Think of a converter transferring the energy in a two-step approach, like a boost or a buck-boost converter: first you store the energy in the inductor during the on-time then transfer it to the load during the off-time. Should a sudden power demand occur, the converter can't immediately react and must go first through another storing-energy phase before answering the demand. This intermediate phase naturally introduces a delay in the response to a change: the current in the inductor has to grow cycle by cycle (it cannot instantaneously jump to the next current setpoint) but this current increase is hampered by the inductor value and the available volt-seconds. The mathematical model of this delay in the response is the familiar RHP zero which appears in the control-to-output transfer functions of the said converters.

Above quote is from user Verbal Kint in this answer on the question: "Effect of right half plane zero on bandwidth of control system"

Why the physical phenomenon describe above by Verbal Kint can be modelized as an RHP Zero ?

Thank you very much and have a nice day !

  • \$\begingroup\$ For an example of interpretation for RHP zero in boost converter (zero of non minimal phase), please refer to my other response here: electronics.stackexchange.com/a/278272/22676 \$\endgroup\$ Oct 25, 2020 at 18:00
  • \$\begingroup\$ Hi, thank you for your answer ! I didn't get where you explained why this phenomenom can be modelized by a zero ? You mean "the inductor is disconected" so the small signal output is equal to zero for a certain time ? This is what you want to mean ? Clearly that is not clear in my mind ... I am sure that it is simple ... Well I will try to read more and more ... \$\endgroup\$
    – Jess
    Oct 25, 2020 at 20:21
  • \$\begingroup\$ Which is weird is that a zero leads the phase where as it seems to lag the output voltage from the control voltage ... \$\endgroup\$
    – Jess
    Oct 25, 2020 at 20:23

1 Answer 1


The right-half-plane zero (RHPZ) in a switching converter featuring an indirect energy-transfer scheme is present because of two things:

  1. the output current delivered to the load depends on \$(1-D)\$. For instance, in a boost converter operated in continuous conduction mode (CCM), the output current is the diode average current expressed as: \$I_{out}=I_L(1-D)\$ in which \$I_L\$ is the inductor average current. You immediately see that if \$D\$ the duty ratio increases because the loop instructs the controller do so (power demand increases), the \$(1-D)\$ term reduces and you actually deliver less current so \$V_{out}\$ drops.
  2. the subtlety which is often overlooked is how the inductor peak current increase after the duty ratio is up can compensate for the shrink in \$(1-D)\$. Look at the below drawing:

enter image description here

This is the instantaneous current flowing in the boost diode. In the left-side drawing, the average current \$I_{d0}\$ is actually the area \$A_1\$ averaged across the switching cycle. This area depends on the peak current \$I_{p0}\$ and obviously the duration \$(1-D_0)T_{sw}\$. When the power demands occurs and the controller asks for more energy delivered to the load, it pushes up the duty ratio a little bit. If the inductor is of small value or if the voltage applied across the inductor during the on-time (the so-called volt-seconds or V-s) is high enough, then the new peak current \$I_{p1}\$ might be sufficiently large to compensate for the reduction in \$(1-D_1)T_{sw}\$ and \$A_2\$ is larger than \$A_1\$. When this is respected, then the output voltage increases as it should despite the increase of \$D\$. Now, if you deal with a high inductance or you operate from a low input voltage, then the available V-s might be too small and \$I_{p1}\$ will not be high enough leading to \$A_2\$ < \$A_1\$: the output voltage drops despite an increase in \$D\$. The control-to-output control law is reversed and the converter is unstable.

The below sketch shows how it works when the duty ratio slowly moves, giving enough time to always ensure \$A_2\$ > \$A_1\$:

enter image description here

There is no problem and the output voltage increases as expected, following the control variable \$D\$. Now, speed up the duty ratio change from 200 µs to 10 µs and the situation differs:

enter image description here

The duty ratio variation is too fast and the inductor cannot increase its peak current quickly enough, engendering a reduction in the diode average current and a drop in \$V_{out}\$. this latter falls before taking off as it should. This is the typical physical manifestation of a RHP zero in a control system.

As a conclusion, you need to purposely slow down the converter so that even if there is a fast demand on the output, it does not propagate and create a fast reaction on \$D\$. The crossover should thus be selected so that there is always enough time for the inductor to build energy cycle by cycle and answer the power demand. Practically speaking, you meet this goal by limiting crossover to 20-30% of the worst-case RHP zero position. Should this RHPZ occur at 1 kHz, you would need to limit \$f_c\$ to around 250 Hz for an acceptable performance. Should you try to push it higher and you may not have enough phase margin.

By going through small-signal analysis (with the PWM switch model for instance), you can unveil the RHP zero in the transfer function. Look at my APEC 2013 seminar, starting slide 65, there is the derivation you want. You could also look at the interesting approach followed by the legendary Lloyd Dixon when he was with Unitrode where he derived the RHP zero position without going through a complete model analysis.

As a final note, whether the converter operates in voltage- or current-mode control, the RHPZ remains at a similar location. In DCM, there is still a RHPZ but relegated to high frequencies and it is generally not a problem unless you need a high crossover frequency.

Additional Edit

You can see the effect of adding a RHP zero to simple 1st-order system featuring a LHP pole and a LHP zero. When you turn the LHP zero into a RHP zero, the time-domain response to an input step first drops below 0 before taking off later on. This is what you observe in the third graph, typical of the RHPZ manifestation:

enter image description here

This video nicely illustrates the time-domain response depending where the poles and zeroes are located in the \$s\$-plane: really easy to follow.

  • \$\begingroup\$ Thank you for your answer. Actually what I do not understand is why as you said this phenomenon "is the typical physical manifestation of a RHP zero in a control system". I completely understand your explanation (as I have already read it in one of your books). Btw, I really appreciate all your books (I read one of them and I'm actually reading an other of your books) . There are very clear and there are many things really interesting especially small signal analysis ! I ve read 2 others books about power electronics and they did not deal with small signal analysis as you do ! So THANK YOU ! \$\endgroup\$
    – Jess
    Oct 26, 2020 at 13:15
  • \$\begingroup\$ The question could be why it is the typical manisfestation of a RHP zero and not for example an LHP zero ? \$\endgroup\$
    – Jess
    Oct 26, 2020 at 13:16
  • 1
    \$\begingroup\$ I spend a lot of time on these books, crafting all the details so that reading is fluid and easy to follow. I am happy if you like them : ) Regarding the RHPZ, you can see how it manifests itself by comparing the time-domain response of the transfer function featuring a LHPZ or a RHPZ. This is well described here but there are no pictures unfortunately. \$\endgroup\$ Oct 26, 2020 at 13:25
  • \$\begingroup\$ I do not know how it could be better ! ;) \$\endgroup\$
    – Jess
    Oct 26, 2020 at 17:15

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