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14

As you have figured out, the gain is only a function of the ratio of the two resistors. Therefore, at first glance, 2 kΩ / 1 kΩ, and 2 MΩ / 1 MΩ are equivalent. They are, ideally, in terms of gain, but there are other considerations. The biggest obvious consideration is the current that the two resistors draw from the output. At ...


10

The op-amp is configured as an inverter and, with +100uV on the input, the output will try to go negative but it can't because you restricted the negative power rail to 0V. So this means your source has to have a negative voltage fed to the input (a possible constraint). The trouble with the LM324 is the input offset voltage is about 2mV i.e. about 20 times ...


8

That's not a capacitor. It's a battery. Capacitors look like the following Notice how the they the plates (horizontal lines) are equal lengths. The value of this battery is 3V. The node (right side of the battery) is Vin.


6

Your first op-amp is the AD8041 and most of the story is revealed in the open-loop gain response: - If you want 26 dB of gain (20 v/v) you can't have a bandwidth greater than about 8 MHz - that's the limits that this device is capable of. This is what GBW means: - Gain x Bandwidth is usually constant (or thereabouts) for a normal op-amp - you can see ...


6

You will need to supply the op amp with a voltage high enough to contain both the output and any headroom the op amp needs to operate in, so that would mean a supply of at least 11.5V or so. If you do not have a supply high enough then you will need to use a boost regulator, charge pump, or voltage multiplier to increase the voltage instead.


6

If nothing is said about it in the data sheet and application notes (as does occur sometimes), leave them floating. If there is direction in the data sheet or application notes, follow it precisely, as @IC_designer_Rimpelbekkie says. I can't recall any exceptions, but that doesn't mean that there are not one or two types of op-amps somewhere that require ...


4

I looked at the TI datasheet and I don't see anything telling the user how to deal with the balance terminals if untrimmed Vos is acceptable. Maybe you can figure it out from the schematic, maybe not. Since TI bought NS that is the official datasheet of the original manufacturer (the go-to document). I think you would have avoided the two downvotes (as of ...


4

To give a really short answer: something in the range of tens of kΩs will probably be good (with most OP-amp models and for most applications). Try 40 kΩ for R1 and 20 kΩ for R2. This is of course not ideal in all circumstances, but it should usually work fine with a reasonable tradeoff between power consumption and noise level. Olin ...


4

As mentioned above, low value feedback resistors have relatively high current which the amplifier must drive. In an inverting amplifier, Rin sets the input impedance, so it is best not to have too low a value because the signal source must drive this. At the other end of the scale, very large resistors not only generate noise (thermal or Johnson noise), but ...


4

As shown in the photos you have neither power supply pin connected. Most ICs work very badly in that state :-). If there is no gnd or VCC on IC then Vout ~= Vin as it is a fancy piece of wire with a 10k to ground and supply. In that case what you see is about what you'd expect. . Note that pin4 is Vcc and pin 11 is ground. Not being rail to rail is fine. ...


4

Since there are a lot of answers already about why that particular part is not suitable, I'm going to mention a few more issues that apply in general, for any opamp: Unless spec'ed specifically, most opamps don't like their inputs to go all the way to the supply rails, so either use a slight negative supply for the opamp or bring the reference up a bit ...


4

Most op-amps should have a spec for supply current in the datasheet. This current flows from the positive supply pin to the negative supply pin, so it loads both of your supplies. In addition you need to consider the current flowing from the op-amp's output pin into the load and the feedback network. If the op-amp is sourcing current, the output pin current ...


3

as "Andy aka" already wrote you've tried to break the GBW limit of an operational amplifier and didn't got the gain you've expected. For your application you need an OpAmp with a GBW of roughly 700Mhz. Well, those exist. They aren't cheap and due to their high frequency they are very picky about PCB layout. Fortunately a different kind of OpAmps exist that ...


3

This is normally (all situations that I've witnessed or read about) a stable configuration. The op-amp would be stable with direct feedback so the question is what does the MOSFET add in terms of gain or phase that might make the circuit unstable. Well, in a source follower configuration the gain of the MOSFET is a little less than 1 so on that score the ...


3

It depends a bit on the op-amp. If it's a bipolar op-amp with high input bias current in a non-precision circuit (referring to the other op-amp) it may be safe to leave the inputs open. Think LM324 in some non-critical circuit. One common practice is to connect the op-amp as a voltage follower and tie the non-inverting input to some voltage within the ...


3

The impedance that the input "sees" determines the amount of noise that will be present. The only possibly noise sources on that net are the output of OA1, the specified input noise of OA2 and anything getting onto the net via the transmission line. Assuming that OA1's output is quiet and there is no external noise, the only noise is the internal noise ...


2

I think preserving the shape and amplitude of the signal is my priority across frequencies I am interested as opposed to having a linear phase difference etc. If the input signal is a sinusoid, any linear filter will preserve the shape of the waveform. If you use an active filter you may need to be concerned about making sure the filter remains linear ...


2

A filter from 1 to 8kHz is not really a bandpass filter. Designing a bandpass so wide will give you a very bad filter - you should probably consider a low-pass at 8 kHz and a high-pass at 1 kHz. uV levels are a pain to filter - it will take very special amplifiers to do that. You might want to consider passive filtering (LC networks) at the input, then ...


2

There is a significant but easily fixable error in your circuit. You have the negative supply connected to 0V and also your op-amp's non-invering input at 0V. This means for positively charged particles hitting the cup you'll see nothing from the op-amp because its output will want to go below 0V (but it can't). Definitely try symettrical supplies such as + ...


2

You are using the AD822 on a single supply with the negative power pin connected to 0 volts. This is not a problem but any signal has to be biased somewhere near halfway between positive supply rail and ground (0V) for correct operation. You are feeding in a logic square wave changing between 0V and 5V AND the output of the op-amp will try and produce a ...


2

XTR110 is a specialty IC designed for use in industrial environments, where 4 to 20mA currents are a common way to communicate analog values between industrial modules. It can be configured for a variety of output currents, including 0 to 10mA. Internally, it's implemented much the same way as an opamp solution, as shown by its schematic diagram in the ...


2

I'm going to suggest that you are far better off configuring your preamp as a non-inverting amplifier instead of inverting. One of the problems is that although most electret mic capsules have about 1k output impedance, some don't. The gain of your amplifier depends on the total impedance from the inverting input to (virtual) ground. Another problem is ...


2

This is a discrete implementation of a classic 3 amplifier Instrumentation amplifier, where the key specifications are Common Mode Rejection Ratio, Input Offset Voltage, input offset current and effective resistor match. Other specifications may be important in a particular application, but these are the dominant issues for standard instrumentation amplifier ...


2

I think he got lucky- that hairball haywired circuit probably had some inductance that caused the oscillation. High currents only require a small layout issue to cause feedback voltage to show up- at the 750kHz that he reported and several amperes it wouldn't take much. In answer to your question, yes there is a valid reason to use such a capacitor even ...


2

The Wiki article is using superposition correctly. It starts by assuming the output is in one state, then assuming the output remains in that state, calculates the comparator input with respect to the circuit input. It uses this relationship to determine the input level (producing zero volts at the comparator) to determine the trigger point at the specified ...


2

The pole formed by the gate resistor and the input capacitance would actually make the circuit less stable. This is because there will already be a pole in the opamp giving a 20db/decade gain change (up to 90 deg phase shift). If you add another pole in the loop you now potentially have 40dB/decade gain change with the phase shift being asymptotic to 180 ...


2

Your methods are fine. What you are looking for in the step response is overshoot or undershoot. What you'd like to see is something approaching critical damping for fast response and good phase margin. Graph borrowed from this SMPS app note. Your circuit has a strong potential to oscillate- I would definitely add a compensation loop around the ...


2

Preamplifier is a role rather than specific topology. It's the first amplification stage, which sits as close as practical to a sensing element (or an antenna). The goal is to amplify the signal before sending it over the cable to the main amplifier. If an unamplified signal straight from the sensing element is sent over the cable, it may get degraded by ...


2

The LTSpice simulation cannot account for circuit items that you have not entered: in this case, your breadboard wiring which is adding a filter (a RLC filter at that). What you are seeing is Step response when you start driving the (almost) square wave into the amplifier. At the point where you initially pulse the input (having been held quiet for a ...


2

Feedback always influences the input impedance of an amplifier. However, it depends on the kind of feedback. When the feedback signal adds with the input CURRENT (example: inverting opamp) the input impedance is reduced by the factor (1+loop gain). However, when the feedback signal is superimposed on the input VOLTAGE (example: non-inv. opamp) the input ...



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