# Tag Info

9

The circuit employs negative feedback and utilizes the very high gain of the op amp. The op amp will try to keep its non-inverting and inverting inputs at the same voltage $V_{set}$ due to its very high gain. Then by Ohm's law $$I_{set} = \frac{V_{set}}{R_{set}}$$ Negative feedback causes the op amp to adjust the transistor base voltage so that ...

5

The opamp is acting as a unity-gain buffer, though it may not be obvious: The rule for opamps is that the output does whatever it has to to keep the two inputs equal, provided that it doesn't clip of course (run into its own supply and stop there). The transistor is used as an emitter-follower, in which the emitter voltage follows the base voltage minus a ...

4

The best way to be able to control arbitrary loads would be to connect the input side of a solid-state relay in parallel with the LED, across points A and B in your schematic. With the appropriate selection of SSR, this can be used to switch AC or DC loads at any voltage or current. Note that this is a poor circuit to begin with. The opamp is being used as ...

4

Probably the easiest way is to change the T1 to an NMOSFET to increase current switching capability. Since a lot of fans are designed for PCs and rated at 12V I would also recommend upping the supply to suit the fan. No other changes would be needed. EDIT NOTE: Have removed connection of pin 12 (carry out - output) from ground. (See EM Fields answer - ...

4

There is no current flowing into the inverting (negative) input, because the input impedance of an ideal op-amp is infinite, therefore the current flowing through the feedback 1k$\Omega$ resistor (let's call it $i_3$) is the sum of the two currents flowing through the two other resistors ($i_1$ and $i_2$). The inverting input is a virtual ground, as ...

4

From a more practical point of view, this is what you have: simulate this circuit – Schematic created using CircuitLab A voltage divider on the input, a unity gain buffer and a voltage divider at the output. Voltage divider output is calculated as follows: $$Vo = \frac{Vi \times R2}{R1 + R2}$$ Where Vi is 7.5V for the first divider.

4

Golden Rules. With negative feedback in place... No current flows into the input terminals of the opamp The voltage at the plus terminal equals the voltage at the minus terminal. Because no current flows into the positive terminal, current must flow from the voltage source, through the series resistors 16 ohm and 24 ohm. So you should be able to figure ...

3

GBP is the to do with the open loop gain of the op-amp. If you have a closed-loop circuit then GBP can help you find where the flatness of the frequency response starts to be eroded. GBP - if it has an open loop DC gain of 1 million and unity gain at 1MHz then the GBP is said to be 1 million. This is useful to know because if you have an op-amp that has a ...

3

The output drive capability is defined in a few places in the datasheet, but not guaranteed except at 2K load. The below graph is based on +/-18V supplies. As you can see you can get 20mA easily out of it, with 15V swing. With a 100R load, the output capability would be a couple of volts or so safely. I don't see why you'd want to follow this part with ...

3

I amplify using LM358 Op-amp and 2 resistors. An LM358 is a really, really bad choice for driving an 8 ohm speaker. Typical output current is 40mA and into an 8 ohm load, the power delivered will be: - $I^2R = 0.04^2\times 8 = 12.8 mW$ Also, you may have wired up your stereo inputs incorrectly and be cancelling out the mono part of the stereo signal ...

3

Is it possible to design a circuit using op-amps such that it's output gives the laplace transform of the input signal? No, it's not possible. For example, $F(0)$ is the integral over all positive time of the input $$F(0) = \int_0^\infty f(t) e^{-0t}dt = \int_0^\infty f(t) dt$$ So, you would have to wait 'to the end of time' to get the 'initial' ...

3

When supplying a peak of 1.4142 volts above midrail there will be a current flow of 62.5mA RMS (1volt/16 ohms) - this current is sourced from a transistor that is connected to 5V hence it, (the transistor), will dissipate a power of possibly over 100 mW. This is repeated on the other bridged op-amp but power dissipation on that side will be slightly higher ...

3

Guitar pre-amps are usually designed with a high input resistance (typically 1M0). Your circuit has an input impedance roughly equal to the value of R3 which I suspect is somewhat lower than R4 probably by a factor of 10. Even at 150k it would be too low. A better approach would be to use a non-inverting configuration for the op-amp such as. The gain ...

2

Actually not the voltages add up but the currents. All currents flowing to the negative input (which is assumed to be high impedance, i.e. "open" and at ground level) must add up to 0. Since currents are proportional to the voltage across resistors, the voltage of the output must be equal in magnitude (but opposite in sign) to the SUM of the voltages at the ...

2

From your schematic, it looks like you're not using the compensation pin for the AD829. The datasheet recommends using up to as much as 68pF on this guy when running with low gain. Since this chip is so fast, you can often have unintended positive feedback at high frequencies due to the layout. I would recommend examining whether or not you should be using ...

2

You are trying to create a detailed model of a sloppy process. For one thing, the gain-bandwidth product is a minimum spec, or sometimes even just "typical". The actual gain-bandwidth can be quite different for any one part, and I wouldn't count on the phase margin either. The raw DC gain can also vary a great deal, as can other parameters. Getting this ...

2

Pin 12 is an output which is high whenever the count is less than 5, and should be left floating instead of being shorted to ground, as it is now.

2

Yesterday I was experimenting with Rigol ds1052e myself and yes you WILL see 50Hz noise with probe without even touching anything. Try to connect ground alligator to the tip of the probe and you will pick up even more noise. This goes away when you unplug probe from BNC. So yes, this is from the probe itself, not necessarily from your circuit. I've seen ...

2

Just to make the point clear and actually put an answer to this question. (see all the comments by Dave, Jippie et al.) They are totally different ICs. The 358 is is dual op amp (it has two separate op amps in the same 8 pin package). The 386 is an audio amplifier. They have different pinouts and cannot be interchanged.

2

It's a typo. The 'Y' axis should be mA not nA! The high resistance (M$\Omega$)applies to differential input voltages less than 1V (where the graph looks to be perfectly flat, at a scale of mA). This is typical of this kind of bipolar super-$\beta$ input op-amp. You should not generally use such an op-amp in applications with significant differential ...

2

The solution set is in error. To see this, simply insert a series voltage source in either input of the op-amp and determine the resulting output voltage- the magnitude will be as Chaudary says. The sign of the result is somewhat arbitrary but normally a positive offset is assumed to be driving the amplifier positive at the output, so I would say the sign ...

2

There are two basic factors to consider when working with an amplifier: the gain, and the power. The gain is basically increasing the voltage of the incoming waveform. The power is the limit to what the amplifier can do. If you think of the output as a simple resistive circuit you can apply Ohm's Law to get a clue as to what is going on. Say you have a ...

2

You can use any amplifier you like. You will get better performance if you use a modern \$5 dual op-amp than a 2-cent/amplifier LM324. Sub-zero means nothing to me. If you want to the amplifier to have guaranteed characteristics below 0°C, then you should buy one that is guaranteed for that temperature range. Below -55°C you may have to qualify them ...

2

The op-amp symbol can sometimes lead one astray as, I believe, it has in this case. Recall that the output of an ideal op-amp is an ideal voltage controlled voltage source which means that the output will source or sink any amount of current. Let's redraw the circuit by replacing the op-amp with an ideal amplifier model: simulate this circuit ...

2

You are imagining it wrong. The feedback loop reduces the gain at he same time as signal is amplified, not after that. Op-amp can be modeled as a feedback control system: The output signal is amplified by $\frac{A}{A+1}$, never more. I agree that with ideal step input, it feels like a chicken-or-egg problem. But in reality, there is always finite slew ...

1

This can not be done because the circuit you are trying to design is non-causal. It has to have nonzero output for all times from negative infinity to positive infinity, even if the input is zero up to some t=0. If you allow an arbitrary delay between the input and the output then, like Scott says, you are talking about making a spectrum analyzer.

1

Not a true Laplace transform, but you could certainly build an analog spectrum analyzer with op amps, if the magnitude of a snapshot of a signal is all you need

1

No - I don`t think it is possible because I see something like a "domain contradiction" in this task. What you are searching is the following: An input in the time domain should produce an output signal that is defined in the frequency domain. How could an opamp perform such a task?

1

To answer your first question about the node between R4 and R3 you have to decide if the feedback is positive or negative. I'll leave that up to you but if you conclude that the feedback is positive then the answer is a certain "no" and both op-amp outputs will be pushed against the power rails in saturation. If you decide the answer is "yes" then, as a ...

1

I would try adding Schottky diodes (eg. 1N5819) from the output of each op amp to ground and +5 (four diodes total). Also put a resistor (something like 1K) in series with the inverting inputs (two resistors total). You should also make sure there is good bypassing on the 5V supply. Perhaps 10uF ceramic in parallel with 1000uF/6.3V low impedance ...

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