# Tag Info

8

The OpAmp is used in an inverting configuration, i.e. when you apply a positive Vi, the output Vo will be negative. So, the current flows through R3 and the upper path (R1, D1), while current through the lower path is blocked by D2. So, for a positive input, Vo+ will be negative and Vo- will be zero. The exact value on Vo+ is determined by the ratio of R3 ...

7

CFA's have been around forever. It's possible to find versions of them in Vacuum Tube designs from the 1930's. Later, there were versions as Chip and Wire Hybrids. Then Comlinear came out with monolithic IC versions, some time in the early 1980's. If you needed a low gain, high slew rate, low distortion amplifier there was no comparison. Parts like the ...

6

This is a typical powered headphone splitter/amplifier. It is a single gain amp to multiple non-amplifying buffers. The 5.1kΩ Resistors represent the load. As it will be in parallel with your actual load, they won't affect much. Notice, no pull-down resistors needed between the Gain Amp and Buffer Amps. This is as generic as it gets.

6

Don't bother using MOSFETs. Instead, use FET-based devices such as the CD4066. Furthermore, if you only need 2 gain settings, you only need 1 switch. As an example simulate this circuit – Schematic created using CircuitLab will give a gain of 10 or 100. (Note that your example, if it worked, would give gains of 11 or 101). Also note that, ...

5

I don't know the specifics of your data board, but in general an amplifier with a gain of 1 (aka unity gain) is useful for its other properties: high input impedance low output impedance These make it a good buffer between various stages so that they don't load each other. And buffer amplifiers are used as the first stage of data acquisition boards for ...

5

Yes - if you can identify the main loop (sometimes there are, in addition, local loops) you have nothing to do than to open the loop at any location for injecting a test signal. However, you have to ensure two important points: (1) You must not destroy/influence/change the dc operating point of the amplifier because often the feedback loop is active for DC ...

4

The pot is connected as a rheostat. It will vary from about zero ohms when the wiper is at the top to about 20K when it is at the bottom. The diodes are connected as a bridge rectifier so that the current through RV5 passes through the meter in the same direction (pin 2 is positive, pin 1 is negative). The meter has some current sensitivity and ...

4

Sketch of a solution : for it to work well, the devil's in the details... A comparator decides which signal is larger. Its output controls an analog switch, aka 2:1 multiplexer, connecting the selected signal (the larger one) to the output. simulate this circuit – Schematic created using CircuitLab You probably want a "make before break" ...

4

There is a simple answer: The bandwidth for the closed-loop gain is determined by the frequency where the LOOP GAIN is 0 dB. In your example circuits the loop gain is not the same - hence, the bandwidth will not be the same. The circuit with the largest loop gain (non-inverter) has the largest bandwidth. Explanation why the Loop Gain (LG) determines ...

4

You're are basically correct. The derivation of the formulas can be found in multiple places e.g. 1 or 2 and the books cited therein, so I'm not going to do it here; you've also done it correctly. In a nutshell, if $f_T$ denotes the Unity Gain Frequency for the open loop opamp and $f_B$ denotes the same for a given circuit with circuit gain $G_0$, ...

4

A 741 op amp cannot swing its output to the power supply rails. There is significant voltage dropped in the output stage transistors. If you look at the 741 data sheet,this will be shown in the graph of output voltage swing versus power supply voltage.

4

This is likely unnecessary architecture. You don't really need (and probably don't want) negative voltages in your application. Your microcontroller ADC can only read positive voltages. Your op-amp can only generate positive voltages. The transducer will only generate positive voltages. The piezo element is a large resistor at no pressure and ...

4

I'm not entirely sure how to answer the why question, but simulation reproduces the same result; first the big picture: But if you zoom in there are tiny little overshoot peaks in there... which are actually pretty close to those measured. TL081 model is from TI, which is actually a very basic one... yet it predicted the overshoot pretty well. Alas ...

3

Rearrange the formula to this: - H(s) = $\dfrac{8336.6}{s^2 + s(189.26) + 8952.6}$ And then note that is of the form: - H(s) = $\dfrac{N\cdot\omega_n^2}{s^2 + 2\zeta\omega_n s + \omega_n^2}$ Numerically derive $\zeta$ (damping ratio), $\omega_n$ and N and try plugging the values into a sallen key type calculator like this one here: - I plugged ...

3

It is actually pretty easy. I'll solve it symbolically and double-check with the QsapecNG result, which I'm also using as schematic to give symbolic names and assign some arbitrary current directions. The symbolic result there (output voltage) is $$V_o = E\; \frac{-R_2(R_3+R_4)}{sCR_1R_2R_4+R_1(R_3+R_4)}$$ So how do we get this by hand? Rather simple ...

3

Forget Sallen/Key for the moment, go back to the equivalent passive implementation of the same filter. (e.g. search "passive butterworth filter" for examples.) You'll see it involves an L and a C per 2nd-order filter section, thus it's basically an L-C resonant circuit, with resistive damping to control the resonance (the height of the knee). Having ...

3

Yes, you can add hysteresis using positive feedback from the comparator output. simulate this circuit – Schematic created using CircuitLab Of course, the amount of hysteresis you add depends on how noisy you expect your signals to be.

3

Whoa, that's a lot of dissipation! At 2.5 V in, there will be 5 A thru the resistor, transistor, and LED. That's 12.5 W into the resistor alone. Overall the supply will have to source 140 W with the LED getting maybe 15 W or so depending on actual forward voltage. Not only is this horribly inefficient, but the FET will take the bulk of the abuse. With ...

3

TI's Analog Applications Journal includes an article about noise analysis for an op amp driving an ADC as in your application. The author uses $F_{\text{L}} = 0.1\text{Hz}$ with the following rationale: When we think about noise at these low frequencies, we may jump to the conclusion that we should take this formula down to a very low frequency, such ...

3

A simple BJT operated in the saturation region is a good start to understanding how a multiplier works: - I'm not talking about the BJT operated in the normal region (where the collector current remains flat for large changes in VCE) - I'm talking sub 300mV between collector and emitter. Look at the pretty colored lines between 100mV Vce and 200mV. ...

3

Use an op-amp precision rectifier, well, actually use two of these: - Shown above is the circuit for one feedback signal - use another identical circuit for the other feedback signal. Then connect the outputs together - highest input dominates the output. Because the basic circuit inverts you might probably want to invert the inputs to each precision ...

3

Let's start with just answering your question: You are going to need a make-before-break solution for the switch from active to sleep to not be disturbed. There's analogue switches and muxes with that ability specified, but they could be complicated to find straight away. If you do not find something suitable, you can use two analogue switches and make ...

3

A summing opamp is the correct way to do this. However, your current configuration is problematic: you have the positive input connected to ground, as well as the negative supply of the opamp. This means that your signals need to be ground referenced (have a center point around 0v), but the opamp can't output voltages below 0v, so half your waveforms will be ...

3

You need to look for "rail-to-rail" op-amps and even then there will only be a few that will do 20mV (with a 1mA sink current) such as the AD8605 - not cheap but pretty good all round and beats the LM324 into pulp on most things providing it runs from less than 6V. Here is linear technology's search engine with a few parameters also selected to give you a ...

2

The implication, since you indicate that the input is a digital signal, is that you want to either turn the LED ON or OFF. If that's true, why don't you ditch the opamp and use something like what's shown below to drive the LED? First transistor inverts the input signal, the second one gets very close to zero volts to hard turn off the third one, which ...

2

The two circuits do the same things, but with different emphasis. The opamp circuit allows the LED current to be changed by altering Vref. It also allows the LED current to be completely independent of the voltage drop across the LED. This circuit would tend to be used for testing and characterisation where the flexibility and accuracy is needed. The ...

2

The ideal current feedback op amp equivalent circuit looks like this: The non-inverting input is high impedance (infinity in ideal model). The 'x1' buffer asserts the same voltage at the non-inverting input, but with low impedance (zero in ideal model). This means it supplies whatever current is required to maintain the voltage it asserts. Depending on ...

2

1 - Quite probably 2 - This search says its quite possible also. 3 - While C1 and C2 block DC they will also be a RC filter, which will hinder low frequency reproduction. You'd need higher values capacitance to reproduce lower frequencies. This calculator can be quite useful to determine such values. Can you try to build the amp first and try it on a ...

2

The LM339 datasheet says this: A basic comparator circuit is used for converting analog signals to a digital output. The output is HIGH when the voltage on the non-inverting (+IN) input is greater than the inverting (-IN) input. The output is LOW when the voltage on the noninverting (+IN) input is less than the inverting (-IN) input. The inverting input ...

2

The TransImpedanceAmplifier (TIA) has a gain (volts out to current in) equal to the feedback resistor. It is 53.6 kohm and take the log of this and multiply by 20 and you get 94.58 dB. Gain bandwidth product is all about voltage gain - you don't have any voltage gain in a theoretical TIA circuit unless you are going to perform noise analysis. If you were ...

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