Tag Info

Hot answers tagged

8

An individual op-amp will have one non-inverting input (usually denoted with a + symbol) and one inverting input (usually denoted with a -). They are very much not equivalent. As their description makes apparent, one inverts it's input value, and the other does not. Now, with regard to the drawn symbol, which is on top is generally a function of what will ...


5

The case of your microphone is probably electrically connected to the flexible membrane inside the microphone, and is responsible for grounding it, while the "ground pin" is only used to provide the return path for the FET preamplifier. In normal operation, both need to be grounded so that there's a complete circuit to drive the gate of the FET. With the ...


5

The offset voltage of each of the op-amps you're using can be as much as +/-3mV at room temperature. So, the difference between two outputs could be as much as 6mV different from the inputs with unity gain. You're seeing 5.4mV which is large, but within specifications and therefore plausible. Since you don't have much gain in the first stage (only 3) you ...


5

First, you should use a comparator if you want to do comparisons. Choose one with a switching time substantially less than what you want the overall switching time of the circuit to be (with 1 ms spec, you should be able to choose almost any of them). Second, choose one that can drive enough output current to pull the FET gate full swing quickly enough. ...


4

Why LED2 doesn't turn on: You should be aware that the LM324 series of op-amps has not got the output stage to source or sink significant current at only 2V margins. If that is not the problem causing your LEDs only very slightly lighting up, it may well be that your finger creates a 200kOhm or even higher resistance, depending on the distance between ...


4

They are regulators as you thought. To properly control the MOSFET on each regulator, the gate needs to rise several volts above the source, probably over 6 volts greater than the source. If the source voltage is 3.3 volts the gate might need to rise as high as 9 or 10 volts when the MOSFET is driving a heavy load current. This means the op-amp MUST be ...


4

The "Dual Decompensated" in the IC's title is biting you. Two things come to mind. The description in the data sheet mentions that that IC is stable in gains of 10 and more, and you are low gain. Also From the data sheet: High Speed Operation When the feedback around the op amp is resistive (RF), a pole will be created with RF, the source resistance and ...


4

R1 and R2 being equal (R) means that the offset error due to input bias current (but not input offset current) is eliminated. If an equal current Ib flows out of each input pin then each pin will be higher by Ib * R and if the impedance at V_in is much lower than R, the two cancel out. simulate this circuit – Schematic created using CircuitLab ...


3

This is called constant gain-bandwidth product but it isn't true for every op amp. It is only true for voltage feedback op amps which use dominant pole compensation for stability. Such op amps can be approximated as a first order system since one pole dominates all others and the others can be ignored. (However, this is not true of current feedback op amps ...


3

Op amps are compensated with a dominant pole. That means the open loop gain rolls off at a constant 20dB/decade vs. frequency. Negative feedback increases the input impedance, decreases the output impedance and increases the bandwidth. Because of the single pole rolloff, the product of noise gain (or non-inverting gain) and bandwidth are constant. Another ...


3

Adding to Andy's answer: Q9 is not fitted, and has been replaced with a short between 2 of the pins If the short is between pins 2 and 3, this would supply 3.3V to the 2.5V output. If the short is between pins 1 and 3, this would make the OpAmp provide the 2.5V output, which would be OK if the current is very very small (note the 10k series resistor, it ...


3

There are a couple of problems with what you are trying to do: The slew rate is not sufficient to pass a 75kHz 12Vp signal. The max rate of change of your input signal is the max of the derivative of 12*sin(wt) or max(12*w*cos(wt)), which is 12*w or 5.6V/us. Look at Fig 11 in the datasheet. The max output signal swing is (typically) the supply rail minus ...


3

Well the input range is clearly 150mV and the output range is found in this sentence: - without exceeding the typical guaranteed linear output range This means that for a normal op-amp (that can swing to within 1.5 volts of the power rails) on a "normal" supply of +/-15 volts, the output swing will be 30 volts - 3 volts = 27 volts swing (p-p). This ...


3

The op amp's positive supply is designed to be biased to the most positive voltage and its negative supply is designed to be biased to the most negative voltage. If an input signal is applied while the supplies are not powered the input voltage may be higher than the positive supply or lower than the negative supply. If that happens a \$pn\$ junction (e.g. a ...


3

You misunderstood the "axiom" An Ideal opamp will do what it can to make the differential voltage between the -ve pin and the +ve pin equal zero. It doesn't state it is zero All it can do is alter its output and thus with negative feedback there OPAMP stands a chance of making the difference zero You then make use of that fact during circuit analysis to ...


2

Vout is not the output of your opamp. If you would add another tag and check the actual output of the opamp you will see it is always set on 0V because you have an inverting configuration without a negative rail so you force the opamp towards the ground voltage which is 0V. So, the output of the opamp can be looked at like 0V, the same as if the diode's ...


2

Almost all ICs are designed with the assumption that the input and (usually) output voltages will remain between the power supply inputs at all times. To ensure this, most ICs have protection diodes between the input pins and the power suppy and ground terminals. If you apply power to an input when the IC not powered, these protection diodes will ...


2

They are not equialent because the op amp has to know WHICH WAY to adjust the output if the pins are not sitting at the same voltage. If you switch them, then the op amp output will simply zip over to one rail or the other and stay put. Op amps are generally designed with a nergative feedback loop. In this case, the outp of the op amp is fed back around, ...


2

First question : What kinds of microphone? Common electret mics may need a bias supply, 1.5 or 3V fed through a 2 k to 10k resistor to bias them. (If you add this, also reverse the input coupling capacitors to their +ve terminal is positive!) Second : what variety of opamp are you using? The +ve input is connected to ground via 1 megohm. The -ve input, via ...


2

With the Laplace transform the capacitors have impedance $$Z = \frac{1}{sC}$$ Use this impedance for the capacitors and analyze the circuit the same way as you would with regular resistors. For an ideal op amp the inputs \$v_{+}\$ and \$v_{-}\$ are at the same voltage and have infinite impedance. Consequently there is a voltage divider at the input: ...


2

What if you use your second (or indeed first, or indeed any of all the other devices) with a device made by someone thinking the same as you with a shared supply in some larger encompassing device? Call the person and ask if the one ground is the same as the other ground? i.e. Is signal ground the same level as power ground? I can tell you, in "cheap" ...


2

If you're trying to output a signal moving a few volts either side of 0V, then you could produce +ve & -ve rails from your +/-24V using a positive and negative regulator, then power any old op-amp of those. If you want enormous voltage swings on the output, then you could use a high-voltage op-amp which you can power directly off the +/-24V rails. ...


2

Your question doesn't show how you're powering the op amp, but I'll take a wild guess that you're not giving it a negative supply, just ground. The chip's input voltages need to be at least a volt away from the supply rails to operate properly. E.g. with a 0V...12V supply, both the inputs need to be between 1V...11V.


2

Here's what the spec of the AD712 says regarding operating either input close to the negative power rail. I say this because you haven't shown the power rails and, like @Rennex, I suspect you are not using a negative rail. Typically exceeding −14.1 V negative common-mode voltage on either input results in an output phase reversal. This was taken from ...


2

It looks like the upper (low pass) 3dB point for 1 of the cascaded filters is 50kHz based on the values R = 47k and C = 68p. In fact it works out at 49.798 kHz. For the same reasoning, the lower (high pass) 3dB point will be 33.863kHz. However, there is a problem. Take a look at this frequency response for a normalized LPF: - Clearly, the 3dB point is at ...


2

There are pros and cons for either. A signal full ground plane has the advantage that signal 0V is a high integrity 0V and can be relied upon but, not when there are currents flowing of any significance. Despite a ground plane being very low impedance, volt drops can still occur when significant currents flow. How much volt drop being a problem depends ...


2

In your lower schema, you are putting a wire between the + and - inputs of the op-amp: This is wrong. There is no \$i_3\$ current, and no current into the op-amp inputs. Just write down the equations: \$V_{A+} = \dfrac{V_2 \times R_4}{R_3 + R_4}\$ \$\dfrac{V_1 - V_{A-}}{R_1} = - \dfrac{V_O - V_{A-}} {R_2}\$ \$V_{A+} = V_{A-}\$ (condition for the amplifier ...


1

They are different packages for the IC. Page 2 of the datasheet shows that CP is a plastic DIP package, and CDR is a small outline package. The R at the end means the ICs come in a reel; if there is no R then they come in tubes. Performance may depend on the package -- you'll have to check the electrical characteristics in the datasheet to see what the ...


1

You need to use a different op amp to drive a \$50\Omega\$ load. Page 8 of the LT1077 datasheet shows the following plot: \$50\Omega\$ load resistance isn't even on the plot, and the voltage gain is quite low even for load resistances in the \$1\$k\$\Omega\$ range. Also, the output impedance is not great (especially considering a \$50\Omega\$ load):


1

You need an opamp with more output current. Search for power opamps. (LM675 and OPA544 are two I've used.) At 200mA you might get away with a TCA0372. Or use some audio driver chips if you don't care about DC. Edit:(more thoughts, comments) Warning the LM675 has a minimum gain of 10. And all the power opamps I've looked at have some cross-over ...



Only top voted, non community-wiki answers of a minimum length are eligible