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I would like to have a buck converter that takes in a reference voltage, and outputs the same voltage.

This is assuming that I'm giving it a Vin that's high enough to support the output voltage that I want.

In other words, I'll give it a Vin of 12v or 24v. I'd like to give it a reference voltage between 0-10v and I'd like it to output the same as the reference voltage. In other words, I want to give it, say, a 5v reference, and have it output 5v.

All of the ones I've seen use a resistor divider hooked up to the FeedBack pin, and compare that to an internal reference voltage.

Are there any parts (buck converter IC's) available that will do this (I require an output of 10 amps).

My other thought is to use a comparator / op amp, and instead of using the resistors, compare the output with my reference voltage and feed the output of that into the FB pin. I'm not sure if that would create an oscillation since it's not expecting that much gain.

Is there a better way to do this?

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    \$\begingroup\$ Do you have access to an MCU? There are I2C/SPI programmable buck converters. \$\endgroup\$ Commented Nov 15, 2023 at 13:28
  • \$\begingroup\$ What maximum current output do you need? \$\endgroup\$ Commented Nov 15, 2023 at 13:33
  • \$\begingroup\$ Maximum 5 amps (I said 10 above to be conservative). \$\endgroup\$ Commented Nov 15, 2023 at 13:35
  • \$\begingroup\$ You need to find an IC or circuit with an option to use an external reference, then you can change it. \$\endgroup\$ Commented Nov 15, 2023 at 13:37
  • \$\begingroup\$ I would prefer RS485/MODBUS rather than I2C, but I’d rather keep it simple if possible. \$\endgroup\$ Commented Nov 15, 2023 at 13:38

4 Answers 4

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Probably more complex than you'd like but the LTC3886 can be digitally programmed (I2C). It has 12-bit digital programmability of output voltage of 2 to 13.8V in 4mV steps and 16-bit digital readback of output voltage and current. You can probably omit the components associated with the second output.

enter image description here

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  • \$\begingroup\$ If I was going to use a microcontroller I would just PWM the output and use a MOSFET, inductor, diode, and capacitor. The point of this question is to be able to control a load with a 0-10v analog signal. \$\endgroup\$ Commented Nov 15, 2023 at 14:53
  • \$\begingroup\$ You can just return the 'ground' side of the reference feedback divider to an op-amp output Vx. Eg. if the Vref is 1.25V and you have two equal resistors your output voltage would be Vo = 2.5-Vx so you'd want the op-amp to output -Vp-2.5V where Vp is your programming voltage. There may be some other details (such as clamping the feedback and finding suitable component values for the full range of output) but stability should not be (directly) affected since the op-amp is not in the feedback path. \$\endgroup\$ Commented Nov 15, 2023 at 15:10
  • \$\begingroup\$ Spehro Pefhany this seems useful but I don’t fully understand. I’ll have to do more research. \$\endgroup\$ Commented Nov 15, 2023 at 17:27
  • \$\begingroup\$ You would need a negative rail. \$\endgroup\$ Commented Nov 15, 2023 at 17:45
  • \$\begingroup\$ @user3338049 If you used an MCU with MOSFET, inductor, capacitor and diode, you'd quickly learn why we almost always regulate in an analogue loop, not a digital one with inherent delays. This IC uses an analogue control loop, with I²C controlled DAC to create a "reference" voltage, as you call it, to avoid those delays and the attendant instabilities. It is not at all the same as an "an MCU with MOSFET, inductor, capacitor and diode". \$\endgroup\$ Commented Jul 17 at 3:54
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Any linear or switching regulator with an accessible feedback can do this.

enter image description here

I didn't draw the regulator, but suppose Vout is the output voltage, R1 and R2 form the feedback divider, and the regulator aims to maintain voltage at node FB equal to its internal reference voltage.

If you inject current into this node, or connect it to an adjustable voltage (node Adj) via a resistor, then this will shift the output voltage accordingly. This has the advantage of not putting the regulator in an opamp's feedback loop, so no special compensation is required. The control voltage Adj will be inverted (Vout goes up when Adj goes down).

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  • \$\begingroup\$ Your solution does not have the converter output track the reference signal, but is only controlled by the reference signal. From the OP "I would like to have a buck converter that takes in a reference voltage, and outputs the same voltage." \$\endgroup\$ Commented Nov 16, 2023 at 1:58
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A circuit based upon the feedback circuit below may or may not suit your needs.

schematic

simulate this circuit – Schematic created using CircuitLab

The output of the feedback circuit \$V_{fb}\$ is given by the formula

$$V_{fb} = V_{out}-V_{ref}+V_{fb0}$$

\$V_{out}\$ is the output voltage of the DC-DC converter. \$V_{ref}\$ is the voltage you wish to track. \$V_{fb0}\$ is the feedback voltage expected by the converter chip when the output voltage is at its target value. \$V_{fb0}\$ is often \$1.250\$ V, but varies by type of chip.

The right hand "circuit" is just present for testing purposes.

It is possible that using this circuit for the feedback of a buck converter may require adjustments to the compensation of the converter's feedback loop. I really don't know the answer, so I am shining a spotlight to the clouds as a sort of Bat Signal in the hopes that VerbalKint sees it and can comment.

An alternative circuit is this.

schematic

simulate this circuit

The output of this circuit is

$$V_{fb} = \frac{V_{out}-V_{ref}}{2}+V_{fb0}$$

It has the advantage that the op-amp is not in the feedback loop of the DC-DC converter, or vice-versa, but is only in the path between \$V_{ref}\$ and the summing node. A disadvantage of this circuit is that it requires a negative supply rail for the op-amp, i.e. the output of the op-amp will be negative if \$V_{ref}\gt 2V_{fb0}\$.

Here is yet a third alternative circuit is this.

schematic

simulate this circuit

The output of this circuit is

$$V_{fb} = \frac{V_{out}-V_{ref}}{11} + V_{fb0}$$

This circuit has the property that the output of the op-amp is not negative as long as

$$11V_{fb0}>V_{ref}$$

Since your original specification is that \$V_{ref} \le 10\$ V, if \$V_{fb0} = 1.25\$ V, you will not need a negative supply for your op-amp.

[In previous answers where I used an op-amp with a inverting gain of absolute value less than 1, I received notes of concern in comments. To answer these concerns in advance, I refer to this discussion, especially the first response from Michael Steffes.]

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You can use a lt7200s. It has dedicated track pins that allow you to configure the internal reference voltage used for the feedback comparator. Simply set a resistor divider so your external reference voltage you'd like to track is within the 0v-0.5V range, then match that divider over to the feedback pin connection to the output voltage.

It's an expensive part, but it has 4 x 5A capable channels that are able to be phased 90 deg apart for a 20A 4 phase solution.

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