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5

"Gain" in the equation: $$GBP = Gain \times BW$$ is a linear gain, not dB. You can't plug dB directly into that equation. Furthermore, bandwidth (in most cases) is defined as the frequency where the output drops by 3dB so it's already accounted for. It's part of the definition. You don't need to account for the -3dB a second time. 20dB is the open loop ...


4

I think I can summarize all the comments from above into one answer: Use an Opamp for this task and forget about the single stage BJT. The Opamp's properties reliefs you from a lot of trouble you would have to deal with when using a single BJT.


3

I = V/R only applies when the source is a voltage source. A current source is not affected by input resistance, at least not until you exceed its compliance range.


3

Your circuit features 3 capacitors but is actually a 2nd-order system (degenerate case). The denominator of the transfer function (whatever it is) follows the following formalized form: \$D(s)=1+\frac{s}{\omega_0Q}+(\frac{s}{\omega_0})^2\$. In this expression, \$Q\$ represents the quality factor and illustrates the losses in your circuit. Because \$D(s)\$ ...


2

\$V_{BIAS}\$ is DC amplified by the factor \$1+\dfrac{R_F}{R_{SHUNT}}\$. So, if \$V_{BIAS}\$ is 1.8 volts and \$R_F\$ = \$2\times R_{SHUNT}\$ the op-amp tries to put a DC voltage at its output of 5.4 volts and this is above the 5 volt power rail. There may be other factors at play but this seems the most likely one.


2

The Emitter Follower's Input/Output impedances however show totally contrasting properties (High Input and Low Output impedance), and from small signal models its quite easy to see that the Emitter Follower does provide a significant current gain (β+1) You need to do a small-signal analysis of the input impedance for signals above the base-emitter ...


2

The resistors between your filter capacitors are counterproductive. They cause extra voltage drop due to load current, and prevent the later capacitors from charging to peak rectified voltage. Here's what happens to the unregulated output when a load of 100mA is applied (simulated in LTspice):- Voltage is very smooth but has dropped to 14V, which is right ...


2

Segue into a design If you find the following an unwanted digression about BJTs, feel free to skip to the design section below. But for some, this may help a little to set the tone of what follows in the design section. I want to start out by stating two things: Design constraints matter a great deal in how one goes about arranging a design process to ...


2

Note: This answer is predicated on the fact that you can verify the clipping by putting a scope across the speaker. And the clipping is happening near the 6.1V mark. If the output of the amp is clipping, then you only have three choices, Increase the voltage to the amplifier (if you can) Reduce the gain of the amplifier if you can. Reduce the amplitude ...


1

The guy in the video misspoke. It's the transistor that's cutting off half of the waveform, because of how it's biased. The blocking capacitor simply eliminates the bias that's already on the microphone signal.


1

This makes it seem like the inverting and non-inverting terminals become common when powered. The action of an op-amp is not to connect the two terminals to each other but to drive the output to bring them to the same voltage (assuming you have connected the output to the input in a negative feedback configuration, which you have via R3). The inputs ...


1

Taking the gain of 10 as a constant horizontal line on the gain/frequency plot (this constant gain is referred to as the noise gain). This constant line will cross the open loop response at 100kHz so the bandwidth is 100kHz. It is the noise gain which determines bandwidth and the noise gain is 10 : Bandwidth = GBW/(Noise Gain) For a purely resistive ...


1

Consider this circuit: simulate this circuit – Schematic created using CircuitLab To calculate the values, the TL431 (ref input shorted to anode) needs 1mA to work, so let's set the current ignoring R3 to about 1.5mA. So R4 ~= 5K, use 4.7K. The resistor ratio R1/(R2||R3) = 0.5 for a gain of 1.50 so let's arbitrarily set R1 = 10.0K We know for an ...


1

One potential problem would be that your circuit shorts the bridge and neck in bypass mode through the capacitors and resistors at the high end frequencies, which would make your blend frequency dependent: simulate this circuit – Schematic created using CircuitLab To use the same parts and avoid this, you could hard-wire your pickup output circuits ...


1

Your circuit will have a voltage gain of unity .Why not config your mosfet as a source follower and enjoy more output swing from the AAA cells .The tone controls are fed from a relatively low impedance of about 1 Kohm now ,They might feel a bit funny .You may want to change this .R9 and R10 are still in circuit loading things down when you are switched on ...


1

I assume your transformer outputs 12 V a.c. This means that after rectification you will get 12 multiplied by square root of 2 minus diode voltage drop. An approximate of 15 V assuming about 1 V drop per diode. This is the peak voltage and in theory it is enough to supply the linear 7812/7912 linear regulators and allow them to function properly (3 volts ...


1

What am I doing wrong or not understanding about this? You're using too low a voltage of transformer. If the transformer ratings are typical, then each output winding is 12VRMS when the AC input is 120V. The output is pulsed DC that does go to around 16V when it is unloaded, but the capacitors have to hold up the voltage between pulses. Even keeping low ...


1

As suggested by many people (thank you all for the suggestions) I have designed a multistage amplifier for the solution using a source follower to buffer the input (due to large source impedance) and then a two stage class a then class b amplifier.


1

Original Post This is the circuit I think you are inquiring about: simulate this circuit – Schematic created using CircuitLab As you can see, there are some specifications. But to be honest, not enough. No idea of \$R_\text{L}\$ or \$R_\text{S}\$ (which cannot be ignored), for example. Suppose \$R_\text{L}=3.3\:\text{k}\Omega\$? Then already you ...


1

Your approach ignores the fact that the minimum voltage that you can get at the output is not zero, whereas the approach you found takes that into account. In other words, your simplified approach would result in a circuit that clips the negative peaks before the positive peaks reach their limit, reducing the symmetrical swing from the maximum possible ...


1

If you simulate the system, you will see that the phase shift created by the motor severely distorts the sinewave and thereby creates harmonics. However, for some purely resistive load there does not seem to be a problem. Clearly, electric motors are not purely resistive load.


1

An emitter-follower is not a (good) current amplifier, although its large/small signal models amplifies current. But the impedances of the emitter-follower disqualify it as a (good) current amplifier.


1

You seem to have almost solved the question. I will try to give you a hint how to continue. You know that distance QA (and QB) is 20uA. So you can see that +/- 20uA variation of Ib leads to ??? variation of Ic, which is the current gain. You can also draw A' and B' which are the corresponding points for (only) 12uA excitation (when we consider this as a ...


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