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20

The diodes keep the bases of the transistors 1.4V apart. This reduces crossover distortion. The resistors are to provide a bias current for the diodes. So the diodes keep the transistors "close" to being on. If they weren't there, as the input voltage went between ~0.7 and ~-0.7V the transistors would be in cutoff and the output voltage would be zero. ...


16

This is a class B amplifier: - Your circuit is a class AB amplifier: - Rv adjusts the bias point of the two transistors so that T1 and T2 are always conducting a little bit of current - this avoids excessive cross over distortion: - See also this article, Crossover Distortion in Amplifiers, for more information. Rv modifies the volt drop across the two ...


11

First, understand this is just a double emitter follower using a darlington on each side. The voltage at the output will be pretty much the voltage at the opamp output. The purpose of the emitter followers is to provide current gain. If each transistor has a gain of 50, for example, then the current the opamp has to source and sink is roughly 50 * 50 = ...


10

Let's start at the given output driver topology: simulate this circuit – Schematic created using CircuitLab On the left is the basic idea, but it is worth some discussion already. \$C_1\$ exists, in part, because a speaker is a complicated device and doesn't entirely enjoy an average DC current flowing through it and because you aren't using a ...


10

There is a simple known circuit which works as a 'programmable zener'. Below is the principle diagram: simulate this circuit – Schematic created using CircuitLab For a real application the variable resistor may be split in three parts to get more accurate control. By varying the resistor you can set the 'zener' voltage between the bases of the two ...


8

The classes were originally arranged to describe the conduction angle for a single quadrant of a power amplifier. In class-A, active conduction occurs throughout all \$360^\circ\$ of the period. In class-B, active conduction occurs for \$180^\circ\$, or one-half of the period. In class-C, active conduction occurs for \$\lt 180^\circ\$ (but more than \$0^\...


7

The purpose of the diodes is to set a bias voltage between the transistors' bases, which sets a small idle current through the push-pull. This makes it work in class-AB and lowers crossover distortion. However the diodes should be thermally coupled to the transistors, to prevent thermal runaway. Also, emitter resistors should be used for this reason. Anyway....


7

The difference between class A and class B is the quiescent current through the last stage. If you make the quiescent current zero then only Q3 or Q4 supplies current when a signal is present. This is class B. If you make the quiescent current so large that for very large signals (even the largest) both Q3 and Q4 never have an Ic = 0 (are never off), we ...


6

Yes it will work. Will it be any good? Nope. The problem with this circuit is around the "zero crossings" which are not at 0 Volt in your circuit but around 6 V since you're using a 12 V supply. The problem occurs when the direction of the output current needs to change. Then the transistor which had been supplying current must be switched off and the ...


6

You have to understand the output topology well in order to know how to create the biasing for it. Although someone did mention that your schematic example has the BJTs arranged in Darlington fashion (with added turn-off speed-up resistors), they didn't tell you that such an arrangement almost always has a better topology. So you'd almost never use that ...


4

The current from an OPA454 is just about limited to a little over 100 mA and on a good day you might get 150 mA so, R3 is chosen to prevent the OPA454 from producing this current i.e. the BJTs take over at a certain point and produce the bulk of the output current. For instance, at 100mA drive through R3 (20 ohms) the Vbe across one of the transistors is ...


4

The voltage gain of this system, which is currently unity is determined by the opamp and feedback network. The push pull amplifier is placed within the feedback loop of the opamp, and is there to provide current, to drive a lower impedance load than the opamp alone is able to. The push pull stage is a pair of emitter followers, and as such doesn't provide ...


3

Actually the class B amplifier doesn't has a base bias. The bias occurs at AB class. But you can bias the base in many ways. If you're using an op amp just like in the image, you could just use feedback. It makes the output equals to the input, just like a buffer but with a power stage. simulate this circuit – Schematic created using CircuitLab You ...


3

I didn't really read the whole question (too long, and didn't get to a clear point fast enough), but this is not what you seem to mean by "distortion": This is very clearly the amplifier oscillating on its own when given a little kick by the input signal. A brief look at the schematic shows why this shouldn't be a surprise. There is no capacitance on the ...


3

What you are seeing is not distortion. It is the amplifier oscillating at high frequency because +20/-20V have no bulk capacitance. This makes it so they have high impedance at high frequency due to the inductance of the wires going to your power supply and the power supply's limited frequency response. Here's a simulation: 1uH is approximately the ...


3

Let as exam the BJT push-pull amplifier along first. Any BJT's need at least 0.5V to 0.7V of forward base-emitter bias voltage before they will go into conduction. In push-pull amplifier both of the transistors will be non-conducting (OFF), when the input signal is in the range +/- 0.5V. And this creates an "deadzone". And also this "deadzone" produce the ...


3

I see a few problems. As Jippie says, C2 and C3 are reversed. The collector of Q1 will be biased somewhere between +50V and ground. The bases of your output transistors should be biased somewhere close to ground. Your preamp biasing looks a little suspect. With a minimum beta of ~75 (from the datasheet), Q1's collector would barely get down to 10V below the ...


3

Class A = both transistors are ON all the time. Class AB = both transistors are ON at idle, then up to a certain output current. When output current is higher than a certain limit, one of the transistors turns off. Class B = either one transistor or the other is ON, but not both. The transistor that is ON is determined by output current polarity. Class C ...


2

The quoted text from the book is: In the quiescent condition, \$V_o = 0\$ and \$V_1 = 0\$ . This is correct in the sense that these are the conditions under which no current flows through the stage, and the output is at 0V. This appears to be from you: What forces \$V_1\$ and \$V_o\$ to be equal \$0\$ when \$V_i = 0\$? The answer is nothing. The ...


2

Brian is right. This is not a complete circuit. Do not bother attempting to analyze the quiescent conditions of this circuit. For the record, as drawn, if \$V_i\$ is greater than 0.6V above \$-V_{CC}\$, then the collector of \$Q_3\$ will be near ground and \$Q_2\$ will pull the output voltage down toward \$-V_{CC}\$. Grey and Meyer are well-respected, but ...


2

Possibly shouldn't be an answer, but... The TL/12 (original, 1947) has separate bias (cathode resistor) for each valve; the others generally used a common cathode resistor and there, you do need reasonably well matched valves. You are correct that none (that I know of) used negative grid bias supply, sorry if I misled you on that. I'm puzzled how you ...


2

Examine the output waveform, then think about what harmonics would be required to reproduce it. For example, here's what adding 66% of 3rd harmonic and 25% of 5th harmonic looks like:- And here's the LTSpice FFT for that waveform:- Just two odd harmonics with the right amplitude and phase were enough to create a waveform very similar to typical class B '...


2

The basic answer to this is to start by making sure all the signal paths have only high pass filters with rolloff no higher than 20 Hz, and low pass filters with rolloff no lower than 20 kHz. There will still be some uneveness in the frequency response, but it shouldn't be drastic. Take a close look at what filters are formed by any caps, and make sure ...


2

Bandwidth and crossover distortion is not the issue with unity loop voltage gain. Your main problem is insufficient current drive. Your Op Amps has a large signal bandwidth of 140kHz of 10V on 600 ohms or 17mA. The GBW os 10 MHz so you can raise the gain from 1 to 100 easily with R negative feedback ratio. The ratio of 600/8 ohms implies base current ...


2

class B is defined as 180 deg conduction angle - so class B is biased to the point of conduction - otherwise its really class C (especially for small signals). The emitter resistors are key to both biasing stability and to allowing each device to turn off during the opposite half cycle. class AB is when the conduction angle is between 180 and 360


2

Whether the class-B stage you propose is a good choice depends entirely on the application. When the output current changes sign, the opamp will have to slew to turn on the other transistor, as you mention. So it will generate a large amount of crossover distortion. For driving a DC voltage into a load, it will be absolutely fine. For audio, it would ...


2

It's a bit more complicated when you drive high currents because of the choice of every component affects the results of output impedance, quiescent current of drivers, harmonic distortion, dampening ratio which affects voltage from back EMF on low frequencies and thus "muddy bass". Naturally from Shockley effects on Vbe vs Tjcn and same for diode, even if ...


2

I just quickly drew this up, the last hour or so. I agree that with wires on a protoboard such as what you are using, it's important to have lots of bypass capacitance right on the protoboard itself. Include that. However, I think you may also have had problems due to the fact that there is lots of capacitance (a few pF) between each nearby hole on the ...


1

Measuring the impedance of a microphone is pretty easy. You need a constant acoustical signal such as a speaker fed by a tone. Mount the microphone in front of the speaker in such a manner that it won't move around as you are doing the measurements. Standard measurement frequency is 1 KHz. Apply the bias voltage to the microphone and measure the ...


1

Now lets use more realistic VDD+ and VDD- supplies, instead of the ideal SPICE. On each of the 2 power lines, install 100,000uF in series with 100nanoHenry to ground{4" of wire}. This resonates at radian freq of 1/sqrt(L*C) or 1/sqrt(0.1F * 1e-7H) or 1/sqrt(1e-8) = 10^+4 radians/sec or 1,590 Hertz, a very audible frequency. To lightly decouple the SPICE ...


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