31

I believe your analysis to be good. I've made sallen-key 4th order filters that cut-off around 3 MHz with absolutely no worry about performance. I don't see that 10 MHz is unachievable. It's all about op-amp choice. For a unity gain stage it's easy to ascertain where the gain starts dropping below (say) 0.99 and regard that as the limiting frequency. On the ...


28

The power delivered to the load is from the Joule heating effect: \begin{equation} P=\dfrac{\Big(\dfrac{R_L}{R_L + 50}G\,V_{IN}\Big)^{2}}{R_L} \end{equation} So from differential calculus we know that a function reaches its maximum or minimum value when we differentiate and equal it to zero. In this case we'll have a maximum value, thus: \begin{equation} ...


24

Yes, you are exceeding the specs. Look at table 6.1 of the DATASHEET on page 3: It clearly states the maximum difference between the V+ and V- pins is 16V. This means you can have +8V and -8V, as the difference is 16V. As you have it right now, your difference is 30V, which is almost double the maximum ratings. Chances are, you have damaged the op-amp and ...


21

Gain-bandwidth product is a small signal specification. It only applies to signals under which the op-amp circuitry remains in a linear regime. That is on the order of 100mV or less for a conventional differential input stage. With the relatively high frequency 2V you are applying you are well into a large signal regime in which non-linear effects take ...


20

The LM324 has a maximum output slew rate of 0.5V/\$\mu\$sec, so it not going to be able to accurately reproduce a 2V amplitude sine wave of more than about: \$f_{SR} = \frac{0.5\cdot10^6}{2\cdot\pi\cdot2} \approx 40kHz \$ To see the unity gain-bandwith directly you can reduce your signal amplitude to perhaps 50mV.


19

Chip suppliers are keen that their users avoid common design errors, shown by application examples in their data sheets. This one is addressed by Linear Technology in their data sheet for LTC6241. It also applies to many other opamps: The good noise performance of these op amps can be attributed to large input devices in the differential pair. Above ...


19

In most circuits, yes. And in your circuit this would be fine. Capacitance is just capacitance and ceramic MLCCs capacitance values have increased quite a bit in the last few decades, leading to a much wider applicability (and the ongoing production shortage). But you must be aware of a couple of caveats that are mostly exclusive to ceramics: In some ...


17

You first idea will not work at all. Your second idea will work, but many OP-Amps aren't going to deliver more than a few mA on their output, which limits the current your circuit may draw from the virtual ground. There are Power-OP-Amps available which may deliver up to a few ampere, but if you cannot get your hands on one, you can use a PNP/NPN transistor ...


16

Intuitively: when you raise the load resistance, you are increasing its share of the voltage (and thus power) versus the other resistance; but you are decreasing the total current (and thus power). When you lower the load resistance, you are decreasing its share of the voltage (and thus power); but you are increasing the total current (and thus power). So ...


15

You are exceeding the amplifiers input common-mode range. It has rail-to-rail output but not rail-to-rail input. The allowable input range is -0.3V to +3.8V when fed from a 5V supply. When configured as a unity gain amplifier the output range will be limited by what's acceptable at the input. Other posters have commented that the data sheet explicitly ...


14

Is there a reason this schematic uses an op amp (AD8031) for the reference voltage of the op amp (AD8544) filter when the voltage divider already brings down the voltage anyways? The usual reason to use an op-amp to buffer a divider like this is to ensure the reference voltage doesn't change if whatever it's connected to sinks or sources current. Can the ...


13

Figure 1. Vmax. This line of the datasheet is stating that the maximum voltage difference between V+ and V- is 16 V. You have probably damaged the op-amp.


13

Is there a reason ... Yes. The two 10k reisistors give the voltage reference an impedance of 5k. This means that if the current drawn from the reference changes by 0.1 mA that the voltage of the reference would change by 0.1m x 5k = 0.5 V. This would be a very unstable reference. The op-amp buffer fixes this. The output impedance of the buffer is close to ...


13

An inverting op-amp inverts the signal; it does not phase change the signal at the output by 180 degrees although, if the input waveform were a sinewave, then it would look like 180 degrees of phase shift.


12

This op-amp boasts input noise of 0.9 nV/√Hz, which is roughly equal to the Johnson noise of a 50 Ω resistor. If you aren't putting resistors smaller than that around it, you're wasting some of this op-amp's performance, and probably should be buying something cheaper. Another useful identity is 1 kΩ ≈ 4 nV/√Hz, there being many ...


12

You can get a small negative voltage by using an LM7705 which produces -232mV nominal output voltage using a charge pump. The advantage of using that part over a typical garden-variety inverting charge pump converter (eg. +5 to -5) or inverting boost converter is that the worst-case negative output voltage generally falls within the maximum negative input ...


12

Capacitors block DC and pass AC. You can use a series capacitor into an opamp with whatever gain you need. Even better might be a simple RC high-pass filter...One capacitor (series) and one resistor (to ground) in front of your amplifier. Like this: simulate this circuit – Schematic created using CircuitLab R2 and R3 set your gain. C1 and R1 set ...


12

Your schematic should be as following: simulate this circuit – Schematic created using CircuitLab Since you're using a single-supply non-inverting amplifier, the non-inv input of the opamp should be biased to a non-zero voltage –ideally to Vcc/2 as in your schematic so that the amplified signal can swing equally. Now let's take a look at the ...


11

The op-amp is going to try its best to keep the voltage between it's plus and minus input the same. In an ideal op-amp, no current flows into the inputs, so the only way that it can do that is by changing its output voltage. In the schematic below, \$v_+ = 0\mathrm{V}\$. That means that the op-amp will try to hold \$v_-\$ at zero, also. Whatever voltage ...


11

The recommendations for you are very simple. Use a comparator for this application instead of an opamp. Select a newer part that operates with orders of magnitude faster response time. It would be the very best thing if the 741 could be eradicated from face of the earth. Here is what can be achieved with the venerable LM393 at 5kHz. The shown circuit will ...


11

The question is not exactly about the wifi, but on how does this operate in full duplex It doesn't; read the more detailed specification on the page you linked: - Operation Mode: Bi-directional, half-duplex, Auto-Switching via carrier sensing. The full detailed spec is here: - Operation Range: 2400-2500 MHz.Operation Mode: Bi-directional, half-...


11

Good start, just one more thing to add... R1 cannot be DC-grounded. It must be AC-grounded with a capacitor. You would choose the capacitor value so that its reactance is equal to (or less than) R1 at the lowest frequency that's important to you. For example, if R1 is 1000 ohms, and you are amplifying audio where 20 Hz is the lowest audio frequency, C1 (...


11

The noise is caused by minute mechanical vibrations of the pot wipers on the rings (the latter are the resistive material.) Since neither the wiper nor the ring material are atomically smooth where they touch, rubbing the wiper on the ring produces slight vibrations. Some of that vibration is perpendicular to the contact patch between wiper and ring. The ...


11

The effective load on the op-amp is likely capacitive. From the LM324 datasheet: Capacitive loads which are applied directly to the output of the amplifier reduce the loop stability margin. Values of 50 pF can be accommodated using the worst-case non-inverting unity gain connection. Large closed loop gains or resistive isolation should be used if larger ...


11

A transfer function of closed loop system is: \$H(s) = \dfrac{F(s)}{1+F(s)}\$ simulate this circuit – Schematic created using CircuitLab Wiki A transfer function of closed loop system with a feedback is: \$H(s) = \dfrac{F(s)}{1+F(s)\cdot G(s)}\$ simulate this circuit Now let's suppose you have a opamp buffer - the output is feedback to ...


10

The datasheet seems to have lost some coherence in the translation from NS to TI. If you look at an original datasheet, it's clear that the recommended voltage is 15V for the LM356N and the maximum is +/-18V. It's not clear what the recommended minimum is, however it's specified at +/-15V. Curves of typical performance go down to +/-5V so probably you're ...


10

The term "zerø-crossover" is a TI marketing term for their solution to achieving a rail-rail input common-mode range. See the Functional Block Diagram from the TI datasheet: In many rail-rail input CMOS op-amps, 2 pairs of input differential pairs (NMOS/PMOS) are employed to achieve a rail-rail input common-mode range. The drawback of this solution is that, ...


10

Put a large resistor from the + input of the left opamp to ground. Say, like 470kohm. The DC bias from the microphone is getting through to the opamp. The TL071 has very high impedance inputs. The tiny bit of DC that leaks through the first capacitor is enough to push that input away from 0V. The amplification then drives the output to one of the rails. ...


10

It's a Platonic ideal of an op-amp; it's not a thing that could ever exist in the real world. So you say to yourself "I'll just assume that my system is stable" (real op-amp circuits aren't, always), and "I'll just assume that my output does whatever it needs to hold the input voltage difference at zero" (real op-amps don't, quite), and "I'll just assume my ...


9

Yes you can use a 741 opamp with a voltage at the inputs between -5 V and + 5 V as long as you give it a supply voltage of at least - 8 V and + 8 V (this is my guess, the 741 has a guaranteed +/- 12V input voltage range when a +/- 15 V supply is used, I re-used that 3 V margin (15 V - 12 V) to get to +/- 8 V) note that that is a symmetrical supply voltage, ...


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