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The RHZ in switch power supply

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huojinsi

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remove boost converter rhz

Hi everyone:
From some papers, i know a RHZ is produced in boost and flyback mode. But i am not clear why the RHZ is produced and how to produce this RHZ ? Pls elaborate this RHZ.
Thks in advance!
 

VVV

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The RHZ is only found in boost and flyback converters operating in CONTINUOUS CURRENT MODE.

I think the easiest explanation of the phenomenon is that given by Pressmann.
In a boost (or flyback, which is a boost-derived topology) operating in continuous current mode, the duty cycle SHOULD NOT CHANGE WITH THE LOAD. That is because the duty-cycle is only a function of the input and output voltages. For example, for a boost operating in continuous current mode, the duty-cycle is:

DC=1-Vin/Vo.

Obviously this does not depend on anything else other than the input and output voltages. (In practice, a small change is observed, since there are load-dependent voltage drops in the transistor, inductor and diode, which effectively get subtracted from thos voltages).

We have established that the duty-cycle should not depend on the load. But a sudden increase in the load current will cause the output voltage to decrease slightly, for the reasons shown above: the voltage drops in the transistor, inductor, diode will all increase with increased current.
This output voltage drop will cause the error amplifier+PWM to take some action and bring the output voltage back to the initial value, although the change may be only small.
But what does the error amp/ PWM combination do in order to increase the output voltage? It will INCREASE the duty-cycle. But this results in NARROWING the current trapezoid during the flyback (reset) portion. But this is WRONG!!!!
This results in a temporary DECREASE in the output voltage.
The situation does not last forever, though. The increased duty-cycle causes an increase in the transistor on time and so the transistor current (and inductor energy) will have more time to increase. Thus, eventually the current will establish itself at the correct level, the output voltage will increase and the error amp will decrease the duty-cycle to the correct value.
Without going into details, as you can imagine this process of the current increasing to the new value takes time and this depends on the inductance, therefore the RHZ depends on the inductance.

So an increase in the duty-cycle results initially in a DECREASE of the output voltage! Exactly the opposite of what is necessary!
This is the RHZ effect: the error amplifier tries to compensate but the effect is "THE WRONG WAY". (In a Bode plot a RHZ affects the gain like left half-plane zero, but it affects the phase the opposite way, like a pole; since the phase is reversed, you can say it goes the "wrong way").

The normal solution to the RHZ issue is to decrease the bandwidth of the error amp. You can understand why by looking at the Bode plot (remember, the RHZ bends the gain curve up by 20dB/dec, but lowers the phase margin just like a pole does).
Or, intuitively, you can see that you do not want the error amplifier to respond very quickly, since you know it will cause the output voltage to change the "wrong way", so you slow it down so that by the time it responds, the inductor current has already risen somewhat towards the new value, so the effect of the RHZ won't be so noticeable.
 

    huojinsi

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huojinsi

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Thks vvv for ur doing much explanation to me!
From ur explanation, i know when the load current suddently increases,the EA/ PWM combination will increase the duty cycle, but actually the output voltage temporarily decreases due to the voltage drops in the transistor, inductor, diode. Is my comprehend right? How to understand this words
But this results in NARROWING the current trapezoid during the flyback (reset) portion. But this is WRONG!!!!
.
Will u give me a wave picture to describe this phenomenon?

In EA compensation, generally which configration is choosed to eliminate this RHZ effect?

Thks in advance!
 

VVV

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Take a look at the picture.
In the first plot, the CCM boost is in steady-state condition. The load current is equal to the diode average current (the average of the red trapezoid).

In the second plot you see what happens when the load current increases. The PWM increases the duty-cycle, which results in narrowing the flyback portion (red waveform). Although there is an increase in height the average drops, because of the narrowing. That means your supply delivers even LESS current than before the PWM took action, resulting in an even greater voltage dip than if the PWM had done nothing at all.
If the PWM had done nothing at all, at least the output current would have been higher right after the load current increased, it would not have gone LOWER, the wrong way. Of course, eventually the duty-cycle will be brought back to normal by the PWM, when the output voltage is restored to its nominal. At that point, the duty-cycle will be remarkably close to the one in the first picture.

By comparison, plots 3 and 4 show the operation of a discontinuous flyback. In this case, the triangular diode current has "room" to increase in width, so the average increases right away, as demanded by an increase in load current. Thus, the average no longer moves "the wrong way".
The duty-cycle will in this case remain larger, though, even after the voltage is restored.
 

    huojinsi

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huojinsi

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Thks VVV !
According to ur second plot wave, i see when the load current increases, the duty cycle also increases, but the diode conduction region(red wave) is narrowing, so the output current inversely decreases, so it the Wrong Way. Now my concept is more clearer than before.
From plots 3 and 4, i can see no RHZ effect is produced in discontinous mode. But i am not familiar with discontinous mode. Now i find the switch power theory knowledge is much and very complex, so i have to learn step by step.
 

VVV

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That is correct. The RHZ only occurs in boost-derived topologies operating in continuous current mode.
 

    huojinsi

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dennislau

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I read some paper which say, in boost-typed converter, there will be RHP zero for trailing-edge modulation while there may be LHP zero (i.e. remove the RHP zero) for the leading-edge modulation.

Could anyone can explain the reasons behind?

Thanks
 

hoonsk

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Hi,
I wish to find out usually how do we usually deal with this RHZ in Boost Converter ? Just put the RHZ out of the bandwidth ? Or there is a technique to cancell the RHZ ?

Thanks !
 

huojinsi

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In my opinion, when we compensate EA in Boost Converter, generally a pole is introduced to cancel this RHZ.
 

VVV

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Unfortunately, we reduce the bandwidth, to keep the RHZ outside it.
Why? Because to cancel the RHZ you would need a right-half-plane pole.
So the usual method is to just reduce the bandwidth.
 

    huojinsi

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huojinsi

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Hi VVV, nice to see ur post again!
When a pole is introduced to cancel this RHZ in BOOST Converter, u say this introduced pole will reduce the bandwidth, why? What relation does this right-half- plane pole have with bandwidth? PLs give detailed explanation.

Thks in advance!
 

dennislau

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huojinsi said:
Hi VVV, nice to see ur post again!
When a pole is introduced to cancel this RHZ in BOOST Converter, u say this introduced pole will reduce the bandwidth, why? What relation does this right-half- plane pole have with bandwidth? PLs give detailed explanation.

Thks in advance!
I think VVV means to keep the RHZ outside the bandwidth, the only way is to reduce the bandwidth. It is because if we want to cancel the RHZ, we would need a right-half-plane pole but the right-half-plane pole is usually or definitely unstable for the system which would not be used in practice.
 

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dennislau got it correctly. You want to avoid a right-half plane pole, since that would make the system unstable.

Therefore, to ensure the RHP zero does not decrease the phase margin too much by the time (read frequency) the gain drops to 0dB, you reduce the bandwidth.
So the gain reaches 0dB at a lower frequency, where the effect on phase of the RHP zero is not too pronounced.
 

    huojinsi

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