# Tag Info

0

This is a curious case – I simulated it and got the same sharp “bandpass” response. I think I know what’s going on. Your transfer equation is correct. It is a high pass filter, and the gain does explode to infinity at high frequencies. This makes sense: the impedance of C1 goes to zero, so the first stage gain R2/0 goes to infinity. But in real life or even ...

0

Well there's a resistor missing from your diagram (see below) then relay switch on and off continue very fast It's quite likely that you need hysteresis to prevent the relay chattering its contacts when the two op-amp inputs become very close in value. Make the "resistor needed" about 10 kohm and make the "hysteresis resistor" about ...

6

Determine the requirements. Determine the topology to use (eg. 2 or 3 amplifiers). Determine the resistor values. Analyze the circuit to see if it meets all the requirements simultaneously. Iterate if required. (optional) Throw away LM741s and use more modern amplifiers or a purpose-built instrumentation amplifier.

1

The one thing I don't understand though is that I thought the way to do it would be just to do a voltage divider followed by a unity op-amp circuit but that is not the case The input voltage range on your INA826 schematic goes from -10 volts to +10 volts and so you need an amplifier that has a negative supply voltage as well as a positive supply voltage to ...

-1

I was asked to delete my obvious ways to design a circuit to do it, since this is homework and there is no actual teacher to help the student.

2

I've heard that opamps are not good at driving capacitive loads "I heard" is practically always an indication that you could have posted a reference, or researched further. However, no doubt, it's hard to drive a capacitive load with anything that relies on negative feedback; if you're not sufficiently dampened, you will start to oscillate. Let's ...

1

The LM324 is an integrated circuit containing four operational amplifiers; this is a common arrangement for cost savings as it allows for less integrated circuits to be used on a board, and lower package costs (which can dwarf wafer fabrication costs in many cases). Three operational amplifiers are needed for this design; a single quad-op-amp chip is easier ...

1

First, I will present a method that uses Mathematica to solve this problem. When I was studying this stuff I used the method all the time (without using Mathematica of course). Besides that the answer of @jonk is excellent. Well, we are trying to analyze the following opamp-circuit: simulate this circuit – Schematic created using CircuitLab When we ...

2

Use the formula below (transposed) to calculate the value of your R3 and then double it to get the value of your R4. Use non polarised capacitors.

2

This opamp has 45 degree phase margin, with Cload of 150 pF. Assume the Fring is 1MHz (near the UGBW). To ring at this frequency, with 150pF Cload, the equivalent output inductance is about 150 microHenries. However, in your circuit you have 10,000,000 pF Cload or 10uF. 150uH and 10uF resonate at 4,000 Hertz. Is that 100 ohm into the 10UF adequate dampening? ...

11

General Form, 2nd Order HPF The general form for a 2nd order high-pass filter is ($K$ is the voltage gain): \begin{align*}\frac{K\:s^2}{s^2+2\zeta\:\omega_{_0}\:s +\omega_{_0}^2}=\frac{K\:s^2}{s^2+\frac{\omega_{_0}}{Q}\:s +\omega_{_0}^2}\label{eq1}\tag{1}\end{align*} Butterworth The next thing is to look up the constants for Butterworth filter. These ...

0

The difficulty in your approach is the lack of link between the PID coefficients and the poles/zeroes positions. In the APEC seminar I gave in 2012, I have re-derived the poles and zeroes calculation based on the existing PID coefficients. Once you know where the poles and zeroes are located, you can determine the components value in your circuit. The ...

0

The circuit belongs to so-called relaxation oscillators in which the voltage on a capacitor "travels" between the two thresholds of a comparator with hysteresis ("Schmitt trigger"). There is no negative feedback in this circuit; there is only a positive feedback during the insignificantly short transition time (I mean the phenomenon and ...

1

Razavi's book is great but I cannot agree that "when v(input)=0 op amp raises Vy such that it barely turns on the diode." OP is right - if the op-amp is ideal, its output voltage will be zero when the input voltage is zero. The diode will be turned on only if the input voltage rises at least a little. There is a lot of philosophy in this specific ...

4

Why doesn't the following circuit have negative feedback? It does have negative feedback and, once the positive feedback (R1 and R2) has done its job, that dominant positive feedback is gradually eroded by the slower negative feedback caused by R and C. After a short while (determined by R and C), the op-amp inputs are equal in value and a very, very short ...

2

The circuit is a Schmitt (after Otto Schmitt) trigger with an added R-C, to make an oscillator. R2, R1, and the op-amp form the Schmitt trigger.

1

Write Uout=A(Uin-Uout)+E where E is how much the output differs from the linear gain output. E of course depends in a complex way on the input and output voltages. In addition E isn't unique, it depends on how big linear gain A we assume. Solve Uout. The formula is Uout=Uin(A/(1+A))+E/(1+A) Uout approaches Uin if A grows and the amp works in a way that E ...

1

Question: Why does the formula which we derived by using small signal analysis apply also for large signal? (1) At first, the input stage of the opamp is a differential amplifier which has a transfer characteristic which mathematically follows a tanh function. For small input signals (millivolt range) this function has a very good linearity. (2) Due to the ...

2

Or is there any other explanation? Forget about op-amps for a couple of minutes and think about a servo control mechanism like this one for example: - You set a "position demand" with the potentiometer on the left and, the error amplifier "calculates" the voltage difference between "actual_output" (from the position ...

0

Imagine that the open-loop gain of the op-amp is not constant, but rather varies with input voltage A = f(Vin), so it is nonlinear. For a good op-amp f(Vin) will be a large positive number, regardless of Vin, though it may vary. Vo = Vi * f(Vin)/(f(Vin) + 1). If f(Vin) >> 1 then Vo ~= Vin. Similarly, an output offset is reduced by the gain of the op-...

0

a larger view of the math is this Vout = VIn * G / ( 1 + G * H) where "G" can be somewhat non-linear IF "H" is a resistive feedback voltage divider. Audio amplifiers depend on this. Transformer_feedback RF amplifiers depend on this (using transformer ratios to set the feedback). Notice the assumption: the H is linear, while the G can be ...

2

Sometimes it helps redrawing the circuit in a different way. Let's start from the fact that the output is pulled down to ground, so vout is zero unless there is some good reason for it not to. For vin<0, the op amp amplifies vplus-vminus and outputs a negative voltage (according to its amplification, it could reach its negative rail if there is enough ...

3

Not sure if this is the op-amp you are referring to, but the OPA2810 typical characteristics contain similar numbers. Interestingly, perhaps, the typical unity-gain bandwidth of the amplifier increases from 75MHz to 105MHz (typ) when the capacitive loading is increased from 4.7pF to 33pF as it makes the amplifier less stable and there is some gain peaking ...

0

I made some assumptions as it comes to the supply of the op-amp, the type of op-amp (picked the Universal one) and came with this simulation. Depending on if you have rail-to-rail supply for your op-amp (+xV to -xV) or single supply (+xV to GND) you get positive and negative feedback or positive feedback respectively.

6

In book it is written that "when v(input)=0 op amp raises Vy such that it barely turn on the diode" . But I don't understand how this is possible because we know for ideal non inverting amplifier has finite close loop gain and if input is zero then output should also be zero (0×finite gain) Well, you're right, the ideal op-amp doesn't need to do ...

6

You are confusing the output of circuit with output of opamp, they are not the same thing. An ideal opamp output does whatever it needs in order to keep opamp input voltages equal via feedback. So in this case the circuit Vout is same as opamp inverting input, and circuit Vin is opamp noninverting input. Thus Vout equal Vin, and opamp output is Vout+Vd. ...

3

As their name suggests, devices with differential input can subtract two single-ended (grounded) voltages. The transistor is a simple example where the first input voltage is applied between the base and ground and the second voltage - between the emitter and ground. The differential amplifier, particularly an op-amp, is a more sophisticated example where ...

5

The short answer is that an opamp has way more going on inside than just the input differential pair. If you recreate the rest of the opamp's internal functions with discrete transistor circuits, you will get the same results.

5

the only circuit with discrete transistors that can add two signals is the differential amplifier with the second input signal reversed before passed to the differential amplifier. That's simply not true. My question is, how does the op-amp add more than two input signals when it is actually made of discrete transistors circuits? Well, a) above statement ...

14

A transistor can add n input signals, a crude example is the following simulate this circuit – Schematic created using CircuitLab You can also do it with a common base circuit.

2

Figured out that the solenoid was the reason why the circuit would not reach 1.5A. The resistance of the coil was 15ohms at room temp, but when it heated up at high currents, the resistance would rise to about 22ohms. So, the solenoid was limiting the current. I confirmed that this was the case by attaching a lower resistance solenoid to the circuit, which ...

0

From your own answer to your question: The problem is that the guitar has it's own resistance from the potentiometer. After measuring that resistance and recalculating R2 using the resistance of the guitar as R1, the amp works as it should. My comment: That means you have omitted the input decoupling capacitor and are running a small DC current through ...

0

One of the op-amps has V+ on pin 7 and the other one has V+ on pin 4. Only one of them is correct.

0

first thing the LM358 op amp is not suited for this application because of the thermopile sensors can give negative voltages, these opamp cannot pickup this.so its better to go with a better rail to rail opamp.I am trying to make a breakout board for a thermopile sensor for the past 2 months.i had tried lm358,AD620 opamp and everthing but it did not give a ...

0

Your schematic has Digit to COM instead of to +5. It ought to be an open collector active low driver to a Common Anode (CA) digit.

1

On the textbook schematic, the A to G pins on the decoder is top to bottom and it is the opposite on your schematic. The proteus model you used seems to be A to G from top to bottom so you have to connect the right decoder output to the right 7segment pins. On your schematic, G is connected to A, F to B, etc...

1

Basically using a single supply, you must ensure the input stays in the Vcm spec range which is >2V or <Vcc-2V. otherwise the input is not operating in close loop linear range. (E.g. Vcc/2 is a good way to bias inputs if possible by AC coupling.) Other fine details, Ringing is most likely from excessive probe ground lead and for <=35ns Tr or >=...

1

Suggestions build the circuit atop a single_sided copper_clad FR-4, so you have a Ground Plane to work with bypass the VDD using 0.1uF from VDD to Ground using 0.1 inch lead. Locate this capacitor at the OPAMP VDD pin. isolate your opamp VDD from the power supply, by inserting 10 ohms in series. This dampened any wire_ringing. phase shifts on the Vin- ...

0

The problem is that the guitar has it's own resistance from the potentiometer. After measuring that resistance and recalculating R2 using the resistance of the guitar as R1, the amp works as it should.

1

I realized that the sense resistor needs to be provided by the user, and is not included in the DRV8301 itself. It's the resistor placed between SNx and SPx.

-1

Circuit is a VCCS, load current is derived from the OP Amp ref input Voltage divided by the Sense Resistor(R4) value, I.E. assume the input ref voltage is 5.0v (pot R1 is in mid position) then the load current is 5.0v/100ohm = 5.0mA.

2

As previously mentioned, the problem is the DC level after the DC coupling. There is no DC path to ground on either end of the 1.2K resistor...except the LT1719 inverting input. Taking a look at the LT1719 datasheet, specifically the I.sub.B "Input Bias Current" specification, 2.5uA of current will typically flow out of that terminal (positive ...

3

This will not work. You would need 4Vp-p at the op-amp output to get 2Vp-p at the output, since RA||RB = 10K, and your Vcc is only 3.3V so you can only get less than 3.3Vp-p. You can do this entire thing with three resistors (such as 10K, 17.5K, 23.33K) if the input signal is low impedance. Maybe you misunderstood your instructor who was pointing out an ...

1

A very simple rail splitter. From the datasheet of TL064 it seems maximum output source/sink current is only about 3 mA (to maintain 2.5 V output at 5 V supply). simulate this circuit – Schematic created using CircuitLab An option to increase output current would be to add a push pull output stage with an NPN and a PNP transistor pair, but that may ...

5

I don't want to use a specific LiPo battery charging IC... If you think its possible can you guys give me some ideas? Yes, it's possible. All you need is a voltage regulator with current limiting. You could do this with a regulator IC (switching or linear) which has a current limit feature, a 3 terminal regulator with a transistor to provide current ...

0

Are you against/avoiding using any digital means of control? A tiny microprocessor would make it easier to control every aspect of the circuit. A set of sensors that measures the LiPO (or any other chemistry that fits in your voltage range) status and proportionally drives transistor biases or PWM signals into a MOSFETs to control charge voltage/amperage ...

6

This kind of op-amp has a bias current that flows out of the input, so opening it will cause the output to go to about +22V. If you connect a resistor (say 100K) from the pot wiper connection to ground you will affect the linearity slightly, but an open wiper will result in an output voltage of only a few mV typically (worst case could be maybe 50mV). ...

5

If you disconnect a pot and leave the top end of R11 floating then the bias current from the non-inverting input will charge up C10 until it reaches one of the power rails. Since you have a voltage following buffer circuit the output will just follow the input.

1

A starting point would be to decide at what value of RTemp you want it to switch (maybe 2K). Then set R1, R2 and R3 all to that value. R4 is the current limiting resistor for the LED and should be set to allow a few milliamps through. Yes, the negative terminal of the LED will need to be grounded for anything to happen.

0

Well, just set R9 to be zero, and all bridge resistors to be 120 and the potential divider produces a ratio of 360/480 and clearly that's wrong because it should be 0.5. At this point, if I were you I'd ignore the document and find another I could rely on. Moving on... just look at the glaring error putting R5 in the numerator when they meant R9 so, if I ...

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