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I'm working on a design that uses two separate buck regulators to generate a positive and negative voltage, and I'm trying to implement current limiting with a current sensor. However, I am running into issues with the way the sense resistor is integrated into the feedback loop as shown in my block diagram below:

schematic

simulate this circuit – Schematic created using CircuitLab

With the way the regulators are set up, the current sense resistors have to be included in the appropriate feedback loops in order to maintain the correct output voltage. Otherwise, the regulator won't take into account the voltage drop, and the output voltages will be less than what is expected.

The issue is with the way the two regulators are configured. When you add the sense resistor to the negative voltage circuit, it seems to create a different 'return node' for the negative voltage. If it wasn't there, both regulators can be tied to the ground node at their returns. What other methods of current sensing can be used for the negative voltage circuit such that the appropriate returns can be tied together?

EDIT: As Hacktastical said, there is a way to have a buck regulator output a negative voltage. I based the reference circuit off of an app note from TI:

Positive to Negative Buck-Boost Converter

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  • \$\begingroup\$ You have much bigger problems. Your circuit is fundamentally flawed. You can't use positive regulators as negative ones. At least half your circuit needs to be reworked with different components Your bottom LM2673 IC is a dead end for current as far as V1 is concerned. You need to review the basics. \$\endgroup\$ – DKNguyen May 16 at 20:47
  • \$\begingroup\$ The circuit is fundamentally incorrect and the question cannot be answered for this circuit. Open a new question about why your negative circuit won't work (or why you cannot wire a positive regulator to produce a negative voltage supply). \$\endgroup\$ – DKNguyen May 16 at 20:55
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    \$\begingroup\$ The only thing that's missing is a tie to GND on negative return. Otherwise it's totally feasable to use a positive step-down as an inverter. See here: ti.com/lit/ml/szzn001/szzn001.pdf \$\endgroup\$ – hacktastical May 16 at 21:30
  • \$\begingroup\$ And here's an appnote for making an inverter from this exact part: ti.com/lit/an/snva022e/snva022e.pdf? \$\endgroup\$ – hacktastical May 16 at 21:37
  • \$\begingroup\$ Please always point the ground symbol towards, um, the ground. i.e., pointing down. \$\endgroup\$ – Transistor May 16 at 22:07
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The schematic you show will most certainly NOT work in the way you want.

You are adding a series resistor into the output and expecting the feedback loop to offset the effect of it. The chip does not have enough response to be able to do this.

The LM2673S already has short circuit protection built in, please read the datasheet.

enter image description here

The current protection is NOT in the output circuit, it limits the current through the switch and therefore the peak current that flow in the inductor. This can be set to provide just enough energy into (c1,C4) and (C2,C3) to allow the PWM to maintain your required output voltage. If your output current exceed a level that can be maintained by the maximum PWM on time, then the voltage will begin to fall.

You must also remember that in the case of a fault current (ie a sudden short circuit on the output) the current is defined by the output capacitors. They will deliver a large pulse of current under these conditions.

Note: You should also include a capacitor from the Vin to -Vout as shown in the application note on using the LM2673 for negative supplies. Without this capacitor, startup will be distorted.

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  • \$\begingroup\$ Thanks for your reply. While the drop in voltage is a good sign that my current is reaching its limit, I am trying to incorporate a visual indicator of sorts such that it would be obvious. That's why the sense resistor is there. I can measure the voltage across them and use that to light up an LED. Perhaps my method is not good in this case? \$\endgroup\$ – BestQualityVacuum May 18 at 18:47
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Without going into too much detail, your approach violates the ESR requirements with an external sense R and also that is redundant with the internal Rs. There is a separate port to provide gain Ri for the OCP threshold with internal current sensing.

If you want to sense current on your own, use a 2~ 10 mOhm max ground sense resistor and high gain balanced Diff Amp with low ESL and a good layout to reject CM noise.

Total ESR on the load cap plus Rsense must be between 10 and 130 mOhm (?) from memory. So your 1 Ohm will result in instability and poor load regulation.

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  • \$\begingroup\$ Thank you for your reply. I see what you mean with the sense resistor. If you don't me asking, what resources can I look into to better understand the relationship between ESR and design? Perhaps this design approach is unsuited for my needs given the output load connection and my need for current sensing. \$\endgroup\$ – BestQualityVacuum May 17 at 3:31
  • \$\begingroup\$ Any search for low side Current sensing ... without violating the datasheet ESR criteria. also note this part is replaced with an improved version with a C suffix. \$\endgroup\$ – Tony Stewart Sunnyskyguy EE75 May 17 at 13:21
  • \$\begingroup\$ Thanks for your reply. I looked at TI's catalog, but I don't see an LM2673-C variant. Considering how the sense resistor is connected to the higher potential wrt their respective regulator, wouldn't they both be high-side current sensing at this point? Putting aside the much needed fixing, given how I want to connect my supplies to my load as seen by my connector, is high-side current sensing not feasible? The GND is shared at my load, and current is flowing from POS_OUT to GND and GND to NEG_OUT on my board based on the potentials. \$\endgroup\$ – BestQualityVacuum May 17 at 14:36
  • \$\begingroup\$ Forgive me, it was another "Simple Switcher" \$\endgroup\$ – Tony Stewart Sunnyskyguy EE75 May 17 at 20:53
  • \$\begingroup\$ No, It's Low Side (0V rail) ti.com/lit/sboa190, For the negative rail, you just swap the inputs to convert to a positive value. I suggest 50mV max for you and ALL current sense R's , but you can go smaller or slightly more in cases where ESR does not matter. \$\endgroup\$ – Tony Stewart Sunnyskyguy EE75 May 17 at 21:02
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I think what you've shown is fine. What will happen is, when there is more IR drop on NEG_SENSE, this will cause Vf to shift down, making the regulator try to compensate by increasing its output voltage. NEG_V increases as a result.

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  • \$\begingroup\$ Thanks for your reply. I posted in my schematic the load connection that I intend to use it for. You can see how it has two pins for positive/negative voltage and one pin for GND. I can't connect the 'returns' of each regulator together to GND for the output load connection. If I connect NEG_RETURN to GND, it ignores the NEG_SENSE resistor since GND is on both sides. \$\endgroup\$ – BestQualityVacuum May 17 at 3:27

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