15

The common mode voltage is a voltage offset that is "common" to both the inverting and noninverting (i.e. "+" and "-") inputs of the instrumentation amp. An instrumentation amplifier is set up as a difference amplifier, so it measures the difference between these two inputs and so rejects any voltage that is common to the two. In other words, if you have two ...


14

The OPA633 has a peak in its frequency response at about 200 MHz of nearly 5 dB when the signal source impedance is 50 ohm. If the source impedance is about 300 ohm that peak is about 1 dB. If you look at graphs in the data sheet you will see this on page 3: - Using 150 ohm appears to be some form of compromise to avoid too much peaking whilst avoiding too ...


14

One of the biggest benefits of the 3 op amp INA is the equal and high input impedance. The op amp's non-inverting pins' input impedance can be up in the \$T\Omega\$ range. I'll leave it as an exercise for you, but if you look at the difference amplifier circuit, the input impedance of the negative input varies with the positive input.


14

The 3 op-amp design has three main advantages over a single op-amp differential amplifier. The input impedance is much higher, since the inputs drive directly into an op-amp input rather than into a resistive divider. The gain can be set by changing a single resistor, so the critical parts can be easily integrated on to one chip (maximizing symmetry) with a ...


12

USB transmits data as a differential signal, but the differential rules are broken in order to create framing signals. Part of the protocol specifies what to do with those combinations of levels. 3.3V/3.3V - signals the device is removed. 0V/3.3V - Low speed device 3.3V/0V - Full speed device 0V/0V - Single ended zero, marks the end of frame, or if held ...


11

It's all about competing or living with noise. A reasonable op-amp will have an equivalent input voltage noise of about 10 nV \$/\sqrt{Hz}\$ and if you implement a 1st order low pass filter having a 3 dB point of 40 Hz, the equivalent noise bandwidth will be about 63 Hz. A 2nd order filter will have an equivalent noise BW of 48 Hz as you might anticipate. ...


10

Most likely a common-mode voltage violation. See Figure 15 of the datasheet. Unfortunately it does not give you the limits for a single power supply, but shows you that common-mode and output voltage limits go hand in hand. Actions that may solve your problem: Use dual power supply. Lift inputs above ground. Use a lower gain setting. EDIT: As pointed ...


10

The problem in your reasoning is that you do not show the complete path of the signal. More specific the impedance level of the signal. You are right in that you cannot have both a high impedance and a low noise. If you want low noise you must keep the impedance low. Simple as that. In the two circuits you have drawn it is unclear what the impedance is of ...


10

In short, with the split and grounded R1 circuit, A1 and A2 amplify both common and differential mode signals by the same gain. simulate this circuit – Schematic created using CircuitLab So their outputs will be easily saturated when trying to extract a tiny differential mode signal superposed on a high common mode. On the other hand the "classical" ...


9

IMO they serve no purpose, and they can be left out. If they were to minimize input offset, then there should also be one in the feedback from the output to the inverting input. Both inputs should see the same impedance. Especially with very high input impedances like FET opamps there seems to be no need for them.


8

When you really need to get to ground on the output so that even a "rail to rail" output isn't good enough, give it negative power. A opamp to drive a A/D input doesn't need much current, so a charge pump should be sufficient. All you need is a consistently toggling digital output. This can drive a NPN/PNP emitter follower pair. That plus two Schottky ...


8

The opamp in a Sallen-Key filter is supposed to be a unity-gain buffer. Yours has a gain of +3, so it isn't surprising that it's oscillating. Wikipedia talks about this. If you need that much gain, you need to do it elsewhere. There's also the Application report from Texas Instruments Analysis of the Sallen Key Filter that explains why the gain of 3 or ...


8

Although it is possible to design a Sallen Key filter with a gain higher than unity, this is rather uncommon for a reason. Any gain in it introduces positive feedback into the structure and leads it towards instability. Particularly when you take the amplifiers own poles into consideration. The OPA177 has a gain bandwidth of ~600kHz, at a gain of 3 you have ...


7

Like The Photon says, they work down to DC, it's just that most customers don't use them for those very low frequencies, so they don't publish data on them. I checked InAmps at Digikey with low slew rates, they will be better suited for your application. The AD8235, for instance, has a slew rate of 0.011 V\$\mu\$s, and the datasheet show graphs for noise,...


7

Instrumentation amplifiers are designed to work down to DC, which is effectively what your application is. You'll basically need to look at all the sources of error (offset voltage drift will be key) and assume they can drift over their full range over one period of your measurement. Typically, your ability to control the operating temperature of your ...


7

Would 3 op-amps be necessary? Not in theory. In theory, you could just connect one end of the thermocouple to ground and then just feed the other end to a non-inverting amplifier. The problem, though, is noise pickup. Thermocouples have long wires, and those long wires act as antennas, picking up all sorts of junk. In most circuits, this wouldn't be a ...


7

Notice the small block diagram in Fig.2 if the AD620 datasheet. Here's a larger diagram from a datasheet of a similar InAmp with a 3-OpAmp topology (LT1920). Notice that VREF is not a high impedance input. In order not to skew the differential output stage, VREF should be connected to a low impedance output. A voltage divider by itself is not a low ...


7

Say a circuit has two inputs, \$v_1(t)\$ and \$v_2(t)\$, we can mathematically decompose this into a common-mode and differential part, making the two circuits below equivalent: simulate this circuit – Schematic created using CircuitLab For these circuits to be equivalent, we need to have \$V_{cm} = \frac{V_1+V_2}{2}\$ \$V_d = V_1 - V_2\$. And we ...


7

A instrumentation amplifier is more than just a opamp hooked up as a differential amplifier. The inamp puts buffers in front of each of the diff amp inputs. This presents a high impedance to the outside, and also eliminates the cross impedance between the inputs of a bare diff amp. In theory you could make a inamp with three opamps, but in practise that ...


7

If you look at section 28.6.3 of the ATmega328P data sheet you will see that it begins to define the zero offset error that you will get when using the ADC. The upshot of this is that it is extremely sensible not to use (say) the bottom or top 10 mV of the ADCs range because you cannot guarantee that the ADC hasn't hit the "end stops": - The above picture ...


7

A common method for generating a negative supply rail for operation- and instrumentation- amplifiers (which don't require a lot of current) is with a capacitive charge pump. A classic part for this is the TC7660. The MAX232 can be re-purposed for this too, because it generates ±10V in order to transmit the true RS-232 levels. There is also a way to get ...


7

Common mode changes slowly, signal changes fast. What you thus need is a high-pass filter, which filters out the DC component. In the easiest case: that's a capacitor in series with your signal source, with a resistor to ground to "short" low frequency content. here's an easy-to-use RC high-pass design tool. Start with something like C=10nF. You seem to be ...


7

Trig theory to extract the phase & magnitude. If you only care about phase... Say you have a signal \$Acos(\omega t + \phi)\$ and you want to extract \$\phi\$. You can use an oscillator of the same frequency to extract this info BUT the issue is the phase. \$V_{sig} = Acos(\omega t + \phi)\$ \$V_{osc} = cos(\omega t)\$ \$V_{sig}V_{osc} = Acos(\...


7

LM358 has very low gain-bandwidth product (1MHz) therefore the Sallen-Key filter topology you used might not work well as a lowpass above a few tens of kHz. What happens is the HF jumps over the opamp through C5, and at HF it doesn't have enough feedback to keep its output pin low impedance, so the HF just goes on to the output. Also as Scott says in the ...


7

The energy stored in a cap is proportional to the square of the voltage -- so a 47uF cap on a 500V rail is storing as much energy as a 4700uF cap on a 50V rail. That's going to be part of it. Anything else is guesswork -- but I suspect that tube amps are less sensitive to power supply variation, and that people are just used to tube amps having a bit of ...


7

What you need to do is add a couple of resistors simulate this circuit – Schematic created using CircuitLab If the instrumentation amp has a gain G, then, since the current through R1 must equal the current through R2, Vin/R1 = G iL Rs/R2, where iL is the load current. Rearranging the terms gives iL = Vin(R2 /R1 G Rs) Note that, strictly ...


7

Look at section 8.10 in the data sheet. The part you have chosen to use has an input offset voltage of \$\pm 0.3\mathrm{mV}\$ typical, \$\pm 2\mathrm{mV}\$ worst case. Your output offset voltage in an amplifier like this is your non-inverting gain times the input offset voltage -- so you're looking at between \$\pm 30\mathrm{mV}\$ and \$\pm 200\mathrm{mV}\$...


7

If you are reading the ADC just once per second then you need to eliminate frequencies above 0.5 Hz to prevent aliasing. If you think your system will have noise at, say 10 Hz, then that noise will contaminate your readings. I recommend that you sample at a much higher rate, perhaps some multiple of the power mains frequency, and perform low-pass filtering ...


6

Is the supply for a differential op amp compulsory or can we operate those without a supply? I just want to amplify the difference of inputs, I'm not in need of any supply voltage. Generally, no, you cannot operate them without an external power supply. In VERY special cases, yes, voltage (but not power) amplifiers that do not require an external ...


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