Why the output voltage remains normal in discontinuous mode?

[chmr]

Discontinuous mode ?

Hi all

I have a question about switching power...

In buck circuit, why the output voltage still remain normal even in discontinuous mode ?

Can anyone explain it in detail ?

Thanks

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chmr

 

[VVV]

Re: Discontinuous mode ?

Because the control loop adjusts the duty-cycle and keeps the output constant.
As long as the duty-cycle can be adjusted over the required range the output will stay in regulation.

 

[Engineer_Bob]

Re: Discontinuous mode ?

I don’t think it can, how are you measuring it?

 

[Alex_TSU]

Re: Discontinuous mode ?

These circuits can operate normally both in continuous & discontinuous modes. The current spikes provided to the secondary are absorbed by the filtering capacitor & the average output voltage is hold within the established range, controlled by the feedback circuit.

 

[Engineer_Bob]

Re: Discontinuous mode ?

With reference to a PWM feedback scheme

As you probably know discontinuous operation occurs when the load is low.
Effectively the current flow in the inductor is stopping or stopped.

Theoretically: If you have no load what so ever the capacitor will hold the output voltage up so the feedback condition will be satisfied. The output will stay constant.

In reality: I found that the capacitor voltage will gradually fall, the feedback will kick the circuit back into life and the transient response will become apparent. If I recall correctly, the output will start to do something called bursting and the SMPS becomes quite unstable. I’ve seen spikes with a continuously repeating period, i.e. the converter stays unstable. (LC circuits make good resonators).

So theoretically the output will stay steady because the charge in the capacitor is satisfying the feedback condition by presenting the correct voltage across the output.

Practically, I’ve connected a scope probe across the output of a buck converter with no load other than the feedback and the probe itself, I’ve seen the output bursting. This takes the form of spikes on the output.

Safety note

To be quite honest I’ve never tried to solve this problem in any way other than adding a minimum load, components are never perfect so I suspect that under these conditions you are getting some kind of initial inrush through the inductor. This initial current spike does present itself as a very high very narrow spike across the output. My concern here is for the capacitor (and the operator), the narrow spike has a high fundamental so thinking about the ESR, if you leave this running you could damage the capacitor or even blow it up! (hay I guess half the fun of power electronics is when it goes pop and faultfinding involves looking around the room for your components, lol)


If you are asking this question simply out of interest then the answer is given above, if however you are building a buck converter I would highly recommend a minimum load is included in your design to keep the SMPS stable.

If you are using a variable frequency scheme, I understand that running in discontinuous mode is desirable, but the filter is a pain to get right (don’t know about this never tried it)

Hope that’s of some help Engineer Bob

 

[VVV]

Re: Discontinuous mode ?

Let's consider the buck circuit in the first picture. Point A is the one you would normally probe with a scope to see if the circuit is working.
Now take a look at the second picture. The diagram on the left shows the buck in discontinuous mode under certain conditions. Applying Kirchhoff's law to the junction point where the load, the inductor and the capacitor meet, you get:
IL-Ic-Io=0 (with the directions indicated). So at any moment in time the sum of the three currents must be zero.

The inductor current is discontinuous. The output current is considered constant, at least compared with the switching frequency. So the capacitor current must be Ic=IL-Io. But the capacitor does not support a DC current. And the capacitor is chosen LARGE enough to present a low impedance in AC, so it will short to ground the AC component. In other words the capacitor current is exactly equal to the AC component of the inductor current.

Since the inductor current also has a DC component, that must be equal to the load current, because the capacitor does not conduct any DC. So the load current is equal to the average of the inductor current: Io=ILave. What would happen if the load needed less current (resistor increases)? The sum of the above DC currents must stay constant. If the inductor current does not change, then the load must continue to take the same current. But with a larger resistance, the only way to do so is with a proportionally higher voltage.
So that is what would happen, the output voltage would increase.

That increase in output voltage is sensed by the control loop which in turn will decrease the ON time and reduce the peak of the current and consequently its average (DC) until the voltage is brought back to the correct value.

The right side of the picture shows a similar situation: what happens if the input voltage increases. Again, the control loop adjusts the ON time so as to mainain the output voltage constant.

As you can see the control loop maintains the output voltage constant by adjusting the ON time and by that the inductor average current= load current. The only time the loop cannot maintain the output voltage is if the PWM duty-cycle cannot be adjusted down to very low values. In all practical situations the duty-cycle cannot be made arbitrarily small, say 0.00001%. There is always some minimum practical limit (2% or so) and below that it just goes right down to zero. Even under those conditions most control loops will maintain the output voltage constant (although with some higher ripple). That happens simply because after a few switching cycles at the minimum duty-cycle the output voltage increases a little too much and the control loop will turn off the PWM until the output voltage drops below the reference. Some PWM's have this feature by design, in order to minimize the current associated with driving the switching transistor, which affect the efficiency at light loads.

Formulas are not hard to derive, but I found that in this case they do not present any helpful info.

(In case you are wondering, the voltage at point A before the ON time is the output voltage, which comes through the inductor. The ringing is given by the inductor resonating with the parasitic capacitance of the diode, switching transistor and PCB).

 

[Engineer_Bob]

Re: Discontinuous mode ?

VVV, regarding your post, I must say I think that’s an excellent and clear explanation.

To anyone in the forum interested in power electronics I’d like to say the following:

When I design and build my own SMPS I usually custom build my own transformers/inductors, build the power section and manually drive the feedback with function generators.
I design the feedback circuits after measuring the performance of the power section. I find this approach works really well.

For anyone interested in SMPS / Power Electronics who wants to learn more about it, then I highly recommend this site.

**broken link removed**

You can even download the entire site in zip format and use it at your leisure.

I haven’t used this site for a few years, but I recall it being really useful when I was a student, and having found it again, I don’t think it’s changed……..enjoy.

Kind Regards

Engineer Bob