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I am using the LM5069MM-2 from Ti for a project. This project requires an over current protection limit. The LM5069 can do this and the data sheet shows how. But I don't understand the process Ti uses to determine the effective sense resistance.

Here is the problem: The application current limit must be 3.8A. To set this current limit a sense resistor, Rsns, is selected. The selection of Rsns is based on the minimum threshold voltage (48.5mV) the SENSE pin of the LM5069 reads before it initiates the over current protection and the desired current limit:

Rsns = 48.5mV / 3.8A = 12.76mΩ

But sense resistors only come in whole value increments 11mΩ, 12mΩ, 13mΩ... So what we can do is reduce the effective resistance of the next larger sized resistor i.e 13mΩ to get an effective resistance of 12.76mΩ.

Here is where the confusion begins to take place:

How are they determining the values of R1 and R2? What theory are they using? The way they are doing it does not make sense to me. They just find a ratio of R1/R2 based of the sense resistor value I need verses the one I have. Then they apply a voltage divider-like equation to get an effective resistance. Can someone please shine some light here?

Here is a snippet from the data sheet that describes their process for finding the effective resistance. enter image description here

Thank you for your time,

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The key here is that 13 mOhm (used by the sense resistor) is very small compared to the resistances used in the divider, allowing some approximations to be made. Additionally, the key is to understand that this resistor network is not a network with equivalent resistance 12.76 mOhm. It's simply a network that will control the LM5069 in the same way that a simple 12.76 mOhm resistor would have.

To start, consider the circuit for some arbitrary current I flowing through the power bus with the sense resistor alone, and perform a conversion to an equivalent circuit. It's easiest to draw it as a Norton equivalent:

schematic

simulate this circuit – Schematic created using CircuitLab

We then perform a Norton/Thévenin conversion. We can approximate the equivalent circuit of the power bus and \$R_{SNS}\$ alone as a Thévenin equivalent with 13 mOhm Thévenin impedance, and with Thévenin equivalent voltage equal to 13 mOhm * I. Here's the circuit converted to its Thévenin equivalent, with the resistor divider attached.

schematic

simulate this circuit

Because Rth (13 mOhm) is much smaller than R1/R2, the voltage drop across it is negligible, and nearly all \$I\cdot13\text{m}\Omega\$ of voltage is applied across the divider. The divider's output voltage is approximately \$I\cdot13\text{m}\Omega\frac{R1}{R1+R2}\$, which is the same as the voltage that you would get if you had a \$13\frac{R1}{R1+R2}\$ milliohm resistor to begin with.

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  • \$\begingroup\$ I really appreciate you taking the time to answer @nanodarad. But I still do not understand how 13mΩ is converted into 12.76mΩ by a resistor divider. Maybe I am missing some core understanding. Thevinin and Norton have never been a strong suite for me. Are these concepts essential to answering my question? \$\endgroup\$
    – Tim51
    May 8 '20 at 20:24
  • \$\begingroup\$ @Tim51 They're essential to the derivation, but not to the conclusion. The point isn't that you make a device that is necessarily a 12.76 mOhm resistor. The point is that you make a circuit, whose voltage is related to the current in the same way the voltage and current are related for 12.76 mOhm resistor, so that it behaves like a 12.76 mOhm resistor for the purposes of designing this circuit. Don't read too far into it "being" a 12.76 mOhm resistor. It's not. It's a 13 mOhm resistor with a few more resistors attached, such that it has the right behavior when used with this particular chip. \$\endgroup\$
    – nanofarad
    May 8 '20 at 20:30
  • \$\begingroup\$ Its the derivation that I would like to understand. What's the difference between using Thev/Nort and just creating a parallel/series resistor network? I thought that's what they were doing at first, but the math never worked out. @nanofarad \$\endgroup\$
    – Tim51
    May 8 '20 at 20:43
  • \$\begingroup\$ @Tim51 The math for parallel/series should work out to give you the correct output voltage across the SENSE and VIN pins. Again, it will not simplify into a 12.76 mOhm resistor, because the statement that Rsns,eff=12.76 mOhm doesn't imply as much as you think it does. It only means that this circuit's current limiting feature will behave as if an effective resistance of 12.76 mOhm was used for the sense resistor. The use of Thev/Nort is simply a way of quickly doing the analysis to find the effective output voltage. \$\endgroup\$
    – nanofarad
    May 8 '20 at 20:46
  • \$\begingroup\$ Interesting. When I performed the series/parallel equations I got 12.99mΩ for Rsns. So I assume that's where you're saying the circuit will behave like a 12.76mΩ resistor is used for the sense resistor but the actual value for the resistor is 12.99mΩ? @nanofarad \$\endgroup\$
    – Tim51
    May 8 '20 at 21:11

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