# Stability - Driving a Capacitive Load

But I have no proper guide/mentor or anyone who can provide me better answers other than users in this site. So, please pardon me for this. I rely on this site for answers which can Solve my queries and help me with my understanding.

If you seen my previous questions, you would notice that a majority of my questions would have been related to control loop stability analysis of a dc-dc converter. I was not able to grasp the concept of zeroes with respect to electrical components. All I understood is that, If I have a pole, it means the electrical component will oscillate (at some frequency) and the system will be unstable. But what happens when a component provides a zero instead of a pole? How does one understand or find whether a component will provide a zero or a pole? What effect will that zero have in the system?

So, I am trying to learn it by watching questions and answers in this site, reading app notes, watching lectures and so on to get clarity.

In this Video, around 3:08 , he says that when the capacitance magnitude matches the resistance magnitude, we get a zero.

I don't understand that. Can you tell me :

1. Is he meaning that at some frequency, the capacitance impedance will be equal to the resistor's resistance? If so, at this point, what will the combination of the resistor and the capacitor connected to ground look like? And when he says zero, what effect will it have on the system?

Please do not think that I am asking multiple questions. If anyone can provide the answer which can help me clear my doubt regarding the poles and zeroes in an electrical circuit (in simple electrical terms for a beginner to understand) and how one can figure that this component will provide a pole or a zero, then that answer would solve all my questions.

Would be really grateful.

• wait a minute, having a pole will not "... If I have a pole, it means the electrical component will oscillate (at some frequency) and the system will be unstable." – jDAQ Nov 21 '19 at 3:31
• you are seeking to understand a difficult topic. Congratulations. Have you examined "Body Plots". I'll give you an answer to assist you. – analogsystemsrf Nov 21 '19 at 3:33
• Based on my previously asked questions and reading other app notes on the web, I have some crude understanding that a pole in the system (due to a component) will make the system unstable (as that component will oscillate about some frequency, maybe its resonant frequency). But I don't know what a zero will do to a system. Most of the answers that the users provide in this site are quite tough to understand for me as I am able to only understand simple terms related to electrical. A little bit detail answer with simple words and sentences or an analogy or example can help me understand – Newbie Nov 21 '19 at 3:44
• As you can understand from my username, I am quite new to the field. Yes, I have examined the body plots. That's where I get a little stuck. I am stuck with the relationship between the crossover frequency and the switching frequency. A little simple detail would help in this regard. Sorry to ask so many questions. Actually, I want to get my basics right. Please let me know if I am wrong somewhere with my basics – Newbie Nov 21 '19 at 3:46

Ok newbie, i've read your questions and comments and have a gauge of your understanding level. what classes have you taken pertaining this subject?

If you're designing a DC-DC converter for stability using bode plots (poles and zeros), you received great answers already. You need to make it a goal of understanding them.

It boils down to what another answer alluded to:

Poles/Zeros zeros have +20db/decade gain and poles have -20db/decade attenuation. You asked why in a comment? At the stage of PSU design you should have learned it, follow the math in my link. Simply, in an RC circuit, as the frequency goes up, the impedance of C will become lower and lower, and the ratio of R - C will drop/increase and this rate happens to be at 20dB/decade. Or thinking in the time-domain instead of frequency domain, the capacitor takes time to charge, so any fast changes in voltage will not be relayed with attenuation.

There is also phase. +/-90 degrees for a single zero/pole (respectively), 180 degrees for double pole/zero. The phase transitions quickly at the R+C or L+C crossover/corner frequency.

Multiple poles/zeros will have Nx20db/decade gain. so 2 pole will have 40db/decade (on average), roughly. With higher order yet poles/zeros other things come into play involving phase, and for real accuracy s-diagrams are needed. This link describes them in great detail: https://web.mit.edu/2.14/www/Handouts/PoleZero.pdf It's not critical to understand this all for PSU design, as it's usually a combination of passive poles or at most a 2 pole LC, but it's good to understand the basics.

With this information you should know how to draw a bode plot. Good link: https://pages.mtu.edu/~tbco/cm416/bodestab.html

Intuitively, this means a pole "slows down" the response of the system, because higher frequencies are attenuated. This often, but not always, has a stabilizing effect. A real-world example of a pole is heating up a pot of water to exactly 90C. You can crank it on "HIGH" (high input power) but the temperature (output voltage) takes like 5 minutes to warm up. The circuits have their own physical limitations that limit their speed, described w/ poles/zeros.

By inverse logic a zero "speeds up" the response of a system, by applying gain to higher frequencies which is more likely to have a destabilizing effect. Intuitively this makes sense: if an "ideal" zero will have more and more gain as frequency increase, then at some absurd 2.4GHz I will have +100dB of gain. So any teeny tiny interference from a wifi-router will cause the output voltage to change by an incredible amount. bad.

Both poles and zeros affect phase and gain, which affect stability.

There's a final point about the gain part of the bode plot: the gain level is usually set externally by an R+C and includes factors like the gain of the amplifier, any R-dividers, and so on. It's a variable that can be changed to affect the stability, where generally lower gain will mean a slower but more stable response.

Stability Critereon Why is this gain/phase behavior of poles/zeros important?
A good read: https://pages.mtu.edu/~tbco/cm416/bodestab.html But basically, one things need to occur for sustained oscilaltions (instability): +When PHASE reaches -180 degrees, there is greater than unity gain ( A > 1V/V, A > 0dB) +Or, described inversely: when gain approaches unity (0.001dB), there is 180 degrees of phase shift

In reality, you don't want to approach anywhere near 0dB at 180 degrees phase shift, so the concepts of "gain margin" and "phase margin" as described in the link arise. As you "approach" this unstable point, you may have "DC stability" where your DC level will be correct, but a sudden shift in the DC operating point (say a step change in the load) will cause decaying oscillations at frequencies near the 0db/180degree point.

Gain margin is how much gain you have to spare when the phase reaches the trechearous 180 degrees. You want it to be well below the 0db threshold, maybe like -20dB. Phase margin is how much phase you have to spare when gain hits 0dB. You want about -45 degrees for damped performance.

Intuitively, this stability critereon makes sense: If you have GAIN while having 180 degree inversion, your system went from negative feedback to positive feedback. Having 180 degree inversion (and thus some positive feedback) is fine if it's gain is <0dB, because it won't accumulate. But having >0dB gain and positive feedback means a positive feedback loop that will grow as it oscillates. Even if gain is only 1.001 or like +0.1dB, it will gain until it reaches some other physical limitation.

To bring it to your original question, why does a C on the output cause instability? A big capacitor right at the output is an R+C zero in some converter topologies, and L+C in others.
It generally has a stabilizing effect on DC-DC converters, but can also cause instability. It depends on what this RC did to the bode-plot and whether it ruined the gain/phase margin of the converter.

Thinking about it in the time-domain, a super large C will cause the output voltage to change SLOWLY. This, like i said, usually stabilizes the output voltage, but you have to consider what the PSU system is doing. Basically, will the PSU patiently increase the current to charge the cap, or will it "accelerate" too quickly and dump huge current into the cap, overshooting. In the 90C pot of water example, you'd hope your control scheme would turn the power down a bit from HIGH as the pot of water approaches 90C, lest we overshoot and accidentally boil the water. This rationale can be applied to any aspect of the PSU, they all have bandwidth/gain--i think of it as inertia as in the pot of water. The output voltage (capacitor) can only gain voltage so fast. The inductor can only increase current supplied so fast. The amplifier only has so much gain at high frequencies. Bode plots allows us to quantify it.

There are also empirical ways to categorize a PSU (or any) control system. By applying a "Step" change to the input voltage or output load, the converter closed loop response can be characterized, and stability inferred.

Let's look at the 2 circuits he presented.

The first is with Rop and C

simulate this circuit – Schematic created using CircuitLab

If we write out the transfer function for this, we get $$H(s) = \cfrac{1}{1+sR_{op}C}$$.

The denominator, D(s), contains a pole. A pole by itself is not bad. It depends. As the video explains, if you cross the 0 dB with a slope of -40dB/dec then it is unstable.

In the example above, you already have a pole from within the opamp (which is not shown). This pole from within the opamp, causes a -20 dB/dec slope. This is seen by the first dot in the bode plot.

Now we are driving a capacitive load, and looking at the transfer function we can see that it also adds a pole. The previous pole, plus this new pole, causes the slope of the bode to decrease by an additional -20 dB/dec. This is now -40 dB/dec. And IF it crosses the 0 dB at this slope, it will be unstable.

So let's look at what he did by adding that additional resitor. As always, lets look at the transfer function.

simulate this circuit

$$H(s) = \cfrac{1 + sR_{add}C}{1 + s(R_{op}+R_{add})C}$$

Now things are different arent they ?

With this additional resistor in a strategic location, we now not only have a pole but we also have a zero. To make things even better, the zero is located after the pole which means, at some point, we have a -40 dB/dec slope (due to the 2 poles) and then shortly after, since the frequency of the zero is not that far ahead, we increase our slope by +20 dB/dec so that our total slope now is -20dB/dec. And when this crosses the 0dB, it's stable.

• Thank you very much. I have two questions with your answer. First, why and how do you say that when a bode crosses the 0dB with -40dB/dec it will be unstable and it will be stable if it crosses the 0dB with -20dB/dec? Second, how do you say, by adding a resistor in the second circuit, you have a zero and a pole? I have trouble at this level of understanding. You answer seems to be very complex to me. I can't even understand why the slope has to be -20dB/dec? Why can't the slope be some other number like -30dB/dec or -35dB/dec ? And how can you directly say that this circuit has a pole/zero? – Newbie Nov 21 '19 at 3:56
• Sorry to say, I am truly a beginner. I have a lots of queries for which I don't get answer on the web or materials. I just want to understand the basics of the concept. I don't necessarily need the answer related to the video I asked in the questions. The video is just a means of getting my basics and understanding of the concepts in line. – Newbie Nov 21 '19 at 3:58
• @Newbie Do you understand decibels ? Do you understand the s-domain and what it means ? If you are not familiar with those, learning control systems is going to be really hard for you. And your questions might have to start at much more basic level. Try this first - electronics.stackexchange.com/questions/374247/… – efox29 Nov 21 '19 at 4:00
• @Newbie A zero is when the H(s) becomes zero. This is done when the numerator becomes 0. A pole is when H(s) goes to infinity. This is done when the denominator goes to 0. In both cases, there is a pole. In the first example, $sR_{op}C = -1$ and $s(R_{op} + R_{add})C = -1$. If those conditions are true, then the denominators of their respective H(s) functions become 0. Hence, a pole. – efox29 Nov 21 '19 at 4:05
• Thank you. See, when you say, that you have a zero or a pole, you are deriving your answer by quoting the transfer function of the system. From the transfer function, I can understand. But how can you say, whether the system has a zero or a pole without looking at the transfer function and only by looking at the circuit. In the video too, he also says we have a zero and a pole without having to write the transfer function. I want to understand more in only electrical terms rather than understanding with transfer functions or math – Newbie Nov 21 '19 at 4:09

[ edited to add simple example of POLE & ZERO in a circuit ]

Lets consider this circuit; the opamp in the bitmap has a RIAA vinyl compensation RCRCR network embedded in the opamp feedback element. For the schematic, examine the left most column, finding the "RIAA EQ" illustration.

The feedback has two poles (at different frequencies) and 2 zeros (at different frequencies). These 4 frequency "points" will show up in the math that describes the system. In the example below, there is no oscillatory behavior, so there is no "pairing" of poles.

Poles will look like 1/( Frequency + J * 2 * PI * 50 Hertz) as you might notice in the gain/phase plot; the amplitude (the "gain" will be dropping as the frequency increases to the right.

Zeros will look like (frequency + J * 2 * PI * 500 Hertz), as you should notice the gain starts to level off, not falling so rapidly as in the 60 to 160 Hz region.

The performance of this system includes a "almost flat" region for +- 1 octave around 1,000Hz, followed by another pole at 2,000Hz whereupon the Gain starts a steep decline once again.

And finally, out near 100,000 Hz the opamp's feedback network (the RIAA Equalization, which I selected from the left-hand menu, causes the closed loop gain once again to level off, forming another zero.

Again, this circuit is stable.

Now letx examine a combined opamp-pole-zero plot.

simulate this circuit – Schematic created using CircuitLab

With the POLE at 160Hz and the ZERO at 100X higher, or 16KHz, the phaseshift almost increases by another 90 degrees (actually about 85 degrees) to be almost -270 degrees near 1,600Hz; above 1,600Hz the ZERO has enough effect on the opamp feedback that the phaseshift reverses the trend and heads back to -180 degrees. And notice the phaseshifts are delays, thus MINUS humbers.

• I really thank you for the answer. But you don't seem to understand what my level of understanding is. I am stuck at a far more basic level. Please look at my questions. They are so basic. Please understand that I am stuck at that level and have not come up to read and understand the answer you have provided me. I have tried reading all possible materials but have not got a clear understanding. I am reaching out to you all, to provide a very basic answer to the zeroes and poles. Please look at the type of questions that I ask. Answers to those questions will help me. – Newbie Nov 21 '19 at 4:05
• are you comfortable with vector algebra? The behavior of a pole --- 628/( J * 6.28 *Freq + 628) produces a flat frequency response for low frequencies; at 100Hz ( 628 radians/sec), the magnitude is down by 1/1.414 and the phaseshift is 45 degrees. – analogsystemsrf Nov 21 '19 at 20:42