# Tag Info

10

Yes you can and I've done with lower power OPAs (in order to preserve low board height etc.) The main gotchas are: You need a way (e.g. series resistors) to limit Voffset induced currents circulating between amps. Series resistors will blow away some power. Their value needs to be evaluated carefully. The higher they are, the more protection the amps have, ...

9

"Vin is 5V, so Vout should be 50,000V." Why? The OpAmp amplifies the the difference between the + and - inputs, not just the value on the + input! OK, you might start with: the output is at 0V, and the input (connected to the + input) is 5V. What you have done is apply a 5V step to the input. Now what happens is that the OpAmp starts to rise the voltage ...

8

Most Basic Interpretation: Here is my intuitive way to understand a given op amp circuit by personification. Picture a little dude inside the op amp. The little dude has a display that indicates the difference in voltages between the + and - inputs. The little dude also has a knob. The knob adjusts the output voltage, somewhere between the voltage rails. ...

7

If you short collector to base and use it as a diode between the opamp output and the inverting input then any base current has to flow from the input signal causing an error, especially at low currents where the HFE drops off. In the arrangement show the base current flows to ground and so the current flowing into the input is slightly less. This could ...

6

It's an inverting amplifier, with a no-load gain of $-\dfrac{R_b}{R_1+R_a}$ at DC. There are low-pass filters on the input and output, with cutoff frequencies of $\dfrac{1}{2 \pi RC}$ where $R = R_1||R_a$ for the input circuit and $R_2$ for the output circuit.

6

There are so many different amplifiers because there are so many different applications, each requiring different attributes. A perfect voltage amplifier, for example, has infinite input impedance and zero output impedance. Neither of these exist in a real amplifier, of course, but there are devices with Gigohms of input resistance (look for a JFET input ...

5

Differential input impedance is the ratio between the change in voltage between V1 and V2 to the change in current. When the op-amp working, the voltages at the inverting and non-inverting inputs are driven to be the same. The differential input impedance is thus R1 + R2. If the op-amp was 'railed' (saturated) then the differential input impedance would ...

5

Presumably the amplifier can drive its output below ground or the diode would not be necessary. The 2N3055 has a Vbe breakdown rating of -7V so we need to be sure not to exceed that voltage. Reverse breakdown of the base is not good for the transistor- it can cause permanent beta degradation over time, and the 2N3055 is already miserably low gain. The ...

5

You cannot analyze the circuit to explain this effect assuming that the op-amp is ideal. The DC open-loop gain of an LM324 amplifier section is typically about 1E5. It has a gain-bandwidth product of 1MHz, meaning the the open-loop gain at 1kHz will be only about 1000. The analysis you need to do will depend on whether your input frequency is a few Hz or ...

5

The inputs of an op-amp with negative feedback are assumed to be equal to each other. Without it, the assumption that $V^+ = V^-$ produces an unstable answer. The circuit you have is using the op-amp with positive feedback and thus $V^+$ doesn't have to equal $V^-$. Instead, it is an analog comparator with hysteresis. When $V^+ >V^-$, the output ...

5

A method that I have used successfully to transfer a linear signal through an optocoupler is to use a second optocoupler to provide the negative feedback. This compensates for the non-linearities. The LED of the second optocoupler is put in series with the main one and the phototransistor form the second one is used in the feedback path of the driving ...

5

Since you are digitizing this signal anyway, I would consider this: Abandon the micro's ADC and use a discrete one. Then isolate the digital lines instead. This gets you much better linearity at (possibly) a slightly higher BOM cost.

5

It doesn't match any op amp amplifier circuits I've seen In contrary, it is one of the well-known controller types applied in control systems: Proportional-Integral-Controller (PI). Here is the transfer function: $$H(s)=-\frac{R_2+\frac{1}{sC_1}}{R}=\frac{R_2}{R}+\frac{1}{sRC_1}$$ Please note that the resistor R is the resistance which is effective for ...

5

The preferred technique for measuring the current of an ion beam is using a transimpedance amplifier (TIA) configuration. I'm think here of gas mass spectrometers but generally it is a good idea because it keeps the target potential at near zero volts. Imagine what happens to your ion beam as the "cup" starts charging up - will it deflect the beam and make ...

4

A lot of comparators just have an open-collector output and this is normally used by connecting a pull-up resistor to a positive supply rail. In your circuit you have shown a TL081 (written since the op edited it to an LM311) and this doesn't qualify as a comparator - it's an op-amp and as such, it doesn't need a pull-up resistor. Here's a circuit of a ...

4

. JUST ALWAYS USE TL071 ! :) Or maybe LM741. Too many devices? The situation is similar with all components. Have you seen the number of 1K resistor types you can buy on digikey? OK, what should we use as a general-purpose hobbyist op amp? For beginners the Op-Amp rule used to be simple: always use Fairchild uA709 op amp for everything, later ...

4

The thing that's usually special about each op-amp is how well they approximate (or don't) the ideal op-amp. In theory if you stick within their specifications the ideal op-amp model works. Don't worry too much, though. If you're doing something which is a few kilohertz and doesn't need to be too accurate (say, less than 1%) pretty much any op-amp will work. ...

4

The equation for the capacitor is rather simple. The capacitor will interfere with frequencies lower than $f = {1 \over {2\pi RC}}$ where R is the input impedance of the next stage. Simply substitute the desired minimum frequency and impedance, solve for capacitance, and pick a larger value.

4

Of course you can try to compensate for the DC offset but why not eliminate the influence of DC offset in the first place ? You can do this if you do not need 10000 times gain at DC. You mention that your signal is 40 kHz, I conclude from that the DC value is irrelevant to you. Then I would just make amplifiers that have 10 x gain at 40 kHz but 1 x gain at ...

4

Trace both current paths from output, and you find they both end up at ground. Via R3 directly, and via R2, R1 and V1. So if the output sinks current I, that current must come from ground. Simple as that... The more interesting question, is what happens to it inside the opamp. From the opamp output there is no direct connection, internally, to ground. ...

4

All three of your questions are related. This serves a dual (triple? see #2) purpose of blocking any DC voltage on the line, and re-referencing the voltage to something the circuit can use. Note that the voltage on the + pin of U1G1 is connected to Vcc/2. If you apply your laws of op amps, you know that in negative feedback mode (where the output is ...

4

These stages are inverting. Your equations are both missing a - in the transfer function. V1 gets inverted twice, that's why it has a positive coefficient. V2 gets inverted once, that's why it has a negative coefficient.

3

Actually, the phenomenon you describe used to be a real problem, way back in the dark ages (1970s). The venerable LM310 Voltage Follower data sheet contains the application hint (bottom of page 2) which recommends a 10k ohm input resistor in order to maintain stability. Also note that your argument can be applied to any op amp circuit, and dealing with ...

3

The equation you have should be correct. I derive it this way: Note that the only assumption I'm making here is that the output swing of the first opamp is symmetrical about zero, ±$V_{O1}$. It doesn't matter whether this swing is related to the supply voltage or not. The voltage $V_{FB}$ at the + input of the first opamp is: V_{FB} = ...

3

The base voltage is prevented from going more than about 0.7 volts negative at the base (with respect to ground). This is sometimes required because BJTs have a limited ability to cope with negative (reverse biased) base emitter regions. The assumption here is that the driver (the triangle) might be powered from positive and negative supply voltages. NPN ...

3

Some of the best performing mixers just use diodes and accurate transformers. (Double balanced mixer) to do four quadrant mixing. They do have loss and need a lot of local oscillator drive but they are low noise so the overall noise figure can be pretty good. Mini-Circuits is one of the most well known manufacturers (Mini-Circuits.

3

What customers don't want to do is find that their op-amp (in their design) has been replaced by a so-called better spec model. Only the naive would assume that the better spec model will not potentially cause problems in their specific design so, you have a proliferation of old crappy op-amps and the ongoing development of new ones. Hardly ever do old ones ...

3

Slew rate limit means that there is a limit to how fast the output can slew- in volts per microsecond. This is a kind of nonlinear distortion. The text is saying that because the output only has to slew 1.4V rather than the negative supply voltage or so, so it will be able to respond to zero crossings faster. If you think about the output, just before ...

3

An opAmp operates in continuous time and not in discreet time. This means that no action can occur instantaneously and actions do not happen in steps. Even if a switch is flipped to connect a voltage to the + pin there is still a transient rise time in the input and the out put continuously follows. This is very commonly described as opAmp action. A spice ...

3

In general the first consideration is what happens if one power amp has a (slightly) different output level than the other one. Will there be a current flow from one output into the other one? If yes, will this damage one or both components? It is usually better to just use one op-amp and use a discreet FET, Transistor or IGBT for the power part.

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