# What happens when a buck converter passes from max load to minimum load

#### HRN88

Joined Jul 21, 2020
24
Hi guys, what happens if a buck converter that previously was designed to operate in CCM at full load (max power output) and then I connect a less load lets say 10% of the maximum load?

I simulated this in LTSpice and I can see that the buck converter changes from CCM to CrCM or even DCM also I can see that the ripple current (delta I) on the inductor has a small variation in amplitud (Ipp). but the DC component level of that current has been moved down doing that the inductor passes from CCM to DCM.

1. Does it's normal this behavior?
2. What are the more stress in the MOSFET and freewheeling diode, in CCM, CrCM or DCM?
3. If a controller is designed to control the driver in CCM what happens when the buck enters into DCM mode?
4. And finally, should say that CCM is the more stress in the switching devices?

I attached my simulation if it helps to my questions or if do you have any comment. (Max output power 60W)
Ty so much friends

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#### Papabravo

Joined Feb 24, 2006
18,981
Every buck converter is different. Depending on the compensation and the control strategy there will be a two-part response, The first part is the transient response that deals with the step change in the load. The second part of the response is what happens in the steady state. Let me try to answer the questions.
1. What you describe sounds like normal behavior. Some things must change to adapt to new conditions.
2. I don't know how to quantify "stress" in a MOSFET or a diode, except to define it in terms of exceeding Absolute Maximum LImits as defined in the datasheet. Power dissipation of a given package is probably the biggest one and when the load is reduced you tend to move away from those limits thereby reducing the stress.
3. This means that the capacitor is more than capable of supplying the load current without an excessive drop in voltage. The efficiency may suffer, but I don't think it stops working. If this was a problem, you could always put a minimum load across the output.
4. I suppose, but proper component selection allows for the expected stress level with considerable safety margin. Nobody enjoys designing so close to the ragged edge that you periodically go over it.
EDIT: So, you have a basic open loop buck converter with no closed loop control. Says me, it cannot react well to a step change in load.

#### Papabravo

Joined Feb 24, 2006
18,981
Since your circuit is open loop, it has absolutely no ability to regulate either output voltage or maximum current or really much of anything in the presence of a load transient because of the FIXED duty cycle.

EDIT: I may have made the load transient go in the wrong direction. I made the load go from 500Ω to 45.45Ω. I can make the load go from a moderate 500Ω to 5000Ω if that was your original question.

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#### HRN88

Joined Jul 21, 2020
24
Since your circuit is open loop, it has absolutely no ability to regulate either output voltage or maximum current or really much of anything in the presence of a load transient because of the FIXED duty cycle.
View attachment 269016

#### Papabravo

Joined Feb 24, 2006
18,981
Increasing the load resistance doesn't do much of anything because, as before the duty cycle is unchanged.

#### Papabravo

Joined Feb 24, 2006
18,981
I did not add any subcircuit. The voltage controlled switch is a component that is built into LTspice. The component name is "sw". You need to supply a "model card" which is an anachronistic way of saying "a SPICE line" on the schematic, which I did, defining the properties of the switch. there are other properties that have defaults as defined in the LTspice Help pages. The purpose of the switch is to provide a load transient. there are other ways to do it, but this one is straightforward and understandable.

I'm not quite sure about "essential jaja". Is that a technical term in your area?

#### HRN88

Joined Jul 21, 2020
24
I did not add any subcircuit. The voltage controlled switch is a component that is built into LTspice. The component name is "sw". You need to supply a "model card" which is an anachronistic way of saying "a SPICE line" on the schematic, which I did, defining the properties of the switch. there are other properties that have defaults as defined in the LTspice Help pages. The purpose of the switch is to provide a load transient. there are other ways to do it, but this one is straightforward and understandable.

I'm not quite sure about "essential jaja". Is that a technical term in your area?
Ok I understand about the load transient, ty so much again. And, with essential, I refer that my control knowledge are basic.

#### Papabravo

Joined Feb 24, 2006
18,981
Did you notice that I changed the diode to Schottky? This is a common thing in DC-DC converters.

#### HRN88

Joined Jul 21, 2020
24
Did you notice that I changed the diode to Schottky? This is a common thing in DC-DC converters.
Yup, I was thinking, is better due to the fast reverse recovery time and low voltage drop?

#### Papabravo

Joined Feb 24, 2006
18,981
Yup, I was thinking, is better due to the fast reverse recovery time and low voltage drop?
What do you figure is your next step?
Have you established criteria for current and voltage ripple?

#### HRN88

Joined Jul 21, 2020
24
What do you figure is your next step?
Have you established criteria for current and voltage ripple?
Mmm, I think there are a lot of things to do to improve this design, for example add a parasitic capacitance to the inductor to see it self resonance, estimate the magnetic looses, a better calculation for the output capacitor and add an ESR, of course implement a closed loop control to control the current and voltage ( I would like to learn how to do this)

For this design I thought these parameters.
Input voltage: 70V
Output voltage: 15-55V
Output current: 2.0 A
Max output power: 60W
Efficiency: 85%
Fsw: 110 KHz

For my calculation I considered:
Ripple current 30% of I output
Ripple voltage 1V

#### Papabravo

Joined Feb 24, 2006
18,981
Mmm, I think there are a lot of things to do to improve this design, for example add a parasitic capacitance to the inductor to see it self resonance, estimate the magnetic looses, a better calculation for the output capacitor and add an ESR, of course implement a closed loop control to control the current and voltage ( I would like to learn how to do this)

For this design I thought these parameters.
Input voltage: 70V
Output voltage: 15-55V
Output current: 2.0 A
Max output power: 60W
Efficiency: 85%
Fsw: 110 KHz

For my calculation I considered:
Ripple current 30% of I output
Ripple voltage 1V
A couple of things:
1. Parasitic capacitance on a small inductor operating at 110kHz won't be much of a factor. You can and should add some series resistance based on the selection of an actual part.
2. Vo of 55V and Io of 2 A is a power output of 110 Watts. You input power will be more than that.
3. Add some DC resistance to the inductor
4. Add some ESR to the capacitor
5. 1V P-P ripple on a 15 volt output seems excessive, Id shoot for 1% or 150 mV P-P
Are you familiar with the following expressions for sizing the inductor and the capacitor in a buck converter?

$$L\;=\;\cfrac{V_0(1-D)}{\Delta i_Lf}$$
and
$$C\;=\;\cfrac{1-D}{8L(\Delta V_0/V_0)f^2}$$
Try them out with the values that you have chosen and the range of possible output voltages, current ripples, and voltage ripples. See if you have chosen wisely. A spreadsheet might be helpful for this exercise. Since you plan of having duty cycles in excess of 50% there are some additional problems you will need to deal with: more on that later.

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#### HRN88

Joined Jul 21, 2020
24
A couple of things:
1. Parasitic capacitance on a small inductor operating at 110kHz won't be much of a factor. You can and should add some series resistance based on the selection of an actual part.
2. Vo of 55V and Io of 2 A is a power output of 110 Watts. You input power will be more than that.
3. Add some DC resistance to the inductor
4. Add some ESR to the capacitor
5. 1V P-P ripple on a 15 volt output seems excessive, Id shoot for 1% or 150 mV P-P
Are you familiar with the following expressions for sizing the inductor and the capacitor in a buck converter?

$$L\;=\;\cfrac{V_0(1-D)}{\Delta i_Lf}$$
and
$$C\;=\;\cfrac{1-D}{8L(\Delta V_0/V_0)f^2}$$
Try them out with the values that you have chosen and the range of possible output voltages, current ripples, and voltage ripples. See if you have chosen wisely. A spreadsheet might be helpful for this exercise. Since you plan of having duty cycles in excess of 50% there are some additional problems you will need to deal with: more on that later.
Yes I know similar equations for the output capacitor and inductor, usually I use these