# Tag Info

9

Because you're trying to have the op amp output a voltage higher than it's capable of. A typical upper limit is V+-1.5V, so the fact that you're getting 3.8V out on a 5V supply is already better than that. Either pick an op amp with a rail-to-rail output or use a supply with a higher voltage.

9

You have to actually read the datasheet before using a part. Here is the relevant snippet from the LM324 datasheet: The middle column of numbers applies to the LM324. With 30 V power, it can only go to 26 V output, meaning it requires 4 V of headroom. The datasheet isn't very clear what the headroom requirement is with 5 V power, but the 1.2 V you see ...

9

As others have already mentioned, the output voltage of a real op amp can never reach (be equal to) the voltages on the power supply rails (+VCC or -VEE). Looking at the schematic diagram for a real op amp, one can understand by inspection why this is. Figure 1 is a schematic diagram I copied from the National Semiconductor LM324 data sheet (dated August ...

8

To get 13V out of the op-amp, you need to have at least 13V as its positive supply (even for a rail-to-rail op-amp). So, just keep it simple and use that instead.

8

For the question in the first paragraph A clarification to my earlier statement. You have to be careful how you explain things. If you assume that "two nodes are always kept at the same voltage by some unknown force", then the problem is simply, no it will not have an effect - because as you stated, the two nodes are always kept at the same voltage no ...

7

You're using the PNP as an emitter follower so no wonder it adds a voltage to the bottom side of the waveform. What you need to do is use an NPN or an NMOS transistor. Then the polarity of the signal from opamp U1 (used as a comparator) is "upside down" but that is easy to fix: just swap the + and - inputs of the opamp. In the case that you use an NPN, do ...

6

Op amps which use dominant pole compensation have a constant gain-bandwidth product. You've cited a 5Hz dominant pole, and the LM301A datasheet gives a typical open loop voltage gain of $160\text{V}/\text{mV} = 160,000$. This gives a gain-bandwidth product of $160,000 \times 5\text{ Hz} = 800,000\text{ Hz}$. You have configured this op amp with a closed ...

5

The dominant pole can be seen only in the open loop configuration. When you add negative feedback, gain is reduced and the BW is enhanced (for the closed loop system)

5

The stable non-saturated solution to an op-amp circuit satisfies equality of voltage across the inputs, but it also satisfies other requirements such as near-zero current into the inputs. If you add a wire, you replace the restriction with a much weaker one -- that the currents at both inputs are equal and opposite. This greatly increases the set of ...

5

The nodes being at the same voltage does not mean that they can be shorted without any ill effects. Consider this circuit: simulate this circuit – Schematic created using CircuitLab Now, assuming the opamp is ideal, the voltage on nodes A and B should be the same (equal to 1V). The current isn't though. The current flowing through R2 is much ...

5

•How is this open-loop gain measured? Do we use some kind of special voltage meter to measure the voltage across the output and the ground? But if so, it is not open-loop at all. It's kind of a dilemma. Or do I derail from the correct path of Electrical Engineering thinking? Measurement of open-loop gain is (in principle) straightforward. You apply a ...

5

In electronics engineering we often simplify our reasoning for "small signal" analysis where we consider that our circuit has lineair behaviour which means that we can sum signal intensities and therefore also multiply them with fixed values. The behaviour for a signal of 1V is then similar as the behavior for a signal of 1mV - we just need to divide by 5000....

5

The input voltage range of the LM319 does not extend to ground or below. The lowest valid input voltage is 1V above ground. Note also that the output is open collector so you do need a pull-up from output to positive rail.

4

There is a difference between OP AMP decoupling and, for example, logic chips decoupling. The purpose of bypass capacitors is to provide sufficiently low impedance on power rails in the whole frequency range of the OP AMP. Different types of OP AMPs have very different frequency range: Gain Bandwidth Product (in fact, it typically defines the bandwidth) ...

3

You can use a voltage inverter like the LMC7660 to generate a negative rail for your op amp. It converts any voltage between 1.5V and 10V into its negative counterpart -- e.g. +5V in, -5V out. Only good for a few mA though.

3

How big a deal is it if the pots are far from the op amp feedback loops? 1) A very big issue. But keep the traces together and avoid digital traces. Cross them at a 90 deg angle if you have to. A 5pf to 22pf capacitor across the output and (-) input of the op-amps will stabilize them if you get ringing or parasitic oscillation in the op-amps. The ...

3

Because the input is only 0.5V (-6db). You should divide the output voltage by the input voltage to avoid such problems or even better use 1V when doing an AC analysis.

3

Figure 1. Internals of the ancient 741 opamp. Source: Wikipedia. Most opamps will have an output arrangement similar to the push-pull arrangement of the old 741. Others will have FET transistors rather than BJTs. In either case if the top transistor (red oval) is turned on the output will be pulled to positive rail. If the bottom transistor (green oval) is ...

3

You did not give us a lot of details, so I don't know whether my solution is really applicable in your case, but the simplest is to AC-couple your 10MHz digital signal using a capacitor: simulate this circuit – Schematic created using CircuitLab On the output, you end up with a signal that moves between about 59.5 and 64.5V (due to the diode ...

3

In an op-amp circuit, a feedback network maintains the two inputs at (very nearly) the same voltage. Usually the feedback network around an op-amp assumes two independent KCL equations, where the input current drawn by the op-amp is negligible (on the order of 1nA). But each node has its own independent KCL in this design. At first, very little current ...

3

Think about this in a reasonable manner. Obviously the opamp cannot output 5000 volts when it is connected to some smaller voltage power supply. So let us say that the opamp is connected to supplies of +5V and -5V. In the case of an ideal opamp this limits the output to not be more than +5V and not less than -5V. In the real world the opamp may get close to ...

2

The main difference is that opamps need to be stable when in a negative feedback configuration. This means that they need to be compensated. Compensation slows your slew rate. For comparators stability is not such a big problem. All you need to ensure is that it saturates and doesn't oscillate. Comparators also have two output states. This means that gates ...

2

Another way to look at it is if you can get the high speed and low sensitivity and IF the device doesn't saturate or slow down if you load it heavily then you need not have to have the log aspect of it. Look at the Analog devices ADN2880 which is a SiGe based 2.5 GHz Integrated TIA. And the HMC6590 (which might be too costly for you) also is blazing fast (...

2

You can use the fact that you are dealing with a linear system. The overall transfer function H(s) = H1(s) + H2(s) where H1(s) = Vout / Vin1, Vin2=0 and H2(s) = Vout / Vin2, Vin1=0 Assuming an ideal OpAmp with infinite gain one of the capacitors is between 0 (=gnd) and the virtual ground of the OpAmp and can therefore be ignored for the calculation of H1 ...

2

The top circuit is the better one by far. The advantage of the top circuit is very fundamental, it's input impedance is low which is the ideal way of measuring currents from a sensor that behaves as a current source. The opamp will keep the voltage at the input zero so parasitic capacitances are kept at the same voltage. In effect, the influence of these ...

2

The current source can be switched to operate either as a current source or a current sink. When it sources a current the capacitor C1 gets charged, when it sinks a current C1 gets discharged. Having a constant current results in a constant slope and produces the triangular waveform. To get a rising and falling edge the current needs to change direction.

2

If your input voltage swings equally positive and negative then it would be expected that your output would do the same. However, with 0 volts as your most negative supply rail this cannot happen. An op-amp can only produce output voltages within the confines of its power rails. This won't be helped by using a rail-to-rail op-amp either. The data sheet has ...

2

The inverting input of an op amp produces a negative output relative to the non inverting input. So if your signal is 2V +-0.3V then you can set the non inverting input's 'ground' reference to 2V and the output will also be relative to 2V. simulate this circuit – Schematic created using CircuitLab

2

First, thank you for the well-written question. Based on your scope captures, I suspect the mid-rail biasing is not working correctly. You should be able to check the output bias voltage by shorting Vo1 and IN- together, then measuring the voltage at Vo1. (Turn your input sine wave off first, obviously.) You should see about 1.6V. If it's more like 0.8V, ...

2

At the very best, in a a RR Output (RR = Rail to Rail) op-amp, the circuit will have been designed to run up to and very close to the rails of the op-amp. Usually however, this is not the case and the outputs will not come close, these values are specified in the data-sheet. For an op-amp to exceed the the rails would mean that it would have to have some ...

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