# Tag Info

11

In outline, it depends on the signal source, i.e. the type of microphone. There are some very low noise vacuum tubes. There are some low noise IC amplifiers, but not many. There are also discrete semiconductors, both bi-polar and JFET, and these are often the best choice for an input stage, possibly using an IC for the later gain and output stages. ...

10

If you ignore the input capacitor and look at the input impedance into your op-amp circuit from the left of the 200k resistor, the input impedance is the 200k resistor. This is because the op-amp is configured as a virtual earth amplifier. In other words there is 200k loading your capacitor and the 3db high pass cut off point is when Xc = 200k ohms (in ...

5

You are confusing 'inverting' with 'negative feedback'. Open loop simulate this circuit – Schematic created using CircuitLab Figure 1: op-amp with open-loop inverting mode. In Figure 1 the op-amp will amplify the difference between its inputs by the open loop gain. Let's say the open loop gain is 1,000,000 and we apply +1 mV at the '-' input. ...

5

Try using negative feedback and not positive feedback. The circuit you have shown is nonesense unless you are trying to make a comparator with rather a lot of hysteresis. When using a simulator it might theoretically "settle" on what seems to be an improbable scenario - If the output is -1V and the input is +1V then the voltage at the non-inverting input is ...

5

Odds are that Q1 is smoked. You forgot to add a current limiting base resistor to limit the current. You should probably add a reverse diode on the base (after the resistor) to protect the transistor. The diode is recommended because you are feeding the base with an alternating voltage that swings above and below zero volts. When it swings negative the ...

5

Offset voltage is unavoidable error in the construction of a opamp. Nobody "places" it anywhere. Manufacturers go to great lengths to reduce it, but of course, can not make it zero. It also makes no sense to ask whether the offset voltage appears on the positive or negative input. The offset is between the two inputs. A opamp ideally does: ...

4

The bottom distortion is trivial to explain - you appear to be running the op-amp between 8+V and ground, you have not created a virtual ground at +4V, and so the signal is naturally clipping where the op-amp has no power to drive it. For a simple approach, add a second battery to provide a -8V rail, if using 8V batteries, and power the negative power pin ...

3

I think that you can get much better than what you are achieving. You say that if you rise the volume the sound becomes acceptable but is distorted... Let's take a step back. You are feeding your speaker with a square wave. That is pretty much the definition of distorted signal. If you want to have a more clean, less distorted signal on the output you ...

3

The drawing shows the LED power supply as 5 volts - that is too low to light four LEDs in series. The forward voltage across an LED depends on the colour and chemistry - common red LEDs are about 1.9 volts, and other colours are higher, up to 3.2 volts for blue and white. The supply voltage must be greater than the sum of the LED voltages, plus a volt or ...

3

Input bias can flow both into or out of an op-amp's inputs, and in certain op-amp topologies may do either. The consequence is that you need to design your circuit accounting for the current, i.e. if your source impedance is 100kΩ and you've got 1pA of input bias current, a non-inverting buffer will have a voltage error of (100kΩ * 1pA) = 100nV. As you can ...

3

You are basically right, except for the choice of 741 to use over the 0-5 V range. The 741 requires some headroom against both supplies. For 0-5 V operation, a CMOS "rail to rail" opamp is a much better choice. Once of the Microchip MCPxxxx series is likely a good fit. The circuit you show (other than, again, poor choice of opamp) is valid and is common ...

3

why they tell to roll off the open loop gain Saying it this way is just a convenience but the practicality is that there will be negative feedback applied that both shapes the amplifier gain to what you want AND corrects for instability. First of all, familiarize yourself with the important features of the open-loop gain. This is for an OPA192: - ...

3

Your input voltage is an AC waveform which oscillates above and below the zero volt / ground line. It should peak at about +0.75 V and -0.75 V. Your op-amp has its negative terminal connected to ground. The lowest voltage it can possibly output is 0 V and most will only get within a volt or two of that. The fix Power your op-amp from a dual supply, +8 V ...

3

'I have checked both data sheets and haven't found anything different' Well you haven't checked very carefully , have you? They seem to be very similar. There is at least one difference, the bias current of the 'A' version is higher than the plain version. Given that the plain data sheet is dated 2012, and the A version is 2016, I suspect that what has ...

3

The coupling capacitor C1 of 10 nF is too small. You made a high-pass filter with -3 dB frequency of 1/(2pi*RC) = 80 Hz where R is the 200 k resistor and C is the 10 nF capacitor. If you make C1 100 nF or more, the problem will be solved.

2

Survey says... simulate this circuit – Schematic created using CircuitLab

2

Since the jumpers have to be set manually, a simple solution might be to use a second jumper that indicates the selected gain to the microcontroller. So the user have to set two jumpers to identical positions. Or just use a rotary switch with two poles and three positions (e.g.: SS-10-23NPE). A completely different approach would be to implement a ...

2

The TL081 is not the best part to use if your trying to gain up your signal. The voltage offset is in the mV range. Your 'new' gain stage has a gain of (1+(50k+49.2k)/20) = 4963. If you gain the Vos=3mV of the TL081 from the bandpass filter by 4963 you get 14.8V, and that is beyond the rails. I would expect this circuit to rail out, but you may have a ...

2

The TL08x series of opamps have a typical input voltage offset error on the order of 3 mV (datasheet here). This offset is indistinguishable from a real input, and it gets amplified the same way. I assume you're talking about U1 in your diagram, which is configured for a gain of about 5000×. This means that the output error could be on the order of 15 ...

2

If the simulator is using a linear model to solve the problem (for example, using the AC simulation mode) it can get easily fooled. The circuit will be basically this: and the simulator will basically write down $$v_o = A_{vol} \left( v_o\frac{R}{R_1+R} - v_i\right)$$ ...and find a solution, independently from the fact that the system is unstable. ...

2

The purpose of this matching is to minimize DC offset due to bias current. If the op-amp has low bias current and the application is not sensitive to offset, you don't necessarily need to match the DC resistance. You can calculate the offset using the bias/leakage current figure from the op-amp datasheet. In this case, if you want to change the impedance ...

2

To start, eliminate V2, C1, R2 and R5. Then the junction of R3 and R4 will have a Thevenin equivalent voltage of 9 volts, and a resistance of 5k. Adding R5 provides a hysteresis of $$\Delta v = 18\times \frac{6k}{1.006 M} = .107\text{ volts} = 9 +/- .053\text{ volts}$$ this will also provide a hysteresis at the R2/R3 junction of  \Delta v = 18\times ...

2

Common mode voltage (DC and noise on whatever it turns out to be) is likely to be your biggest problem. A shunt is a very low impedance source and is easily filtered if you don't need fast response. You don't need a twisted pair in that case, though it won't hurt. If you filter the shunt voltage well at the ADC end and use a differential amplifier with a ...

2

As has been indicated, this is not a good circuit, but I'll try until I get tired. Let's start with function. Apparently you want to do the following: 1) Run each battery through a difference amplifier to produce a (more or less) 3.7 volt level. 2) compare each level to 3.3 volts with a comparator 3) if any cell in a 4-call pack is low, turn on an LED ...

2

We're not here to give you the solution for homework, we're here to help you solve problems, or more importantly clarify confusing steps in work put into solving problem. So instead of answering a question of two problems, I'll show you an approach for these kinds of problems. Circuit Analysis of Op-Amp using KCL We're analyzing here for small signal ...

1

A non-inverting amplifier always has a voltage gain greater than one- the formula is Vout/Vin = (1 + Rf/Rz). But you can attenuate the signal before the op-amp sees it, and have a gain as little as one (for example Rf = 0, Rz = open). The op-amp will have a closed-loop output impedance much less than 2.5K, so your requirement will be satisfied.

1

Option 4: Use a voltage converter IC to generate a higher and a lower voltage from your 5V. For example I use this little circuit to power OpAmps and comparators from 5V without any problems: This generates a voltage of approximately 9.5V at the VA+ terminal and -4.7V at the VA- terminal from just a 5V supply. If you use this to power opamps and use ...

1

The cheapest solutions, assuming you only have one supply, are to redesign the circuit so it will work from a single supply or generate the negative supply. All monolithic op-amps that I know of will actually work on a single supply- very few actually have a ground pin, so they don't know the difference between +/-5V and a single 10V supply. They do know ...

1

The LTSpice Yahoo group is the place to go. You can download models of many common devices. Also keep in mind that while models of real op-amps are pretty good, they are far from perfect, and may not simulate all the parasitics in a real circuit. If you are very concerned about parasitics, there's little you can do other than building the circuit and ...

1

The closed loop output resistance should be very low, less than 1 ohm. 75 ohms is more-or-less what you'd expect for AC output impedance. The output resistance is only low for currents that the op-amp can supply. It will go into limiting mode if the output current exceeds a couple tens of mA roughly. It also cannot drive close to the rails with a heavy ...

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