2
\$\begingroup\$

I am currently designing a circuit which should be able to take 12-28V DC power in and output 12V DC @ 1.5A.

The question is: is this possible with a 100% duty cycle buck converter (like TPS54302)? What will happen if the input and output voltage are nearly the same?

If this is not possible what options do I have? If it would be ok to drop up to 1V when the input voltage is below 13V, will this make things easier?

Would it be possible to detect a low voltage and bypass the converter if the voltage is too low?

\$\endgroup\$
  • \$\begingroup\$ "No" voltage drop would require a device to be constructed exclusively from superconducting materials, because it implies a resistance of zero. I don't think I have yet seen the announcement from the Nobel Prize Committee about that, so I suspect that with most devices, a small amount of resistance is still the best we can get. Thus any practical device will be likely to have a small voltage drop. Note that duty cycle is unrelated to voltage drop; an electrical space heater can be designed for a 100% duty cycle but drop quite a few volts. \$\endgroup\$ – a CVn May 18 '17 at 14:24
  • \$\begingroup\$ @MichaelKjörling I briefly thought about putting an "almost" in the question, but thought it is self-evident that no circuit has no voltage drop. Is the voltage drop really totally independent of duty cycle? Won't the converter have a hard time maintaining a high output voltage under load if is not capable of turning the switch on at least most of the time? Or is it just a matter of how much current is drawn? \$\endgroup\$ – Karsten May 19 '17 at 12:33
5
\$\begingroup\$

Remember a typical buck converter configuration:

schematic

simulate this circuit – Schematic created using CircuitLab

The product page indicates that duty-cycle can be %100.

If duty-cycle is %100 then the MOSFET (M1) will not act as a switch (because it's fully on). Instead, it will turn into a resistor with a value of \$R_{ds-on}\$.

So a small voltage of \$V_{DS} = R_{ds-on} \cdot I_{LOAD}\$ will drop across this resistor. For example, according to the datasheet, on-resistance is 85mR. And for a load current of 3A, the voltage drop will be about 270mV. Note that there will also be a small voltage drop across the inductor due to its DC resistance.

Would it be possible to detect a low voltage and bypass the converter if the voltage is too low?

This is called Brown-Out Detection and a lot of chips on the market have this feature. But instead of bypassing, they tend to turn themselves off to protect the converter from drawing excessive currents.

\$\endgroup\$
  • \$\begingroup\$ That's what I thought, but I am unsure if the regulation will be stable, because the design tool Texas Instruments provides for there components do not allow input and output voltages to be the same and recommend bigger and bigger components the closer the values are together. This may be because - as you said - there will always be a voltage drop and this may cause the regulation to become unstable. This is what I don't know. \$\endgroup\$ – Karsten May 18 '17 at 8:18
  • \$\begingroup\$ I don't know much about the buck controller internals and therefore don't know what it will do if the output is consistently lower than the set voltage. Will it just permanently enable the transistor? \$\endgroup\$ – Karsten May 18 '17 at 8:20
  • 1
    \$\begingroup\$ If I were you, I would use the EN function of the controller instead of worrying about duty-cycle. For example, build a comparator that pulls the EN pin low when VIN is lower than or equal to, say, 12V. Also put a relay between VIN and SW pin so that you can also bypass the the controller and direct the input to the output by shorting these pins as long as the EN pin is held low. \$\endgroup\$ – Rohat Kılıç May 18 '17 at 8:32
4
\$\begingroup\$

Reading your question it sounds to me that a "buck-boost" converter might be more suitable for your needs. These converters can produce a fixed output voltage with inputs that range from significantly below to significantly above the output voltage. Check out offerings from Linear Technology and Texas Instruments.

This method is a lot simpler than trying to bypass the buck circuit to ensure adequate operation around 12 volts.

\$\endgroup\$
  • \$\begingroup\$ I thought about that, too. The problem is that these controllers incl. BOM seem to be much more complex and expensive than simple buck converters. \$\endgroup\$ – Karsten May 18 '17 at 9:03
  • 1
    \$\begingroup\$ What price for performance and functionality? \$\endgroup\$ – Andy aka May 18 '17 at 9:36

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.