New answers tagged

2

Hopefully you know that this is an intercomputer data transmission line interface, it sends and receives serial data stream which is inputted and outputted by a serial data port of a computer. The datasheet http://pdf.datasheetcatalog.com/datasheet/nationalsemiconductor/DS006750.PDF claims several kilobits per second are transferrable, but the actual ...


1

As others say, shielding might not be necessary... But watch out for voltage drop on the ground wires, if the current in them changes. For example if something else on your board turns on, that current step will transfer to all microphone outputs. Large power supply capacitors on the board will help. Separating power ground and signal ground, and using ...


0

Your input highpass filter and feedback lowpass filter have simple gradual slopes so they actually reduce the level of your 1500Hz signal and also do not make a good bandpass filter. A multiple-order Sallen-Key highpass and lowpass filters make a good bandpass filter. A Wien Bridge, Multiple Feedback or Twin-T bandpass filter has a high Q at its peak with ...


0

The unshielded short cable is 300 ohms so it probably will not pickup much interference. The voltage follower opamp does nothing and should be removed. The input resistor of the inverting opamp can use the 1.6k resistor then the 0.1uF capacitor feeding it cuts frequencies below 1kHz. The 2.2uF output capacitor will cause the inverting opamp to oscillate. To ...


1

My opinion only: so long as the output impedance of the preamp is fairly low, you should NOT need to provide shielding between the output of the preamp on the card containing the microphones and the other card containing the ADC. Several factors affect noise pickup. First is the actual level involved. The higher the desired level sent down the conductor,...


0

How is it possible that it behave like a resistor (which is what will happen in triode region)? You can’t equate it to the triode region unless you are holding Vgs at some fixed value and, you can’t do that because Vgs = Vds when wired as a MOSFET diode. may be a diode connected Mosfet useful for an amplifier? If it’s connected as a diode there are ...


0

The crosstalk, even in a 50 ohm system, may be too high for a SPDT in one IC. I suggest you implement "T" switches: two SPST in series, with the center_node either "shorted to Ground" or floating.


0

If the opamp inputs are bipolar transistors the operation is impossible because one transistor cannot feed the mandatory few nanoampere bias current for the other. Both base current directions cannot be right at the same time. If the opamp has fet inputs the gate voltages must be defined in relative with the fet substrate. That means some connetion is ...


0

The difference is that it is not balanced with respect to GND. You mean: not referenced to ground. And that is actually the issue, for example if you would do this: simulate this circuit – Schematic created using CircuitLab then the Vfloat battery only sets the voltage difference between the opamp's inputs. The voltage (or potential) at the opamp's ...


1

Since no current flows on RL, are we authorized to replace it with a short circuit, or with an open circuit, and then solve the resulting circuit? Yes you can do this on this particular circuit but be careful about expanding that idea; if you did it on an op-amp circuit (because you assumed that the voltage between Vin- and Vin+ was zero i.e. a virtual ...


0

How to transform a set of analog voltages from V to ppm/K? The basic way is to take your 1st graph and draw red lines on it like I have: - The red diagonal line represents the average slope between -40 °C and +125 °C. I've also added horizontal and vertical lines that you can use to pick-off the numbers. So, the voltage changes 10 mV in a temperature ...


0

When you calculate the derivative, the resulting units are the units of y divided by the units of x. So here, it would be \$V/K\$ (deg C and K are the same unit when considered differentially). If you divide the derivative by the typical/nominal voltage, the resulting units are then just \$1/K\$. The values will be fairly small, so you often use a unitless ...


3

Your system is unstable because you have a pole in the right half plane, and the result is that the output grows unbounded. Any non-ideal op-amp has a finite DC gain \$\ A_0 \$ and a finite bandwith (crossover frequency \$\ \omega_c\$ below infinity) such a non ideal op-amp can be approximated as a single pole low pass filter, with the following transfer ...


0

You can do something like this: simulate this circuit – Schematic created using CircuitLab I have included both a relay output and an opto output, depending on what you want to use. R3 is for hysteresis for clean switching. Because R3 must connect from output to non-inverting input for hysteresis, the reference divider must go on the non-inverting ...


0

If the signal is relatively clean, 150mV might be fine to work with. Sample it with Analog.read on A0, spit the number to the serial port, and view it in the Serial monitor in the Arduino IDE. This might be exactly what the Analog.read example in the Arduino environment does. In the Serial monitor, sample with the detector signal off for a while. You ...


1

You are wanting to detect Changes in Voltage ,not absolute voltage .This means that you can capacitively couple to remove the nominal 18VDC component .You can easily amplify the AC signal that represents changes in voltage .A simple transistor or opamp will do to give a gain of say 10 .This nominal 1.5 Volt peak signal could go into an analog input for ...


2

Charge injection may or may not be a problem, depending on the performance required of the PLL. Using two devices switching in antiphase, and making the assumption that an on-going device generates the opposite charge injection to an off-going device, gives you a first order cancellation of the injection current. Uncancelled charge injection can be ...


1

This design should help you a good way on towards your aim although it won't provide the full +&- 5V, maybe +& -4V. I chose the common op amp for its rail to rail output capability and low input bias currents. It's a Wien Oscillator followed by a variable gain power stage. R6 & R7 linearise the J-Fet. EDIT This bootstrapping technique should ...


0

If you are familiar with the question 'an opamp has zero voltage between the input pins, so how can it produce an output?', then you have the answer. A PLL is a feedback system, with infinite gain at DC, large gain well below the loop bandwidth, gains around unity around the loop bandwidth (in fact unity gain defines the loop bandwidth, the loop filter ...


1

The output of the phase comparator is the same : the output of the "loop filter" is not. To achieve zero phase error, you need infinite gain at zero frequency; i.e. a component of the loop filter is an integrator. An easier way to think of the loop filter, in this case, is as a PI controller, with the P (Proportional) term providing fast tracking of phase ...


5

Small Signal voltage gain from input to output at 400kHz (no LO) It will be \$\frac{1}{2}\frac{Z_L}{R_{E1}}\$ Small Signal voltage gain from LO to output at 400kHz (no signal at input) It will be \$\frac{Z_L }{ 2 {r_{e_{ac}}}}\$ where \$r_{e_{ac}}=\frac{0.026} {I_{CQ_1}}=\frac{V_T}{I_{CQ_1}}=\frac{k_BTq}{I_{CQ_1}}\$ where \$k_B\$ is Boltzmann ...


0

You need some way of shifting the input signal to the new ground level. The easiest way is to use an isolation amplifier. How practical that is depends on any accuracy and cost constraints you may have and whether you can produce a supply rail relative to the Vee (you could always use an isolated DC-DC converter. There are probably commercial modules that ...


1

I've been reading passages from ADI's OP AMP Handbook and it's really helped me with designing and analyzing precision op amp circuits. https://www.analog.com/en/education/education-library/op-amp-applications-handbook.html It's all free online. The parts about noise analysis are very practical and applicable to my current work.


0

I didn't get how the O/P terminals above is connected to the wind wave. Doesn't such a potentiometric wind vane have 3 terminals? Which terminals are connected to the O/P in this case? The wind vane is configured internally as a variable resistor, rather than a ratiometric divider, hence why there are only two terminals. We can't be certain why it has ...


1

Why they choose one or other method is uncertain - but the current source driven method should work well enough. Change 741 opamp to LM324 (quad) or LM358 (dual) The transistor provides a current source of about 0.5 mA- see below. The windvane is connected as a variable resistor (2 wire connection) and not as a potentiometer (3 wire connection). Both ...


1

From the document you linked: - What could be the reason they are using current source? To convert resistance (linear with wind direction) to voltage linearly. Doesn't such a potentiometric wind vane have 3 terminals? Not according to the document you linked. They are using it like a rheostat.


0

Is there any reason one might want to measure both the primary and secondary coils on an LVDT? To get better accuracy is the usual reason. The amplitude is somewhat subject to noise perturbations so, if you also measure the phase angle between primary and secondaries you get a more reliable result. To do that you need to know what the primary wave form is....


0

The circuit analysis you described indeed assumes that the presence of the feedback capacitor does not cause any oscillations and the circuit is absolutely stable (defined below). If the circuit is stable, then phase margin can be used to measure the relative stability. But usually resistively loaded circuits are always stable and the test for absolute ...


5

the range of the signal is such that the 10 bit ADC value is moving from 200-500 (1V-2.5V) on a scale of 0-1023(0V-5V). You are right at the point where it’s tricky to get a decent benefit without significant attention to detail. Currently you occupy a range of 1.5 volts (1 volt to 2.5 volt) out of 5 volts but, you don’t really have a useable range of 5 ...


0

It lowers the device input impedance, which is not good for a voltage amplifier. Firstly it’s an audio amplifier and any input impedance above 10 kohm is generally accepted as being fine except when used as an input stage for a guitar amplifier (which it shouldn’t be). And, because it’s an audio amp that drives a speaker, having both inputs lightly DC ...


2

It provides a path for the DC bias current to the input transistor so that the signal can be AC coupled with a capacitor. Note that in the application diagram (fig 13 onwards in the data sheet) the two inputs can be connected to ground - it is not required to provide a bias voltage. The value of the resistance, 150k, is high enough that it does not have ...


0

An unstable closed loop system can be made stable in two ways: - Lowering the open loop gain or, Ensuring that the open loop phase change does not reach 180 degrees before the open loop gain drops below unity. Clearly, for people who make opamps, they like to sell you something that has a massive DC open loop gain so, if the overall gain is dropped (to fix ...


8

Constant current sources ... there are no such devices in nature; almost all of sources are (constant) voltage sources. So we have to make them. I think you are correct. Due to physics chemical batteries (DC) and electrol-magnetic (AC) almost always (wriggle room!) create voltage sources. Note, however, that if the load resistance is very low in comparison ...


5

I will expose my philosophy in a few consecutive steps each of them illustrated by a colorful picture. The left part a is a conceptual electrical circuit illustrating the basic idea; the right part b is an exemplary electronic circuit based on this idea. The quantities do not have exact values. The representation by voltage bars is approximate but the ...


3

You can use a TLC551 and a MCH3383 p-channel MOSFET. It will work down to 1V. Follow the astable multivibrator equations in the TLC551 datasheet, and drive the MOSFET gate with the output. MOSFET source to +1.2, drain to motor, other side of motor grounded. Reverse biased 1N4148 or BAT54 across the motor. Done.


0

You can't do this with purely passive components, period. You probably should use a dedicated timer IC. They're extremely cheap and ubiquitous, and perfect for exactly your application. If you absolutely must go 'strictly analog', you can do this with an op amp circuit called a Wien bridge oscillator. You will have to increase your supply voltage to make ...


2

Be aware that small-signal parameters assume linear operation, which usually means that the intended operating parameters vary over a small range. It is apparent that the intended application of this device is as a switch, operating over a very large range. Deriving small-signal parameters from this data sheet will likely result in gross approximations that ...


1

That sounds like a resistive touch screen membrane, A long time ago a company called "Koala" did a resistive digitiser pad for "Apple ][" PCs and other 8-bit machines. but these days your best bet is probably to re-purpose a resistive touch-screen digitiser.


4

The video cable from camera is also used to transmit pan/tilt/zoom control data back to camera if it contains such a feature. The logic level TXD output data from the video receiver chip is buffered by the transistor which drives the video cable with AC coupled data signal.


0

It looks like the impedance is about 1K-2K through your band of interest. I don't know what you mean by "much more," but 20K input impedance for your preamp should probably be easily achievable, and 200K wouldn't be much of a challenge. Note that your internal resistance is significantly lower--almost negligible--but the impedance will still affect both ...


0

Looks like a tester for large coils in the 5H range to resonate at 50 Hz or so near Q=1 according to my RLC nomograph. What did you want to know?


0

Looking at the block diagram, the input jacks are unbalanced, with the third terminal being a switch contact to the tip. The switch contacts connect to RCA jacks on the rear panel, which will be disconnected when a plug is inserted in the front jack. The tip contact is the "V" at the end of the top line of the jack, and the switch contact is the upward-...


-1

They are mono, look on the user manual page 3, there is a block diagram showing how the mixer works. Besides, normally mixer inputs are mono.


0

The input impedance is frequency dependent because of C4 in your circuit. Add an additional stage on the input without a capacitor in its emitter circuit. R10 & R11 in your circuit have a bearing on the output impedance and they are too high in value. Reduce these values but this will reduce the gain as it will reduce the input impedance of the last ...


0

You're confusing two slightly different things. When you're using an amplifier with a certain output impedance then you can extract the maximum power from it when the load impedance has the same value. Example: a signal generator (like this one, they're used in electronics labs), these have "50 ohm" outputs and only when you load them with 50 ohms will you ...


0

In a power amplifier the designer is typically trying to present a perfect, undistorted voltage signal to the speaker. The output voltage should simply be the input voltage multiplied by the gain of the amplifier. You get this in part by having your source impedance be much lower than the load impedance. Without this buffer there would be some voltage ...


0

Think about it this way. If you had a 12 volt light bulb that has a resistance of 12 ohms and you applied it to a 12 volt power supply, it would produce a certain amount of brightness based on the power consumed. Then, you insert a 12 ohms resistor in series with the bulb and looked at the bulb’s brightness. In which scenario is the bulb the brightest? ...


0

simulate this circuit – Schematic created using CircuitLab Sorry you have to find someone else to make this.


2

The engineer method would be a mux or digital switch or output from the IC8 Synthe Chip on the schematic to a bluetooth audio IC. Proper grounding, decoupled analog and digital power etc. A hacker way would be to tap the IC11 D/A output at TP4 and TP3 to get the audio before the filter IC, and route that to an internal bluetooth audio module. But the ...


1

You can short circuit the input to ground, this gives you a straight line at 0 V. You can position this with the offset knob at a position you like - this will be your zero volt position. Now you can switch the input from ground to a DC coupled input. You will see your waveform, the AC as well as the DC. The DC content is the offset of the middle of your ...


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