### Unit 1 - Power amplifiers

### Power amplifiers

**Power amplifiers**

with introduction and classification of power circuits

### Amplifiers for analog electronics

In typical analog circuits (as in operational amplifiers) the amplifier must pay attention to the following aspects:

**Signal amplification (voltage or current)**

**Linearity**of the output signal delivered to the load

**Frequency range**of the output signal

**Input and output resistance**of circuit

the power gain between input and output signals is not the main goal for the power stage (except for RF power amplifiers)

e.g. a radio system that needs to amplify the signal from the antenna, but the load is another circuit that processes the signal.

Too much power transferred to the load could also be detrimental

### Review of basic of power amplifiers for analog electronics

In typical output stages (as in operational amplifiers and audio systems) the power amplifier that drives the load must pay attention to the following aspects:

- \(\textbf{Power conversion efficiency}\) $\eta = P_L / P_S $

defined as the ratio between the average power (in a cycle) given to the load and the one taken by the power supply (always less than 1). Often given as a percentage ($<100 \%$)

**Power dissipation**on the active device

- the
**power gain**between input and output signals is now relevant

### Classes of operation of power amplifiers (and power circuits)

The operation of the power amplifiers is defined in different classes according to the way the active devices in the circuit are operating during the period of input signal (sinusoidal input exemplify the typical working condition). For analog amplifiers we define two main classes of operation:

**Class A **where the active device is conducting during the entire time period of input signal waveform \((2\pi \, rad) \)

**Class B**, where the device is conducting for about one/half the time period of the input signal waveform \((\pi \, rad) \) Two devices are required to obtain a good output linearity.

For the switching circuits we can define a **Class D** operation where the device is made to commutate between full conduction (on) and interdiction (off) states (we will discuss it later on)

### (Analog) Power amplifiers

**(Analog) Power amplifiers**

Class A amplification

### Class A operation

In class A operation the device (here a BJT is assumed) is biased at the middle point – Q(Io, Vo) – of the load line, and the operating point is driven by the input signal along the load line to a max current less or equal than Imax and min current larger or equal than 0.

The output power is max when the operating point reaches Imax (ideally, when the Vce(saturation) is neglected) and 0.

### Class A power efficiency

Max power efficiency: assuming a linear operation up to the limit values one has:

Power absorbed from the positive supply voltage: \(P_{S+} =V_{CC} \dfrac{1}{T} \int_T i_C(t) dt = V_{CC}I_O\)

Power absorbed from the negative supply voltage: $P_{S-}=V_{CC}I_O$

Total power from the supply: $P_S =2V_{CC}I_O$

Power provided to the load: $P_L=\dfrac{V_PI_P}{2}$ ($V_P$, $I_P$ peak values of the a.c component)

Max a.c. peak values (for maximum efficiency) : $V_P=V_{CC}, I_P=I_O$, $P_{LMAX}=\dfrac{1}{2}V_{CC}I_O$

The max power efficiency is then: $\eta_{MAX}=\dfrac{P_{LMAX}}{P_S} =\dfrac{1}{4}=25 \%$

The power absorbed by the supply is always constant and equal to $P_S$.

The efficiency is linearly dependent on output power $P_L$.

### Power balance in class A amplifiers

The max power efficiency is $25 \%$

The power balance of the circuit is: $P_S=P_{Dev} + P_{pol} + P_L$

Where:

$P_{Dev}$ is the power dissipation on the power device

$P_{pol}$ is the power dissipation on current generator

$P_L$ is the dc power dissipated in the load

$P_{pol}$ is constant: $P_{pol}=V_{CC}I_O$ The max power dissipation on the device is obtained for zero power on the load and is 50% of $P_S$.

$P_{Derv-max} = P_S - P_{pol}=P_S -V_{CC}I_O = V_{CC}I_O $

Let us consider the meaning of these results:

- To obtain a (controlled ) power output of
**50 W**one need a supply power of at least**200 W**

- the device must dissipate
**100 W**in the steady state to transfer a**max power of 50 W**to the load!

### Power balance in class A amplifiers

**Exercise:**

Which is the power dissipation on the device when the maximum power is transferred to the load?

Simple calculation shows:

$P_{Dev} = \dfrac{V_PI_P}{2} = P_L$ ($V_P, I_P$ peak values of the a.c component)

**Exercise:**

Typical impedance of a 50W loudspeaker is 5 Ohms.

Which is the peak voltage and current the BJT needs to handle?

Simple calculation shows: Vce_max>45V, Ic_max≈4.5A

### (Analog) Power amplifiers

**(Analog) Power amplifiers**

Class B amplification

### Class B operation

In class B operation the device (here it is assumed a BJT) is biased at zero current point – $Q(0, V_{CC})$ – of the load line. As a result the power dissipation in the quiescent state is zero. Two devices (and two power supplies) are needed to obtain an output signal analog to the input one. The NPN device operates as an emitter follower for positive signal swing, while the PNP device operates as an emitter follower for negative signal swing.

### Class B power efficiency

**Max power efficiency: **

Assuming linear operation up to the limit values, and sinusoidal input.

The power $P_S$ absorbed from the supply, assuming the following plot for the currents given by the two supplies, is: $$P_S=\dfrac{2V_{CC}}{T} = \int_0^{T/2}I_{MAX} \sin(\omega t)dt = \dfrac{2V_{CC}I_{MAX}}{\pi}$$

The load power $P_L$ is again: $P_L=\dfrac{V_PI_P}{2}$ with $V_P=V_{CC} \qquad I_P=I_{MAX}$

The maximum output power on the load is: $$P_{LMAX}=\dfrac{V_{CC}I_{MAX}}{2}=\dfrac{V_{CC}^2}{2R_L}$$

The max power efficiency is then: $$\eta_{MAX} =\dfrac{P_{LMAX}}{P_S}=\dfrac{\dfrac{V_{CC}I_{MAX}}{2}}{\dfrac{2V_{CC}I_{MAX}}{\pi}} = \dfrac{\pi}{4}\approx 78.5 \%$$

### Power balance in class B amplifiers (1/2)

Indicating with $P_{DEV}$ the power dissipation on both active devices, with $P_S$ the power provided by the supply and with $P_{L(a.c.)}$ the a.c. power transferred to the load, the power balance of the class B circuit is: $$P_S=P_{DEV}+ P_{L(a.c.)}$$

The power dissipation on the devices is null for zero power on the load.

Let us compute $P_{DEV}$ as a function of the output current: $$P_{DEV}(I_P)=P_S(I_P)-P_{L(a.c.)}(I_P)=\dfrac{2V_{CC}I_P}{\pi}-\dfrac{I_P^2R_L}{2}$$

This is a second order function in $I_P$, and the max is located somewhere between $0$ and $I_{MAX}$. It can be found as: $$\dfrac{dP_{DEV}(I_P)}{dI_P} =0 \rightarrow \dfrac{2V_{CC}}{\pi} -\dfrac{2I_P^*R_L}{2}=0\rightarrow I_P^* =\dfrac{2V_{CC}}{\pi R_L}$$

### Power balance in class B amplifiers (2/2)

Substituting $I_P^*$ we get the maximum power dissipation on the devices.

$$P_{DEV}(I_P^*)=P_{DMAX}=\dfrac{2V_{CC}^2}{\pi^2 R_L}$$

and we obtain the following ratio between $P_{DMAX}$ and $P_{LMAX}$ $$\dfrac{P_{DMAX}}{P_{LMAX}}=\dfrac{2V_{CC}^2}{\pi^2 R_L} \Big/ \dfrac{V_{CC}^2}{2R_L} =\dfrac{4}{\pi^2}\approx 0.4$$

### Power balance in class B amplifiers

From the previous results on power efficiency and power dissipation it comes out that:

- The max power conversion from power supply to load is 78%
- the total power dissipation (on both devices) is 40% of the max output power: then each device must dissipate 20% of the max output power

Let’s consider the meaning of these results:

- To obtain with class-B a power output of
**100W**we need a supply power of at least**130W** - To transfer a
**max power of 100W**to the load, each device must be able to dissipate**20W**(at the IP* rated)

**Conclusion:** Class B is better than class A in power conversion, (we pay this with some degradation in linearity), but this is still not sufficient if we need power conversion above several kW.

For a **10kW** output power we need a power dissipation on each device of more than**2kW** and this is not feasible with usual power packages, as we will see later.

### What is the ‘Package’

It is the plastic, metallic or ceramic box that surrounds the piece of silicon (chip) in which the power device is manufactured.

The package provides: mechanical reliability protection from environmental humidity, chemicals, dust, and various pollution

### (Analog) Power amplifiers

**(Analog) Power amplifiers**

Class D amplification

### Class D operation

To **increase** the power available at the output of a power circuit, one must decrease the power dissipation of the active devices (the circuit)

This is to increase the efficiency and to afford less expensive and bulky power devices since device power dissipation is upper limited by the package size and material (we will come back on that point later). The best way of reducing the power dissipation on the device is to let it operate into two limit operating points: a) OFF state, where the power dissipation is zero because the device current is null. b) ON state, at the minimum voltage drop allowed by the operation of the device (often indicated as saturation voltage) This is the Class-D operation: the device operates as a switch, that is either open (OFF state) or closed (ON state). In this way, the device, driven by input pulses capable to bring it either in ON or OFF state, can operate at a power much less than the available output power, thus increasing both the power output and the power efficiency.

### Class D operation

To **increase** the power available at the output of a power circuit, one must **decrease** the power dissipation of the active devices (the circuit)

This is to increase the efficiency and to afford less expensive and bulky power devices since device power dissipation is upper limited by the package size and material (we will come back on that point later).

The best way of reducing the power dissipation on the device is to let it operate into two limit operating points:

a) **OFF state**, where the power dissipation is zero because the device current is null.

b) **ON state**, at the minimum voltage drop allowed by the operation of the device (often indicated as saturation voltage)

This is the **Class-D** operation: the device operates as a **switch**, that is either open (OFF state) or closed (ON state). In this way, the device, driven by input pulses capable to bring it either in ON or OFF state, can operate at a power much less than the available output power, thus **increasing both the power output and the power efficiency**.

### Class D operation

Ideally the power dissipation of the device working in the ON-OFF state is

OFF state: $P = V_{CC} \cdot I_{leakage} \rightarrow $ Negligible

ON state: $P = I_{max} \cdot V_{ON} \rightarrow $ is a fraction of $V_{CC}$

The power dissipation is small ON-OFF and OFF-ON **transition**: depends on the power device. The slower the transition the higher the power dissipation (more time spent far from ON and OFF states)

In class D operation the chosen power device can easily handle a significant power as the limitation is NOT on the maximum static power (there is a limitation on the dynamic power however, but this is less stringent) but on the maximum current and voltage.

**Example**: designing a circuit with 50V-2A supply voltage and load current (100W)

In Class A operation need to chose a power device with the ratings of 50V, 2A, and 50W of maximum power dissipation.

In Class D operation the maximum power dissipation can also be as low as 10W since VON is about 2V.

### Class D operation

In other words, as schematically shown below with reference to a BJT device, the operating load line (red line) can overcome the max power dissipation locus (green hyperbolic line), because in the ON state (point B) the dissipated power is much less than the maximum power dissipation $P_{DMAX}$, and in the OFF state (point A) is almost zero (assuming negligible the leakage current)

### Class D operation

However, one must pay attention on the time required by the device to switch between ON and OFF states: we can define an average $\textbf{steady-state power dissipation}$ $P_{DS}$ and an average $\textbf{dynamic power dissipation}$ $P_{Dd}$

$P_{DS}$: average power dissipation in the ON state (assuming negligible the one in the OFF state)

$P_{Dd}$: average power dissipation during the switching transitions $\Delta T1$ and $\Delta T2$ between ON and OFF states

$$P_{DS}=\dfrac{T_{ON}}{T}I_{ON}V_{ON} \qquad \qquad P_{Dd}=\dfrac{1}{T} \int_{\Delta T_1 + \Delta T_2}i(t)v(t) dt$$

The total power dissipation is the sum of $P_{DS}$ and $P_{Dd}$.

### Class D amplifier

In class D amplifier, the information content of the signal cannot modify the amplitude of the pulses, because these latter are of constant amplitude, but it can be transferred to the output by a **modulation** of the width of the pulses.

In other words, we need a **Pulse Width Modulation** (PWM) to drive the device and to transfer this information to the (amplified) output, i.e. to the load.

The simplest PWM modulation technique is done by using a signal comparator to compare the analog signal with a triangular waveform.

The output will be made of a pulse train having an amplitude equal to the supply voltage of the comparator, and ON (OFF) duration defined by the time interval where the triangular waveform is lower (higher) than the one of the modulation signal.

### PWM Modulation of Class D amplifiers

An example of PWM modulation, made by a sinusoidal signal $f_{Sig}$ using a signal comparator and a triangular waveform $f_M$, is reported in the following plot

### Example of analog Class D amplifier

### Class D power block (1/2)

To reconstruct the output signal after the class D operation we need to demodulate the signal with a low pass filter that removes the modulating frequency $f_M$, while leaving unaltered the signal up to frequency $f_{Sig}$.

The filter is made with inductors and capacitors to minimize the power losses.

### Class D power block (2/2)

Assuming that the L,C components of the circuit are lossless (in real cases at least the inductance has a series resistance that dissipates power), the power efficiency of the circuit is linked to the power losses of the semiconductor devices for the switching elements.

The power efficiency is defined as: $\eta = \dfrac{P_L}{P_S} = \dfrac{P_S - P_D}{P_S} 1 -\dfrac{P_D}{P_S}$

where for each device the steady state and dynamic power is calculated as previously reported.

$$P_{DS} = \dfrac{T_{ON}}{T}I_{ON}V_{ON} \qquad \qquad P_{Dd}=\dfrac{1}{T} \int_{\Delta T_1 +\Delta T_2}i(t)v(t)dt$$

We will use these expressions to evaluate the power efficiency of the switching power circuits.

### PWM Modulation

The filter cut-off frequency $f_F$ must be:

$f_F << f_M $(modulation frequency) to suppress the modulating signal

$f_F >> f_{Sig} $(signal bandwidth) to leave unaltered the signal

Then $f_{Sig} << f_F << f_M$ - Thus the ability to work with large bandwidth signal and the constraints on the filter are relaxed if it is possible to increase the modulating frequency $f_M$.

This requires power devices with **high operating frequency** and **low switching times**.

### Circuit simulation of a class D power Amplifier

Below the schematic of a push-pull power amplifier operated in class D using a PWM modulation with a voltage comparator, an LC filter at the output, and two complementary Power MOS

### Circuit simulation of a class D power Amplifier

Top plot: Vin and Vout

Middle plot : Gate to Source NMOS waveforms vs. Vin

Bottom plot: Power dissipation.

### Measurements through the simulation

Below the instructions for the measurement of the data shown in the previous slide.

### Discussion on the LC filter

The pole of the LC circuit is at 7.9kHz: $f_F=\dfrac{1}{2\pi \cdot \sqrt{LC}}$

The signal maximum frequency, $f_{Sig}$, is 5kHz while the modulating signal, fM, is at 50kHz For analog amplifiers this kind of dimensioning of the LC circuit is adequate.

As we will see in the following, when dealing with power converters or power supply circuits, it is more convenient to see the LC tank as an energy storage circuit taking care of providing constant voltage and current to the load.

### Power Circuits

The role of a power circuit is providing electric power to an electronic system (personal computer, processor, lighting system, etc.) or an electric motor (air conditioning, automotive, railway, etc.).

Differently from power amplifiers, they do not have a variable input signal that has to be amplified. The input is an electric power source that can be AC or DC and is not regulated. The output is a regulated DC or AC power.

The power circuits share with power amplifiers the need to have high power efficiency and the use of power devices that the designers would like to be as ideal as possible.

### Power Circuits : classification

The basic power circuits are:

DC/DC converters, that control the d.c. power on the load, by variable control signals

DC/AC converters (Inverters), that generate a regulated a.c power from a d.c. power supply, and control the a.c. power delivered

DC/AC converters (Inverters), that generate a regulated a.c power from a d.c. power supply, and control the a.c. power delivered

### Power Circuits

Apart from the above circuits, we can consider also the power circuits that transform AC power (usually the one of the grid main supply) into a DC one (unregulated) : these circuits are named Rectifiers.

AC/DC converters (Rectifiers), that generate an

(unregulated) d.c power from a a.c. power supply

These circuits are usually made of transformers and diodes. They are needed if the electronic system is supplied by the mains; in that case the rectifier is used to give the unregulated DC input power that is the input for the power circuits presented above.

### Risorse della lezione

- Power amplifiers
- Rectifiers
- DC/DC converters
- DC/DC and DC/AC converters
- Power device ratings
- High Voltage PN Junction
- The Power Diode
- The Power MOS
- Power bipolar transistor
- The family of Thyristors
- IGBT
- Snubber circuits
- Isolated DC/DC converters

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