# Tag Info

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You are correct. at t=0-, Vout=0, and there is no current, because there no voltage supply. At t=0+ (juuuuuust after you turn the power on), the capacitor acts like a short, so all of the current passes into the capacitor as it starts charging. As the voltage level in the capacitor builds up, a potential across the R1 is formed, and you get a current ...

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Exactly correct. You understand it perfectly. If you then turn off the power supply, the capacitor would be discharging through the lower 1MΩ resistor, following the dotted blue line shown here:

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The current direction on the diagram is supposed to be the right answer The current direction in the diagram is the WRONG answer. If you combine those two resistors into a single 100kohm resistor, current flows from the higher potential to the lower potential and the amount is the potential difference (20V-10V) divided by 100kohm i.e 100 micro amps ...

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First read datasheet of the module, then ask something. Here is datasheet: http://www.hanssummers.com/images/stories/ultimategps3/skm61.pdf

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No, there are no conventions. If putting the switch in the wire connecting battery to ground does the job then so be it. Not all battery-powered circuits are grounded and, those that are, may have positive or negative terminals of the battery grounded. I'm mentioning this because the question assumed a negative ground. If a battery powered circuit has a ...

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one may put switch on any supply line which it is NOT grounded. If it is a floating circuit stray voltage may damage sensitive devices. prior to selection of grounding supply pl. ensure, especially with input instruments, power supply which supply terminal is grounded.. select the same polarityfor grounding your circuit and put the switch on the other ...

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You should place it between the positive terminal and the circuit supply, because if you place it between the negative and your circuit ground, even if the switch is open, you circuit is "live" because if anything else ground the circuit before the switch it will turn on. Another way of saying it, placing it on the positive insulate you circuit from the ...

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It's an organizational thing, basically. It takes a multi-bit bus and splits it out into individual bits. Just a change in representation in the schematic. There is no logical function to a splitter, only interconnect. You would implement it on a breadboard with wires.

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Doublers are possible using digital circuits, the following diagram is one such example. The trouble with it is that it relies on propagation delays in delay chains in order to generate the doubled frequency. This means that you don't get a guaranteed duty cycle, it will vary depending on how long your chain is. The clock will probably also be fairly ...

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Not really. You need some sort of analog circuitry to do this. A PLL is generally the proper solution in this case. The idea is to use a voltage controlled oscillator to generate a new frequency that has a precise mathematical relationship to the reference frequency. There are some other techniques that involve delays (possibly RC circuits or inverter ...

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See the line that dips in the middle of the board that cuts of the two sides as such the resistor thats supposed to be connected to the RED LED is not since there is a gap between it extend the resistor over the gap OR move the LED up and connect a wire from the LED to the voltage rail (i think you connected it to 0v ?)

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It looks like you are misunderstanding the way a breadboard is wired internally. Check out this link for information. Here is an image, taken from that link: One issue you have, as an example, is the resistor connected to your green LED. Both leads are plugged into a single node, and therefore it doesn't act to bring power to your LED (or do anything ...

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Take a look at this section of your first picture: - Red and green LEDs are unconnected.

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simulate this circuit – Schematic created using CircuitLab This is a simple circuit ( which I think will mostly serve the purpose). The rules are simple : The capacitor holds the charge and based on the amount of charge held by the capacitor, you can decide which floor you want to go on. If you want to go on floor 5, press SW1. Entire voltage ...

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An elevator controller is constructed from logic gates, typically in a microcontroller (or in the old days from a set of electromagnetic relays). Using an op-amp as your only active device is possible but impractical because op-amps are intended to handle analog not digital signals. Consider the controller as a state machine that has inputs from these ...

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Here is a picture and description: - Taken from here. Below is a better view of the circuit taken from another website: - From here

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Its very simple, For example: In case of three phase induction motor 1. You can interchange or reverse any of the two input phases of voltage supply.

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You can run multiple 780x style voltage regulators in parallel if you - as you've suggested - connect their outputs via a diode. The voltage drop across the diode will reduce the total output voltage. Fortunately you can compensate for this drop by adding another diode into the ground path of the voltage translator like this: The total output voltage ...

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It's in the Proteus Help file: SCHEMATIC CAPTURE HELP (F1) -> MULTI-SHEET DESIGNS -> Hierarchical Designs See the section "External Modules" - you should be able to define the child sheet as a .MOD file, then use the Make Device command on the parent component and use include the .MOD file there. I've never done it before, but I'll try it now. Sounds like ...

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When you say "no volt out on the motor" do you mean that the LEDs don't light? Or do you have a meter which is telling you no voltage across the motor? If the latter, you need to measure each motor lead with respect to ground, and see what that tells you. If the former, you desperately need to buy a cheap DMM and learn to use it. To begin with, over on the ...

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Inspect visually, check the supplies, check the voltages at the MOSFET gates (with the 10K pull-ups I think they should be okay). Check the input voltages at each op-amp pin including power. The schematic looks okay to me, except those diodes appear incorrect. They should presumably parallel the MOSFETs. This kind of error will generally not be picked up ...

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It's impossible to read the supply voltage on the LM324 but, assuming it is 12 volts i.e. common to the voltage rail on the H bridge, you'll never properly turn-off the top MOSFETs because, the LM324's output can not swing higher than about 10V (on a 12 volt supply). Other than that, there does not appear to be a logic supply on the AND gates.

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If the requirement is 8 bit by three words, it sounds as if the word size is 8 bits (the data), and there are three individual location for that data (the address).(This address range is unusual in practice, but theoretically there is nothing to prevent it). Using your 2 four bit by three word devices, you must 'address' the words (you will have some ...

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It is the same as any other capacitor. If you put DC through it, current will flow, but not for long; charge accumulates on its plates until no more can flow. At the very beginning, it acts as a short circuit. But as the charge on it increases, the voltage across it increases, to the point where no more current flows. So when using a capacitor as a bypass ...

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The watch runs either on special ASIC (if volume is very high), or on MCU (microcontroller), often the supercheap version of MCU which is mask programmable and which uses very low clockrate (less than 1MHz) They do have memory, but often very little of it -- a small MCU might only have 16 bytes of memory (as registers), while ASIC would only have enough ...

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The number of storage locations in a memory chip is 2 raised to the power of the number of address wires. Your 4 bit x 3 word chips therefore contain 2^4 = 16 locations (addresses). You want an 8 bit x 3 word design. This will have 2^8 = 256 addresses. You must provide: 1. A 16-output binary decoder for 4 of your address inputs. 2. Quantity 16 of 4x3 chips. ...

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The ADC will handle +/- 80 mV signal BUT needs the proper power supply to do so. One "naughty" but entirely valid way of getting a small negative supply is to place a diode in the -ve return power supply lead. System ground including all regulators etc are at the true ground = anode = most positive end of the diode and - 0.6V or so for a Si diode is at ...

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if they are labelled 12 volt they will be the type with a built in resistor . so they are fine to put in parallel with no extra resistors needed. Its not unknown for standard 2v or 3.5 v LEDs to get into the wrong packet so I would sacrifice just one first on 12v to prove they are correctly labelled .

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For $V_D=8\rm{V}$, the quiescent current (through the MOSFET) is $I_D=\frac{16\rm{V}-8\rm{V}}{1.2\rm{k\Omega}}=6.67\rm{mA}$. Knowing that the transistor must be in saturation, you use the equation for saturation drain current: $$I_D=\frac{k}{2}(V_{GS}-V_{TH})^2$$ Of course, transconductance parameter $k$ and threshold voltage $V_{TH}$ must be known. ...

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If you rearrange your formula, you get $$(U_m - 2U_T) = I_m (2R_d + R)$$ which looks like $$V = IR$$ where $R$ is the sum of all of the resistors in the loop and $V$ is the voltage across that summed resistor. Kirchoff tells you that the voltage across this resistor is the same as the voltage looking at the rest of the circuit, which is $U_m - ... 2 Yes, they are the same. In a Moore machine, the outputs are associated with the states, while in a Mealy machine, they are associated with the edges. This means that a Mealy machine can often have fewer states than the corresponding Moore machine. In this specific example, your states s2 and s3 could be combined, since they have the same set of output edges ... 3 There are three problems. 1 - The biggest is that you have no excitation. That is, your simulator will provide a perfectly balanced set of operating points, and the circuit will have no reason to start oscillating. Try adding a circuit to provide a brief pulse into the - input just after startup. Adding a noise generator to the input to simulate real op amp ... 1 This problem needs just two nodes, and hence two simultaneous equations, to solve. Since Vout is an output and not a source, and also since we are solving for Vout/Vin, we can consider the node equations at X and Y only, where X is the junction of R5/C1/R1 and Y is the junction of R4/R3/R2. Also note that R5=R3. First, simplify the notation a little: let ... 3 The easy way to solve this is, as has been mentioned in comments, is to add buffers amplifiers in each branch before the potentiometer (R1, R2) output. The hard way, which isn't really all that hard since there end up being 3 nodes, is to write the node equations and solve for Vout/Vin. If the nodes are defined as: Junction of Vin and R5. Junction of ... 0 Here is my guess, this would be a lot of algebra: 3 Ok, let's get started on this. Unfortunately you can't analyze the two circuit halves separately, but I believe there is a fast way to compute at least the poles of your transfer function. First of all you can split R5 and C2 and connect the input to both of them, this is always valid. Now if you could ignore the fact that the two circuits are coupled ... 6 I'll try to discuss the circuit without going too much into all the details, cutting a few corners here and there to make it easier to understand rather than being 100% accurate. Important to realize is that amplification actually consists of two things: voltage gain current gain (often forgotten) I think the black magic is really in the DC bias of the ... 6 None on them are "simple resistors". If they were then you'd find a resistor there instead of a transistor. Every transistor in that circuit is in some way involved in amplifying the signal (although there could be some argument about Q2, since that one is in the feedback loop). No sane circuit designer is going to use a transistor to do the job of a simple ... 0 You said "My PC and softwares i used to read/write data from/to sim-card could't recognize my sim-card reader" What did you expect? It is a very simple terminal circuit using a RS232 interface to send and receive data. There is nothing to recognize, e.g. no reader ID... Another thing you have to consider: the data line is bi-directional, i.e. there is one ... 3 I would start by saying that Vcc is not high enough for the amount a gain you are looking for with an amplitude of 0.010V in your input signal and your gains we have 0.010V x 3 x 30 x 30 = 27V of amplitude for your output signal but yet your Vcc its only 15V. The first 2 stages work because 0.010V x 3 x 30 = 0.9V which in Vcc is able to provide 0 You do not give nearly enough information to answer the question, so here are common problems from website's FAQ http://www.ladyada.net/make/simreader/faq.html" 1. You have the wrong COM port selected (make sure you have this right) 2. Your serial cable is damaged or is the wrong kind (say its a null modem cable, which is not correct) 3. You have a ... 1 In these solution i have shown RL and RC Transient Responses,We saw that the currents and voltages in RL and RC circuits decay exponentially with time, with a characteristic time constant t, when an applied current or voltage is suddenly removed. In general, when an applied current or voltage suddenly changes, the voltages and currents in an RL or RC ... 5 This is to confirm what @David wrote. (+1 to him, by the way.) A terminal that has no internal connection and that can be used as a support for external wiring without disturbing the function of the device, provided that the voltage applied to this terminal (by means of the wiring) does not exceed the highest supply voltage rating of the circuit. JEDEC ... 8 No connect means just that -- don't connect the terminal to anything. They often appear on devices because the manufacturer uses a standard form factor that has more terminals than are needed for that particular device. 2 Do$2^N$first. For$2^N$you need a 4 bit to 1 of 16 decoder. You can make one using 2 74LS138 3 bit to 1 of 8 decoders.$2^{2^N}\\$ can be done by just re-labeling the outputs!

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When you nodal analysis, you have to specify the direction of currents before you even begin. In your example you have specified that the current goes from the right side of your schematic to the left. Not a problem. Conventionally, we think of current going from a positive voltage to a negative voltage. If you look at your circuit, you have a 120V source ...

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did a LTSPice simulation for circuit below D1 at 9mA and D2 at 127uA

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Assuming both diodes have the same forward voltage then only D2 will be forward biased and conducting most of the current. D1 will just have leakage current from .2 volts below its forward bias. Vx = 0.2V - Vforward, assuming no leakage in D1. Schematic below can be used to simulate the dc levels. simulate this circuit – Schematic created ...

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Your original battery of 7.2V might be a 2-cell Litium based cell or a 6-cell NiMh. I don't know. The important thing to remember is that written voltages of the cell packs are "nominal". This is explained in a Texas Instruments application note SNVA533. Here is a picture of the nominal voltage, here called MPV (mid-point voltage) from the application note: ...

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Cars and big rigs have a lot in common but one's a car and the other's a truck. Here's a cut/paste from another forum (elec-tech-online): if i were to design a discrete op amp, it would be very similar to a power amp design, except for the output stage, which would be just two small transistors for the op amp, but as many as 3 pairs of successively ...

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