14

Your schematic should be as following: simulate this circuit – Schematic created using CircuitLab Since you're using a single-supply non-inverting amplifier, the non-inv input of the opamp should be biased to a non-zero voltage –ideally to Vcc/2 as in your schematic so that the amplified signal can swing equally. Now let's take a look at the ...


12

You can get a small negative voltage by using an LM7705 which produces -232mV nominal output voltage using a charge pump. The advantage of using that part over a typical garden-variety inverting charge pump converter (eg. +5 to -5) or inverting boost converter is that the worst-case negative output voltage generally falls within the maximum negative input ...


10

Let's look at a simplified diagram: simulate this circuit – Schematic created using CircuitLab Let's set \$R_f=68\:\text{k}\Omega\$. You know that \$1+\frac{R_f}{R_{th}}=3.5\$, which is your desired gain. So it follows that \$R_{th}=27.2\:\text{k}\Omega\$. If the input is \$4.5\:\text{V}\$ then the output is \$3.0\:\text{V}\$ and \$V_{th}=5.1\:\text{V}...


9

When selecting a comparator, it is always best to choose a part that is specifically designed for the job. Op amps can be used as comparators, but really, it is usually best to just buy a comparator! As for the 741 IC, this has many features that are undesirable for a comparator, as a lot of people will tell you. However, we can focus on your specific ...


8

A voltage is a difference in potential measured between two points. Traditionally, we label ground as 0V and measure from there, but you can choose any point on your circuit to call zero. If you choose your negative supply, then just by changing your labels, it's back to a single supply! So yes, it is always possible. That said, take a bit of care when ...


7

You are right, when the circuit is unpowered the CT output could violate the absolute maximum ratings of the opamp. This is not good. The protection diodes will be forward biased, and the power rail partially powered. This might be OK, depending on what is connected to the opamp power and what exactly the datasheet says about this. It might be acceptable ...


7

In comments you added this information, this is a single run of the ac analysis This method won't allow you to measure the input or output (especially the output) impedance accurately. You need to test the output impedance by applying a source to the output with the input zero'd and vice versa. You will need two separate runs of the simulator to do this....


7

In the circuit, you have posted the Opamp dos not work in a linear region. More about the linear operation you can find here: Op-amp: Virtual ground principle and other doubts But instead of this, the opamp is working as a voltage comparator with hysteresis (positive feedback). Here you can see a voltage comparator without positive feedback. As you ...


6

The job of R2 and R3 is to create a voltage that is roughly half the supply. Lower values give you lower impedance of the output voltage, but also increase current. In this case, the voltage only needs to drive a opamp input, which is high impedance. The 50 kΩ output of the divider is plenty low enough, and 45 µA of current was deemed ...


6

The 741 is generally not able to go from power rail to power rail, usually falling short by a couple volts. Page 6 of the rev G data sheet shows this. For a +/-15v supply, the max guaranteed output is only +/-12v. If you really need a wider swing, you need a different part, or a follower stage of some kind.


6

Getting a precise analogue of a digital moving average filter is somewhere between very difficult and impossible using analogue circuitry, as it involves delaying a signal for a long time. There are various approximations you could use, which might (or might not) be acceptable to your particular application. The first is a simple lowpass RC filter. Instead ...


6

The gain of an op-amp is set by the feedback network. If you have an amplifier compensated for minimum gain of 5 and apply feedback that results in less gain, phase margin will be inadequate and you run the risk of oscillation, especially at unity gain. For example, the LMH6624 is decompensated so a minimum gain \$|A_V| \ge 10\$ is required for stability. ...


6

The output stage of op-amps under saturated conditions may have a bit lower output current or significantly higher supply current than the datasheet promises as a maximum. The information on behavior with saturated outputs is not typically shown directly in the datasheet, but here is one reference. Having the op-amp biased so it cannot balance could also ...


5

If you redraw your schematic using the normal triangle op-amp symbols it will be easier to understand your circuit. simulate this circuit – Schematic created using CircuitLab Figure 1. Upper circuit redrawn. The idea of a schematic diagram is to show the schema of the circuit. Yours is a very bad example as it it quite difficult to identify the ...


4

You need to determine if you want your detector to operate in photovoltaic or photoconductive mode. If the former, simply get rid of R1 and reverse your PD. simulate this circuit – Schematic created using CircuitLab If the latter, reverse the postion of R1 and the PD, like so simulate this circuit In theory, photoconductive is faster than ...


4

You need to look for "rail-to-rail" op-amps and even then there will only be a few that will do 20mV (with a 1mA sink current) such as the AD8605 - not cheap but pretty good all round and beats the LM324 into pulp on most things providing it runs from less than 6V. Here is linear technology's search engine with a few parameters also selected to give you a ...


4

The values of R2 and R3 are chosen so that the unwanted voltage \$V=G*I_{in}*R/2\$ (Where \$I_{in}\$ the maximum input current of the opamp and \$G\$ is the closed-loop gain) remains acceptably low. This voltage will appear as error on the output. C1 is chosen so that the low-pass filter formed by R2||R3 and C1 provides the desired PSRR in the target ...


4

Any leakage in C2 would only be corrected by the output of the opamp moving positive to supply the leakage current through R2. If this moved the output close to the positive supply rail it would reduce the output swing capability or even saturating the opamp. The actual effect will depend upon the value of R2. The caution seems to be somewhat exaggerated. ...


4

The answer is "no" and "yes". simulate this circuit – Schematic created using CircuitLab Figure 1. (a) Voltage follower. (b) Voltage divider and follower. (c) Inverting amplifier. (a) Since the IN signal is connected directly to the op-amp non-inverting input it is not allowed to exceed the input voltage limits. Usually that means keeping within the ...


4

I want to use an inverting amplifier to linearly shift and convert a signal with voltage range -10...+10V into 0 ... 3.3V for an ADC of a microcontroller. That's doable without much trouble. Instead of the 40 Hz source I would input the -10...+10V supply and instead of -15V I want to use a -12V supply. That should probably read: Input signal: -10 V to ...


4

You could generate a small positive voltage, and use it as a virtual ground. Since you selected a differential ADC, its large common mode rejection can allow you to get away with a very simple way of generating that 0.2V reference voltage. simulate this circuit – Schematic created using CircuitLab


4

Let's assume there's no trivial error, but something which needs a little math to be spotted. I guess the hysteesis is too wide. Put a 15kOhm resistor in the place of the current 10k between the output and the +input of the leftmost opamp. The output swing is currently too narrow to cause the Schmitt trigger to flip to both directions.


4

Data sheet, page 5: Vicm Common mode input voltage range VDD + 1.15 to VCC - 1.15 V You're trying to make it work VDD + 0 to VCC - 0 V; that won't do. This is just what is known as Latch-up behaviour. You have a rail-to-rail output opamp. What you'd need to have is a rail-to-rail input and output (RRIO) opamp for this to work.


4

The LM741 (datasheet here) has a "common mode" input voltage range of not within about 3V of either supply. In your example, with a 9V supply, you must have an input voltage in the 3V-6V range (0+3, 9-3) for the opamp to function correctly. The absolute maximum input common mode range varies with supply voltage. It is usually not well covered in data ...


4

I know it's acting as a current limiter to the LEDs but confused about its role in feedback. No, it's not acting as a current limiter. That's why you're confused. R77 is part of a current controller. Opamp U8A operates to keep the voltage across R77 equal to the voltage on its non-inverting input. It will drive Q4 gate to whatever voltage is needed to ...


4

Solving for V2: Vout=Vin * R2/(R1+R2)= Vin* 0.3299 So at 10V at V2 you get 3.3V output. simulate this circuit – Schematic created using CircuitLab Solving for I2: Vout = I2*165 Vout = 20mA*165 = 3.3V simulate this circuit In order to use this circuit with 3.3V supply, you should use a rail to rail opamp in order to maximize the output range and then ...


4

Your design goals are to have \$V_{out} = 3\$ when \$V_{in} = 4.5\$ \$V_{out} = 10\$ when \$V_{in} = 6.5\$ Solving for a linear equation, you want \$V_{out} = \frac{7}{2}V_{in} - 12.75\$ An often used Op-Amp topology for amplification and level shifting looks like this. simulate this circuit – Schematic created using CircuitLab To find the resistor ...


4

Does anybody have an idea on what's going on and how to deal with it? Thanks It's called signal inversion (also phase reversal) and happens when the input goes beyond the upper or lower limit specified in the data sheet. Basically, the output polarity flips and things become fairly meaningless: - 2nd picture from here. My advice is to add a small negative ...


4

Oldskool linear transformer-diode-cap power supplies work great. However, ironically, the reasons behind their excellent short-term overload tolerance are the same reasons why you don't want to use them: they're huge, heavy, and therefore expensive. In a linear power supply, if the transformer is sized for average power with a thermal cutout, it'll handle ...


4

Your symbol pinout order does not match the model's pinout order. Here is the relevant section in your OPA379.LIB file: * END MODEL FEATURES * * PINOUT ORDER +IN -IN +V -V OUT * PINOUT ORDER 1 3 5 2 4 * .SUBCKT OPA379 1 3 5 2 4 * Q21 25 26 24 QNL R77 27 28 200 R78 29 28 200 R79 30 26 100 Notice how the order is listed as: +IN -IN +V -V OUT ? Well, ...


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