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14

Yes - it has a name. In control theory this circuit is known as a PD-T1 unit. It has a proportional-derivative behaviour with a certain delay term T1. In filter terms, it works like a first-order high-pass with a superimposed constant gain. The transfer function is H(s)=1+[sR1*C/(1+sR2C)] This device is used to enhance the phase (for stabilizing purposes) ...


13

I'd just call that a non inverting amplifier. Calculating the transfer function is quite easy if we can consider the op amp ideal. In DC C1 is open, so you don't have current in R1 nor R2, so the op amp is in the buffer configuration and the gain is 1. When f gets very big C1 is closed, the gain of the circuit is the usual 1+R1/R2 leading to a 2 gain for ...


7

With two voltages (input, output) I would expect that it is likely a "voltage follower". More than that, are you sure about the part number? When speaking about classical voltage opamps the real closed-loop gain for the unity gain amplifier (follower) is \$Acl = \dfrac{Ao}{1+Ao}\$ with Ao: Open-loop gain. Fort a typical value \$Ao=10^5\$ (100 dB) we ...


4

The 1uF capacitor forms a simple low-pass filter with th 4.7K to filter the PWM square wave from the microcontroller. The 100nF cap in parallel with 4.7K does not do much- it prevents erosion of the op-amp phase margin due to input and stray capacitance on the inverting input.


4

You can use the ICL7660. It takes a positive voltage (up to 12v) and generates the inverse: Very easy to use, I've used it in several commercial products. If the ICL7660 chip is difficult to find, you can also roll your own" charge pump inverter" using a 555 timer, which should be available anywhere, and a few extra components: There are lots of ...


3

A 741 cannot do what you ask. Find a 741 data sheet, and look for "output voltage swing". You will see that, for instance, at a power supply of +/- 15 volts it will allow an output of ~+/- 12. In other words, the lowest output voltage specified is never anywhere near the - supply. In your case, the - supply is ground, so you cannot expect the 741 to give you ...


3

I'm assuming you're using a non-inverting configuration. First, your DC offset. When you looked at the output, you had no input at all - not even a resistor to ground. So the input floated and the output went to the rail. Try connecting your input to ground with, let's say, a 10k resistor. If, for some reason, you are using an inverting configuration, ...


3

The diode in the picture isn't a clamp rather it glues the negative portion of the signal to about -0.7 volts whilst the positive part of the signal rises freely. In effect it tries to self bias the average output after the capacitor to the p-p voltage divided by 2. Setting the "clamp" voltage to be a tad higher can be done by using a forward biased diode ...


3

This doesn't directly answer your question, but I have a few suggestions to improve the design which should make it unnecessary for you to use the clamping diode at all. simulate this circuit – Schematic created using CircuitLab Move the 10k/10k resistor-divider to your opamp's non-inverting input to give your input signal a DC bias point of ...


3

Your signal is all positive, so you can amplify it around 0 to get what you want. The issue is that not all opamps work with their inputs at 0, or can drive their output to 0. What you want is a rail to rail opamp, or at least one where the common mode range extends to the negative rail, and the output range also extends close to the negative rail. ...


3

There is no actual convention as to a number of transistors- the designers themselves don't particularly care. What matters is the performance, the manufacturability, the processes used (extra layers, certain types of parts such as good PNP transistors, CMOS, higher voltage-handling ability, and precision resistors require special or extra process steps ...


3

The error will be composed of the input offset voltage and change of input offset voltage with input common mode voltage (CMRR) plus error due to finite gain- so a small offset and a small gain error. The error due to finite gain with a precision op-amp is usually pretty low at DC, but increases with increasing frequency. Noise will generally be higher ...


3

The error is essentially the input voltage offset of the op-amp. This can range from microvolts to millivolts. Traditionally, op-amps with \$V_{os}\$ below 1 mV have been sold as "precision" op-amps. And you'll find this as one of the main categories of op-amps on most vendors webpages (Analog, TI, Linear, ...). If this parameter is critical, you should ...


3

You already pretty much described the answer. A voltage divider will follow the battery voltage and maintain the virtual ground at about half the battery voltage. You would also add either a bypass / bulk-storage capacitor or an op-amp configured as a follower (or both). The op-amp buffer makes the virtual ground "stiff", meaning that it doesn't bounce ...


3

The precision resistor divider is a better way to go than the zener diode. However a lot depends upon what kind of load impedance the divider will have. Too low of load impedance means lowering the divider resistors and consuming a lot of current. My favorite way to solve this is to use the divider resistors that consume low current but then buffer the ...


3

The zener method will generally have a much lower source impedance, so if you're demanding that the virtual ground source or sink more than leakage current you might have more inadvertent coupling through the virtual ground than is tolerable. For example, if you are willing to use 20mA from a 12V supply to create the virtual ground you can use two 300 ohm ...


3

Not a bad idea. I disagree with the other comments about using the MCU to control this. Why waste the quiescent current? Periodic brief wake ups do have a real effect over time. Just add some hysteresis to the comparator so that it's not oscillating. You could consider having the output of the comparator connected to an interrupt pin on the MCU if ...


3

For the main part of the open loop response a typical op-amp acts like an integrator and therefore shifts the phase angle by about 90 degrees. The curve that looks most like 90 degrees is the phase response. Here's the response for an OPA192: - Taken from this blog. And below is a generic op-amp's bode plot: - Taken from here and below an op77: -


3

Try something more like this- you may have to fiddle with the value of the capacitor to optimize it if you need really high frequency response. simulate this circuit – Schematic created using CircuitLab


2

There is no generic name for it I believe. It has unity gain at DC and, at some point in the spectrum the gain will have risen to 2. This is dictated by: - High frequency gain = 1 + R1/R2 The frequency where the gain is nearly 2 (the 3dB point) is determined by R2 and C1. When the reactance of C1 equals R2 this is the 3dB point and the reactance of the ...


2

Sure can. It's the 10k to V+. As the circuit powers up, assuming the set and reset lines are held at the same value, the - input to the op amp will reliably be held higher than the + input. During the early stages of the power-up cycle, while V+ and V- are low, the output transistors in the op amp output stage are not able to produce much current. As a ...


2

What if you used another switch from the same package as a dummy SW1 and calibrate out the effect? Then you'd be down to the difference in resistances. Maybe you could increase R1 as well. That could well get you the 30:1 improvement you want.


2

This does not answer the original question exactly but demonstrates a means by which the startup-reset state can be assured and shows another low cost and effective "Heath Robinson" means of achieving the desired aim. I have built many circuits using this approach with excellent results. As an option to using an op-amp you could consider the use of 2 x ...


2

Many have told you to beware input offset voltages, which will be your main error. You may also see slew rate limits, which may distort your output at fast transiotions. Also, many op amp outputs can not reach closer than a few volts below the power rails, so you may have saturation problems.


2

You can move the input filter capacitor to the other side of the switch (leave it connected to the input power) and add a small capacitor (maybe 100nF) on the actual input of the regulator. You might want to add a fuse to the input (either a polyswitch resettable fuse or a one-time fuse).


2

You can't run the OPA657 at unity gain - it oscillates - even when configured as a TIA you can run into trouble with it. Here's the open loop gain and phase (red): - The op-amp's phase margin is such that at a closed loop gain of about 6 to 8dB the negative feedback will be positive feedback i.e. open loop phase is about -180 degrees at about 850MHz and ...


2

In general there's no simple formula to calculate the total cut-off frequency. You would need to consider the total transfer function (i.e. the multiplication of the transfer functions of the individual filters), and from this transfer function compute the 3dB cut-off frequency.


2

This answer assumes that each stage of the multistage filter is unaffected by stages before or after it i.e. the stages can be thought of of having buffer amps between them thus they are isolated from nuances of other stages. Take the bode plot of stage 1 and mathematically add it to the bode plot of stage 2. Continue until all the filter stages are plotted ...


2

This is a dominant-pole compensated op amp, so the gain curve starts out with high gain at low frequency and rolls off with a slope of -20dB/decade. The phase starts off at low frequency with a -90 degree shift and stays relatively constant until the other poles (due to transistor Ft etc.) start to influence it. So the top curve at 10kHz is the gain.


2

There are a couple clues you can look at to determine which line is gain and which is phase. To get near-ideal performance, the open loop gain should be very high -- sometimes as high as 120 dB (1,000,000x). The gain wouldn't start out at 20 dB (10x). Meanwhile, when the phase response is flat, it should be sitting at a multiple of 90 degrees. Phase doesn't ...



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