The actual output voltage of the SEPIC circuit does not match the theoretical value.

xiaxingxing

The actual output voltage of the SEPIC circuit does not match the theoretical value. [Copy link]

 

As shown in the figure, Vin input is 7~12V;

R3=9k ohm, R6=1K ohm;
power chip XL6008 (Xinlong), FB pin voltage VFB=1.25. (Measured VFB=1.248V);
theoretical output Vo=(1+R3/R6)*1.25=12.5V, but the measured output voltage is indeed 13.7V;
SW pin oscillates when no load, but does not oscillate when connected to a load. The output ripple is not large, only a few hundred mV;
When R3 and R6 are set to other resistance values, the output voltage also has a similar situation, and the actual output is much higher than the theoretical value.
What is the reason for this? Thank you all!!!

qwqwqw2088

Where does the simulated SEPIC model of XL6008 come from?

xiaxingxing

qwqwqw2088 posted on 2021-1-14 17:31 Where does the simulated SEPIC model of XL6008 come from?

There is no simulation, I actually did a board test.

g6g6

I still need to keep learning, the basics are not enough

qwqwqw2088

xiaxingxing posted on 2021-1-14 17:50 There is no simulation, I actually did a board test.

Misunderstood

It seems that the reality is inconsistent with the theory

I thought it was a simulated sepic circuit

maychang

13.7*3/22=1.868, but you actually measured 1.248V.

Either your resistor is inaccurate or your multimeter is inaccurate.

qwqwqw2088

Then keep asking.

Where does the theoretical value come from?

xiaxingxing

maychang posted on 2021-1-14 20:02 13.7*3/22=1.868, but you actually measured 1.248V. Either your resistor is inaccurate or your multimeter is inaccurate.

Senior, 13.7*3/22=1.868, how did you get this calculation?

Isn't it calculated as follows:

Because VO=(1+R3/R6)*VFB, so VFB=1/(1+R3/R6), and R3=9K, R6=1K, VO=13.7V,

So VFB=13.7/(1+9)=1.37V.

Thanks!

xiaxingxing

qwqwqw2088 posted on 2021-1-14 20:03 Then let me continue to ask, where does the theoretical value come from?

VO=(1+R3/R6)*VFB, and R3=9K, R6=1K, VFB=1.25V.

So VO=12.5V, is this the calculation?

maychang

xiaxingxing posted on 2021-1-15 09:06 Senior, 13.7*3/22=1.868, how did you get this calculation? ? ? Isn't it calculated as follows: Because VO=(1+ ...

"And R3=9K, R6=1K, VO=13.7V,"

The first floor diagram shows R3=19k and R6=3k.

maychang

xiaxingxing posted on 2021-1-15 09:06 Senior, 13.7*3/22=1.868, how did you get this calculation? ? ? Isn't it calculated as follows: Because VO=(1+ ...

If R3 and R6 are 9kohm and 1kohm respectively as described in the text, then the 13.7V and 1.248V mentioned in the first post do not conform to the voltage division ratio of these two resistors. So it can be said that "either your resistor is inaccurate or your multimeter is inaccurate". Of course, it is also possible that both are inaccurate.

xiaxingxing

maychang posted on 2021-1-15 10:45 xiaxingxing posted on 2021-1-15 09:06 Senior, 13.7*3/22=1.868, how did you get this calculation? ? ? It's not based on...

Well, thank you, senior. I will go back and confirm it.

I have a few more questions about the negative voltage generated by SEPIC. I would like to ask the seniors

As shown in the circuit diagram below, I use XL6008 to generate negative voltage. The problem is as follows:

1. The datasheet of XL6008 does not mention the VFB voltage when XL6008 is used to generate negative voltage. However, ADI chips all mention it, and the VFB when outputting negative voltage is half of the VFB when generating positive voltage and then rounded to the negative. So I guess the VFB when XL6008 outputs negative voltage is -1.25/2=-0.625. Is there such a rule? (VFB of XL6008=1.25V)

2. According to the parameters in the circuit diagram, my output voltage is calculated as follows: VO=-0.625*(1+19/3)=-4.58. But my output is much larger than this, about negative ten volts.

3. Is my circuit diagram correct? My circuit requires an input voltage of 12V and an output voltage adjustable from -1.25 to -40V. (In actual debugging, R10 is replaced with an adjustable resistor)

Thank you senior!

maychang

xiaxingxing posted on 2021-1-15 19:35 Well, thank you, I will go back and confirm. In addition, I have a few more questions about the negative voltage generated by SEPIC. I would like to ask the seniors for the following circuit diagram, ...

I don't know whether XL6008 can use the same FB pin as the feedback pin for both positive and negative output voltages, because I haven't read the chip datasheet carefully. According to the second line of the page you posted, it can.

However, the connection between diode D2 and inductor L4 is a Cuk circuit, not a Sepic circuit.

xiaxingxing

maychang posted on 2021-1-15 19:59 I don't know whether XL6008 can use the same FB pin as the feedback pin for outputting positive voltage and outputting negative voltage, because I haven't read the chip carefully...

The output voltage of the CUK circuit is calculated in the same way, VO=(1+R10/R14)*VFB, right? ? ?

I now set R10=8.2K, R14=1K, C11=10nF, the test output voltage VO is -8.34V, and the multimeter tests the voltage of the FB pin to be -0.63V. The following is the oscilloscope waveform.

My question is that the theoretical output voltage VO = (1 + R10 / R14) * VFB = (1 + 8.2 / 1) * (-0.63) = -5.8V, but the actual output voltage is -8.34V. What is the reason for this? Is there a current flowing in/out of the FB pin?

Thank you, senior! (The specific resistance value of the voltage divider resistor for the positive voltage output in the previous few floors has not been determined, but the test of the negative voltage output is definitely correct. I have tested it many times.)

maychang

xiaxingxing posted on 2021-1-16 12:04 The output voltage of the CUK circuit is also calculated in this way, VO=(1+R10/R14)*VFB, right? ? ? I now set R10=8.2K, R14=1K, C11=10nF, ...

"My question is that the theoretical output voltage VO=(1+R10/R14)*VFB=(1+8.2/1)*(-0.63)=-5.8V, but the actual output voltage is -8.34V. What is the reason for this? Is there current flowing in/out of the FB pin? "

I haven't found the datasheet for XL6008, so it's hard to say. It's possible that the input impedance of the FB pin is not high enough and a certain current is required.

xiaxingxing

maychang posted on 2021-1-16 13:02 "My question is that the theoretical output voltage VO=(1+R10/R14)*VFB=(1+8.2/1)*(-0.63)=-5.8V, but the actual output voltage is -8.34V. This is...

I have uploaded the specification sheet of XL6008. Could you please take a look at it for me? Thank you!

xiaxingxing

maychang posted on 2021-1-16 13:02 "My question is that the theoretical output voltage VO=(1+R10/R14)*VFB=(1+8.2/1)*(-0.63)=-5.8V, but the actual output voltage is -8.34V. This is...

Senior, I just measured that there is 875uA current flowing out of the FB pin. What is the reason for this? Is it an unreasonable circuit design or a problem with the chip itself?

maychang

xiaxingxing posted on 2021-1-16 14:05 I uploaded the specification of XL6008. Please help me take a look. Thank you!

In the datasheet of this chip, the second item of Feature does say Programming with a Single Feedback Pin.

maychang

xiaxingxing posted on 2021-1-16 14:05 I uploaded the specification of XL6008. Please help me take a look. Thank you!

However, its Absolute Maximum Ratings clearly states that the FB pin voltage is -0.3 ~ Vin and must not be exceeded, otherwise it may be damaged.

xiaxingxing

maychang posted on 2021-1-16 15:31 However, it is clearly stated in its Absolute Maximum Ratings that the FB pin voltage is -0.3~Vin and cannot be exceeded, otherwise it may be damaged. ...

I see, now it makes sense. I didn't notice the input voltage range of the FB pin. Now it seems that the FB pin voltage can only feed back the positive output voltage. Thank you!