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The MC34063 Application Note lists the equation for calculating minimum inductor size as follows:

$$L_{min} = \frac{V_{in} - V_{sat} - V_{out}}{I_{pk}(switch)} t_{on}$$

But this implies that as Ipk(switch) (eg, the maximum switch current) decreases, the minimum inductor size increases. This is backed up by interactive calculators such as this one, which show the same effect.

Why is this the case, and does it imply that the regulator will only work as designed if running at peak load, and I thus need to increase the inductor size if I want to handle smaller loads?

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A more theoratical explanation:

The current through the inductor of a SMPS is like a triangle. The average current of this triangle is equal to your load. The peak-to-peak value is determined by the various input and output voltages, switching frequency, duty cycle and inductor.

Buck converter Waveforms for a buck converter

The first figure shows a buck converter. The second shows the waveforms of the buck converter. It shows the switch S, the voltage across the inductor and the current through the inductor. When the switch is closed, the voltage across the inductor is Vin-Vout. When the switch is open, the voltage across the inductor is -Vout. The diode is assumed in this ideal and therefor has zero voltage drop. A buck converter has a rule that Vin>Vout, so you have a positive voltage 'charging' the inductor, and a negative voltage 'discharging' the inductor. The rate of change in current is dependant on this voltage and inductance. If you want a stable output, the upramp should be as 'high' as the downramp. Otherwise you get a falling or rising average. There is an equilibrium. In maths, this comes down to this:

Equation for buck converter

The first term of the formula describes the upramp, and the second term describes the downramp. As you can see the switching frequency and duty cycle have been simplified to t_on and t_off. The duty cycle is only dependant on the ratio between the output voltage over the input voltage. The duty cycle will not change with varying load.

The level of upramp and downramp 'speed' will only change if you change the input/output voltages, inductor value or the switching frequency. Increasing the switching frequency will lower the up and downramps, but it's not always possible to increase the switching frequency (maybe you're already operating at the maximum). The input/output voltages are to remain constant, that's the application you're dealing with. If you increase the inductor then the change in current through the inductor is going to drop. That's the only tool you have available.

Why is this a problem? Well, in the waveforms I've shown the converter is running fine. The minimum current through the inductor does not reach zero. What happens if the average current drops so much that the inductor does reach zero?

The converter would need to resort to discontinuous mode. Not all converters can do this. This sometimes requires the converter to skip cycles. If the converter opens the switch for a minimal amount of time, a certain amount of energy is transferred. This is stored in the capacitor, but is not consumed fast enough. This will influence the output voltage, which makes the converter unstable. If you skip cycles the converter basically waits before the output voltage drops far enough before it requires another cycle.

A higher value inductor will mean that the minimum current will closer to your average current , possibly avoiding discontinuous operation. This also implies why you calculate the minimum inductor through the datasheets. You can always use a bigger inductor, but smaller may cause issues on low loads. However if the SMPS is also designed to deliver high power in situations, the inductor may be too bulky and expensive..

A converter capable of switching to discontinuous mode is pretty much trouble free with this and you don't have to go through this. The MC34063 is a fairly old and generic chip, so it's a bit more tricky.

If you can't fit a bigger inductor.. add a minimum load yourself.

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    \$\begingroup\$ The MC34063 datasheet doesn't explicitly say if it handles discontinuous mode, but the application note implies it in its description. If that's the case, it sounds like I should be fine, albeit with possibly increased output ripple at lower loads. \$\endgroup\$ – Nick Johnson May 16 '12 at 22:34
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Think of the reverse. A larger inductor builds up current more slowly when the same voltage is applied accross it. Therefore, if you need a lot of current you have to use a smaller inductor to build up current more quickly, or leave the switch on longer to build up more current.

For smaller output current, you don't need a larger inductor necessarily. However, there is a limit to how short it is reasonable to keep the switch on, so there is some minimum current buildup in the inductor each switch cycle. That minimum current causes some minimum voltage increase on the output when it is dumped there. Therefore, switching power supplies designed for high current will have larger output ripple voltage than those with tighter maximum spec, all else being equal.

If output ripple isn't a big concern, you can use discontinuous mode with a pulse on demand control scheme and get as little average current as you want. Most SMPS chips are designed for continuous mode since they use high frequency to keep the physical inductor size down. They aren't going to go into all the design tradeoffs, and will make some assumption about what you want the output characteristics to be. This is usually low ripple and fast transient response. With these considerations, there is some limited current range where the characteristics will be "good". By chosing parameters just enough for the highest current case, you give yourself good performance down to the lower current levels.

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    \$\begingroup\$ So a regulator rated at 100mV ripple at 1A will cause more ripple at lower loading? What if I pick a larger inductor than the minimum value? Your writeup implies this is a bad idea, but the app note definitely indicates the inductance as a lower bound, not an upper one. \$\endgroup\$ – Nick Johnson May 16 '12 at 12:33
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    \$\begingroup\$ @Nick: Depending on the design, the 1A switcher might have more ripple at 10 mA. Or it could have less ripple at 10 mA if rated for only 100 mA. There are a lot of tradeoffs and control schemes. With canned switcher chips, many of these have been made for you often without elaboration. Every part with built in switch has a lower limit on inductance. This is because there will be some minimum switch on time, which requires some minimum inductance to not exceed the maximum switch current. \$\endgroup\$ – Olin Lathrop May 16 '12 at 12:38
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    \$\begingroup\$ Thanks. I'm mostly trying to validate my assumptions, namely: 1) That I can spec the regulator based on maximum current (and voltage ripple at that current), and expect regulation to be maintained within reasonable bounds of that at, say, a third of the max current, and 2) That I can pick an inductor larger than the minimum given for a particular max current for convenience, without compromising the design. If I understand your answer correctly, both of those statements are true? \$\endgroup\$ – Nick Johnson May 16 '12 at 12:44
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    \$\begingroup\$ @Nick: Probably, but you can't know all the tradeoffs that went into the design of any particular switcher chip. Only the datasheet can tell you for sure what the valid range of part values are. \$\endgroup\$ – Olin Lathrop May 16 '12 at 12:51
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Lighter loads require more inductance to stay in continuous conduction mode (CCM).

The app note equation you ref gives an inductance Lmin which puts the converter on the boundary between CCM and discontinuous conduction mode (DCM). If you use the max load current in this calculation, the resulting converter will drop into DCM at anything less than max load, where its dynamics will change. (DC regulation will remain good.) Instead, base the inductance calculation on the anticipated min load, so the converter remains in CCM over the load range.

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I'm in a similar boat as you with this chip. From what I understand (and to reiterate whats been said above) you want to set your average current such that your peak to peak current ripple through the inductor is always above 0 amps. If you look at the chart with average current, voltage and switch state you want to make sure i_min can never hit 0. To accomplish this shrink your current ripple and this will allow your average current to go down as well....approaching 0.

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