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7

The concept of the "op-amp virtual earth" is very important to understanding why the op-amp is used as a mixing/adding circuit: - I've stolen the OP's picture and marked in red where the virtual earth is. Anything directly connected to the -input is called "virtual earth". OK that doesn't explain why so, here goes... If the voltage at the +input is 0V ...


3

Try this circuit with a dual op-amp. It uses the suggestion by @horta to move the voltage away from the rails. U1A produces a stable voltage at +3V. The input is capacitively coupled to that voltage. Since your pulses are -22.5mV, the threshold is set at 3V - 11.25 mV approximately (set by the ratio of R9 to R8 approximately) +/- 2mV of hysteresis (set by ...


3

The impedance of your 100pF coupling capacitor is too high at 5MHz (about 318 ohms). in comparison to the 52 ohm resistor. The total impedance of the series combination is about 322 ohms (they add in quadrature) so you're getting an ideal gain of 1.58 from the first stage and 10 from the second stage.


2

You need to create a new pulse from the original pulse. The new pulse should start from zero and rise to a positive voltage when t = t0 + 2. At t= t0 + 6, the pulse should complete and return to zero. This allows you to use an integrator between the start of the new pulse and t = t0 + 4. At t = t0 + 4 the integration process should be suspended (or halted) ...


2

Vref is the new 0V and this is where you are getting confused. With 1V going in with a gain of unity (Rf = Rin), the voltage out is: - \$V_{OUT} = V_{REF}-(V_{IN}-V_{REF})\cdot\dfrac{R_F}{R_{IN}} = 2.5 - (-1.5) = 4 volts\$ This is an offset from 2.5V of +1.5 volts and remember your input voltage of 1V (relative to ground) is actually -1.5V relative to ...


2

You are using an LM358 as a mid-rail generator - you are hoping it will provide a steady mid-point voltage and be rock-steady in the presence of a 5 MHz signal that it gets inflicted with via the 100kohm resistor R8. R8 is big in value but your gain is also large - the LM358 has a unity gain bandwidth of 1 MHz and it cannot possibly hope to keep its output ...


2

As far as I'm aware the ADC on an arduino requires 0V to 5V as the range of input and because you have a signal that is -1V to +1V, you need to centre it about 2.5V. If the signal is purely AC then that is easily done with a potential divider and capacitor: - Note C1, R1 and R2. C1 blocks the average 0V level of the input signal and R1 and R2 set the ...


2

A few thoughts: - The recommended supply voltage range is up to +/-15V not 16 volts as you have in your circuit 8.2pF capacitors are probably totally ineffective as supply de-couplers - try 100nF Breadboards are probably not the best way of connecting the device up and can cause large instabilities. What is your shut-down pin connected to? All the specs on ...


2

Looking at the topmost op-amp and ignoring the \$100 \Omega\$ resistors, write by inspection: $$v_{X+} = v_{OUTX} + v_{X-}$$ For the bottommost op-amp, write $$v_{X-} = v_{X+} - v_{OUTX}$$ Thus, $$v_{X+} - v_{X-} = v_{OUTX}$$ So, this circuit converts a single-ended input signal, \$v_{OUTX}\$ to a balanced output signal; it's an active 1:1 ...


2

The most effective filter for receiving the magnetic part of the EM spectrum with a loop antenna is a tuning capacitor right across the input coil. It: - Minimizes unwanted large signals being amplified Improves the signal to noise ratio because you can "peak" the tuning right at the input Works at rejecting above and below your desired 5 MHz operating ...


2

Without an input, pin 3 will find its own level and the output could be anything between 5mV and about 7V from a 9V Vcc. OK that's based on the spec and you say +7.5V - I'm not surprised. When you connected the input to 1.5V, the gain of the op-amp circuit being about 8.7 will still mean the output voltage is end-stopped at 7.5V. With a 741 - you can't use ...


2

There is no guarantee what a opamp will do with one of its inputs floating. Your schematic shows a gain of 2, but your description implies a gain of 8.7. If the gain really is 8.7 (this kind of ambiguity is annoying in a question since we have to waste time dealing with it), then putting in 1.5 V should result in 1.5 V x 8.7 = 13 V. The opamp can't ...


2

As pointed out by Gsills, there is no negative power rail for it to go negative. The LM358 is a pretty old, and for general purposes where you would 'expect' rail-to-rail ( 0V -> VCC) outputs it doesn't cut it. Your comment of using an MCP6002 is a good idea, it will definitely get the job done. Just make sure it's slew rate and in general it's bandwidth ...


2

It is a 10uF non-polarized electrolytic capacitor. Here's a similar ECG front-end circuit that uses the AD620 as an electrode amplifier (from this site, en Francais): In that case, the designer used a 2,2uF film capacitor with a 2,2M resistor to form a 0.033 Hz high pass filter (\$ f_C = \frac {1}{2\pi RC})\$. As has been pointed out, the symbol is ...


2

Omitting the supply rails and decoupling capacitors is a common shorthand in circuit diagrams. It's admissible, when supply rails are not important for the discussion. 0.001 μF in the original post is not a decoupling capacitor. It's a compensation capacitor between pins 1 and 8 of the CA3130 OpAmp. fig.8 on p.9 in the datasheet. Decoupling ...


2

I know analog guys don't like to hear this, but a small microcontroller could easily meet all of your requirements. It would be easy to modify its operational parameters even after the hardware is built, and it would have by far the lowest overall parts count. A built-in ADC could digitize the audio (as well as the voltages from your control knobs), ...


1

It would be wiser to amplify the signal far above the supply noise then use a Schmitt trigger or comparator with positive feedback as required. One solution uses a current sense chip on either 3.3 V rail or 6V rail, whichever is cleaner. http://www.maximintegrated.com/app-notes/index.mvp/id/1180 using a gain of 50 on a MAX4372.


1

I'll assume you want "average reading" equivalent to RMS for a sinusoidal input. If you want "true RMS" you'll have to add a more complex conversion circuit. This is the same principle used in non-"true-RMS" multimeters. The transformer gives you (60A/7000) * 47 = 402.8mV RMS AC full scale (full scale of the transformer) So you need to convert an AC ...


1

You can use a current sense ampflifier that has an external sense resistor. TI offers several, for example. Then you can choose the sense resistor value as low as you like to suit the current levels you need to measure. With high current, for accurate measurements you'll want to be sure that your sense resistor has a low TCR so that self-heating doesn't ...


1

The first thing to do when looking to design a filter for a signal is to obtain the spectrum of the signal. This can be done in different ways. For example, you can measure or capture the signal with a scope. Most modern scopes can calculate an FFT of captured data and give a spectrum. For a design though, it is nice to make an idealized waveform and then ...


1

Well, you have a gain of 1+1000/130 = 8.7. So connecting the input to either 9V or 1.5V will rail the output. The LM358 allows single supply operation and sensing near ground, but if you leave the input floating you are also likely to have the output go to the rail. (You might be better off with dual supplies.) Did you tie the non-inverting input to ground ...


1

It depends on where the currents are flowing. In general, if current flows from the power supply rails (your +10V/0V) through the load resistor to your virtual ground, I'd use three capacitors, with the one across the 10V supply before the rail splitter and two equal capacitors from the +5 and -5 rails to the virtual ground. From the datasheet figure, you ...


1

How do I wire it all up? Towards the back of the data sheet it gives you a typical circuit to use and this shows where GND comes in: - Note the incoming power arrives on the circuit at the top left (red rectangle). The ampere rating of your power supply depends on how the speakers are configured and what supply voltage you decide to go for. This ...


1

The allowable input voltage range for a 741 is 2 volts inside the power rails. This means that on a +/-15V supply, the inputs will still adequately function from -13V to +13V. If you are trying to tempt performance from a 741 with a grounded negative rail then don't expect the inputs to function lower than 2V above ground and worst case 3V. Let's look at ...


1

At first I thought the circuit was a Differential Howland Current Pump. Similar to this one here. I thought maybe the cross-coupling makes the current sources share the available voltage. But I did a simulation since the analysis didn't indicate that was possible.. With no load, the (-) output is a virtual ground and the (+) output equals the input ...


1

Just because the battery voltage is 3V doesn't mean there isn't some sort of switching regulator to boost the voltage. However, even if there isn't it is possible to achieve using just a single 3V rail, 2 op-amps, a couple of resistors, and a photodiode. You just need an op-amp which can operate on low voltages. You will either need to bias the ...


1

It's a H bridge amplifier: - The top output has a voltage that is \$V_{IN}\cdot (1 + \frac{R_1}{R_2})\$ and the bottom output is the inverse of this. Clearly I'm talking about AC voltages because the input amplifier removes DC/low frequencies.


1

Just to move things forward I'm posting an idea (not an answer): - I believe that the waveform shown above is current into a device containing a bridge rectifier, charging capacitor and load. It's too similar to the stereotype to be anything else (within reason) and the only conclusion of this is that the OP is incorrect about the sampling rate because ...


1

The main part of the filter (sectioned in red below) is a modified MFLP filter and not a Sallen-key filter: - I've used Mr. Okawa's web site for doing the analysis and it tells me that the cut-off frequency is about 3.2kHz and there should be no peaking because the Q (or damping ratio if you did the math) is about optimum for flat response ...


1

Maybe something like this: You will have to look around to find a opamp that can run from a single 5 V supply and has suitably low noise. We can't tell if this meets your noise spec, because "low" is no spec at all. This will have a voltage gain of -10, which gives you the 20 dB you are asking for. It also looks like a 2.5 V source with 3.3 kΩ in ...



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