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There are a range of methods which can be used to provide offset voltage compensation. The best method to use varies with the application circuit, but all either apply a variable current to a circuit node or vary the voltage of a node which a circuit element connects to. The methods described below can easily be applied to your circuit by Adding a ...


11

Capacitive coupling has been suggested, but this has two big disadvantages: Your signal is no longer a square wave It will only center your signal around 0V if the duty cycle is 50%; you'll see the signal go up and down if you play with the duty cycle A good function generator will have a potmeter to set an offset to the signal. One way to do this ...


7

This is very easy: R1 and R2 must be equal in your case. The overall values of R1 and R2 depend on things you haven't told us, like the source impedance and the maximum allowed impedance of OUT. If OUT is a signal going into something like a microcontroller A/D, then start with 10 kΩ for R1 and R2. A better answer will require a better question.


7

What you need to do is simply remove the DC offset all together, not supply a negative one. This is known as AC coupling. If you run the output of your square wave generator through series capacitor, it should do what you need. This will however be at the expense of making the square wave less square. An example circuit is shown below for you: And the ...


6

Several comments: Add component designators to your schematic. It is difficult to talk about the circuit otherwise. You appear to be a little confused about resistor dividers. You have two resistor on the input, but only one is doing anything. The resistor to ground is just a load on the input but otherwise has no bearing on your circuit. It sortof ...


6

You are close to having a working solution. I attempted to edit your schematic but I don't see a edit link below the schematic in the preview window. So I'll draw a new one here. simulate this circuit – Schematic created using CircuitLab R5 & C2 filter noise from your 5V rail. You may need to increase the value of C2 if you are getting ...


5

Of course you can try to compensate for the DC offset but why not eliminate the influence of DC offset in the first place ? You can do this if you do not need 10000 times gain at DC. You mention that your signal is 40 kHz, I conclude from that the DC value is irrelevant to you. Then I would just make amplifiers that have 10 x gain at 40 kHz but 1 x gain at ...


5

The equation for the capacitor is rather simple. The capacitor will interfere with frequencies lower than \$f = {1 \over {2\pi RC}}\$ where R is the input impedance of the next stage. Simply substitute the desired minimum frequency and impedance, solve for capacitance, and pick a larger value.


5

You would need a resistor from the output to Ground to remove the offset. Without a resistor to ground, leakage in the capacitor will hold the output voltage near the voltage at the bottom of R1.


5

If you have a linear circuit where Vout = f(Vin) and f(0) is ideally 0, it's easy, just null the input offset with 0V applied and apply something close to full scale and adjust the gain. No iterations required. For more complex situations where f is nonlinear and/or f(0) is not zero you may need to adjust the gain and offset both before and after the ...


4

The answers above are both unsatisfactory in some ways. Andy's has incorrect assumption and calculation, while "placeholder"'s essentially tells you nothing concrete can be said... which is not the case. Andy's error is to assume that in the numerical example the PSRR is to be considered at 1kHz, but it actually needs to be considered at DC given the ...


4

You need to use meg instead of m for the resistor value. You have specified a 0.007\$\Omega\$ resistor, which will tend to reduce the gain to 1. Your amplifier has a GBW product of 18MHz meaning it will only have an open-loop gain of about 500 at 40kHz, not the 14,000 you appear to be expecting. However it will have extremely high gain at DC so any offset ...


4

I don't know what weird results you are getting, it could be that you didn't check the input common range. The input common mode voltage (average voltage of the inputs) has to be about 0.2V above the negative rail. As pointed out by Carloc, a detailed description of the behavior of the comparator over the input voltage range can be found in the datasheet. ...


4

I think that the historical definition of amplitude comes from nature. When Earth was young, most things that were wavey tended to oscillate sinusoidally about a middle point. Think sea waves or a tree swaying in the wind. When man emerged he invented violins and tuning forks and they too oscillated symmetrically about a mean point. They called it a sine ...


4

You can define a straight line as y = mx + c. There are only two unknowns, the gain or slope m, and the offset c, so you only need to make two measurements to calculate them both. Make two measurements y0 and y1 for two inputs x0 and x1, then solve for m and c using simultaneous equations see wikipedia, or as two equations is quite simple, trial and error ...


4

Should first the gain error be compensates or the offset error? That's up to the particulars of how exactly the compensation circuits get in there and tweak things, and where in the overall process these compensations are applied. You are basically asking what is better:     Y = mX + b     Y = m(X+ b) While different values of m and ...


4

Actually the 1nA bias current of the AD620 is pretty high for this application. The probe can be 100-1000M\$\Omega\$ output resistance. With your gain of 12.2 the output-referred offset with a 200M probe resistance would be about what you are seeing with a 1nA bias current. Normally we would use a low input bias current amplifier for this application, pA ...


3

First of all you should really get better resistors. 5% with a strain gauge is just ridiculous. To "trim and calibrate" your measurement you should just leave your gauges be, i.e. leaving them in a known state, measure the output and save it in a variable in your micro. This value should then be subtracted from all subsequent measurements. If you can you ...


3

I'm assuming you're using a non-inverting configuration. First, your DC offset. When you looked at the output, you had no input at all - not even a resistor to ground. So the input floated and the output went to the rail. Try connecting your input to ground with, let's say, a 10k resistor. If, for some reason, you are using an inverting configuration, ...


3

If you are stuck with being only-able to replace the op-amps then.... Just use much better op-amps. You can get zero drift op-amps that have less than 10uV offset voltage - altogether that's 11.11mV at the output with regular DC coupling and all the offsets pointing the same way. 40kHz is a breeze for a lot of them - gain-bandwidth-product required is ...


3

The offset voltage is increasing as you have increased the gain and you have the non inverting input set above "mid point". With the top circuit there is actually attenuation, so the opamp doesn't have to drive it's output very hard to keep the inverting input at 2.5V. When you increase Rf to 10Meg, the output has to drive it's output to the necessary ...


3

From page 142 of IC Op-Amp Cookbook 3ed., Jung, you can come up with the following circuit. Note that the offset also is affected by the gain resistors, so that the output offset is Vo_offset = V+ * (1+R1/R2) A couple of notes for implementing this circuit: The resistor values are for reference only. You should scale them to your circuit impedance vs power ...


3

You did not give us a lot of details, so I don't know whether my solution is really applicable in your case, but the simplest is to AC-couple your 10MHz digital signal using a capacitor: simulate this circuit – Schematic created using CircuitLab On the output, you end up with a signal that moves between about 59.5 and 64.5V (due to the diode drop). ...


3

There's no offset. It works exactly as it should work. Your integrator charges at the positive half cycle of your V1 and discharges back at the negative half cycle. this is the ideal operation. In practice the integrator drifts due the leakage currents or opamp offset voltage. It can get stucked to max or min output voltage limit or it can gradually forget ...


3

I do not know the circuit you are working on, but if you are putting a DC blocking capacitor in the input, then the capacitor will create a highpass filter with the input impedance of the circuit. For optimizing the value of the capacitor you need to know your input impedance. The highpass filter will have a cut off frequency. This frequency is given by: f=...


3

You should have a capacitor between the LM324 output and the middle of your voltage divider, which is also going to be the node that is the input on the ADC. The capacitor will allow the DC offset, while at the same time allowing the middle voltage of the divider to be pushed around by the audio signal. The divider provides the bias voltage (presumably 2....


3

If you connect A, B, C and D to ground (through the four voltage sources), you're putting R3 and R4 in parallel, and also R5, R6 and R9. You end up with a noninverting amplifier with a gain of 3 (defined by R1, R3 and R4) being fed by a voltage of 5V / 4 = 1.25V (defined by R5, R6, R8 and R9). So naturally, the output voltage is 3.75V, more or less. But to ...


3

When you run a simulation, in most circumstances, the default starting voltage across a capacitor is 0 volts. This means, that at the instant of beginning the simulation, if 4.5 volts is forced onto one plate of that discharged capacitor by the op-amp output, 4.5 volts is also seen on the other plate. Without a discharge resistor (such as R7), the OUT2 node ...


2

Yes, low impedance is important. The differential input impedance (looking into either input) is 696 ohms when the diffamp is in balance, so if you're looking for 0.1% accuracy you need the source impedance to be well under an ohm. You can use another of the same kind of op-amp (LT1818) to buffer the 1.5V reference, which could be derived from a higher ...


2

look into this amplifier, by putting 3v on one of the input pins you can filter that out, you have the option to invert the signal to (depends on which pin you place the 3V) http://en.wikipedia.org/wiki/Differential_amplifier#Other_differential_amplifiers


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