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It looks to me like a reed switch (I can't be really sure though because the image is not clear enough).
A reed switch is an electrical switch operated (usually) by an applied magnetic field. A common example of a reed switch application is to detect the opening of a door or windows, when used as a proximity switch for a security alarm.
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Since you seem to not be too familiar with control systems, electronics, and wiring things, I recommend a simple threshold detector that controls a relay. The relay then switches the heating element on or off.
Unless the IR thermometer has a output signal for this purpose, it won't be useful as the sensing element. However, there are various devices that ...
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Anisotropic etching is a linear process, like welding, so presumably the units would be distance per time. Unlike welding it is at a small scale. See this slideshare. In this case units are Angstroms per minute. One Angstrom is $10^{-10}\ \textrm{m}$, indicating the process is on an atomic scale. Since it appears to be used for integrated circuit fabrication,...
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Firstly, by temporal I think you mean spatial. Secondly if you consider acceleration, the units are m/s^2. Seconds-squared are not spatial, but they are temporal. But they needn't be either. Consider E=mc^2. c is the speed of light.
Your formula comes from these two formulas:
P = I V
In other words, the power consumed by a device is equal to the voltage ...
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R2 and RL are in parallel so the voltage across them will be the same so can be treated as a single resistor when thinking about the voltage across them.
So
Req(lower) = (1/R2 + 1/100kohms)^-1
Req(upper) = (1/R2 + 1/200kohms)^-1
Vs/(R1 + Req) = I
Hopefully this gets your started, you're looking to produce a system of 2 equations that you can use to ...
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The simple answer is that you can get off the shelf modules which does exactly what you want.
They are often called 'split charge modules' and are designed for installations like boats and motor-homes which have batteries charged off an engine alternator but prevent the starter battery form being fully discharged.
This is one example http://www....
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The capacitors serve two distinct functions
They serve as "reservoir" capacitors. If the regulator is somewhat
remote from the energy source then sudden current peaks will cause
$V_{in}$ to dip due to V = I x R drop in the wiring. $C_{in}$ serves
to provide current for such peaks. $C_{out}$ acts similarly. It is
charged to Vout_regulator in normal operation ...
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My understanding is that when you have a circuit with that's connect to a battery they has to be some resistor in the circuit of some kind to ensure the current on the wire is small enough to power the device (e.g a bulb).
The bulb is the resistor! The power supply (battery) will provide the power the load requires subject to the battery's capability limits....
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In my experience getting data from a rotating component is tricky without resorting to some kind of radio transmitter or a passing the data through a slip ring, which would probably require amplification of the data signal due to the noise generated by the slip ring... it can be done though.
I'd be cautious about using an IR thermometer for anything other ...
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My input is DC from a small DC motor which I use as a generator.
My input current is 300 milliamps and I need 1200 milliamps as output.
Is there some way to increase the output to 1200 milliamps ?
With out using external energy input you can transform voltage or current but you can not increase the power level. As power = voltage x current, that means,...
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If it was a microwave oven then all metals can damage the oven if they are placed in the oven & the oven is operating.
If the oven is is thermal oven that is heated by an electric element, gas or wood, then there will no issues with using steel or stainless steel items in such ovens,
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One good place is to start is LTL-307EE datasheet. From the datasheet take look at current profile for the device. Below is the forward current and voltage characteristics.
Also take look at absolute maximum rating for the device.
Now you have to size R1 resistor such that no more than in this case 30mA of current flows through the LED. R1 is called a ...
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The issue is certainly NOT that the chassis ground is at -100V, it's absurd. The chassis ground by definition is at equal potential for power supply and load, you can treat it like one big wire that's large enough to take all the current you could throw into it. If you put a voltmeter between a car's chassis & earth, I promise it will never read -12V.
...
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CircuitWorks doesn't simulate the electronics; it just integrates between some common PCB design software formats and SolidWorks for mechanical information. For example, maximum component height, location of heat sinks and mounting holes, keep-out areas, etc.
It sounds like what you want is a plug-in that drives the mechanical constraints in SolidWorks (...
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IBM/Lenovo ThinkPad line is basically the only manufacturer/brand of mobility devices that includes circuitry and software that allows one to vastly prolong battery life of laptops used as stationary workstations by making sure to never charge the battery prior to it being discharged to below 96%, and having options to start charging only when the battery ...
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You start by making A-B an open circuit. This implies that the current entering or leaving node A from the load is 0 Amps. This implies that the current in the two 1 $\Omega$ resistors have the same current flowing through them.
The open circuit across A-B also implies that 0 current is flowing into or out of Node B. This means 0 current is flowing ...
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You are right in that $V_x$ can't be 10 V. Fortunately, 10V for $V_x$ simplifies the circuit considerably, making it particularly easy to analyze.
Since $V_x$ is the same voltage as the supply on the left, both ends of the 5/11 Ω resistor are at the same voltage. That means the current thru it is zero. That means the current thru anything directly ...
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First off, neural networks are only very rarely done in hardware, as actual summation of electric signal. Much more often they are just programs that run on standard computers, simulating a network with numerical inputs and outputs, and when you want superior speed and size, you implement it in hardware of FPGA, but still with numerical input and output. ...
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If you trying to measure voltage, connect the black connector probe (Terminal) to COM. Move the rotatory switch 20 DCV. The dot should point to 20 DCV. Then place the black exposed probe to negative and red expose probe to positive button on the battery. You should read somewhere around 1.5V for the D size battery.
Checkout some of the links below. This ...
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I'd use a video camera and attach a striped pattern to form a moire pattern which is easy to see in the video. Here's a link: [http://www.unappel.ch/people/emin-gabrielyan/public/070804-multi-ring-moire-indicator/]
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$R_{23} = {\frac{1}{\frac{1}{R_2}+\frac{1}{R_3}}}$ Resistors in parallel
$R_{23} \approx 9.018KΩ$ Plugging in the values
$V_{out} = V_{in} \cdot {\frac {R_{23}}{R_{1} + R_{23}}}$ Voltage division
$V_{out} \approx 0.9895 \cdot V_{in}$ Plugging in the values
$k \approx 0.9895$
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Partial answer. More later - maybe!
Figure 1. OP's circuit.
There are a number of problems with your circuit.
Your op-amp VCC+ is fed through a 5.6 MΩ resistor. The datasheet says that the quiescent current, ICC, is 0.7 mA. From Ohm's Law we can see that you would need a supply of V = IR = 0.7m × 5.6M = 3920 V just to power up the op-amp. With a 12 V ...
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Without a sketch your setup is difficult to visualise. A typical industrial solution would be to use an inductive proximity switch. These are non-contact, fast switching, water-proof, etc.
Figure 1. A selection of inductive proximity switches.
You seem to suggest that the switch and circuit will rotate. This means that you need to consider those forces ...
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To solve this issue, you need to be familiar with Euler's formula:
$$\sin(wt) = \frac{e^{jwt}-e^{-jwt}}{2j}$$
You definitely started on the right track with this issue, only finding the right value for $K$ and $K^*$ Might be hard. So lets continue from there:
$$K\left(s+\frac{1}{2}\sqrt{2}-\frac{1}{2}\sqrt{2}j\right)+K^*\left(s+\frac{1}{2}\sqrt{2}+\frac{1}{2}...
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The final value theorem is for a signal, not a transfer function.
Use the transfer function to express the output signal
$$ V_{\mathrm{out}}(s) = \frac{1}{RCs+1} V_{\mathrm{in}}(s),$$
with input $V_{\mathrm{in}}(s)$. Now, I assume that your input signal is a step-function
$$ v_{\mathrm{in}}(t) = \begin{cases}0, \; \mathrm{for} \; t < 0 \\ 10, \; \mathrm{...
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Try using a replacing a 9V VDC voltage source with a VAC voltage source. Pay attention to the frequency of the VAC voltage source. This should address the issue.
With regard to determining the time value for a particular voltage, review PROTEUS DESIGN
SUITE Getting Started Guide. In the section VSM tutorial (Graphs) read around page 230.
You might have ...
answered Dec 22 '20 at 13:12
Mahendra Gunawardena
7,0755 gold badges23 silver badges62 bronze badges
2
Try this.
Figure 1. A simple snubber indicator.
Pressing SW1 will pass a current through L1. D1 will be reverse biased and LAMP1 will remain off.
Releasing SW1 will cause L1's top terminal to switch to a negative voltage, D1 will become forward biased and current will flow through the lamp.
You can estimate the pulse duration time constant from $ \tau = ...
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If the switch is ON then the OR gate output will turn on. The NOT gate will not affect the OR gate.
If the switch is turned OFF (having been on) then OR will turn off which turns NOT on which turns OR off again and the cycle repeats.
The speed of oscillation will be determined by the propagation delay of the gates.
Would ... the output end up being analog?
...
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