of the op-amp behavior. (It controls the output always in a way, that the voltage difference between + and - is zero.)
- This means, at the input side, we have a simple circuit consisting of a voltage supply (U_HV) and a resistance which consists of all series resistances.
- We can calculate the current in this input circuit: I = U / R = U_HV / (sum of all series resistors).
- Using this current, we can calculate the voltage drop on the feedback resistors: U_feedbackResistor = I * R_feedbackresistor.
- And add the reference voltage, to get the output voltage.
Github: http://github.com/uhi22 --- Patreon: https://www.patreon.com/uhi22