I am designing an adjustable power supply. Input voltage is 12V from adapter, and the output should be ranging from 0.5V to 12V.

I'm planning to use LM2575, but apparently its minimum output voltage is 1.23 based on datasheet. My concept is instead of using potentiometer to adjust feedback voltage from the LM2575, I use DAC MAX522 to create feedback-like voltage (this is possible right?).

Here is my current schematic enter image description here

The question is:

  1. is there any recommendation about the DC-DC step down regulator with minimum output 0.5V?
  2. To avoid wasting resource, will this schematic works?

Thanks in advance

  • \$\begingroup\$ What output current do you need? \$\endgroup\$
    – AngeloQ
    Commented Mar 21, 2017 at 3:53
  • \$\begingroup\$ I think in theory your concept could work, but I fear your response time will be too slow (ADC + MCU processing + SPI + DAC) and your regulator bandwidth may be too low. If you don't need good transient response it might be OK. I would probably add a cap on the feedback pin. There is probably a better way to get 0.5V output. But that's why asked what current you are looking for. \$\endgroup\$
    – AngeloQ
    Commented Mar 21, 2017 at 4:00
  • \$\begingroup\$ Just an idea on how to do this - instead of a DAC, you can PWM some pin of the MCU, then put two or three stages RC filter groups after it, in essence getting an analog signal. Or you can get a MCU with integrated DAC. Anyway, I whare AngeloQ's worries that your response time will be very slow. \$\endgroup\$
    – anrieff
    Commented Mar 21, 2017 at 4:13
  • \$\begingroup\$ The output current should be up to 1A. About the response time, I don't mind if its slow, because the voltage output will be controlled by push button (manual). As for the DAC chip, I'm using it in order to decrease the size of the PCB \$\endgroup\$
    – duck
    Commented Mar 21, 2017 at 4:22
  • \$\begingroup\$ @RohatKılıç on p.2 fig.1 of LM2575 datasheet, it seems that the feedback is on noninverting mode. I can't wait for the schematic, thanks! \$\endgroup\$
    – duck
    Commented Mar 21, 2017 at 4:47

1 Answer 1


First, it should be noted that there's a comparator inside the LM2575 and its purpose is to compare a portion of output voltage (the voltage on FB pin) to a reference voltage (1.23V for LM2575) and generate a signal for the control circuitry. So, what the LM2575 does is to keep the voltage on FB pin at 1.23V. That's why minimum output voltage is given as 1.23V.

Consider the following schematic:


simulate this circuit – Schematic created using CircuitLab

The key point here is the current into/out of the resistors connected to FB pin. And we have 3 equations:

1) \$I_{R2}\$ is constant: \$I_{R2} = I_{R3} + I_{R1} = V_{ref}/R2\$.

2) \$I_{R3}\$ varies with DAC output voltage, \$V_{DAC}\$: \$I_{R3} = (V_{DAC} - V_{ref})/R3\$.

3) And finally, \$V_o = V_{R1} + 1.23V\$, where \$V_{R1} = I_{R1} \cdot R1 = (I_{R2} - I_{R3})\cdot R1\$. If we make \$I_{R1}\$ negative then we can get output voltages lower than 1.23V.

But selecting resistors needs a two-unknown equation to be solved.

For your needs: \$V_{o-min} = 0.5VDC\$ and \$V_{o-max} = 12VDC\$

Let's select R2 = 1k2.

\$I_{R2} = 1mA\$

For \$V_{DAC} = 0V\$ (zero code);

  • \$I_{R3} = (0 - 1.23)/R3 = -1.23/R3\$
  • \$I_{R1} = I_{R2} - I_{R3} = 1 - (-1.23/R3) = 1 + 1.23/R3\$
  • \$V_o = 1.23 + I_{R1}\cdot R1 = 12V \Rightarrow R1 = \frac{10.8}{1+\frac{1.2}{R3}}\$

For \$V_{DAC} = 5V\$ (full-scale code);

  • \$I_{R3} = (5 - 1.23)/R3 = 3.77/R3\$
  • \$I_{R1} = I_{R2} - I_{R3} = 1 - (3.77/R3)\$
  • \$V_o = 1.23 + I_{R1}\cdot R1 = 0.5V \Rightarrow R1 = \frac{0.73}{\frac{3.8}{R3} - 1}\$

From \$R1=\frac{10.8}{1+\frac{1.2}{R3}} = \frac{0.73}{\frac{3.8}{R3} - 1}\$, you'll get R3 = 3k5 and R1 = 8k.

If you crosscheck, you'll see that the minimum output voltage (with full-scale DAC output) will be 0.51V and maximum output voltage (with zero-scale DAC output) will be 11.94V.

With this method, any output voltage range can be obtained.


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