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9

The gate capacitance of a MOSFET is a more complicated topic than a lot of people realize. It depends very strongly on the operating conditions of the device. This makes sense — the capacitance we're talking about has the gate itself as one plate, which is a fixed physical structure, but the other "plate" is not just the source, drain and substrate ...


8

If the output load is primarily to ground, then two capacitors. If it's to either supply, then one capacitor will suffice. The purpose of bypass capacitors is to provide a low-impedance close to the chip (bypassing any series inductance to the supply rails). Since most op-amps do not have a ground pin the internal circuitry does not care about the ground ...


7

I'll interpret your question a bit more broadly, since the results with one or both inputs floating are unlikely to be working usefully. In general there are two categories of op-amps, bipolar ones which have significant bias current that is of guaranteed direction, and FET or MOSFET-input types which have smaller bias current of undetermined (and possibly ...


7

One capacitor for each supply, connected to ground. Consider that, if you have dual supplies, and each supply produces a positive spike of equal amplitude, a capacitor between the two supplies will see a constant voltage, and will attenuate neither spike.


5

What intuitions or rules of thumb would faithfully guide me on when to consider an OTA instead of a "regular" op amp; perhaps illustrated by any "classic" applications where an OTA would be preferred (and why)? You can't really compare an OTA with a regular OpAmp. OpAmps are simple building blocks that you'll usually "configure" to do one fixed ...


5

The gate resistor isn't helping anything (that I can see) so get rid of it! The circuit should be reasonably stable (without the 10K resistor) so long as your MOSFET source resistor doesn't get too much lower than it is. If you need lower than about 50 ohms then you'll need to add compensation. Be sure to check stability with a relatively small square ...


5

One parameter of the opamp you are violating is the input common-mode range. For the LM358 and many other opamps the inputs must not be with a certain voltage of the positive or negative supply rail. The LM358 was one of the first that allowed the inputs to go down to the negative rail and is often referred to as a single supply opamp. Even the LM358 does ...


5

Outcome Report Based on the accepted answer by LvW, I removed C1 to see if that would do the trick. Unfortunately no dice, as the waveform just changed to a different capacitive response: However that was when I was sure LvW was right, and that using capacitors for compensation had to go. Who knew that using such a conventional compensation strategy ...


5

I would go with Hypothesis 2. I think, you have simulated the loop gain correctly. The input impedance at the inv. input seems to be much larger than the combined source impedance of the remaining network. I recommend to make another test and select another point for placing the test signal source: Between the opamp output and the common node of C2 and ...


5

You can definitely do it with op amps. Some things worth investigating A 'T' feedback network in the op amp Be advised of the issues with that much gain in an op amp. You will need to account for things like Dynamic range Signal to noise ratio Current/Noise voltage (i.e. you will probably want most of the gain in the first of several stages as the ...


5

You can ground the source, connect the drain to the desired bias voltage (with a large capacitor - maybe 1uF ceramic) across drain-source) and directly measure the gate capacitance with a battery-powered meter or LCR bridge. The Vishay datasheet says around 0.7nF at 30V and 1nF at 2V Vds (for Ciss). If you don' t have a C meter, a reasonably small value ...


4

You need to respect the amplifier input common mode range (and output range). If your amplifier (I'll assume it's an AD8221 inamp) had large power supply voltages your bridge configuration would be fine. You're putting about 0.7mA through the RTD element, which is reasonable for a Pt100. Your full-scale span will be 100mV, so you need sufficient gain in the ...


4

The purpose of the resistor R2 is to eliminate the DC offset caused by the op-amp input bias currents. If the bias currents are exactly matched, then the voltage drop across each 1K resistor (R1 and R2) is the same and the output voltage is equal to the input voltage. Fine, but if the op-amp has significant input capacitance or there is a lot of stray ...


4

While I generally agree with Spehro, there are a few things I think you should pay attention to. First, you MUST add some decoupling to your power line. A 9-volt battery is not going to have the performance you need. Try about 10 uF, tantalum, as close to the amp as you can get. From the picture, it looks as if there may be an electrolytic serving this ...


4

You only need to consider the phase around the feedback loop. Since you don't have feedback from the output of the second to the input of the first you can treat the two independently. If you wanted to feedback from the output back to the input you would have problems. You have rather low feedback resistors - is there a reason for that? They will consume ...


4

A opamp in this application makes no sense. It seems you want to do ordinary digital signal level translation, so get a level translator chip. Even a TTL logic chip run from 5 V will work since 3.3 V is well above the guaranteed logic high level for a TTL input.


4

EDIT - IGNORE C1 - unfortunately this is the only circuit that I could find with the Rout component shown within the feedback loop so, please ignore C1 - it is not meant to be there. This technique is used (with care) a fair amount: - Ignore C1 and just analyse how much current the op-amp can supply to a load in the place of C1. If the maximum output of ...


3

Can you please explain what "mismatches in the differential input stage" is? All op-amps have an input voltage error term. It's just like a small millivolt battery being placed in series with one of the inputs. If you build an amplifier with a gain of ten, the output voltage from the op-amp becomes offset from where you would expect it to be by 10 x ...


3

To ensure the two circuits have the correct intended gain, a strong voltage source should feed node 1 and node 2. In other words the output impedance of the voltage source has to be very low or the amplifiers will not have the gain as expected. Because of this, it's reasonable to say that any loading (due to adding the number of amplifiers in parallel) will ...


3

It's nominally a noninverting opamp configuration with unity gain, although with a 10M feedback resistor, the feedback is not going to be very effective because of the opamp input's bias current and capacitance. So yes, the opamp is running essentially open-loop with a feedback path that introduces enough phase shift — along with the phase shifts ...


3

I would go with Hypothesis 1. The LT technique works when your injection point has a high input impedance on one side, and a low source impedance on the other side. I tried to get that technique to work where my feedback network was pretty high impedance (yours is too), and ran into similar disagreement between measured circuit oscillation (I had negative ...


3

Can an op amp work with both the inputs floating? If nothing is connected to the inputs, the op-amp doesn't care but the output will be nonsense and, by "working" this is meaningless. can an op-amp work with any one of the inputs floating ? Again this is a nonesense question because for an op-amp to "work" it has to do what it is intended to do ...


3

At a guess, your PA341 is bad. Try disconnecting the buffer, and just close the loop with about a 10k load resistor. EDIT - Since the PA341 is good, a closer look establishes your problem. The app note clearly requires a depletion-node FET, and you are using an enhancement-mode part. They do not work the same.


3

This circuit has problems. It is doubtful that M1 ever becomes active. Vgs is clamped by D5 and D6 to about 1.4V, but needs to be over 4V to get over the Vth needed. So, M1 doesn't conduct, that also means no current for Q2. PA341 supplies buffer output current through D5 and D6, then through D1 and D2 until it hits current limit at ~48mA, as set by R4. ...


3

This is indeed an interesting problem, because of the variation of effective load capacitance with the load resistance due to Mr. Miller, and your need to not overcompensate it. I suspect a biased push-pull BJT output driver would work fine- maybe 4 small BJTs (2 connected as diodes) a couple bias resistors plus maybe a couple ohms each of emitter ...


3

Your calculation is correct - and as you say the exact gain is 1001. The capacitor is to provide a high gain at DC for the second stage but a lower gain at higher frequency to keep the overall circuit stable. If the capacitor was not present the gain of the second stage would be 1/10 so any output voltage would mean that there was a large non-zero output ...


2

There are a few issues with your implementation but first I think you need to clarify your intent. Do you wish to reduce the dynamic range of the signal, i.e. keep it more constant, or expand the dynamic range? Is it for audio use? The usual way to implement AGC is with feedback not feedforward as you have. To reduce the dynamic range take the input to ...


2

In addition to the application areas mentioned already I like to add the following: OTAs - in comparison to voltage opamps - can be easily realized as integrated circuits (simpler design, less stages). Hence, it is possible to realize, for example, fully integrated filter ICs (OTA-C filters, gm-C filters, switched-capacitor filters). As another advantage, ...


2

If you must use a current sense amp with comparator rather than an eFuse or something like that, TI makes quite a few devices that could work for your application. Here are a couple of examples but you can search the website for more: http://www.ti.com/lit/ds/symlink/ina202-q1.pdf http://www.ti.com/lit/ds/symlink/ina300.pdf http://www.ti.com Looks ...


2

This should do what you are asking - it takes the audio in (AC-coupled) and amplifies it (inverted, but that won't matter) with a variable gain of up to about 5.5 (use R1 to adjust gain). It also centre-biases the signal around 2.5V (analogRead = 512 on the Arduino). simulate this circuit – Schematic created using CircuitLab However, please do ...



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