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I am hoping to use 9V batteries, and realize I might have to run two or more in series? This is how to make bipolar power supply using two batteries. And for reverse engineering the board - check to which connector terminals goes pin4 and pin8 of the IC. Then connect batteries correspondingly +Vs to Vcc+, -Vs to Vcc- and GND to GND. Image taken from ...


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The circuit comme from : "https://hal.archives-ouvertes.fr/hal-01856768", the Nauta configuration has a very low CMRR. Best Regards


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A simple way to get the frequency response is to measure the impulse response. The impulse response is related to the frequency response. The frequency response is just the Fourier transform of the impulse response as described on the DSP pages.


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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 ...


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The 12AU7 needs a much higher plate voltage than your circuit is supplying. I have a bass rig with 12AX7A input stage and the tube is supplied from 250V through a 100K plate resistor. I don't think you have a chance in hell of making the circuit work nicely with anything less than 75V, just guessing. You should study the characteristic curves here: https://...


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Significantly lower the values of R19,R20 & R21


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For part 2, one possible approach would be to consider each element of the desired function as an independent circuit. Consider the formula as the following block diagram: If an opamp is used to build each block, it is clear that we need a basic amplifier, an integrator, and a summer, so the circuit will be a bit more complicated: simulate this circuit &...


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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 ...


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For #3, notice that the gain has to be 2 because 7-2 = 5 and 10-0 = 10. You can shift the output with a summing circuit. Something like below: simulate this circuit – Schematic created using CircuitLab


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Not sure why you need the middle voltage follower opamp. For #1, something like below should give \$v_o(t) = 3v_i(t) - 2\$: simulate this circuit – Schematic created using CircuitLab


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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 ...


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That's Kirchoff's voltage law combined with the assumption that the opamp does its job i.e. the output settles to such Uout that the voltages at the inputs are forced to be equal. The changes in the voltages are assumed to be so slow that the speed of the opamp is high enough to retain the equality at the inputs. So, the plus input is at 0V => The minus ...


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The current which flows through C1 and R1 is the same, and equals: $$i_2=C_1\cdot \frac{dU_{c1}}{dt} = \dfrac{U_{out}-U_{c1}}{R_1}$$ Rearranging and solving for Vout: $$U_{out} = U_{c1} + R_1 \cdot C_1\cdot \frac{dU_{c1}}{dt}$$


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Hint: If voltage across capacitor is \$u\$, then the current through it is given by \$C\dfrac{du}{dt}\$


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There is already a similar question, however the design procedure is still misssing there. Now to your questions: I can't really find characteristic curves for transistors on datasheets which makes it difficult. Although there is no output characteristic curve in the datasheet, you can find specific biasing points which might already fulfill your needs. ...


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That 2.2uF is connected the wrong way around. The +ve should be towards the higher voltage level. With no signal from the mic all three terminals of the pot are at 0V. With a mic output signal the signal at the top of the pot wiggles about 0V and the wiper signal also wiggles about 0V but is a reduced amplitude version of the signal at the top of the pot. ...


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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 ...


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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.


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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 @...


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Just draw the small signal model of the circuit to calculate the impedance seen at the emitter. Assuming a resistance \$r_b\$ is connected to the base, the impedance is calculated as: simulate this circuit – Schematic created using CircuitLab Apply KCL at the emitter to give: $$\beta i_b + i_b +i_x = \frac{v_x}{R_E}$$ Apply KVL in the emitter-base ...


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Well, we have the following circuit: simulate this circuit – Schematic created using CircuitLab When analyzing a transistor we need to use the following relations: $$\text{I}_\text{E}=\text{I}_\text{B}+\text{I}_\text{C}\tag1$$ Transistor gain \$\beta\$: $$\beta=\frac{\text{I}_\text{C}}{\text{I}_\text{B}}\tag2$$ Emitter voltage: $$\text{V}_\text{E}=\...


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I think that expression should actually be:- ((R1//R2//Rs)/Beta) + re where Rs = signal source resistance In the ideal case of Rs = 0 the left hand term disappears and the equation reduces to re.


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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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Looking from the emitter terminal, collector is a current source whose current cannot be changed. It offers infinite resistance and acts as an open circuit: \$R = \dfrac{\Delta V}{\Delta I} = \dfrac{\Delta V}{ 0} = \infty\$ Thus \$R_C\$ doesn't affect the impedance seen from emitter side. You can obtain the resistance of a block by measuring the voltage ...


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If your wires can be as long as needed to have an antenna you could detect some really long waves, those which would be hearable if an AC voltage with the same frequency was connected to your speaker. In this case the AC voltage would be harvested by your wires. 1uF capacitor is not especially high reactance at 10kHz, so you could well hear something (=...


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Click and pop suppression is a standard feature in an amp driver these days. Since the amp you have doesn’t handle it properly, consider a different amp. That would be less work than trying to retrofit this one.


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With those values, you will receive pretty much nothing from RF. If it were resonant at an audio frequency, then you could (possibly) pick up something from the wires carrying the driving signal to a really high powered loud speaker (say, several thousand watts.) The real speaker would be so loud, though, that you wouldn't need your "receiver." The ...


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simulate this circuit – Schematic created using CircuitLab Figure 1. Simplified schematic of amplifier output and proposed switching arrangement. The "pop" occurs because the amplifier is running on a single-ended supply. To provide drive for positive and negative excursions of the speaker the amplifier is biased to half-supply when quiet, so, on a ...


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After replacing the 1N4148 diodes with the much stronger HER305 diodes, the clipping disappeared, even though the 1N4148's seemed in fine working order when tested with a multimeter. Still not sure why this worked, but it did. I was able to push the output wave up to 13.7V @ 3 MHz. And for those who were asking, this is what my amplifier looks like (...


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Q1 emitter-base short circuit. Check out Q1 or replace it! The effect is also apparent in the absence of 15 volts of the Q1 collector.


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I'm not sure why you AC-coupled your input and output, given that you have balanced power supplies, which makes it unnecessary. The point that the commentors are making is that your bias network doesn't do what you think it does. You're assuming that the junction between the two emitters is close to zero volts with no signal. But in fact, the actual voltage ...


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1A is a bit wimpy for an audio amplifier; 1A into 8 ohms is only 8W. I would concur with Mattman that these were probably 3A (1N5403 rings a bell, or 5405 for higher voltage) in that case shape. A Google image search for "3A diode" shows several in this case shape (merely illustrating my confirmation bias; "1A diode" shows some too, but fewer I think) ...


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Put in the highest peak-repetitive forward current diodes you can. And install a small fan. The only way for heat to exit is by the PCB foil, and there is very little of that.


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Those were a very common diode at one time. My very strong recollection is that they were rated at one Amp. 1N4005 should replace them just fine. Be sure to replace all four diodes.


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Replacing the amplifier part of the radio with a completely different amplifier will not fix the damaged radio parts. Did the radio make much smoke when it was damaged by the AC power supply? I think the entire radio and its power supply needs to be replaced.


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You might want to look into using monolithic Class-D Audio Amplifiers for this. I am not an audio person but there are a few things you need to look out for from a power electronics perspective. If you pass a sinewave to the gates of your FETs, you're not going to get the same sinusoud out because the FETs wont switch on properly. You will also need to ...


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This gives class_C operation, the best efficient RF mode. just follow the FET with PI matching.


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Older power MOSFETS were temperature stable and did not self-destruct. Newer power MOSFETS (this is according to the automobile industry and to NASA) can self-destruct. Thus your question may need to reflect the generation of FET.


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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, ...


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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 ...


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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.


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If the signal source has a low output impedance then the HP filter is -3dB at 314Hz and the LP filter is -3dB at 4.0kHz. Telephone or AM radio sounds. Maybe it was used many years ago for an Electrik Geetar.


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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 ...


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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" ...


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A tale of two transistors Generic NPN transistors often contain no parasitic capacitances, and often have other "ideal" properties. They are useful to see how parasitics affect frequency response. When a real transistor model is substituted, capacitances and other non-ideal effects become apparent. In LTspice, the generic, default transistor (called NPN) ...


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That amplifier has two cascaded high pass filters (assuming you add a load resistor). The filters cut-off frequencies, fc = 1/(2*piRC) Where, for the input filter R = Rs + (((B+1)R4)//R1//R3) say B = 100 and Rs = source resistance = 0 Ohms. And for the output filter R = RL + R2 I haven't done the maths but I would expect an overall high pass cutoff ...


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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.


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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" ...


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At the moment, you're directly applying a "hard" voltage to the base of Q1. That is unrealistic and can never be done in a real circuit. Try adding a resistor of 1 kohm in series with C2. Normally in this circuit the collector-base capacitance of the transistor is the main factor that limits the bandwidth. This capacitor is often referred to as a Miller ...


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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.


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