# Tag Info

## New answers tagged analog

0

Another thing you need to watch on this device is that the output voltage, over the range of specified input voltages is just about proportional to input voltage - this means that if your input voltage is 5.5V (i.e. 10% high) then your 6V output will be 6.6V. I've been tripped-up by this on a similar product from Traco. It's all in the table called "output ...

3

The converter is operating within its specifications. See the "Tolerance Envelopes" at the top of page 3 of the datasheet. The output is allowed to be as much as 10% high at 10% load, which could be as high as 6.6 V in your case. Try other loads, up to 100%, and see if it generally follows the load line that they show.

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I'm not sure what your second diagram is trying to show, but it together with all the stuff following (I tuned out after the first diagram) suggests that you are making this complicated. It is good to see that you have thought about this problem and put some effort into solving it, but it's easier to just solve it than to try to figure out exactly what you ...

2

No, that's not what he does. He feeds each microphone through an amplifier then through an output impedance that feeds the line through R14 and C4 shown in thick red circle below: - I've shown the wire that is the actual line used by the intercom in a thin red/brown circle. Because of the output impedances of several units connected all at once there ...

3

I don't use Arduino, so I don't know what is happening in those routines, but there are few common ways of taking an ADC measurement with AVR. One of the great things about AVR chips in the ATtiny and ATmega lines is that a lot of peripheral registers have the same (or very similar) names in different chips. I'm going to be using the ATtinyx4 (datasheet) in ...

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The nice thing about AVR is code ports very easily from processor to processor. I have a set of functions on file for setting up the peripherals. Here's my ADC initialization function: void init_adc(void) { sei(); ADMUX = 0b11100000; ADCSRA = 0b10001100; ADCSRA = ADCSRA | (1<< ADSC); } The sei() function serves as a global ...

3

Your professors analysis and lecture is very good. Your question above is out of context and as such you've really kneecapped anyone who is going to analyze it without them understanding that this is really just part of an analysis and rationale for how you do small signal analysis. He is taking you through replacing the non-linear elements and ...

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Here is a great piece of reference material for understanding instrumentation amplifiers. I have a paper copy on my desk so I know it's good. It's produced by Analog devices. Don't read it yet.... The circuit you have shown may have problems and these are to do with common mode AC rejection. The top amp in your diagram will have a different common-mode ...

1

An instrumentation amp is essentially a difference amplifier. The output is directly proportional to the difference between the two input (essentially, which is what you need) and so any common mode signals, ie, those which exist in both inputs, should in principle be eliminated. It looks like a fairly straightforward application of an inamp to me. Simply ...

1

No, a wire fed axially through the centreline of a solenoid would not induce anything into the solenoid coil. The magnetic fields (if both are energized) will be at right angles. There would be no transformer action. At much higher frequencies there will be circulating eddy currents induced in the solenoid winding (from the straight wire) but these won't be ...

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In order to avoid this drift in my gyro, I added if statements to filter out the drift. Something like this: if(abs(degrees(gyroRate[XAXIS]))>0.5) rollAngle += degrees(gyroRate[XAXIS])*timeDiff/1000.00; if(abs(degrees(gyroRate[YAXIS]))>0.5) pitchAngle += degrees(gyroRate[YAXIS])*timeDiff/1000.00; if(abs(degrees(gyroRate[ZAXIS]))>0.5) ...

2

simulate this circuit – Schematic created using CircuitLab You're going to run into some accuracy issues such as: noise from the 5V line self heating possible voltage drop (due to length of wire) I would just go with a one wire thermometer like the DS18B20/DS18S20. Using a digital sensor, less immune to noise. You can even run it at a much ...

1

It may not seem a very wide range of digital numbers but have you considered avaeraging (say) ten readings. Noise (if generally above the bandwidth that you are needing to measure) can be used to your advantage by a process called dithering. In effect the randomness of the noise will give you a spread of readings for a fixed temperature and averaging several ...

3

You must use the basics of your ideal opamp assumptions and derive the KVL and KCL equations for Vout/Vin. If you recall ideal opamp assumption "1", the inverting terminal voltage is equal to the non-inverting terminal voltage, and so that node, where R2 and C2 connect, is at ground (they call it "virtual" since its actually the opamp forcing this node to ...

3

Tips: (1) The voltage across C2 is -Vout. Thus: $I_{C_2} = -sC_2 V_{out}$ (2) The voltage across R2 is: $V_{R_2} = I_{C_2}R_2 = -sR_2C_2 V_{out}$ (3) Due to the virtual ground: $V_{C_1} = V_{R_2}$ This should give you a good start.

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Since the current through R2 is zero, Vo = Vx. You can calculate Vo (and Vx) just using the known expression of a voltage divider: $$V_o = V_x = \frac{\frac{1}{sC}}{\frac{1}{sC} + R_1} V_i= \frac{1}{R_1Cs + 1} V_i.$$

4

If you had the code for the micro doing the PWM, this would be a trivial firmware change. However, it seems you don't have access to the internals of the "control board". What you need therefore is a inverting amplifier. You want 500mV-4.5V to map to 5V-0V. Or put another way, you want a gain of -1.25 centered around 2.5 V. This is easy to do: The ...

0

So I had to re-read through the ADC document and ask for some help, but I think I've found the way. My previous iteration used a method of defining certain parameters, but I ended up just doing direct assignment to the registers eventually. For anyone else with the same problem, the following is my code: Configuration for analog: DDPCONbits.JTAGEN = 0; ...

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CircuitLab solves the circuit because it doesn't simulate effects like junction temperatures reaching beyond the limit, so that semiconductors melt. A diode is not a fixed voltage drop. Current through a diode is related to voltage by an exponential equation. That exponential equation goes on forever: for any imaginable voltage, you can find a current. ...

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The circuit as shown is not viable - or you could analyze it in two phases, if you must: Phase 1: Each 1n4148 diode is rated for 200 mA continuous, 450 mA peak repetitive current. When wired as indicated, each diode will drop approximately 1 to 1.5 Volts (Fig.3 in datasheet) before the current exceeds absolute maximum rating As the supply voltage is 5 ...

1

The best way to do this is to being the signal into a counter/timer channel of the microprocessor. Configure the counter/timer to operate in counter mode. If you can count the pulses for a set interval of say one second would give you a reading of pulses/second from the motor. There will sometimes be a situation where waiting one second is intolerable ...

2

Most of these classic tables come from the work of Wilhelm Cauer on network synthesis. Such filters are known as Cauer Networks or Cauer Ladders as a result of this work. There is also the Cauer filter which is a special case of such a network. A good book on circuit theory should cover some of this and, if I remember correctly, it involves long division of ...

3

Power on reset circuits are used to trigger the point that a microprocessor can start executing its code without the need to worry about whether the voltage supplies are within their prescribed limits for successful operation. The basic principle is based on charging a capacitor. The capacitor charges at a rate intentionally slower than what the power ...

2

Rarely does a chip startup need to be overly precise as the design will likely be simulated/designed for a range of temperatures, Voltages and process corners. What is more likely is the need for certain ranges in dV/dt voltage ramps to ensure the proper sequencing of states during the initial start up phase, but typically there are brown out detectors and ...

2

It depends on how your ADC is configured, and what the reference voltage is. Assuming the number is read directly as a N-bit integer that's justified to the LSB (i.e., the most significant bits beyond ADC resolution are zero), $$V_{in}= V_{ref}* \frac {ADC \ Reading}{(2^N-1)}$$ Sure, there are nonlinearites and noise considerations, but this should be ...

3

There are two possibilities here: That the uC has a fully calibrated ADC with precision circuits and temperature correction. In which case you just need to read through the 418 page *.pdf and find the relationship or .. It is uncalibrated and you can only get an approximation. SO that means that you have to calibrate the input vs. binary # ahead of time ...

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I'm not quite sure exactly what you are trying to do here, but: If the input at +UB is +12VDC, then you can ignore the capacitor and just calculate the divider voltage: $12V \cdot \dfrac{R19}{R18+R19} = 12V \cdot \dfrac{3.3k}{13.3k} = 2.98V$ The diodes will limit the voltage range from just over 5V to just under 0V. For the digital representation, ...

4

This is not a great circuit for a variety of reasons. It is measures the current that leaks between the two probes with a reasonably constant voltage applied. For reference, here is the circuit again (this is what I'm answering even if the question is edited later): The opamp is used as a inverting amplifier. It looks like the intent to hold the + ...

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The circuit measures the resistance of the soil between the probes. The size and shape doesn't matter so much for a rough reading, so spikes or plates will work. If you want to try and calculate the resistivity of the soil, then two square plates would probably be best. The potentiometer can be used to adjust the gain, and set the zero point to whatever ...

2

It depends on the application. I'll take two examples, the first is that of the wired telephone lines from the local exchange to the Customer Premises equipment and the second being the use of an Amps for AC amplification. But before that, I will assume that you are comfortable with the entire concept of Negative Voltage. Simply put, it's the reference ...

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Using a positive and negative power supply at the same time yields a total potential voltage double to either of those voltages on their own. For example, using a +12V and -12V allows for a differential of 24V. Still, other circuits will split a signal into positive and negative values to transmit them, comparing the results at the receiver to eliminate ...

2

Yes, when you have a negative power supply current flows from supply ground to the (negative) supply output pin. (But the electrons disagree, they flow from negative output to ground.) Negative supplies were the norm in the days of PNP germanium transistors. Later the dominant transistor was the NPN silicium version, and positive supplies were the norm. ...

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