# Tag Info

8

Maybe look at it like this: - I've removed components that get shorted or open circuited by the PTT switch contacts In transmit mode, the speaker becomes the mic and the op-amp has a large gain determined by R3 (100k) and the small impedance of the speaker. The Rocking Armature's impedance is 150 ohm so for a rough estimate, the gain is 100,000 ÷ 150 = ...

7

A few problems with the op amp circuit: Depending on your motor, its highly likely that the op amp cannot source enough current to drive it. The op amp is running open loop. The output will almost always be saturated at the positive rail because the inverting pin is pulled to common. You may have a few tenths of a millivolt of control, depending on the ...

6

Some points: In the non-op-amp circuit, as you turn the motor near maximum voltage, a lot of current flows through a smaller and smaller segment of the pot. So it's a good idea to think of something better, indeed. 100 ohm resistors to ground are wasteful of energy; you don't want on battery power. If you have the op-amp in place, the potentiometer value ...

5

Yes. In general all the circuitry on a board that uses the same voltage has the same supply. The only exception is when you have (for example) "digital" 5 V that will have noise on it and "analog" 5 V that is cleaner. I am assuming that the batteries can supply enough current to power your several op-amps. 25 mA is fine. Higher currents will result lower ...

4

To make the new circuit work the same way as the existing one, you'd need to replace Q1 with a p-channel MOSFET, not an n-channel one. However, you would not save any power dissipation by doing this. Because you wouldn't operate the MOSFET fully switched, you'd be operating it as a variable resistor. You can see that there's no way to significantly ...

3

Your op amp is oscillating because your open-loop gain is larger than 1 at a frequency at which the phase shift is 180°. The op amp in your circuit is driving an almost entirely capacitive load - the MOSFET's gate. There are many possible ways to correct this using just a well-placed resistor or a capacitor. It might be best to use a series resistor or a ...

3

Wow, it is impressive that you would ask this question, it shows admirable courage. Loop Stability Analysis in the Real World. "How does one develop a Bode-plot for circuits such as this using non-ideal op-amps that contain important poles in addition to the ones created by my passive components?" Two questions should be kept in mind while ...

2

The Zener, in combination with the 1K resistor, is forming a stable voltage source for the comparator to compare against. If your 12V supply is stable enough, you can divide that down and use that for reference instead. But not the way you have shown. The Thévenin impedance of the original circuit is ~0 + 10k, or 10k ohms, but the Thévenin of yours is 10k ...

2

The gains of the two op-amp circuits you refer to are DC gains and, in simple circuit configurations apply across a range of frequencies up to a certain "limit". The "limit" is usually (but not exclusively) the point where the op-amp can no longer sustain the desired gain and this may be due to parasitic capacitance on the circuit board, intentionally ...

2

If your 9V supply is actually a split supply, you will probably be fine. Use caution if you intend to make more than one of these. As far as changes, you won't need to make many. Put voltage dividers on + and - pins to scale their voltage to $\pm20mV_{max}$. You will get a more linear response this way. If you do this, eliminate the diode bias ...

2

Note that the LMC6772 chip you are using is not an opamp; it's a comparator that has an open-drain output. Since it can't actively drive its output high, the Schmitt trigger circuit won't work as designed. The datasheet says, "Refer to the LMC6762 datasheet for a push-pull output stage version of this device." However, look at how you have the optocoupler ...

2

The underlying problem description suggests a possibility of a low-tech solution rather than the high tech path. This is not to say that the mentioned integrator based approach won't work, of course. Basically, how about two current limiting mechanisms in series: One that works well within the 40 microsecond time, but triggered only at close to the ...

2

Assuming that the problem is the capacitive load (gate of the MOSFET) some ideas are: In audio amplifiers, the classic approach for defending against capacitive loads is the inclusion of an output inductor, often in series with a resistor. Just an idea to keep in mind: don't forget inductors as a way of isolating from capacitances. Ever notice how the data ...

2

This is really simple - use an N channel FET and have it as a source follower. You can even use a BJT. The one below has gain due to the 3k3 feedback and the 1k to ground from -Vin. If you don't want gain connect the output directly to -Vin and omit the 1k. A unity gain buffer on the output of an op-amp is either an emitter follower or a source follower. ...

2

Here is the vital part of your question: - This tells me that you'll need two cascaded 2nd order high-pass-filters. Cascading the filters is the same as the multiply in the middle of the bottom equation. Here's a sallen-key high-pass filter (remember you'll need two cascaded): - The transfer function for it is: - Now you need to convert your ...

2

No, that's not what he does. He feeds each microphone through an amplifier then through an output impedance that feeds the line through R14 and C4 shown in thick red circle below: - I've shown the wire that is the actual line used by the intercom in a thin red/brown circle. Because of the output impedances of several units connected all at once there ...

1

Note that the 20mv error is a measure of how close the output can go to 0v (using your supplies) against a 100 kilohm load pulling the output to "mid-supply" i.e. 2.5V. Without that pull-up, as in your application, the output will go considerably closer to 0V. However if you need to inject offset; the place to do it is "ref" i.e. pin 6 - whatever voltage ...

1

As has been pointed out, the output impedance of the LM358 is interacting with $C_{\text{iss}}$ of the FET to place a pole at about 20kHz. Since the loop still has lots of gain there it oscillates. You could try: Lowering the output impedance of the amplifier (a lot) by adding an emitter follower buffer at the amp output. Isolate the ...

1

Your opamp is not stable probably because you are driving a capacitive load (gate capacitance). Remove C10 and lower the value of R15 to tens of ohms. You can also try using a different opamp. The datasheet of LM358 says: Capacitive loads which are applied directly to the output of the amplifier reduce the loop stability margin. Values of 50 pF can be ...

1

The sky really is the limit when it comes to op-amps - it is definitely possible to find audio bipolar op amps that have low offsets, tiny input bias currents, and low noise. A quick Google search turns up the LM4562, which has better input bias current specs than the NE5532 by two orders of magnitude. It also costs about 5 times as much. Whatever ...

1

Rather than using a current sensing resistor, I would suggest that you instead monitor the voltage drop across the MOSFET and compare it with a reference signal that indicates how much the MOSFET is allowed to be dropping at any given time. When the MOSFET is not enabled, it should be allowed to drop the full supply voltage (indeed, that would be expected). ...

1

Generally a guitar amp would have about 100k (or greater) input impedance - this is because the tone controls and volume controls are about that sort of range. Yours has 10k input impedance. I'd make R8 100kohm and R7 1Mohm Your 2nd stage is not needed - it has unity gain for the relevant frequencies and your filter cap might just as easily be placed across ...

1

Many problems: You forgot to power the opamp! This is not a rail to rail opamp, so the power scheme needs to be thought about carefully. You say you have a non-inverting amp followed by a active filter, but that is not what the schematic shows. Both stages are clearly inverting. You mention a 3.3 V supply, but then the schematic shows a battery used ...

1

The lower circuit, RC only works when: The output impedance of Vin source is 0Ω and at the same time: Vin is much larger than VC . Ideally the capacitor voltage is 0V, but that defeats the whole function as an integrator. The voltage across the capacitor influences the current through the resistor. and at the same time: The load of the circuit ...

1

U1A is a unity gain buffer, given by R5/R2. C2 is there to put the non-inverting inputs of U1A and U2A at AC ground, because the power supply isn't symmetrical: it's not a low-pass. U2A has gain of 100, given by R10/R7, i.e. 220k/2k2. If you consider the OP-amps one at a time it becomes clearer.

1

Referring to the schematics in the question, there are unlikely to be any advantages to using an op-amp, and quite possibly a number of disadvantages. First, your op-amp schematic is drawn without feedback and with the negative input grounded. Due to the high DC gain of the op-amp, the output will most probably be saturated at either the positive or ...

1

The "buzz" is the usual mains hum, picked up by the body as an antenna and transferred when you touch it. Breadboards tend to be very good at picking up ambient noise anyway, and capacitatively coupling signals that should not be coupled. Try running the whole thing off a battery; that will tell you whether PSU noise is a problem. (Are there two power ...

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