Miscellaneous Talk on Miller Capacitor

maychang

Miscellaneous Talk on Miller Capacitor [Copy link]

 

Miscellaneous Talk on Miller Capacitor

Let’s first look at Figure (1).

figure 1)

This circuit can be said to be the simplest non-pure resistance circuit, with only one resistor and one capacitor.

If the input (IN) signal is a step function (called Heaviside function in some textbooks), that is, the voltage is zero before time zero and 1 after time zero (of course, it can also be a certain constant value, which can be done by multiplying a constant), and the capacitor does not store charge before time zero, then the voltage across the capacitor gradually rises. Roughly as shown in Figure (2). In Figure (2), the red curve is the input signal (step function), and the blue curve is the output signal, that is, the voltage across the capacitor. As can be seen from the figure, the voltage across the capacitor gradually rises, getting closer and closer to the amplitude of the input signal over time. Mathematical analysis shows that the blue curve is an exponential curve. In actual circuits, this RC circuit is often used to delay signals. If the input signal is a sinusoidal signal, then the output signal is still a sinusoidal signal, but the amplitude is reduced. The amount of reduction is determined by the resistance value of the resistor and the capacitive reactance voltage divider of the capacitor. The capacitive reactance decreases with the increase of frequency, while the resistance value does not change with the frequency, so the ratio of the voltage across the capacitor to the input voltage decreases with the increase of frequency. In other words, low-frequency signals can easily pass through this RC circuit and are reduced very little, while high-frequency signals cannot easily pass through this RC circuit and are reduced more. Therefore, this circuit is a low-pass filter circuit. In addition, the output signal is not out of phase with the input signal, but lags behind the input signal. The higher the frequency, the more the phase lags, up to 90 degrees. Since the differential equation of this circuit is a first-order differential equation, this circuit is also called a first-order low-pass filter circuit.

figure 2)

In this circuit, if the voltage at the right end of resistor R rises by a small value, such as 1 millivolt, the charge stored in the capacitor will increase by a small value, and the value of the charge increase is proportional to the capacitance of the capacitor. This is determined by the definition of capacitance C=q/u. Obviously, the more charge is added to the capacitor, the greater the capacitance is from the perspective of resistance. In actual circuits, capacitance C is unchanged and is a constant.

What will happen if we connect the end of the capacitor that was originally grounded (the voltage is always zero) to the output of an inverting amplifier, as shown in Figure (3)?

image 3)

Assuming that the voltage gain of this inverting amplifier is A, then the voltage at the junction of the resistor and capacitor, i.e., the input terminal of the inverting amplifier, rises by a small value such as 1 millivolt, then the output terminal of the amplifier will drop by A millivolts. If A=100, the right end of capacitor C will drop by 100 millivolts. We can see that the charge stored in the capacitor will increase to 99 times that in Figure (1), close to 100 times. This is because the voltage change across the capacitor is 99 times that in Figure (1). Then looking from the resistor to the right, the capacitance of the capacitor is not C, but 99C. Because the charge stored in the capacitor must flow into it from the resistor after all. It is as if the capacitor in Figure (1) has become larger and has become a capacitor with a capacity of 99C. This phenomenon is called the Miller effect. This 99C capacitor is of course virtual and is generated by the voltage change at the output terminal of the inverting amplifier, and is called the Miller capacitor. Approximately, it can be considered that the capacitor C is amplified by as many times as the voltage gain of the inverting amplifier. If this circuit is used to delay the input signal, then approximately the delay time will also be increased by as many times as the inverting amplifier voltage gain.

The difference from Figure (1) is that the output in Figure (3) is inverted with respect to the input.

In an actual inverting amplifier circuit, such as a common emitter amplifier circuit composed of bipolar transistors, or a common source amplifier circuit composed of MOS tubes, if a capacitor is connected between the collector and the base, as shown in Figure (4), since the output of the common emitter amplifier circuit is inverted with the input, looking at the capacitor from the input resistor to the right, the capacitance is also approximately amplified by A times, where A is the voltage gain of the common emitter amplifier circuit.

Figure 4)

The voltage amplification factor of the bipolar transistor common emitter amplifier circuit is related to the characteristics of the transistor, such as the current amplification factor of the transistor, the load, etc., so the amplification factor of the capacitor C is also related to these factors.

It should be noted that even without external capacitor C, the collector junction of the transistor still has distributed capacitance. This distributed capacitance from the collector to the base will also be amplified by A times in the common emitter amplifier circuit, but it is not shown in the schematic diagram. This amplified distributed capacitance from the collector to the base is also called Miller capacitance. This capacitance is an important factor affecting the high-frequency characteristics of the common emitter amplifier circuit.

If the bipolar transistor in Figure (4) is replaced with a MOS tube, because the common source amplifier circuit is an inverting amplifier, the drain and gate signals are inverted, so the Miller effect still exists, and the capacitor across the drain and gate will also be amplified by A times, where A is the voltage amplification factor of the MOS tube common source amplifier circuit. Similarly, even without an external capacitor, the distributed capacitance between the drain and source of the MOS tube will be amplified by A times. The Miller capacitance, which is not shown in this electrical schematic, will have a considerable impact on the switching power supply, limiting the switching speed of the power switch tube of the switching power supply. It can be seen from Figure (4) that the smaller the resistance R, the smaller the impact of the Miller capacitance. Therefore, the switching power supply always requires that the driving source of the power switch tube can output a sufficiently large current, that is, the driving source is required to be close to an ideal voltage source. In this way, the impact of the Miller capacitance can be reduced and the power switch tube can have a sufficient switching speed.

maychang

This post is an explanation of the role of capacitors C1 and C2 in the above figure.

@S3S4S5S6.

se7ens

Good article, mark it first

S3S4S5S6

maychang posted on 2023-10-30 17:37 This post is an explanation of the role of capacitors C1 and C2 in the above figure. @S3S4S5S6.

"The smaller the resistance R, the smaller the influence of the Miller capacitance. Therefore, in a switching power supply, it is always required that the driving source of the power switch tube can output a sufficiently large current, that is, the driving source is required to be close to an ideal voltage source. In this way, the influence of the Miller capacitance can be reduced and the power switch tube can have sufficient switching speed." In this figure, the resistor R3 cannot be too small, but should be as large as possible to limit the base current of the transistor, so that the charging time of C2 becomes longer, thereby achieving a delay effect.

maychang

S3S4S5S6 Published on 2023-10-31 08:41 "The smaller the resistance R, the smaller the influence of Miller capacitance. Therefore, in the switching power supply, it is always required that the driving source of the power switch tube can output a sufficiently large current...

The requirement for the power switch tube of the switching power supply is not to turn on and off slowly, but to turn on and off quickly, which is exactly the opposite of your slow turn-on requirement.

MrCU204

Figure 4 shows this usage in a sawtooth wave generator.

maychang

MrCU204 posted on 2023-10-31 22:25 Figure 4, this is used in the sawtooth wave generator.

[This is used in the sawtooth wave generator]

Now that op amps are so cheap, it is often possible to generate sawtooth waves using an op amp rather than discrete components.

MrCU204

Cut off Rc, connect an ammeter in series to the base and then add the signal.

If the signal frequency is gradually adjusted higher, then at a certain frequency, the ammeter reading will increase with the increase of frequency, and the collector capacitance will become like "Rc".

maychang

MrCU204 posted on 2023-10-31 22:36 Cut off Rc, connect an ammeter in series to the base before adding the signal. If the signal frequency is gradually increased, then at a certain frequency, the ammeter reading will be...

[Cut off Rc, connect an ammeter in series to the base before adding the signal]

If the collector load is "cut off" (short-circuited), the Miller effect will no longer exist. The capacitance from the base to the collector is equivalent to being directly connected to the positive terminal of the power supply. If the collector load is "cut off" to make the resistor open-circuited, the transistor collector will have no power supply and will not work at all.

MrCU204

The Miller effect or other defects can be corrected by discrete circuits for each active device, but the compensation of the op amp can only be done by external plug-ins (large loop feedback).

There seems to be no good solution to the Miller effect except using a sufficiently powerful constant voltage source for hard charging, unless a cascode architecture is used. In fact, even if the Miller effect is eliminated by those parasitic capacitors, the loading effect on the tube itself and other risks still exist.

MrCU204

maychang posted on 2023-10-31 22:51 [Cut off Rc, connect an ammeter in series to the base before adding the signal] If the collector load is "cut off" (short-circuited), the Miller effect will no longer exist...

It is not used for operation, but to show the effect of parasitic capacitance Ccb.

Only add the signal, do not connect Rc and power supply, let Ccb completely replace Rc, this "Rc" will decrease as the frequency increases.

MrCU204

These junction capacitances not only cause the Miller effect, but can also lead to parasitic oscillations. The solution is to use neutralizing capacitors.

maychang

MrCU204 posted on 2023-10-31 23:19 It is not used for work, but to show the effect of parasitic capacitance Ccb. Only add signal, do not connect Rc and power supply, let Ccb completely replace Rc, this "Rc" will...

The Miller effect will only appear when the amplifier is in an approximately linear operating state. If the transistor or MOS tube is in a nonlinear region, such as the saturation region of the transistor, there will be no Miller effect.

Therefore, when it is not used for work, the effect of parasitic capacitance Ccb cannot be demonstrated.

maychang

MrCU204 posted on 2023-11-1 23:52 These junction capacitors not only cause the Miller effect, but also cause parasitic oscillation. The solution is to use neutralizing capacitors.

Neutralizing capacitors suppress parasitic oscillations, which is only effective when the input and output of the tube are both LC circuits. If the junction capacitance of the tube in the large loop feedback causes parasitic oscillations, other methods must be considered, and "neutralizing capacitors" cannot be used.

lkh747566933

I often see the introduction and sharing of Miller platform. However, I have not used MOS to this extent yet.

elec32156

Hello, teacher! I have some questions about the Miller platform that I have not figured out yet. I would like to ask you for help. I am sorry for taking up your time.

1. After entering the Miller platform, the Vgs voltage remains basically unchanged, and the MOS tube is in the saturation region. At this stage, the current Id increases, causing Vds to decrease. However, from the output characteristic curve, this relationship does not seem to correspond, because in the saturation region, when Vgs remains unchanged, Id does not change with Vds. What is the reason for this?

2. When Vgs is greater than Vgs(th), it continues to increase and enters the Miller platform at a certain voltage point. What factors determine the size of the Miller platform voltage? I am a little confused. The actual simulation found that when Vgs is larger, the platform voltage is also higher.

3. After the Miller platform ends, the linear region begins. At this time, Vgs increases again, and Id remains basically unchanged. Is the voltage beyond the platform redundant? Because the higher the voltage, the longer the switching time.

For the convenience of explanation, the figure below is the schematic diagram and waveform of my actual simulation, and the gate drive voltage is 4V.

maychang

elec32156 posted on 2024-3-8 18:47 Hello teacher! There are some questions about the Miller platform that I have not figured out. I would like to ask you and take up your time. 1. Enter Miller...

【After entering the Miller platform, the Vgs voltage remains basically unchanged】

[Basically unchanged] It is not completely unchanged. It is basically unchanged because the Miller capacitor absorbs most of the current output by the signal source, and a considerable part of the driving voltage is dropped on the internal resistance of the signal source, but in fact the gate voltage is still rising. You can think about it this way: if Id does not change, then Vds will not change either, the Miller capacitor no longer exists, and Vgs must rise.

maychang

elec32156 posted on 2024-3-8 18:47 Hello teacher! There are some questions about the Miller platform that I have not figured out. I would like to ask you and take up your time. 1. Enter Miller...

【 After the Miller platform ends, it enters the linear region. At this time, Vgs increases again, and Id remains basically unchanged. Is the voltage beyond the platform a bit redundant? 】

It is redundant. But it is necessary for the switching power supply. On the one hand, as Vgs increases, the on-resistance of the MOS tube can be reduced, so the conduction loss of the tube can be reduced. On the other hand, the dispersion of the tube is quite large, and it is difficult for us to ensure that the gate drive voltage stops rising just after the Miller platform ends, because the Miller platform of different tubes is different in height.

elec32156

maychang posted on 2024-3-8 19:18 [After the Miller platform ends, it begins to enter the linear region. At this time, Vgs increases again, and Id remains basically unchanged. Is the voltage beyond the platform a bit too much...

OK, I understand the problem.

elec32156

maychang posted on 2024-3-8 19:13 [After entering the Miller platform, the Vgs voltage remains basically unchanged] [Basically unchanged] does not mean completely unchanged. Basically unchanged because the Miller voltage...

I understand. But Vgs changes very little after all, and in this Miller platform area with very little change, Id changes a lot. What I mean is that if we look at the output characteristic curve, the change trend in the saturation area does not match. Is it that the output characteristic curve does not take the Miller effect into account at all? Or is there something wrong with this comparison method itself?