Does the voltage gain follow linear relationship for boost converter?

And when you change the switching frequency , does the voltage gain still follow linear relationship ?

  • \$\begingroup\$ Most boost converters have voltage feedback so there is no voltage gain. \$\endgroup\$ – Bimpelrekkie Sep 11 '15 at 11:51
  • \$\begingroup\$ For a boost converter with feedback the output voltage is constant so the output voltage is independent of frequency (within practical limits). \$\endgroup\$ – Bimpelrekkie Sep 11 '15 at 12:34

Almost all boost converters have a feedback loop to control the output voltage. This means that there is no voltage gain, the output voltage is independent of the input voltage (as long as it is withing practical limits).

Perhaps you are thinking of this type of circuit:


simulate this circuit – Schematic created using CircuitLab

The voltage gain of this circuit without any load is infinite ! Why is that so ? When S1 is closed the electrical energy from the battery charges the inductor L1. Yes an inductor can be charged with current ! This build up a magnetic field in the inductor. When S1 opens the inductor wants to keep the current flowing. So the current flows through D1 into C1. Since L1 behaves as a current source when it is charged but S1 is open, the voltage at the anode of D1 will simply reach the value that is required to make the current flow. So with ideal components the voltage will reach an infinite value.

Switching frequency has nothing to do with this, a lower switching frequency means that S1 opens and closes faster meaning L1 is charged less per open-close cycle.

  • \$\begingroup\$ For ideal, yes. But practical inductor (typically ferrite core) reaches a limit when the core saturates, the Li-squared energy storage limit. Faster switching requires better inductor, higher-mu core and lower-resistance windings. \$\endgroup\$ – MarkU Sep 11 '15 at 20:29
  • \$\begingroup\$ I suspect he's talking about the error amplifier gain in a voltage mode control converter or something like that. Reason not to pander to poorly phrased and undeveloped questions. \$\endgroup\$ – Fizz Oct 23 '15 at 8:18

When you're changing frequency - optimal parameters for the output LC-filter also changing. If you're increasing frequency - optimal parameters will become lower and there is no troubles (until L and C can work on such frequencies). If you're decreasing frequency - parameters are going up and you can move from CCM (continuous current mode) to DCM (discontinuous current mode). In DCM transfer function for steady state will depend from load value and will become non-linear.

So for your case answer is - yes, when you're increasing frequency it will be linear until parameters of circuit allows it (if inductor doesn't designed for 1 MHz switching frequency - then it's your troubles).

Also you must consider that increased frequency will cause increased dynamic power losses (if we're talking about hard switching then it's increasing linearly with frequency).

  • \$\begingroup\$ To me, what you state seems more relevant to buck converters instead of boost converters. \$\endgroup\$ – Bimpelrekkie Sep 11 '15 at 12:32
  • \$\begingroup\$ In fact for both. When you're increasing frequency or output current of buck or boost - values of LC-filter goes down. And both types do have transfer function which is independent from frequency (in steady state). \$\endgroup\$ – Looongcat Sep 11 '15 at 14:12
  • \$\begingroup\$ * in steady state in CCM \$\endgroup\$ – Looongcat Sep 11 '15 at 14:57

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