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I am testing the control loop stability of a DC-DC Converter.

One of the Pass-criteria of the DC-DC Converter states, "Cross over frequency should not be higher than 1/8th of switching frequency"

Which cross-over frequency is this referring to with respect to the DC-DC Converter Switching frequency and could you explain the concept behind the 1/8th factor?

Thanks.

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  • \$\begingroup\$ I believe the Z-transform methods are involved in sampled-system stability. \$\endgroup\$ – analogsystemsrf Sep 30 '19 at 5:11
  • \$\begingroup\$ you can google for "stability analysis of switching converters" get the knowledge of all the terms. Cross over frequency is the frequency where Gain plot reaches 0 dB.1. quora.com/… 2.electronics.stackexchange.com/questions/126568/… \$\endgroup\$ – user19579 Sep 30 '19 at 5:12
  • \$\begingroup\$ Yes, my confusion is there are 2 cross over frequencies. Gain Crossover Frequency and Phase cross over frequency. Which crossover frequency is that statement referring to and why the other crossover frequency is not considered? That's my doubt \$\endgroup\$ – Newbie Sep 30 '19 at 5:48
  • \$\begingroup\$ Gain margin and phase margin are both used in assessing relative stability of the closed loop, based on open loop metrics. The gain margin is the negative value of gain (dB) at the frequency where the open loop phase angle is -180 degrees (= phase crossover frequency),and the phase margin is (180 + phase angle) at the frequency where the gain is 0dB (= gain crossover frequency). \$\endgroup\$ – Chu Sep 30 '19 at 7:30
  • \$\begingroup\$ Yes. I understand. But if someone says just "Cross over frequency", should I take it as "Gain Crossover frequency" or "Phase Crossover Frequency". That's my question. \$\endgroup\$ – Newbie Sep 30 '19 at 10:07
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Which cross-over frequency is this referring to with respect to the DC-DC Converter Switching frequency

It's more than likely referring to the the LC resonant frequency (\$f_c\$) of the energy storage components within the DC-to-DC converter. See L and C below: -

enter image description here

Cross over frequency should not be higher than 1/8th of switching frequency

The L and C form a low pass filter that below resonance, barely alter the phase angle between input (the switching waveform) and output (the smoothed DC voltage). However, as resonance (\$f_c\$) approaches, the phase changes dramatically from near 0° to 180°. That change of phase is unavoidable and can turn a stable circuit into an unstable oscillator. For the LC filtering to be effective, \$f_c\$ has to be some way below the switching frequency. The further below the switching frequency, the lower the output ripple amplitude.

Simulation

Using an on-line simulator, consider L = 10 uH and C = 10 uF for the energy transfer components and, look at the green trace (phase response) below: -

enter image description here

Link to Interactive calculator

Slightly below \$f_c\$ = 15.9 kHz (referred to as \$f_n\$ in the picture above), the phase shift is quite close to 0° and this poses no threat of introducing loop instability. However, slightly above 15.9 kHz, the phase has shifted nearly 180° and this can really "shake the ground" when it comes to stability. This is why compensation circuits are added within the PWM control block (see top picture) to retard the 180° phase shift and prevent this oscillatory condition arising. The compensation is a counter-measure to unwanted oscillation.

Low output ripple vs faster closed-loop control

To achieve adequate filtering of switching voltages on the output, you need to keep the resonant frequency (\$f_c\$) of the energy transfer components (L and C) significantly below the switching frequency. The further you go below the switching frequency, the better the result i.e. lower output ripple voltage. The LC is a great low pass filter for this and, in the above picture, you can probably see that if the switching frequency were at 159 kHz (\$10\times f_c\$), the attenuation of the switching voltage will be 40 dB compared to DC. That's a 100:1 reduction

Example: if the switching is 10 volts p-p, the resulting 1st harmonic on the output waveform will be 100 times lower at 100 mV p-p. However, you also want to keep the resonant frequency high so that your closed-loop control system can react quickly to load and supply changes.

These two requirements are in opposition so a compromise is necessary.

Why 8:1? Why not 10:1? It's a rule of thumb and like most rules of thumb, you can choose to push the rule this way or that way depending on your most dominant needs.

Hopefully, the information above will allow you to see that the choice of LC cross-over frequency is a compromise based on juggling these somewhat opposing constraints: -

  • Good loop response to load and supply voltage changes (\$f_c\$ needs to be high)
  • Ensuring the compensation circuit is effective at the resonance (\$f_c\$ "right")
  • Minimizing output ripple voltage (\$f_c\$ needs to be low)
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  • \$\begingroup\$ Could you please explain the relation/difference between the switching frequency and the cross-over frequency \$\endgroup\$ – Newbie Jul 24 at 14:39
  • \$\begingroup\$ DC/DC converters have a switching frequency and they have analogue components (an inductor and capacitor in most cases) which have a cross-over frequency. One is an active frequency by which the transistors are switched and the other is the result of an LC circuit transfer function. \$\endgroup\$ – Andy aka Jul 24 at 14:50
  • \$\begingroup\$ Thank you very much. The analog components which you mention LC, are they the output capacitors and inductors which are present outside the IC in the case of TPS54620 IC? And why should this cross-over frequency be lower than the switching frequency (do you have any simple analogy for this? Please help to understand these two doubts. Thank you \$\endgroup\$ – Newbie Jul 24 at 14:53
  • \$\begingroup\$ Yes the external components and, my answer fully goes into explaining why the "cross-over frequency be lower than the switching frequency". \$\endgroup\$ – Andy aka Jul 24 at 14:55
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    \$\begingroup\$ Thank you very much for the clarification \$\endgroup\$ – Newbie Jul 26 at 12:08
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As mentioned by others, the crossover frequency refers to Gain Crossover frequency. A converter's control specification is mainly

  1. Steady state accuracy

  2. Transient overshoot & settling time.

The second part has to do with 0dB crossover frequency of the loop gain & the phase margin.Typically this frequency is set to 1/10 to 1/5 of the switching frequency, as the speed of the system response to load transients is governed by it. If wc is the crossover frequency then we can estimate the settling time to be 3/wc to 4/wc.For an acceptable transient overshoot, the phase margin is taken more than 45 degrees.

Higher the crossover frequency (i.e frequency at which loop gain = 1) faster the load response, But this should also be low enough to accommodate attenuation of switching noise.

In my opinion 1/8 is not a rigid number to go by, it is usually anything between 1/10 and 1/5 of the switching frequency.

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  • \$\begingroup\$ Thank you for the answer. I have a doubt. When you say,"Typically this frequency is set to 1/10 to 1/5" - What is this frequency are you referring to and how can it be set? And another doubt, in the bode plot, the gain curve represent that gain of which circuit? \$\endgroup\$ – Newbie Jul 24 at 6:31
  • \$\begingroup\$ You want to set your 0dB crossover frequency to 1/10-1/5 times switching frequency.This is the frequency that you finally want; if it's not already achieved by open loop, then you need to go for closed loop control. \$\endgroup\$ – SM32 Jul 24 at 7:05
  • \$\begingroup\$ You'll first need to see the open loop bode plot (or calculate it)to see the phase margin,gain margin,gain crossover etc.Then after you design your compensator(I like to do this in parts)check this bode plot and compare with the open loop one to see if you've made any progress in the right direction. \$\endgroup\$ – SM32 Jul 24 at 7:07
  • \$\begingroup\$ Thank you very much for the comment. Just 2 questions - 1. "You want to set your 0dB crossover frequency to 1/10-1/5 times switching frequency" - When you say this, how do I set this crossover frequency? Do I do it by selecting and tuning the output capacitors values or is it part of the internal IC Compensator circuit (say lets take TPS54260 buck regulator) or how? And 2. "This is the frequency that you finally want" - When you say this, - Where is this frequency measured in a buck converter (lets take TPS54260 for example) output stage regulator circuit? \$\endgroup\$ – Newbie Jul 24 at 8:27
  • \$\begingroup\$ In my answer,I assumed the compensator design would be done using basic R,C and OPAMPs,(which is fundamental in understanding how compensation works anyway).I design in 2 steps-a lead-lag compensator & a PI compensator.This has a method actually,when you need to choose wc.For example,if G(s) is a second order system with a complex pole pair,then choose the zero of the lead lag = w0,and pole at 10*w0. As for using TPS54260,you might want to check out section 8.2.1.2.11 & 7.3.20 till before 7.4. \$\endgroup\$ – SM32 Jul 24 at 9:25
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"Cross over frequency should not be higher than 1/8th of switching frequency"

Which cross-over frequency is this referring to...?

Such a statement is ambiguous considering that there are two crossover frequencies, i.e. gain crossover frequency and phase crossover frequency.

Ask the person who made the statement what exactly they are referring to. You could make an assumption they meant the gain crossover frequency, but how do you know for sure what they meant?

Ask them for clarification.

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Edit: In non mathematical terms, the crossover frequency can be likened to the bandwidth of the control system. The reason the crossover frequency should be less than 1/8th of the switching frequency is to avoid switching noise and ripple interfering with the controller. The controller should respond to the average output and 'ignore' with switching ripple.

Detailed answer: The crossover frequency is referring to where the magnitude loop gain bode plot of the system intersects with the 0dB axis. Referring to the first figure below, the transfer function of the system is given by:

$$ \frac{V_o(s)}{V_i(s)}=\frac{1}{1 + G(s)H(s)}$$

The loop gain is G(s)H(s) and is basically the point at which the gain of the controller plus converter is 1. If you measure or plot this, you will get a bode plot. Take a look at the second figure below for an example. The crossover frequency in the second figure is denoted by fc.

For information about how to measure the loop gain, take a look at this link. This presentation explains the loop gain in detail also.

enter image description here enter image description here

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  • \$\begingroup\$ Could you please explain the relation/difference between the switching frequency and the cross-over frequency? \$\endgroup\$ – Newbie Jul 24 at 14:39
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In answer to the first part of our question: The voltage output control is basically an inverting amplifier that monitors the output voltage for changes due to increasing or decreasing load and sends a signal back to the converter to increase or decrease the conversion to keep the output voltage constant. Like any amplifier it does not have constant characteristics from DC to light. The gain cross over is the frequency where the amplification drops from amplifying to attenuating and the phase cross over is where the feedback goes from being negative to positive. If the phase cross over frequency is lower than the gain cross over there is a range of frequencies at which positive feedback is amplified and the circuit becomes an oscillator. So the control loops are always designed to have the gain cross over be lower than the phase cross over. But that is internal oscillation.

The switching is a powerful external oscillation that feeds into the control loop. Since it is external it is independent of the internal phase angle and so not critical of the phase cross over frequency. But the gain at the switching frequency is important since it controls how much the switching oscillations are amplified or attenuated. So the cross over frequency they are referring to is the gain cross over.

The reason for the factor of 1/8 is a rule-of-thumb that the control loop should attenuate the switching oscillations by at least 3 octaves. With even a simple first order filter in the control loop limiting the high frequency gain response this gives 3 factors of 1/2 for a total of 1/8th gain (attenuation) of the switching oscillations. If this rule-of-thumb is approached in the design then the external components will have to filter out the remaining 1/8th.

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For any 2nd order low pass filter, 8x this breakpoint frequency results are 3 octaves with an asymptotic slope of -12 dB/octave or -36 dB at 8x the frequency where the fundamental switching frequency may be considered.

The crossover frequency for gain=1 is used for amplitude attenuation.

Shown by simulation below enter image description here

The filter affects the slew rate at which the loop can be corrected and ripple attenuation of switching spectrum are design tradeoffs.

The better design spec and test for stability is to define at least 2 step loads of different sizes and directions for better observation of overshoot stability. (e.g. 50~100% 90~10% step current) . The choice is yours depending intended design load applications. Don't expect all DC-DC supplies to have the same stability.

AC ripple must be measured into an AC coupled 50 Ohm termination at DSO to exclude false noise.

There are better stability criteria tools but beyond the scope of this question.

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  • \$\begingroup\$ Thank you very much for the answer. But I am a beginner. I just want to understand certain things electronically rather than mathematically. I understand that math is inevitable. Just a more intuitive explanation would help me lot on how the control loop stability is done and why that frequency values should be on where they should be \$\endgroup\$ – Newbie Jul 26 at 15:45
  • \$\begingroup\$ Design is often about ratios than absolute numbers. So 1/10th is based on attenuation ratio of about 1% at 1/10th f (second order squared) 1/8th is just a little less attenuation \$\endgroup\$ – Tony Stewart Sunnyskyguy EE75 Jul 26 at 15:48
  • \$\begingroup\$ beware that the ESR of the cap has a big impact on ripple \$\endgroup\$ – Tony Stewart Sunnyskyguy EE75 Jul 26 at 15:51

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