0
\$\begingroup\$

For example my calculation for MP1584 inductor is 10uH, but i could put only 47uH to it. It is a problem? I'm doing this wrong?

\$\endgroup\$
2
  • \$\begingroup\$ I'm doing this wrong? Probably. You cannot expect to use a 4.7 x larger component value and still get the same performance. In the datasheet there's a table stating you can use 33 uH when Vout is 12 V. If you use 47 uH one or more of the following might happen, the converter: does not work, becomes instable, output current is limited, excess ripple on the output voltage. If you don not fully understand the consequences of using 47 uH, then simply don't. \$\endgroup\$ Sep 4, 2016 at 13:15
  • \$\begingroup\$ Try it, and then let us know... \$\endgroup\$ Sep 4, 2016 at 15:19

3 Answers 3

3
\$\begingroup\$

Let me get this straight. The datasheet gives you a formula for inductor value. You crank thru this, and discover that the inductor should be 10 µH. Then you decide "Eh, what do they know, I'll ignore all that and use something off by 5x from what it says.". Now you're seriously asking if you're doing something wrong!?

Using a component nearly 5x off from what the datasheet specifes is a bad idea. This really should be obvious. Inductors have 10% or 20% tolerance. That kind of error is most likely built into the formula and the compensation network. But 5x is way beyond reasonable error.

Some datasheets tell you the tradeoffs with different inductor values, but most don't. You don't know what considerations went into designing the compensation network and guaranteeing stability while living up to the ripple and transient response specs. If the datasheet doesn't give any guidance about different inductor values, then you have to assume the values they specify are requirements within reasonable inductor tolerances.

\$\endgroup\$
1
\$\begingroup\$

For a given switching frequency, when voltage is applied to the inductor during the "on part" of the duty cycle, current will ramp up at so many amps per second. This rate (di/dt) is determined by the voltage applied across the inductor and the value of the inductor, given by the formula V = L di/dt.

So, if inductance is too large, the current rate of rise is too shallow and the final current at the end of the "on part" of the duty cycle won't be enough to impart sufficient energy into the inductor (energy = \$LI^2/2\$).

The energy needed is determined by the load and the switching frequency so, if the load requires more energy per switching cycle than what the inductor has stored, it won't be enough to keep the output voltage in regulation.

\$\endgroup\$
2
  • \$\begingroup\$ In a heavilly loaded buck converter the inductor current never drops to zero. So the inductance of the inductor does not directly limit the power that can be passed through the converter. \$\endgroup\$ Sep 4, 2016 at 16:45
  • \$\begingroup\$ @peter I was really debating about the possibility of this but decided to keep it simple for the OP to understand but you are quite eight, a higher average current and lower rate of change can yield enough energy to make heavier loads get enough energy per cycle. \$\endgroup\$
    – Andy aka
    Sep 4, 2016 at 17:13
1
\$\begingroup\$

The inductor and capacitor on the output side form an LC lowpass filter that smooths out the voltage and current through the load. For these,

\$F_{cutoff} = \frac{1}{2\pi\sqrt{LC}}\$

and

\$Z = \sqrt{\frac{L}{C}}\$

So increasing the inductor value will lower the cutoff frequency and increase the characteristic impedance.

The main effect you will see is decreased efficiency and worse behaviour when current demands change suddenly. I recommend load step testing your converter.

\$\endgroup\$

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.