This thread has been locked.

If you have a related question, please click the "Ask a related question" button in the top right corner. The newly created question will be automatically linked to this question.

  • Resolved

LM358: Grounding issue on my rectifier design

Intellectual 305 points

Replies: 18

Views: 137

 Part Number: LM358

Hello 

I'm trying to implement a rectifier and AC voltage sensor. The rectifier part works just fine. I'm able to power my device without any issues.

Please look at C2 in the screenshot (Circuit schematic). This connector is connected to AC voltages. Vin is a rectifier and the DC part is to power the device. AC-1 to AC-5 are AC voltages that are sensed at ADC-0 to ADC-4 respectively. AC-6 is the neutral line of all the AC voltages connected to the board.

When AC (1-5) are connected to AC voltage, the device works just fine. No issues at all. But when AC-1 and AC-2 are connected to 120 VaAC while AC (3,4 and 5) are not connected to anything, ADC-0 and ADC-1 read the right value but ADC (2,3 & 4) read 7.5V while they should read 0 V.

I connected all the -ve terminals of the bridge rectifiers to have a common ground but when there is no AC voltage, the respective +ve terminals of the bridge rectifiers reads 58V (+ve is floating while -ve is grounded). Is there any way to fix this? cause my ADC is reading the incorrect value of 7.5V when it should read 0 V.

I was told to use an isolated amplifier with unity gain. Unless there's isn't a better solution, Can I use the following Opamp with unity gain to accomplish this:

https://www.digikey.com/product-detail/en/texas-instruments/LM358P/296-1395-5-ND/277042

If not do I have to use a specific isolated opamp like the following:

https://www.digikey.com/product-detail/en/texas-instruments/AMC1100DWVR/296-46067-2-ND/6571602

Thank you 

  • Hello User,

    The common neutral line does make it through the rectifier from that AC input to the + output even if the other AC input is open (not connected). This happens because your common ground is 'hot'

    If the common ground was the common neutral (AC6) instead and use a single diode (anode to AC IN and cathode to each resistor divider output) to get positive half of AC input then unconnected channels would be zero. The downside is that you are only looking at the input positive half of the waveform. This may be good enough.

    Alternatively you could lose the 5 input common grounds and measure the difference voltage between + and - for each of the five channels.

    Regards,
    Ronald Michallick
    Linear Applications

    TI assumes no liability for applications assistance or customer product design. Customer is fully responsible for all design decisions and engineering with regard to its products, including decisions relating to application of TI products. By providing technical information, TI does not intend to offer or provide engineering services or advice concerning Customer's design. If Customer desires engineering services, the Customer should rely on its retained employees and consultants and/or procure engineering services from a licensed professional engineer (LPE).

     

  • In reply to Ron Michallick:

    Hello Ron, 

    Thank you so much for your response. I considered half-wave rectification but had 2 issues. The capacitor needs to hold the voltage longer than before (twice as long). I could get by this issue, but I was told that isolation is necessary from the AC line. Since the isolation amplifier is mandatory, I thought I go with full-wave rectifications. I'm not pulling any current. It's just to read the voltage via ADC. If isolation is optional, then I could go with half-wave rectification. I agree that it's good enough (no isolation though). 

    All those AC lines are coming from the same module. They all have a common neutral line. I can't change this. 

    Thank you 

    Varun R

  • In reply to user6202586:

    Varun,

    What is the goal of the circuit. What part of the AC voltage do you need, average, peak, RMS (hardest), instantaneous value?

    Where does the output go (to other circuits that can share a potentially 'hot' ground or not)?

    If the output goes to something external or to a person then isolation from neutral and line will be required.

    Regards,
    Ronald Michallick
    Linear Applications

    TI assumes no liability for applications assistance or customer product design. Customer is fully responsible for all design decisions and engineering with regard to its products, including decisions relating to application of TI products. By providing technical information, TI does not intend to offer or provide engineering services or advice concerning Customer's design. If Customer desires engineering services, the Customer should rely on its retained employees and consultants and/or procure engineering services from a licensed professional engineer (LPE).

     

  • In reply to Ron Michallick:

    Hello Ron, 

    The goal of the circuit is to read the voltage in the line via the ADC's of my IoT device. This device is powered via the top part of the circuit and consumes 0.3-0.4 at 12V. Whenever the ADC's change by more than 10 V, I programmed it to send data via its modem to our a could server. 

    The +ve of the device is connected to the +ve of the bridge rectifier while the -ve is connected to the -ve of the bridge rectifier. That's why I had to connect all the -ve's of the bridge rectifiers to have a common ground for the device. 

    There is a module that is powered by the AC mains voltage. The 5 AC lines which go to the ADC come from different parts in the module. They turn OFF and ON depending on various scenarios which is what I'm trying to monitor using this board. Then this board is connected to the IoT device (Which I think is external relative to the board). So I'm guessing isolation is necessary?

  • In reply to user6202586:

    Varun,

    Would it be good enough to digitally detect (on or off) for the 5 AC lines. That could be done with a simple optocoupler that includes isolation and won't need external power.

     

    Regards,
    Ronald Michallick
    Linear Applications

    TI assumes no liability for applications assistance or customer product design. Customer is fully responsible for all design decisions and engineering with regard to its products, including decisions relating to application of TI products. By providing technical information, TI does not intend to offer or provide engineering services or advice concerning Customer's design. If Customer desires engineering services, the Customer should rely on its retained employees and consultants and/or procure engineering services from a licensed professional engineer (LPE).

     

  • In reply to Ron Michallick:

    Hello, 

    I thought about that. But my boss told me that it won't work. Hence I have to do it the hard way. Since you asked me this, I'm assuming I definitely need isolation. 

    I already have a few LM358 opamps with me to test. I could buy this one: https://www.digikey.com/product-detail/en/texas-instruments/TL074BCDR/296-14999-1-ND/562659

    It has 4 channels. I will use a combination of LM358 and TL074 to isolation all the AC inputs. I'll first perform half-wave rectification, then reduce the voltage via a voltage divider, then pass the voltage to the opamp. 

    I'll simulate it this weekend and share the results. Please let me know if I'm on the right track. 

    Thank you 

    Varun R 

  • In reply to user6202586:

    Varun,

    The half wave rectification and reduction sounds good. The GND point will be AC6 (common neutral). 

    Regards,
    Ronald Michallick
    Linear Applications

    TI assumes no liability for applications assistance or customer product design. Customer is fully responsible for all design decisions and engineering with regard to its products, including decisions relating to application of TI products. By providing technical information, TI does not intend to offer or provide engineering services or advice concerning Customer's design. If Customer desires engineering services, the Customer should rely on its retained employees and consultants and/or procure engineering services from a licensed professional engineer (LPE).

     

  • In reply to Ron Michallick:

    Hello 

    Just to be clear, there is no isolation required?

    I simulated the circuit in Tina with an opamp and got the result I wanted. I can power the opamp using the output of the transformer which itself has isolation. I have attached the circuit and simulation results to this message. 

    If you think the opamp is not necessary, let me know. I'll get rid of it. 

    Thank you 

    Varun R 

  • In reply to user6202586:

    Varun,

    Isolation will be needed. The neutral AC line ground must be treated as if it was hot.

    If you want average line voltage at a gain of 1/10 (for sine waves) then this would work. 120VAC in is 12VDC out

    HW1.TSC

    You also should distribute the line voltage across multiple resistors to lower the voltage each resistor has to endure. This lowers power dissipation and voltage modulation.

    Regards,
    Ronald Michallick
    Linear Applications

    TI assumes no liability for applications assistance or customer product design. Customer is fully responsible for all design decisions and engineering with regard to its products, including decisions relating to application of TI products. By providing technical information, TI does not intend to offer or provide engineering services or advice concerning Customer's design. If Customer desires engineering services, the Customer should rely on its retained employees and consultants and/or procure engineering services from a licensed professional engineer (LPE).

     

  • In reply to Ron Michallick:

    Hello Ron, 

    I ran into a problem when I practically implemented the circuit. I don't have any DC voltage here. The challenge is to power the opamp. I didn't think this would be a problem because, in my first schematic that I shared here, I have a rectifier that is working great. I thought I could power the opamp from this rectifier.   

    However, there is a grounding issue. The opamp power supply is to the ground of the bridge rectifier which is isolated from the AC mains by a step-down transformer. When I implemented the circuit using 24V DC, the opamp is not giving the voltage sensed by the voltage on its + pin. 

    In order for this circuit to work in the real world, I should get rid of the opamp. Or find a different way to isolate the circuit. I have attached the full simulation schematic to this message. 

    Please let me know if I missed something. 

    Varun Rfullwaverectifier.TSC

This thread has been locked.

If you have a related question, please click the "Ask a related question" button in the top right corner. The newly created question will be automatically linked to this question.