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6

I would say the output was offset at 1 volt and this is approximately 100 times higher than the offset on the input waveform. Given your amp is configured to have a gain of about 100, there is no surprise here. The triangulated output is due to the poor gain-bandwidth-product of the op-amp you chose. You want a gain of 100 and you are feeding it 18kHz and ...


6

This behavior is called phase inversion. It happens in certain op-amps when you exceed the input common mode range limits of the op-amp (for example, when the input voltage is below the most-negative supply voltage). It's explained at the circuit level in a TI presentation I found (warning: Powerpoint link), and also in an EE Times article. Notice the ...


5

You need to actually read the datasheet: Note that a negative input voltage is out of spec. When you violate a spec, there are no longer any guarantees what the part might do. So yes, there is nothing unexpected about it inverting the signal onto the output, or doing anything else that might appear strange. When you don't follow the rules, you can't ...


5

I remember seeing a similar question here (but couldn't find it). The component seems to be a transmission gate, or analog switch. (http://en.wikipedia.org/wiki/Transmission_gate) Google-ing it helps you find the schematic: https://www.google.hu/search?q=transmission+gate+symbol


4

You're applying inputs that are well outside the common-mode input range of the chip. See the datasheet. You should not use this graph either, but use the guaranteed range, which means you should not expect proper operation with less than about 2V applied to the inputs (relative to -Vs) or closer to +Vs than about 1.5V.


4

When you're measuring the voltage direct with the DMM you basically have this arrangement: simulate this circuit – Schematic created using CircuitLab From that you can calculate what the total resistance in the lower half must be in order to get 5V at the junction. When you have calculated that total resistance, you can then separate out the ...


4

Model the circuit without the voltage follower as so: simulate this circuit – Schematic created using CircuitLab This is still just a voltage divider, but you have \$R_2||R_{\text{DMM}}\$ as the lower resistor instead of just \$R_2\$ as in the unloaded case (and the case with the voltage follower, which ideally has infinite input impedance). ...


3

You can write their transfer function from input to the output. The first is a active filter, assume a ideal Op amp: $$ H(s)=\frac{-R_{2}}{R_{1}} \times \frac{1}{1+R_{2}C_{2}s} $$ The second it's basically passive RC filter followed by a voltage follower.: $$ H(s)=\frac{1}{RCs+1} $$ Then, you can see some differences, such as the first one has negative ...


3

First, please check LT1007's datasheet, section "Unity-Gain Buffer Application (LT1007 Only)", it has BJT input with input voltage clamp diodes. So if the output voltage goes beyond 1V + 0.7V, the input clamp diode will conduct, try to use another OP without input clamp, such as LF412, it has JFET input. Another problem is just as others say, if your input ...


3

The problem is that your bit inputs must be at either a logical 1 (Vcc) or a logical zero, (GND) as they would be if they were connected to the outputs of, say, a counter like an HC191. You've shown them floating when the switch is OFF, which won't work properly if you're using an R-2R ladder. This works, and here are the LTspice files so you can play ...


3

Why do we need dual supply voltages in Op-Amp circuits ? You need dual supplies if your signal has a negative voltage portion and a virtual ground (DC offset superimposed on your signal) is not suitable. Which parameter of Op-amp causes to make a square signal's shifting between 0-1,1-0 slower? The slew rate defines how fast an op-amp can change ...


2

I went to a break out session titled "Top 10 Op Amp Mistakes" at a conference not long ago, given by an Analog Devices FAE and this scenario came up. The first option with bypass caps from both power and the ground plane to the virtual ground will be better. Ideally, you'll want to have several caps each, say 10uF, 1uF, 0.1uF, even better, in progressively ...


2

The LM386's internal circuit is similar to an opamp, but optimized for high power audio. To get the best out if it you should follow the recommended circuit shown in the datasheet. Your circuit actually does work, but not very well. The main problem is that you don't have a capacitor on the output, which is required to block DC voltage from getting to the ...


2

Your R2 is supposed to be a potentiometer connected between +5v and Gnd, and with its wiper connected to pin 3 of your op-amp. What you have is a single fixed resistor from +5v to the op-amp instead. The function of R2 is to adjust the DC offset level out of that op-amp (at pin 1) but the way you have it connected that one will never produce anything other ...


2

Slew rate limits will cause distortion at high output frequency and amplitude. If your amp is slew rate limited, putting in a periodic waveform (sine, square, etc.) will result in something that looks a bit like a sawtooth wave. This can create frequency harmonics that are not present in the original signal, especially when the source signal is a pure ...


2

For my opinion, the simplest solution makes use of the classical feedback formula from H. Black: V2/V1=H(s)/(1-LG) with: H(s)=H1(s)*H2(s)=Forward transfer function for an open loop (in our case: H1=V3/V1 for R2>>infinite and H2=V2/V3.) Loop gain LG=Product of all three transfer functions within the loop (with V1=0 or R1>>infinite). Note that H(s) is ...


2

Strictly speaking, you don't need dual supply voltages. You can use a single supply connected to the op amp's rails. Dual supplies are generally much easier to work with, though, because you can easily set the op amp's common mode to ground. If you only have a single supply then you may need to generate a common mode voltage (e.g. at half the supply ...


2

It's not the oscilloscope probe behaving like an antenna -- it's YOU. You are capacitively coupled to noise in the environment. The cap prevents DC signals from reaching the amplifier, not AC signals. If you don't want to see the 50Hz noise, hook the input up to the signal you're trying to amplify, and don'd touch the input pins.


2

Batteries are a special case because their internal resistance increases as they discharge. Probably something like 100uF/16V in parallel with 1uF/16V ceramic. Here's what the impedance vs. frequency of a 100uF capacitor with 5 ohm ESR in parallel with a 1uF ideal capacitor looks like from 100Hz to 5MHz: So, it's pretty low over the entire audio range ...


1

This is a bit of a complicated design question as these parts have very different environmental specs as well. First of all, maximum allowable common-mode voltage. The input offset voltage is a complicated beast, but one of the major contributions to it are input FET bias current asymmetry. This is the current that is being drawn by the individual internal ...


1

Here is a method that uses only six components (3 resistor networks, 2 quad op-amps and a gain-setting resistor R5). simulate this circuit – Schematic created using CircuitLab


1

Op amps can be used for many purposes. Generally, one will want to have a slew rate which is fast enough that the op amp will never be slew-rate limited while processing a "continuous" AC signal. On the other hand, if an op amp will be used to process a discontinuous signal which represents a number of DC levels in sequence, the output of the op amp will ...


1

If that image is accurate, then I would suggest simplifying your board layout. When I was a TA for Electronics lab classes I often saw students get confused by a maze of criss-crossed wires on the board. Starting with the IC is definitely a good idea. Some of your resistors can connect directly from the LM324 pin to +5V or ground. R7 is a prime example, and ...


1

Question 1 (biasing): It is no surprise that the output signal contains a dc part because the input signal is centered upon a dc voltage. However, the dc ouput level cannot be verified simply by amplifying the input dc voltage because the form of the output signal has changed. Question 2 (form): There are two different effects which influence the output ...


1

Overall, with the mods discussed for stability, I think this is a fine circuit. You need an op-amp with rail-to-rail output (or close) but everything else is pretty non-critical. I approve of the use of a 180W-capable MOSFET in this (linear) application. You could certainly use a BJT or a Darlington (or a Sziklai pair) but there's not a lot of reason to at ...


1

0-1 V does not sound like valid PECL levels. Even with 2.5 V supply, PECL levels should be something like 0.7 and 1.4 V. If your PECL device has a VBB pin, that's there specifically for the purpose of re-biasing ac-coupled signals --- and it's tied to any Vcc variation to avoid duty cycle distortion. Remember that PECL logic are referenced to VCC rather ...


1

There are high current op-amps available, such as the OPA547 from Texas Instruments. You could use that (or similar) to apply fractional gain (attenuation) to a larger voltage (say 0-8V at 10:1) from a trim-pot (8V is the minimum supply voltage, so would be convenient for 10:1 attenuation). The op-amp has a 5mV offset to take into account. That would give ...


1

If you have a negative supply available, you can use an LM317 or similar adjustable regulator, with the bottom of the voltage divider connected to the negative rail, rather than to ground.


1

Guitar can put out 2-4 Vpp if you really bang on it just right. But if you're tuning, the range would be limited to no more than 200-400 mVpp I think. Anyway, I might use something like the following circuit (just a sketch of course): simulate this circuit – Schematic created using CircuitLab This uses a bandpass input with an inverting op amp ...


1

Neither one of those op-amps can reliably turn that particular MOSFET 'off' (by swinging within a few hundred mV of the positive rail), though one is better than the other. I would look for a guaranteed swing within 200mV of the positive rail, since the -400mV minimum Vgs(th) has a temperature coefficient and you probably don't want it to go overvoltage on a ...



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