Electronic – Input voltage for feedback pin on LM2596 Voltage regulator

It's really hard to do this with a generated voltage without degrading the voltage accuracy of the regulator.

But if you have a handful of digital outputs that you can dedicate to this, you can use a resistor digital-resistance-converter (DRC?) like this. I've represented the regulator as an op-amp.

^{simulate this circuit – Schematic created using CircuitLab}

The four trim resistors are switched out of the circuit by setting microcontroller pins to inputs and they are switched into the circuit by setting microcontroller pins as digital output lows.

When all four pins are inputs, the voltage divider is R1 and R2. When some pins are set, the divider is R1 and the parallel combination of R2 and the trim resistors.

I would suggest setting the lowest voltage with R1 and R2, the select the trim resistors to pull the voltage over the range that you want. Do not choose all four resistors the same value. Instead you will want something like a 1k,2k,4k,8k sequence so you can get 16 different values out.

For extra credit:

• If your microcontroller has switchable high-side outputs (open-drain) you can do that instead of switching input to output, then
• If your microcontroller has outputs as bits in a PORT register, arrange the resistors with the largest value on the least-significant output bit. Then with careful resistor value choice, you can write 0-15 to the port register to get monotonically increasing voltage.

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.