# Tag Info

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In the example circuit, R1 has virtually no effect on the speed of the flashing, since it is so much lower in value than R2. Therefore, if you replace R1 with the ALS, changing the light level will not change the speed of the flashing. And you can't put the ALS where R2 is, since the circuit relies on current flowing both ways through R2, but the ALS won't ...

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A nice guide for abbreviations used in electronics you can find here Includes some of the appendix A of ANSI/IEEE Std. 991-1986 which is the standard

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If you use level-triggered flip-flops (without a two-phase, non-overlapping clock) then all of the flip-flops will be transparent at the same time. The input to the first flip-flop will propagate all of the way to the output of the last flip-flop whenever the clock input is asserted. By the way, I think it is better to call a level-sensitive storage element ...

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A linear circuit obeys the principles of proportionality and superposition. A linear circuit is an ideal approximation of a real-world physical circuit. All physical circuits have some non-linearity. Linear analysis is useful when the real-world non-linearity is small enough that you get answers that are close enough to your required accuracy. Ideal ...

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For very quick solutions (and under legal issues) I would suggest to use COTS hardware like programmable ISM Band wireless transceiver modules, which you can easily program using, e.g., C. eZ430-RF2500 AT86RF230 etc. You have just to wire the button and the LED to the boards, write a small program and that's it. All the interesting high frequency stuff ...

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A lot depends on whether you are doing a commercial PCB to be manufactured and sold, or a few one-off boards for your own use. If the former, then a ground plan is more important to keep radiated emissions down to meet FCC (Part 15, unintentional radiators), CE and other regulations, although I've seen plenty of commercial products with one or two-sided ...

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Short answer: No, you can't use a resistor to reduce the voltage seen by the switch in your circuit. It's the voltage difference across the component's terminals that matters, not the voltage at the component. In your circuit, when the switch is closed the voltage across it is (close to) zero: When the switch is opened, it will see the full voltage of ...

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Idealy you should always use a ground plane. But for different kind of reason it is not always possible or suitable: 1 and 2 layer(s) PCB: For 1 layer, I think the reason is quite obvious. You can fill the unrouted PCB with copper to GND but that's not a real ground plane. For two layers, you don't always have enough space for routing on one layer and a ...

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First, you need to make sure that your assumptions are valid. For example, if your circuit is connected to two pieces of equipment powered by switch mode supplies, there will be a common-mode high frequency current flowing between those units. The current will flow through your device, generating "noise" voltages on the impedances of the traces in your ...

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The first op amp sets up a reference at Vcc/2 for single supply operation. The resistors on the input limit current to the body and set up a body reference at or near the common mode voltage on the electrodes (using big noisy resistors). The instrumentation amp takes the differential signal, provides a gain of 5, and spits it out as a single-ended output, ...

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I=Vs/(R1+(R2∗Rl)/(R2+Rl)) is correct. VRL=I∗Rl is the wrong part. VRL=I*Req; Req=RL||R2=RLR2/(RL+R2);

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You are right about open circuit, the wires are disconnected. In case of short circuit as well as closed circuit, the wires ARE connected but the difference is that in case of short circuit, the resistance between the connection is extremely low so very high current flows as per ohm's law, whereas in case of close circuit, the connection offers considerable ...

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Your equation for I is correct. The next step is Vo=I*(R2||RL)

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It looks like you're implementing a relatively straightforward ripple-carry adder. It looks like it would probably work, though the wiring is a bit of a rat's nest. An alternative approach also using five DIPs, but without such a rat's nest, would be to use four 74HC153 chips and a quad XOR. The 74HC153 chip contains two four-input multiplexers with ...

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If R$_L$ wasn't present the voltage output would just be: - $V_{S}\times\dfrac{R_2}{R_1 + R_2}$ So, adjust R2 mathematically to account for R$_L$ being in parallel with it. Can you take it from here?

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You can probably use smaller resistor values, or none at all, with the LEDs. Most CMOS gates, operated at 5 V, will not be able to supply 20 mA anyway so there is little risk of damaging the LEDs. However, it is essential that you add a pulldown resistor on the right side of each switch (the side connected to the gates). Otherwise your gate inputs will be ...

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The admittance of this circuit can be written as : Y = $\dfrac{1}{sL} + \dfrac{1}{R + \dfrac{1}{sC}}\ = \dfrac{CLs^2 + CRs + 1}{Ls(CRs + 1)}$. Substituting $s = j\omega$, multiplying the denominator by its complex cojugated and simplifying into real and imaginary parts gives us: $\dfrac{R}{\frac{1}{C^2 \omega^2} + R^2} + ... 1 Multi-layer boards and huge ground planes are your enemies. Your friends are a powerful iron and this strategy: crush, melt, or otherwise break the connector so you can extract each pin individually (if all pins are on the perimeter, preferably snip them free of the plastic body) heat the pin, not the pad, just enough so it can be wiggled free with a ... 4 Usually, there is one pin that is making it hard to remove. That is usually the ground pin, and it will be attached to the ground plane. The ground plane will try to suck away the heat as fast as you can apply it. This is where you need a high wattage iron, so you can get it hot quickly. On a good day, you can remove the solder, then wiggle each pin ... 3 If you've already removed some of the solder, but find the pin is still stuck in the hole. What I've found helps is adding a little extra solder back onto the pin to make sure you get a nice reflow up through the hole and then I use a solder sucker instead of a braid. Also, you might check for any kind of glue under the part. Circuit parts that are ... 9 Add more solder. Use tin-lead solder or special low-temperature solder (ChipQuik). If you can make a solder blob over all five pins you can just lift the jack out. The solder is easy to clean up with braid afterwards. 2 Logic level converter: - This one works with two Vcc levels from 0.8V to 3.6V. 1 You can also use a bi-directional voltage-level translation device. Here are some from ti, analog and maxim. 2 The$s$-domain or frequency domain is not really what is special. It is the Laplace transform that is special. With appropriate assumptions, Laplace transform gives an equivalence between functions in the time domain and those in the frequency domain. Laplace transform is useful because it interchanges the operations of differentiation and ... 1 You could use a RC delay circuit. A simple one is made out of a resistor and a capacitor. For example, charge the capacitor, attached to ground, through a pull-up resistor R1. Then discharge through another resistor R2 and the "long-press" reset button to the ground. The trick is to calculate the values of C, R1 and R2 to make the timings right, so that ... 3 Your question makes little sense. You haven't even specified the type of transistor. For bipolar transistors, it is current, not charge (electrons or holes), that turn on the transistor. A small charge thru the base will allow a larger charge to flow thru the collector, but this is a one-time thing and wouldn't make the transistor be "on", whatever that ... 0 You can calculate the number of electrons required to turn on a discrete MOSFET from the gate charge datasheet numbers. It is not a small number. For example, looking at figure 10 of the 2N7002 datasheet we can see that 0.8nC is about enough. The charge of an electron is$1.6\times 10^{-19}$C, so it takes ~$4 \times 10^9electrons. If you're ... 0 Use a microcontroller that receives the button signal and drives the reset and on/off lines accordingly. Even the tiny PIC 10F200 can do this job easily. It comes in a SOT-23 6-pin package, with one input line and 3 I/O lines. You'll even have a extra I/O line left over. The first thing the micro needs to do is debounce the button. This creates a ... 1 Well, to consider your basic question: what is the minimum number of full adders required? You have initially twelve partial products (bits), while your result has six. Each fully utilized full adder will remove one partial product (bit) and therefore you need exactly12 - 6 = 6 full adders. Then, given that you do not use half adders, there may in practice ... 2 It could also be, more specifically, a norator. I've never used this symbol before, but I saw it in a paper that mentions a nullor equivalent circuit. _Dependent Current Source_ _Independent Current Source_ _____Norator_____ 2 A 60W incandescant light bulb requires 500 mA @ 120 V to light it up to full brightness. You can easily pass 10 mA through it — enough to operate a control module like Lightwave — without causing it to light up at all. This is why such modules are often specified to work only with incandescant (resistive) loads. CFLs and LED bulbs will often ... 0 There are a few misconceptions in your question, and it might help you understand things if we clear those up. For a) and b), replace "there is" with "there may be". The way you have defined an "open circuit" means that there is effectively an open circuit between any two points in a circuit that are not connected together by an ideal wire. When there is not ... 1 simulate this circuit – Schematic created using CircuitLab 0 I may be wrong since I did not study this in English, but I see a major difference between b) and c). From a very practical point of view, a closed circuit is good, a short circuit is very bad ! Short circuit is for instance connecting a wire directly between the poles of a battery or power supply. Whereas a closed circuit is just a "normal" load between ... 1 For (a,b,c) that's more or less correct. In general, there doesn't have to be a voltage/current just because there is a short/open, there just can't be any voltage in a perfect short and there can't be any current in a perfect open. Another way to re-word these two terms is that a short circuit has 0 resistance (R=0), and an open circuit has infinite ... 1 Yes. The component that does this is called a transformer. However, keep in mind that transformers are subject to the law of conservation of energy. This means that power in must equal power out. Electrical power is the product of current and voltage:  P = IV  If power in and power out are equal, then:  \begin{align} P_{in} &= P_{out} \\ I_{in} ... 1 The law of Energy Conservation say that energy must be conserved; the Law of Entropy says there will be some loss. So, if you want a 3333:1 gain in current, you need to see at least a 3333:1 loss in voltage. So, if the input is 10V, the output will be less than 3 mV. In general, passive conversion of current or voltage of an AC signal is done with a ... 1 The formula is derived by practical experiences and not from mathematical 1st principles. It is unprovable other than by being practical and thinking what a diode detector has to achieve. Firstly, the formula states that RC has to be equal to or greater than\dfrac{1}{\omega_c}\$. If the RC time constant were too short there would be significant levels ... 1 I have tried it too, it's due to Q4 connected to Q3, instead connect the emitter of Q3 to the base of Q4. Download the Falstad Java simulator and import this, it will make a simulated circuit showing how it works. (http://www.falstad.com/circuit/) Falstad Code:$ 1 5.0E-6 78.57719942274176 85 5.0 50 R 368 128 368 80 0 0 40.0 9.0 0.0 0.0 0.5 r 368 128 368 ...

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Yes, R1, R2 & R3 are parallel to each other & hence, resultant resistance is 5 ohms. While finding equivalent resistance between any two points, imagine that you're travelling from one point to the other. Let's consider here that we're travelling from A to B. One end of R2 is short-circuited to A & other end is short-circuited to B. Same is the ...

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simulate this circuit – Schematic created using CircuitLab Try redrawing the circuit with the 'A' connections in a bar at the top and the 'B' connections in a bar at the bottom (or use left/right). They've deliberately made it confusing. If you're redrawing, here's a set of guidelines that Olin wrote up: Rules and guidelines for drawing ...

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Yes, you should consider that the initial current is V/2R, as inductors behave as short circuits in the steady state. However, without doing any calculations, there are many things that tell me your answer is incorrect: The dimension for your current is Voltage/Inductance, which is wrong, should be voltage/resistance. The dimension of your exponent is ...

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Although the electron velocity is very low, which is propagated almost instantaneously is the electric field. This causes the effect that all the electrons in the wire to start moving simultaneously (almost).

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An analogy Imagine pressing 1 cm ball-bearings into a 10 m long horizontal pipe (1.1cm internal diameter) which is full of identical ball-bearings. You take 1s to slowly press one new bearing into your end of the pipe That 1 cm ball bearing travelled at 0.01 m/s As you press a new ball-bearing in at your end, another ball bearing gets pushed out the far ...

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Maybe this analogy will help. Imagine you and a friend are holding two ends of a very long rope, and you're holding it tight, without any slack. If you pull on the rope, your friend will feel you pull almost instantly. The rope hardly has to move, and could move quite slowly. Similarly, an electron from a battery doesn't have to move very fast or far for ...

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Imagine one electron drifting slowly from the negative pole of your battery into the wire. At nearly the same moment, an electron from the beginning of the wire drifts to a bit further on in the wire, replacing an electron that is (at nearly the same moment) drifting yet further ..... and at the end of the chain an electron from the end of the wire drifts ...

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IF the system is linear, then you are right, it can be characterized solely from its impulse response. It can be thought of as convolving the input by the impulse response. Often the impulse response is difficult to meaure. One way around this is to measure the step response, which is usually easier, then differentiate that. That may have a lot of noise, ...

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As @Andy aka said, cost is a big issue. There's no reason to use a part that is more expensive or takes up more space than a better alternative. One other thing to consider is EMC. Remember capacitors store energy in the form of electric fields and inductors store energy in the form of magnetic fields. Therefore, if you have a circuit that emits a great ...

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If you want to block dc but let frequencies above 20 Hz through largely unaffected (as in a simple audio amplifier), the time constant of your circuit will need to be about 0.01. Additionally, if you want an input impedance of at least 1 kohm (ball-park for audio) you will find that the inductive or capacitive Reactance at 20 Hz needs to be also 1 kohm: - ...

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For an electronic switch you can use the circuit below : simulate this circuit – Schematic created using CircuitLab R5/R6 create a voltage divider that keeps M2 gate at about 9.5-10v (if the 24v can be much lower you can use a zener instead of the divider). R4 is a pull down resistor keeping the source grounded when there is no 12v supply, ...

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