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9

The ADG708 can only handle negative inputs if it's in the -2.5 to 2.5V dual supply configuration. If you've got it hooked up to 0 and 5V, then that makes complete sense. There will be protection diodes in the ADG708 that will clamp to -0.7 or so below it's lower rail (VSS).


8

There are two problems with having the -Vs of a 741 connected to ground. First, as others have said, the output cannot go below ground. In fact, with a 741 it can't even get very close to ground, only to within maybe 1V or 1.5V of ground. This is called the output swing and is usually rated with some kind of a load resistor. Take care that the load ...


8

The basic differential front end of a opamp works like this: On a single chip, the resistors would be current sources, usually a current mirror. Either way, think of how this works. Q1 and Q2 are built to match as closely as possible. Assume they are identical, and consider what happens as IN+ and IN- are moved up and down together. Being identical ...


5

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). ...


5

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 ...


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


5

The gain of an op-amp is usually very high (in the order of tens or hundreds of thousands at DC at moderate to low frequencies so, if the op-amp output is not end-stopped against the power rails, it's a reasonable assumption to say that the input voltage difference is Vout/100,000 for example. If |Vout| is maybe 10 volts or less, the input voltage ...


5

The differential amplifier is what gives you the common-mode rejection. Differential gain implies common-mode rejection. The high input impedance also comes from the input stage. If FETs are used, the high resistance is a property of the gate of the FET. BJTs multiply the resistance of the bias current source, which is pretty high to begin with. There's ...


4

The input bias current of the AD8051 might be as high as a couple of micro amps and this doesn't sound much but, through a 1Mohm feedback resistor, this will generate a dc error of 2 volts on the output. Note that with the +input seeing a DC resistance of about 5kohm, the same bias current will produce an error of about 10mV. I would suggest equalizing ...


4

I cannot speak to the digital question, but the analog is pretty straightforward. You should use a TIA with a 1000 lux sensitivity, about 8 kohm / full scale volt. Use about a 5 volt bias, and expect a photodiode capacitance of 4 or 5 pF. I'd suggest that you don't need a PGA, rather, use a pair of x10 amplifiers in series and an analog mux to select the ...


4

Use the relaxation oscillator to make a ramp or triangle wave and then compare that wave's voltage to a DC voltage that you can control (for example with a potentiometer).


3

The problem is that from a term "Series-Shunt" it is not clear what comes first: "in" or "out"? I have discovered that different authors handle this subject differently. For this reason I prefer, for example: Voltage-controlled current feedback. Examples: Voltage-controlled voltage feedback: Non-inverting opamp, Voltage-controlled current feedback: ...


3

The main thing that is almost always mis-stated or mis-understood in all these examples is the fact that the inputs being close or even identical to each other is NOT a effect of the amplifier but is an effect of negative feedback. The amplifier is assumed to be ideal to simplify the discussion. An ideal op-amp without negative feedback will NOT have the ...


3

Why in the world are you worried about Johnson noise? Consider that the LM833 has a power bandwidth less than 1 MHz. Assuming your ADC has a 3 V input and 16 bit resolution, 1 lsb is ~46 µV. A 1 kΩ resistor with a 1 MHz bandwidth has a Johnson noise voltage of ~4 µV, and noise scales as the square root of the resistance. It would take a 529 kΩ resistor to ...


3

There is a pretty much general method for finding the Thevenin equivalent circuit: Find the open circuit output voltage. This means take the circuit you are trying to find the equivalent of, replace the load with an open circuit, and find the output voltage. In your circuit this means replacing everything to the right of node "B" with an open circuit. So ...


3

Without seeing the input signals, I'd say the circuit is working just fine. That is, it's working as it should given the topology. Unfortunately, it won't work as an instrumentation amplifier, or even as a differential amplifier. And I'm pretty sure a differential amplifier with gain is what you intended. First, of course, is the fact that you're feeding an ...


2

It is worth pointing out that fundamentally the transconductance of a BJT is much higher than for a MOSFET. i.e. the current varies with the exponential of the applied voltage in the case of a BJT, whereas it only varies with the square of the voltage for a MOSFET. Ideally all systems would be a mix of BJT and MOS, but that is not how the world works. So ...


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 ...


2

What you were told is not theoretically correct. You have to change the way o thinking what really happens. It is the feedback loop that can force IN-=IN+ , not IN-=IN+ that makes the loop work. The condition IN-=IN+ des not happen magically. Start thinking that \$Vout=A_0(IN_+-IN_-)\$ where \$A_0=10^6\$ This is the IN\Out relation of an open circuit ...


2

Offset current is just the difference between bias currents on the two inputs and this can be regarded as fairly constant. For non inverting configurations, input current is bias current plus a dc current proportional to input voltage due to non infinite input resistance and, for higher frequencies, there is a noticeable ac current due to non infinitesimal ...


2

Let's get you started on just the first filter for now. The first filter is just a simple inverting op amp amplifier. For such a circuit with input impedance \$Z_{I}\$ and feedback impedance \$Z_{F}\$ the transfer function is simply $$A(j\omega) = \frac{v_{O}}{v_{I}} = -\frac{Z_{F}}{Z_{I}}$$ An ideal op amp can force the output to any voltage necessary ...


2

How can I extend this circuit to a 1st or 2nd order low pass filter, with a corner frequency at about 20kHz. 1st order solutions: - Put the same capacitors across R6 and R8 is one method (maintaining R6 and R8 at the same value). Another method is put a 1k resistor in series with the output and shunt the 10k with a capacitor (it works more perfectly ...


2

This circuit is a classic diff-amp. The output is V2-V1. One way to analyze this circuit is to think of the affect from each input to the output separately. Start by grounding V2 and thinking about the response from V1 to the output. With V2 grounded, the + input is just held at 0. Now you have a simple inverting amp with a gain of -1 from V1 to Output. ...


2

Of course, both resistors act as a voltage divider. However, you most consider the fact that you have TWO voltage sources at the same time (Vin and Vout). Hence, you must apply the superposition rule for calculating the voltage at the midpoint between both resistors: \$V_{n1}=V_{in}\dfrac{R_f}{R_i+R_f}\$ and \$V_{n2}=V_{out}\dfrac{R_i}{R_i+R_f}\$ with ...


2

"Wish I didn't know now what I didn't know then.. " (an Ohrwurm for you) The OPA227 is a bipolar circuit so latchup is not an issue. Usually, the voltage limits are not as important as the current limits, but there is no spec on those. The datasheet specifically says: The inputs of the OPA277 series are protected with 1kΩ series input resistors ...


2

When you solve positive feedback circuits like this, you need some initial values. We can say that \$V_{sat+}\$ as the upper limit to what the opamp can drive to and \$V_{sat-}\$ as the lower limit. If we make an initial assumption that \$V_{out} = V_{sat+}\$ then you will get $$ V_+ = V_{sat+}\dfrac {R_1}{R_1+R_2} $$ $$ V_{out} = A_v(\dfrac{R_1}{R_1+R_2} ...


2

On reflection, the answer is to heed the advice "any time you can't find an opamp with two critical characteristics, use two opamps". A voltage feedback amplifier like the LMH6657 configured as a unity gain differential amp, followed by a high slew rate current feedback amplifier to produce the required voltage swing can together produce the result I need.


2

It looks like the problem is the values of R6 and R7. The voltage divider ratio is \$\frac{R7}{R6 + R7}\$, which works out to \$\frac{1.8k}{18k + 1.8k} = 0.091.\$ That gives you about 45.5mA, which is close to what you see. You actually want a ratio of 0.1. Add a 220-ohm resistor in series below R7 and you'll get much closer to your ideal value.


2

Assuming your schematic is really how you've hooked it up, I'd hazard a guess that all your problems (except the pot burning up) are caused by the fact that you have the power and ground reversed. +5 goes to pin 8 and ground to pin 4, not the other way around. Once you fix this, the question of offsets may or may not become relevant, but replacing Rsense ...



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