New answers tagged analog
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You can use an external DAC IC and have the PIC talk to it using SPI, I2C, or some other interface. As you mentioned, you would use a lookup table in the firmware. To change the frequency, just change the rate at which you update the DAC with a new sample from the lookup table. Alternatively, you could keep the update rate the same, but skip entries in ...
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The default reference should be 5V, fed from Vcc to Aref by way of an internal reference select multiplexer. If might be less than 5V if the board is powered from USB.
Measure your 5V source - is that varying as well?
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It's usually fairly simple.
Most PLC analog inputs are voltage inputs with an optional shunt resistor to convert 0 - 20 mA to 0 - 5 V or 0 - 10 V. A 250 Ω or 500 Ω resistor will do that job.
simulate this circuit – Schematic created using CircuitLab
Figure 1. a) A two-wire sensor connection. b) An externally powered sensor connection.
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pull up or pull down is for prevent of float state( micro could not realize pin is 1 or 0 ! )
new micro controller have internal pull up or pull down...
if your micro controller did not have , you must do it.
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You're thinking about this backwards. What you're describing is a full-matrix mixer, in which each of the outputs can have a mix of any of the inputs. In other words, treat each output independently as an N-channel mixer, where N is the number of inputs. Then build as many mixers as you have outputs.
You might need buffer amplifiers (unity gain, low output ...
1
I am stuggling to find a component which can assign one signal input to 0..8 (or 0..16 if available, for stereo sound) outputs.
Because that doesn't typically exist – you'd use a system of analog switches to achieve that. So, basically, for every single connection you'd want, you'd have to have a switch that's either open or closed.
In essence, each of ...
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Since a barometric sensor is used to sensing pressure in gases, which is exactly what you want to do: yes. That's the type of sensor you're looking for.
5
Bias current and offset are both DC phenomenon. The way to deal with a capacitor for DC analysis is to treat it as an open circuit (and in the unlikely event that you have any coils in an op-amp circuit, you treat them as short circuits, or as the coil's DC resistance only).
In that event, the given expression for \$R_{comp}\$ is correct.
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This looks like a Series-Shunt circuit topology. The way I learned it was that after you identify the feedback circuit, trace the path from the input to that circuit. If you passed through an active device, it is series, otherwise it is shunt. Repeat that process for the output. Then if it is input series, and output shunt it is called "series-shunt."
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Since Vbe =0 it cannot operate.
Since Vce= 0 it cannot operate.
To invert input voltage with respect to output collector voltage (which is always the case for relative changes in voltage being both positive with respect to emitter or 0V, when DC biased properly.
Furthermore, the NPN collect Vce must be >= Vce(sat)for some linear operation. Thus the 10k ...
0
Another logical perspective is when the inverting output applies an opposite voltage such that the differential voltage is always zero, (regardless of voltage gain).
You can see Vin+=0V so Vin/Rin=Zin you input impedance. Even if there was a level shift on input to Vcc/2 and same with signal! The Zin is the same the “virtual ground “ on Vin.
This only ...
1
As the OPAMP is taken to be ideal thus the input impedance of basic amplifier becomes infinite.
I am unsure what you mean by "basic amplifier" Maybe you mean: the opamp has inputs with an infinite input impedance. It simply means that no current flows into the opamp's inputs. This does not say anything about the circuit which is opamp + feedback network.
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1
When you do the feedback calculations, you need to do the calculation on the amplifier with feedback. This means that the input impedance you use is the input impedance of the amplifier with the feedback network added. So the raw amplifier has infinite input impedance and zero output impedance, but as it's used in circuit, the amplifier has an input gain ...
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Some microcontrollers have a sperate analog and digital ground to control the ground return currents from the microcontroller. If the microcontroller is switching large amounts of current (as it changes state) the currents exit through the ground pin. The small amounts of resistance and inductance in the pins and PCB can create small voltages that may create ...
1
It looks like I found a solution highly dependent on the device parameters.
Low leakage parts are the critical factor.
Take careful selection of Ids @ Vt and Vgs
Two inversions are done: The 1st must have a gate threshold of << 1/3 to 1/2 of Vbat and not 2 to 4V. This is what causes decay on the 1st time constant.
hi-side mechanical switch,
low-...
1
Your question is in error. You have fixated on the Theta term, but that is not your problem.
You present two circuits. The first is a simple RLC circuit. You present a circuit analysis of that simple circuit which neglects initial conditions (in particular the initital current through the inductor).
THEN, you show a second circuit which is not equivalent ...
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I don't know if your question received a proper answer and the problem is solved or it is still actual.
Anyway, your Fostex already appears to have an interesting bias loop control, due those parallel FET working as controlled resistors by a tension coming from an upper stage that is not full displayed. It is possible that this circuit to do a kind of HX ...
1
You have in an unknown way got a solution formula where the unknown angle and the max charge in C are the freely selectable parameters. Physically they can be derived from the circuit current and capacitor's charge at t=0. We call generally them "initial conditions" and both are needed.
Your cryptic program code doesn't at least produce right plot for a ...
1
The takeaways you should get from this answer for an underdamped sinusoidal 2nd order response are;
there was no "error" in the question for the values or slopes or initial conditions or his calculations. It was his expectation that was perplexed.
The formula with some phase offset almost looked like the response but in fact, there is no initial phase ...
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This design is going to deviate from expected performance due to insufficient GBW requirements.
With a gain of 3.3 in each stage with a Q of ~14 in each stage at staggered frequencies to achieve a net BW of 10% fc= 10kHz, if one truly expects a Butterworth response 1 octave up, means the minimum GBW product is >> 50 MHz. A GBW of 10MHz reduces the output ...
1
If you want to spread out the multiplexed outputs among the six inputs, what comes to mind is six 8:1 analog multiplexers, one for each Arduino input. Two inputs are unused on each chip.
To select the best part, you need to know the output impedances of the sources, the input impedances on the Arduino inputs, and the maximum frequency of interest in the ...
3
The circuit you show is also called a "data slicer" and is used effectively in extracting data from the output of an FM demodulated signal used by several not-so-cheap 434 MHz RF transceiver circuits you see quite often.
Yours is a variation but probably doesn't need R102 because R101 will perform that function as far as I can tell but there's no harm ...
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Below shows a high pass sallen key filter on the left, and a low pass sallen key filter on the right. The resistors and capacitors are swapped. (Both high and low pass filters form a band pass). If you want to convert one sallen key configuration to the other, then swap resistors and caps
Source:
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Probably white and black, although it shouldn't really matter, assuming that the earbuds are passive, as the speakers in the earbuds are basically just a coil and there is no directional preference in which direction the current flows.
It would be best to test them briefly, by connecting to an audio source and seeing which way around the music sounds better....
2
Looking in the datasheet I see:
note that these are not the maximum ratings but the recommended operating conditions!
Which is good, basically TI says: please use it under these conditions.
So we can use the inputs in the whole supply range.
So your SITUATION ONE and SITUATION TWO will give the same result. This ADC can simply handle both situations ...
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Frequency is related to rate of phase change in a same way than the speed of a car is related to rate of change of its position.
2
They're not equal because of anything, they are equal by definition.
We often write a sinewave as as function of time as \$v = \sin(\omega t)\$
Every time the argument \$\omega t\$ increases by \$2\pi\$, then \$v\$ comes back to what it was before, defining one cycle of the sinewave. \$\omega\$ is known as the angular frequency, in radians per second.
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Phase can be measured in many units. Volts, radians, degrees and converted by some constant for some application.
Consider 3600 rev per minute
3600 RPM = 3600 revs * 360 deg / 60 s = 3600 * 6 deg/s
Next consider for each 1 Hz of difference frequency there is a slew rate change of 360 deg per cycle per second.
When a PLL mixer uses a phase detector to ...
2
A sine wave can have a relative phase to another sine of the same frequency or be described as having a specific phase at time zero. This is not the type of phase they are discussing here.
They are referring to an instantaneous phase. The frequency is the derivative of the phase, ω = dθ / dt. All units are rads, of course.
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This is a two-part problem. The first one is easy: How to get the difference between two input voltages: Use a instrumentation amplifier. As you did not specify what the common mode (i.e. average voltage of the input voltages) does, I would recommend to go for the 3-amp instrumentation amplifier. There are fully integrated versions available from e.g. Analog ...
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The way to find differences between two voltages in the analog domain is with a differential amplifier.
Source: https://www.electronics-tutorials.ws/opamp/opamp_5.html
There are plenty of other ways to mimic a neuron, however. If you want one with the traditional sigmoid output -1 to 1 then this circuit might be most valuable.
Source: https://www....
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High impedance inputs are prone to EMI.
Current transducers 4~20mA at 0mA may not read 0 so your reading indicates an open fault.
I wonder if a sensor short circuit would be any different.
I don't see any reason why a flowmeter EPROM fault should affect a current transducer or an RTD.
2
It's common in control system design to want to indicate a faulty sensor in a positive (unmistakable) manner. In an analog system the only way to do that is to send a signal that's well out of the normal range of measurement. An input of zero (faulty sensor) could otherwise be confused with a valid input.
In the case of temperature transmitters used in ...
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Small negative values close to zero are essentially noise in the absence of a signal. Larger negative values may arise if the plc has been calibrated with a gain and offset for that input. If the offset would be negative then with a zero signal the calibration would return a negative value.
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After looking into this a bit more, I think I have a better answer than my comment above.
Since the LM148 is (as the datasheet says) "a true quad 741", I had a look at the actual schematic of the famous 741 opamp. Which you shouldn't be using, read here why. Of course for educational purposes, studying the 741 (or LM148) is fine.
This site has an excellent ...
1
You have a temperature sensor with analog voltage or current output.
Different operating modes, but one is the basic process controls 4-20mA two-wire current loop meant to be fed into a remote <600Ω load resistor. 3.5mA or 23mA (NAMUR NE43) are used to indicate errors. And 0mA indicates an open circuit or unconnected sensor.
Not the same, but relevant ...
4
I gather you are talking about Figure 12.12 from the 5th edition. Here's the schematic redrawn, just slightly:
simulate this circuit – Schematic created using CircuitLab
You can see a very similar circuit to the portion of the above circuit within the right box in an answer I provided here. Note that I didn't include the equivalent of \$R_{17}\$ in ...
2
I agree with all the answers so far but would like to add a different perspective.
The purpose of Q6 as a Q7 current limiter is to make the AC impedance of Q7 so high that it does not introduce any significant AC nonlinearity. This permits the next 4(?) stages of NPN/PNP Emitter Followers to buffer the AC load impedance scaled by each hFE.
This results ...
2
Quite a lot going on in these 3 transistors.
View Q6/R113/R114 as a high-gain common-emitter amplifier; the addition of C5 serves to cut the ACgain in half, but greatly boosts the AC Power Supply Rejection. Common-emitter resistive-load stages have gains of Vdd/0.026 at the most (when Vce is very low, as in this case); assuming 50 volt rails, the gain is 50 ...
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Q6 and Q7 form a constant current source of about 10.3mA at Q7's Collector.
Q5 is simply a relative mirror of 7mA using the Base voltage of Q7 as its reference. Q5 has nothing to do with the Q6/Q7 CC source.
1
A little more than a Q6 VBE across R18 turns Q6 on, which then shunts away the bias coming through R13, R14 to Q7. So Q6 and Q7 servo the voltage across R18 to be roughly constant. The voltage on Q7 base is then its VBE more than this, so after Q5's VBE, the voltage on R10 is the same as that on R18.
I'm happy with Q5 and Q7 VBE's tracking, but R18 voltage ...
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I believe what they're telling you is that although the supply voltage is nominally 24V, you'll never see more than 18V at the output terminals, even if you leave them open-circuit.
1
Two transistors Q1,Q2 are identical
Unfortunately you do not describe in which context this statement is made.
I have seen such statements before in the context of analog IC design. On a chip we can make transistors that are "identical" in the sense that they have very similar transistor parameters like \$\beta\$ and what \$V_{BE}\$ you get under certain ...
3
The common mode input voltage range for the OPA1642 is: -
(V–)–0.1 (V+)–3.5 V
This means that if you have a power supply of 0 volts and 5 volts, the input range is -0.1 volts to +1.5 volts i.e. it will heavily distort the input signal because it is biased at 2.5 volts DC.
OPAMP is OPA1642 rail-to-rail type
No it isn't. The output is specified as ...
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simulate this circuit – Schematic created using CircuitLab
ths one should turn oun at around 1.5v in
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For your purposes, adding a current sense resistor would be easiest.
You can put it anywhere in series with the load, as long as it is thermally isolated from the load. I'd put it on the MOSFET source to make the voltage measurement single-ended and therefore a little easier.
Then use an instrumentation amplifier with a low pass filter to measure the ...
2
Why is it that in every triac analog dimming circuit I find that the triac triggering circuit is fed through the load and not straight from the mains?
In most countries there is no neutral available in the wall switchbox where the dimmer is mounted. Arranging the circuit to find a neutral path through the load avoids having to rewire the switchboxes with ...
0
You don't have to use voltage divider there. The resistance that base sees is R1||R2 (they are parallel according to base).
It is a basic switching circuit. Simply, the idea is to make the Hfe*Ib bigger than expected collector current to keep it in the saturated region to keep the colector voltage as low as possible.
Simple but not without some glitches. ...
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You are using the transistor as a low speed switch, not as a linear amplifier. This makes things much simpler because you don't need to control the bias voltage then. The transistor is always driven into saturation, which is actually good because it limits the power dissipated inside the transistor to very low values. No heatsink required.
Drop R2 ...
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