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schematic

simulate this circuit – Schematic created using CircuitLab

This circuit lets me adjust the output voltage of a LDO further by varying V2 via a DAC. I am trying to understand how it works and why this works. Also, how do I come up with the equation that will determine my output voltage based on the value of V2. I am using the LT1965 that has a 1.2V adj.

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  • \$\begingroup\$ V2 biases the resistor network. \$\endgroup\$ – Ignacio Vazquez-Abrams Jan 12 '17 at 0:57
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I will leave the math to you as this is a great exercise in understanding the basics of power electronics. But, I'll give you the flow of what's going on.

The LM317's regulation loop is going to attempt to maintain some voltage (Vref) at the ADJ pin. It does this by comparing this voltage against an internal reference. For the time being, perform the analysis without R3 and V2. The part's job in life is to provide whatever voltage at OUT it needs to ensure that Vadj == Vref. Knowing the regulation voltage of the ADJ pin and the value of R1, you know the current flowing in R2. Vadj + Ir2 * R2 = Vout.

Now, understand that V2 and R3 are just providing some amount of current into (or out of) the ADJ pin's node. If V2 = Vadj, then the circuit performs just as if there were no R3 connection. If V2 goes above Vadj, then current is added into the ADJ node from this connection. Since that current also goes through R1, the control loop has to decrease Vout to accommodate this additional current. If V2 goes below Vadj, current is pulled out of the ADJ node, and Vout must increase to provide the additional current through R1. As you write the nodal equations, just remember that the feedback loop of the regulator is just trying to hold the ADJ pin at Vref.

Note that this is just the first order analysis and there can be stability and transient issues with approaches like this, but this exercise should greatly increase your understanding of this circuit.

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  • \$\begingroup\$ ah make sense, so I did the node analysis. I see that because the current that is going through R1 is consistent based on the value of R1 I choose. R2 current adjust based on extra or less current from V2. Therefor the voltage at the output has to increase to account for the extra current needed or decrease for the extra current when V2 is sourcing it. \$\endgroup\$ – J. Jones Jan 12 '17 at 17:06

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