I am in the middle of a project where i want to use a mcu (ST32F3028U6 with DAC of 2.4V - 3.6V) to control the V_out voltage of a LT8494 SEPIC DC/DC Converter. I understand the basic operation of the converter and the purpose of some of the components in the application example.

I have also included a snippet from the LT8494 datasheet

In my case V_in is from a 1S Li-Po, so approx. 3.0 - 4.2 V

I understand completely the functions of and the external components belonging to pins V_in, SWEN, SS, PG, GND.

What i am having trouble with is the circuitry in the upper area of the picture. Hence, the question of this post is: What is the purpose of the circuitry in the upper section of the snippet (including components D1-D4 etc), what is the purpose of BIAS and FB pins and how do i choose the correct resistor values R1, R2, R3 (in the picture 1M, 78,7k and 26.7k) to be able to control the output voltage with the DAC of my MCU

My current understanding of the circuit is as follows: Please correct me if I am wrong at any of the following aspects/explanations

When the NPN in the SW pin is enabled the current from Vin is stored in L1 and when the NPN is disabled the current is discharged through C3 to the output, generating the desired output voltage using R1, R2, R3. I dont understand the effect of L2 and the rack of diodes and capacitors.

Sorry for the long post, i was trying to be thorough in my explanation of the problem. Any help is greatly appreciated

EDIT: My desired output voltage would be in the range of approx 1.3V - 12V

Typical Application


"FB" stands for feedback and is an absolute necessary part of a practical boost or sepic converter. The voltage at the FB pin when the device is in perfect regulation is between 1.178 and 1.230 i.e. about 1.2 volts. Read the data sheet!

Therefore, if you want an output voltage of 5V you choose a resistor divider that gives you ~1.2 volts at the FB pin.

Now, if you want to control the output voltage with some form of demand voltage, all you are realistically allowed to to is "add" a voltage or "subtract" a voltage to the voltage that is fed back. This then effectively alters the resistor division ratio and forces (or maybe you could use the word "cons") the device into believing it needs to produce a higher or a lower voltage on the output for perfect regulation.

You don't need a cockcroft diode multiplier. If you only need 12V use this basic circuit: -

enter image description here

You can see it produces 12V from an input voltage as low as 3 volts and, producing a lower voltage is fairly easy. You do this "perfectly" by injecting a DC current into the node formed by the 1 Mohm and 110 kom resistor. As the circuit stands a nominal 1.2 volts sits across the 110 kohm. This produces a current thru the 110k of 10.9 uA but, if you force fed that node with (say) 10uA, only 0.9 uA needs to be pushed thru the 1 Mohm resistor and this means the device will regulate at 1.2 volts + 0.9 volts i.e. 2.1 volts.

If you force fed that node with 10.8 uA the regulator's output would be 1.2 volts + 0.1 volts i.e. 1.3 volts.

You've got to be careful though - it is a sensitive control input so don't hang long wires from it.

| improve this answer | |
  • \$\begingroup\$ Thank you very much for the simple and understandable explanation to overcome my confusion towards this problem. I will accept your contribution as an answer to this question. If you could be so nice to explain the exact purpose of L2 it would be ideal. Also, in the LT8494 the NPN is rated at 2A am I right to say, that i cant draw more than 2A from the battery. to ensure the survival of the NPN? \$\endgroup\$ – Martin1 Oct 23 '15 at 23:53
  • \$\begingroup\$ L1 and L2 are coupled windings and there is no simple explanation other than look up how a flyback regulator works. It's basically a flyback circuit with a capacitor that bridges primary and secondary windings. If you have problems understanding flybacks you'll have problems understanding sepics! Basically L2 is the seondary winding of a flyback converter. The NPN current mx depends on the duty cyle as far as I can tell so it might be as low as 1.3 amps - see pg 3 of data sheet. \$\endgroup\$ – Andy aka Oct 24 '15 at 0:01
  • \$\begingroup\$ Thank you again, you have been a huge help. I am rather new to analog electronics as you could see, but thanks to you i now understand how that system works - will look into flyback converters too. \$\endgroup\$ – Martin1 Oct 24 '15 at 0:04

You don't say what output voltage you require.

EDIT - with a range of 1.2 - 12V you don't need the multiplier, you can use the circuit at the top of page 19 of the data sheet. Change the feedback resistors (316k) to match your voltage needs.

This circuit goes beyond a simple SEPIC converter which wouldn't have D1-D4, D7 and D8 etc. Normally the output would be from C2. If you don't need as much as that you may be able to get away with fewer diodes or no multiplier at all.

Those other components have been added to increase the output voltage up to 60V. They form a DC multiplier (sometimes called a Cockcroft-Walton multiplier).

One common reason for using the SEPIC configuration is that it is tolerant of a short-circuit on the output - the normal boost converter has a direct path from Vin to the output so a short will probably destroy the inductor. In the SEPIC configuration the signal passes though the capacitor C3 so no DC can flow. The inductor L2 provides a DC path for the output current to flow through the diode D5 while allowing the AC voltage from C3 to couple to D5.

Another advantage of the SEPIC is that the output can be higher or lower than the input voltage, a normal boost circuit can only produce voltages higher than the input and a buck only produce voltages less than the input.

R1, R2 and R3 are to provide the feedback signal to the FB pin of the IC. The value needs to be low enough so that it is not affected by the input current of the IC (only 20nA from the data sheet) but high enough so the power consumed by the resistors is not significant relative to the load current.

The FB pin will be at 1.25V when the circuit is in control of the output voltage, if the voltage goes higher than that it will try to reduce the output, if it goes lower it will increase the output. The DAC input to the resistor network will modify the voltage at the output to achieve that balance. Just treat it as a resistor network to determine the range of control that you need. Increasing the voltage from the DAC will cause the output voltage to reduce. At the specified output voltage the output from the resistor network should equal 1.25V.

It looks like the BIAS input pin is used to power the internal circuitry for situations where the input - it is explained on page 14 of the data sheet. This is specific to this device.

| improve this answer | |
  • \$\begingroup\$ Thank you for the answer. I added the desired output range as and edit to my original post. Now, given the desired output, will i need the DC multiplier? If i would treat it as a resistor network, would i then just add a "1.2V battery" instead of the FB pin, and a "battery" as desired maximum voltage and then choose resistors accordingly, to produce the output desired with modifying the DAC? Also, what is the purpose of the inductor L2? Thanks again. \$\endgroup\$ – Martin1 Oct 23 '15 at 23:05

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.