4

the purpose of the 0.1 µF input capacitor and the 200 kΩ resistor is not a HP filter, but an AC input coupling; if it was standalone, it would be a high-pass filter with a ~8 Hz cutoff frequency, indeed. But here, I'd really call it "AC input coupling" A decoupling capacitor is something different. This is an AC coupling capacitor What you label "LP filter" ...


4

Bear with me, the schematic editor wasn't really made for this... simulate this circuit – Schematic created using CircuitLab In the circuit on the left, we can see 10pF feedback capacitance, carrying a fraction of the anode voltage back to (and in antiphase with) the input voltage on the grid. This capacitance arises naturally from two conductors in ...


4

It makes just about zero sense to build an Instrumentation Amp from discrete components if you really need the CMRR of an instrumentation amp. Because of resistor mismatch and temperature drift, there is almost no chance that you can match the CMRR performance.


4

But is the only reason one would use a difference amplifier over an instrumentation amplifier cost? No, one great advantage that a difference amplifier provides is substantial "beyond the rails" operation. Consider the ADI difference amplifier below: - Look at +IN - it sets the scene for the voltages at the actual inputs to the internal op-amp. If "REF" ...


4

I don't think that's a dual power supply, I think it's a single power supply with a chassis earth terminal between the two output terminals. You need a supply with positive and negative outputs or two supplies that you can connect in series (the minus terminal must not be grounded for that to work). You can verify by measuring the voltage from -V to GND ...


3

The main problem you have with your design (other than power supply issues) is that there is no DC feedback to the inverting input and your op-amp is therefore going to be hard against one of the power rails due to input offset voltages causing a small input offset to be magnified by the open loop gain to some ridiculous figure. It seems that you got the ...


3

Gain is the output voltage change divided by the input voltage change. In the small signal hybrid pi model the collector current changes proportional to the change in base voltage (relative to the emitter) so there is a voltage change at the collector proportional to the load resistance (parallel with ro). If you add an emitter resistor, the base-emitter ...


3

It's useful to think of a transistor operating in three modes, cut off, linear, and saturated. When cut off, the collector current is zero. When saturated, the base current is high, and VCE is small. These are the two modes when using a transistor as a switch. You appear to be thinking of the saturated mode in your question. Wikipedia is talking about the ...


3

It is not a job of the power amplifier module to do anything else that to amplify its input and drive the speakers. Normally you would put the volume control before the input of the power amplifier.


2

People used to routinely hook pens up to galvanometers to draw lines on paper. The paper moved with constant velocity. These devices were called chart recorders.


2

The 47k component appears to be a potentiometer, using a somewhat nonstandard symbol. The center/side connection going to the capacitor is the wiper. The capacitors are being used as DC blocking capacitors, preventing any DC link between the stages of the amplifier and decoupling the microphone's bias voltage from the signal. As @BrianDrummond and @...


2

Rightmark Audio Analyzer is a free tool for performing frequency and distortion measurements on audio equipment using a sound card (or any recording device that can save PCM). If your goal is only to measure frequency response, almost any laptop would be sufficient. Hook up the line out from the laptop to the input on your device under test. Then take the ...


2

As @beccaboo mentions, this is possible with a single op-amp, and in this particular case, the numbers fall out nicely, making it relatively simple. Where this circuit fails in practice will be its input impedance being set by the opamp itself rather than the resistor network. Of course, the same concept would apply if we were to add a resistor divider on ...


2

Lets approach this (stability) question another way. The photodiode has some capacitance. That Cdiode in conjunction with the feedback resistor, forms a delay and also a phaseshift and a pole. 1) the circuit already is inverting, thus provides 180 degrees 2) the Cdiode + Rfeedback is another 90 degrees (above some frequency) 3) the opamp provides another ...


2

In a common-emitter amplifier, think of the transistor as a variable resistor controlled by the base current (\$i_b\$): simulate this circuit – Schematic created using CircuitLab The output voltage (and thus the voltage gain) is determined by the voltage across the variable resistor (i.e. \$V_{ce}\$). Now put a resistor between the low-end of the ...


2

The emitter resistor causes negative feedback. This can be shown using a block diagram based on the classical formulas for describing the input-output relations. It is a well-known fact that negative feedback reduces the gain factor. This is - in most cases (opamp) - a desired result of negative feedback. Note that the output current Ic is determined by the ...


2

You won't find it if it's not explicitly listed (which is rare). Typically Zo of the order of a hundred ohms for that type of op-amp. Closed loop output impedance, of course, is almost zero.


2

You can easily find opamps with internal Rnoise of 1,000 ohm; such a relatively low noise would show up in the opamp datasheet as "noise density" of 4 nanoVolts / rtHz. In 100 Hz bandwidth, this would produce 4nV * sqrt(100) = 4nV * 10 = 40 nanoVolts RMS noise. if your signal is 60dB (1,000X higher in voltage) bigger at input to the opamp, you have 40nV * ...


2

Here is a very common audio amp topology, from Douglas Self: It has the usual 3 stages : Input stage is transconductance, converts error voltage into differential current which is then turned into push-pull current by current mirror "VAS" or "voltage amplification stage" which is both a current gain stage and a transimpedance stage ; at frequencies above ...


2

I just expected but I could turn it down with some sort of master volume pot. Am I wrong? For this one, yes. In a stereotypical amplifier the volume control is placed between the pre-amplifier or in the pre-amplifier itself. simulate this circuit – Schematic created using CircuitLab Figure 1. Volume control placement for a power amplifier. A ...


2

I think, the explanation is as follows: Voltage opamps (if they are unity-gain stable) are internally compensated, which means: Their open-loop gain has a pretty small 3-dB cutoff frequency (20...200 Hz). As a consequence (with 20dB drop per decade) and - let`s say Aoo=100 dB - the transit frequency is app. (2...20) Mhz. This is necessary because the ...


1

Using this small-signal diagram We clearly see a voltage divider at the input. Therefore $$\frac{V_{B1}}{V_S} = \frac{R_1||R_2||(\beta_1 +1)r_{e1}}{R_S + R_1||R_2||(\beta_1 +1)r_{e1}}$$ Next, let us find the first stage voltage gain (Q1). This stage is working as a CE ( common-emitter) amplifier. $$V_{B1} = I_B\cdot(\beta_1 +1)r_{e1}$$ $$V_{C1} = -...


1

Reduce the Rsource from 10,000 ohms to 50 ohms. If the bipolar runs at 1mA, if the diode (emitter-base junction) "ideality factor" ==1, and if the bipolar beta is 100, then the input resistance is 26 ohms * 100 = 2,600 ohms. With Rsource of 10,000 ohms, a 14 db (5:1) attenuation exists.


1

The current through the base-emitter junction of a transistor is governed by the diode equation: \$i_e = I_S \left(e^\frac{v_{be}}{n\ V_T} - 1\right)\$. At room temperature, \$V_T \simeq 26\mathrm{mV}\$. So unless your peak-peak input voltage is significantly smaller than \$26\mathrm{mV}\$, you'll get distortion. On the bright side, you have lots of ...


1

The 2.2uf is WRONGLY installed. god eye for catching that, Jeffrey. Since the VDD is Positive, the base of Q2 will be about 0.6 volts above ground, thus the 2.2uf cap needs its (+) terminal to the base. ============================= Unfortunately, during the power-on transient as VDD quickly rises to +12 volts, the top pin of the volume control will ...


1

The two capacitors are DC blocks, conducting the (AC) signal while blocking the DC so that different circuits can work at different bias levels. The pot is just a volume control.


1

consider the RCA 3N170 dual-gate MOSFET, of era 1970s. Miller Effect was greatly reduced, and the precious RF energy could be used in lower-capacity-higher-inductance high-Q narrow-bandwidth amplifiers that had many dBs more gain. And, as others indicate, the S-param S12 and S21 change dramatically, usefully altering the stability-circles.


1

The 200k is superfluous. The 100k feedback resistor together with the inverting input stage makes a 'virtual ground' input stage, with essentially zero input impedance. The impedance then seen by the external signal is 5.1k. Any bias current required by the input stage is met through the 100k. Your input HP filter is therefore the 100n + 5.1k, which has a ...


1

To answer your questions: Gain is not really relevant, you can cascade as many amplifiers as you want. What matters is NEP, that sets when your SNR is limited by detector noise vs. shot noise. What NEP do you need? Sure, amplify as much as your want. It does not help SNR. Specs are not well defined yet. Can't specify if it is possible. APDs have lower NEP, ...


1

Now that it seems certain that you have a dynamic microphone, let's have a look at what your DRA818V module needs at the microphone input. On page 3 of the datasheet, there's a table of radio characteristics: Down towards the bottom is this line: Sen_MOD Modulation Sensitivity @1KHz at 2.5KHz Fdev. 10mV That's telling you that a 10mV peak to peak ...


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