Use the microcontroller DAC to output analog voltage and control the output voltage of the DC-DC circuit so that Vout=2*Vset. Please help me design the circuit

mars4zhu

Use the microcontroller DAC to output analog voltage and control the output voltage of the DC-DC circuit so that Vout=2*Vset. Please help me design the circuit [Copy link]

Recently, I am designing a DC-DC control circuit with adjustable voltage. I use a DAC of a single-chip microcomputer to output an analog voltage Vset to set the adjustable voltage output Vout of the DC-DC.

Basic Overview of DC-DC Power Supply Fixed Voltage Output Circuit

For a general DC-DC power conversion circuit, there is a pin Vfb for feedback output voltage Vout, which is compared with the internal reference voltage Vref of the DC-DC chip.

And there are the following feedback control laws:

(1) When Vfb < Vref, the DC-DC circuit will adjust the duty cycle to increase Vout;

(2) When Vfb>Vref, the duty cycle is adjusted to reduce Vout;

(3) When Vfb is positively correlated with Vout, the DC-DC circuit will eventually reach a steady state, the Vout output remains unchanged, and Vfb = Vref.

Usually, a series resistor voltage divider is used, so that Vout = (R1+R2) / R1 * Vfb = (1+R2/R1)*Vref. The ratio of the voltage divider resistor R2/R1 is used to achieve a fixed voltage output of the DC-DC circuit.

Implementation of adjustable voltage circuit:

In order to realize the adjustable voltage output of the DC-DC circuit, a Vset voltage is used to set the output Vout of the DC-DC circuit so that Vout = N*Vset.

A feedback circuit is designed as shown in the figure. Using Vset, Vout and a fixed bias voltage Vbias, the feedback voltage Vfb is output to the feedback pin of the DC-DC.

The Rp resistor value and the relationship between Vbias and Vref need to be calculated so that Vout = 2Vset.

Calculation process:

make

S1 = S2 = S3 = S4 = 1/10KΩ

K = Sp/S1.

Since the operational amplifier circuit is a negative feedback circuit, the concept of "virtual short and virtual break" is used:

Vp = Vn

Ip = 0, In = 0

According to Kirchhoff's current law, we get:

0 = (Vp - Vout) + (Vp - Vbias) + K*Vp

Vp = (Vout + Vref) / (2+K)

0 = (Vn - Vset) + (Vn - Vfb) + 0*Vn

Vn = (Vset + Vfb) / 2

Simplified to:

Vfb = (Vout + Vbias) * 2/(2+K) - Vset

Vbias and Vset are both kept at fixed values. When Vout increases, Vfb increases, and the DC-DC circuit will adjust the duty cycle according to the increase of Vfb as mentioned above to suppress the increase of Vout. Therefore, the circuit will enter the negative feedback of the voltage control closed loop. In the final steady state, Vfb must be = Vref.

The conversion equation is:

Vout = (2+K)/2 * (Vset + Vfb) - Vbias

To meet the requirement of Vout = 2Vset, we must have:

(2+K) /2 = 2

(2+K)/2 * Vfb - Vbias = 0,

thereby:

K = 2, i.e. R5 = 1/2 * R1 = 5kΩ

Vbias = 2*Vfb = 2 * Vref

Please give me some advice, great masters;

Jacktang

I mean, let the masters verify the calculation again.

h472438470

No follow-up? Looking forward to the actual demonstration of the master? Learn more and improve every day. Thank you

Orima

Learn the solution...

MrShangPING

The DAC output should use a follower, and then have FB feedback to control the duty cycle or turn off the output

Gen_X

Agree with the above proposal!

whuer

MrShangPING posted on 2021-9-13 08:23 The DAC output should use a follower, and then have FB feedback to control the duty cycle or turn off the output

DAC does not need a follower. The DAC has a built-in voltage follower or a 2x in-phase proportional amplifier.