New answers tagged

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Assuming a purely resistive load. Put your left hand on the the positive terminal and your right hand on the negative terminal of an 800-volt Porsche Taycan battery. Now switch hands. The current flow through your now fibrillating heart went in different directions but the power, either (volts)(amps) or (-volts)(-amps) was the same. We can conclude that in ...


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It is true that if you assume that all of the diodes are reverse-biased then their anodes must be at 5V. However, that would imply that the forward voltage across the bottom diode is 4V. We usually assume that any normal diode would in fact be forward-biased with 4V across it. So, what we conclude is that the assumption that all diodes are reverse-biased can ...


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You are misreading what it's about. Imagine you have a battery of 9 volts connected to two series resistors of 5 Ω and 4 Ω. Clearly there will be an amp flowing (a side issue) but, of relevance, we can say this: - $$\text{9 volts} = V(R1) + V(R2)$$ Now, if you rearrange that formula into this: - $$\text{9 volts} - V(R1) - V(R2) = 0$$ You have to respect the ...


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If so, why is the diode resistance in parallel with the current source? My intuition tells me it should be in series A resistor in series with an ideal current source still produces exactly the same current through its load regardless of the voltage across the terminals. So its behavior is exactly the same as the ideal current source with no resistor. ...


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You can use the below simple circuit, which uses one diode and one resistor for each General-Purpose Input (GPI) pin on your MCU. When the switch is on, LOAD pulls D1's cathode well above 3.1 V and reverse-biases it. GPI is pulled logic HIGH by R1. When the switch is on, D1 conducts and clamps R1 and GPI down at approx. 0.2 V. GPI is a good logic LOW. The ...


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Question : <<< I'm trying to find the output voltage using the superposition theorem for this circuit configuration when RG is connected. >>> For helping a little ... here is the literal result for output ... (E&OE). Can be perhaps simplified. If all resistors = 10 kOhm, result is : \$Vo = -4 * VB + 4 * VA\$ Simulation. Two cases. <...


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Assuming the 12V load pulls some current when S1 is closed, then you can use a current sensing amplifier to detect the closed state. Choosing a large gain like the INS212 is good since you only want to saturate the output to V+ (3.3V) and detect when current is passing through Rs sense resistor. You need to calculate the Rs value such that it meets the Vih ...


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The NE5532 needs a MINIMUM supply of 9v. It is a very good and very low noise op amp but the lowest output is about 2v from ground and highest output is about 2v from the positive power rail. Even with 9v, you'll only have ~4v peak-to-peak. If you really need a 5v supply, you might want to look for a rail-to-rail op amp.


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I have experience with electrical design, but never worked on any motorcycles in that regard. So take this answer with a grain of salt. Also, what more would you like to see on the dash? Your MT09 is nice and modern. If you don't like the way the dash looks they make aftermarket ones, which will be cheaper and faster to get than what you are trying here. You ...


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From what I can see from your pictures, in the first two pictures you are measuring the voltage across resistors, and if you have a current through them then you surely get a voltage. But, in the last picture I can see that you are measuring voltage across a wire, and since a wire doesn't have a resistance (at least not in low frequencies) you won't read a ...


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Best thing? Thermal camera if you can find one, the components that are using up the most power (and usually the most current) will heat up the most. It's especially useful for finding shorts or other issues with one singular component that is bringing down the whole power rail.


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Often distributed high currents come from oscillations in CMOS. Using a short 10:1 probe as a loop antenna, you can sniff around to verify if this is the case, then determine is supply ripple is the cause then what controls that ripple , the unexpected load from ESD damage or the source caps. Comparing with a KGB (known good board ) helps alot.


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The difference in power consumption is about a watt (assuming 5V power). That's enough to get a reasonable size component noticeably warm (or perhaps remove a bit of skin if it's a small part such as an SOT-23 or SOIC-8). So look for charring or smell with the power off. If safe and feasible you might be able to probe the circuit with a finger you don't need ...


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I do not know how you are measuring "resistance," but if doing so with a multimeter (while circuit powered off), then you cannot measure the resistance of an IC (or rather, it has nothing to do with how it behaves once powered.) In your place, I would look for any hot components in your bad circuit. Be careful, depending on the faulty component and ...


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What I tried so far is, to compare the resistance of the ICs in both of them to notice if there is any difference. Won't work. Putting a resistor in front of your broken device will then highlight the resistor as the broken thing. Also, you're assuming that your measurement device is even capable of correctly assessing the voltage across a point. That is ...


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The device in the video does not literally deliver an electrical shock. It is merely a vibrator (a DC motor with an eccentric mass attached to the shaft) wired in series with a N.C. microswitch contact and a battery. It’s possible to make a very simple electromechanical circuit to literally shock people, but I don’t think that’s very funny so I won’t bother ...


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Some 3 phases breakers and overload relays don´t work properly if one phase is not connected (unbalanced current), so this is a usual way to be sure that current is balanced and the device will trip at setup value. Also used in DC current to reduce arcing. Depends on size, some 2 phase breakers are difficult to find or are more expensive than same current ...


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There is a type of circuit breaker known as a 'double break'. It has a rotary contact that disconnects the circuit at two points simultaneously. This is to help reduce contact arcing and give greater isolation than a single set of contacts. What they might have been trying to do here is get the same sort of functionality by using two sets of breaker contacts ...


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Metal detectors which detect metals and other conductive materials near the coil are most often based on the fact that AC current in the coil (or as well pulses) generate a changing magnetic field. The changing magnetic field induces eddy currents to the conductive piece. Eddy currents also generate a magnetic field which invariably cancels a part of the ...


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Just think about what I've drawn below your picture: - I'm not explicitly telling you how to look at this; I'm giving a hint: Is the current through the lower left hand resistor affected by either of the two resistors above it? Here's another hint; when you see a circuit that might confuse you; redraw it.


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Forget you ever heard that KVL and KCL were useful for solving problems like this. They're just ways to state the obvious. KVL says the sum of voltage changes around a loop is zero. Of course, it is. otherwise, the voltage at a node could have two or more values. KCL says the sum of the currents entering a node is zero. Again, of course. Otherwise, there ...


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To become a proficient fault-finder on electronic circuits it is largely about studying, over a lengthy period of time, to learn about how individual electronic components perform and about how they can be interconnected to implement the many circuit building blocks typically utilised in practical circuits. Fault-finding is then often about studying the ...


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Not really an answer, but not enough room in 'comments' to post a long reply. Please know I don't mean this to be flippant..... Troubleshooting electronic stuff well isn't something you can learn from reading a few articles. It's not like working with mechanics, where you can SEE what is happening (usually) and things work as per your intuition. (if the ...


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Sorry for my late answer (I was travelling for 2 days): I have calculated a Thevenin solution with the given values and for a fixed termination Rphi only: In this case, the Thevenin voltage source has the value Vth=[10/(4+2kOhms/Rphi)] Volts with a series resistor Rth=Rphi||0.5kOhms. Example: Assuming Rphi=2kOhms we have Vth=2V and Rth=0.4kOhms. The ...


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Is this a sensible analysis, or am I missing something here? You are no longer missing anything and your waveform looks nearly correct (more below). Always look for negative feedback when assuming an op-amp circuit. For comparator circuits either no feedback is used or a little bit of positive feedback. The positive feedback is useful to avoid the output ...


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Ignore all the number and look. consider the extremes. Vin at GND and VDD. what happens ? start deviation from the extreme..How long until something happens? make a sketch of 2+3. fill in the numbers.


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Try this : the output is used to charge and discharge the capacitor.


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When any amplifier saturates the gain is zero. When analyzing the small-signal performance of a comparator, we assume it is not saturated and thus operates in the linear region, which is rather difficult as it is not unity gain compensated to do so but may perform well in an open loop with low offset. This means comparators are not designed for linear ...


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The comparator isn't 'active' when the inputs are more than a few mV different in voltage, it's saturated to one rail or the other. The important parameters, propagation delay and input offset, are a function of how the comparator amplifier behaves when in the linear region.


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Yes, that appears to be the right technique. You removed (open circuited the current sources) and correctly split the remaining driving current by the correct ratio. In other words, when removing the 6 amp source you are left with 24 amps flowing into a 4 Ω resistor that is in parallel with 2 Ω in series with 6 Ω. One-third of the current (24 amps) will ...


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I think choosing the fuse, you should account current which flows during your 0.75 seconds ON period. Effectively assuming your sine wave is there all the time. So, I think Normal Operating Current = Vrms/R. Again during this 0.75 seconds. RMS or Peak to Peak voltage was discussed thoroughly here: Does the rating of an AC fuse point to the amplitude or rms ...


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Seeing the forest behind the trees After these comprehensive but specific answers, I will try to summarize what has been done and why it has been done in this way (ie, I will try to guess what were the intentions of the author when they assembled the circuit). My personal belief is that without this concluding "philosophical" part, we cannot claim ...


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When the precision rectifier's input goes negative, its output goes positive. When the precision rectifier's input goes positive its output also goes positive by the same amplitude. So during the positive input half cycle, when the precision rectifier's output also goes positive there is virtually no voltage drop across either R1 or R2. This is why a buffer ...


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You should analyze it in two conditions, due to the diode non-linearity. Using the same value for both resistors, since the current though them is the same (ideal op. amp.) so is the voltage across them. In the negative input semi-cycle, Vout rises, the diode conducts and, due to the negative feedback, you have the inverter you are expecting. But in the ...


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This is called a precision rectifier. Imagine the SIG going negative by 1V, the opamp wants to make the voltage at the inverting input the same as at the non-inverting input, which is 0V, so it needs to output a voltage that will cancel out that -1V taking the feedback components into account. So assuming R1=R2 it will need to make the output -(-1V)+Vd, or ...


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I'm having issues with the first amplifier. Although it seems very simple (just an inverter), can't see clearly how the first diode rectifies the negative part of the SIG signal. It's an inverting amplifier and, the op-amp tries to keep the voltage at its inverting input to be the same as that on the non-inverting input. This is all about negative feedback. ...


0

The transfer function is as follows:Vout = (((Vin/R1)/K1)xK2)xR2) Due to equal K factors and identical resistors of 51kOhm, the transfer function is simply the voltage divider output. Vout = Vin x (7500/7500+499000+499000) Therefore, BRG_COPTO = 0.00746 x ACREF


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So seemingly the shape of the denominator of the transfer function H(s) depends strongly o[n] the input signal. No. Transfer functions are useful for linear circuits, which satisfy linear differential equations, at least within the sphere of electrical engineering. So, pretty much by assumption, when the transfer function of a circuit is referred to, it is ...


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Another way of looking at the problem. simulate this circuit – Schematic created using CircuitLab Because the voltage is the same across each resistor, current will flow through each resistor. Thus, the total current in the above example is: $$ I_T = {V \over R1} + {V \over R2} + {V \over R3} $$ Using Ohm's law, we can find \$R_T\$. $$ R_T = {V \over ...


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It's a lot easier to relate to parallel resistors if you flip the units. Meet conductance, which is the inverse of resistance. The unit is siemens, and is 1 / ohms. Say our resistors are 2, 4 and 6 ohms. What are their conductances? 0.5, 0.25 and 0.16667 siemens. When resistors are in parallel, conductance simply adds. 0.5 + 0.25 + 0.166667 siemens = 0....


2

If you turn all of the faucets in your house on at the same time, does water only flow out of the faucet with the easiest path for the water? Or does it instead flow out all of them at varying rates depending on the resistance to the water flow (pipe size, faucet size, bends in the line, etc)? Electricity acts very similar. Water pressure -> voltage water ...


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simulate this circuit – Schematic created using CircuitLab Figure 1. Three parallel paths. Consider three roads or water pipes or corridors (or wires). R2 has a resistance half that of R3 so twice as much traffic / water / people (or current) can flow through it. R1 has one third of the resistance of R3 so three times as much traffic / water / people (...


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Step by step: Combine C29 and R49+C37 by parallelism (A is the new impedance name) Then you can use a PI-T transform between A, R54 and C41 Parallelism combination between C33 and (C49+R65+R69) PI-T transform again between (R61, R53, C45) In the end calculate the gain between Vin and V1, by Vout I mean the point 1, then you will have the potential in 1 ...


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If you want a rapid check of your maths when calculating parallel resistors, the following graphic method works: So, given a base of any length, obviously greater than zero and for accuracy I chose a value around half the value of the sum of the two resistors, draw 1 resistor with the number of squares to match its value; scale as you wish. Do the same for ...


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There are two fundamental electrical properties; Conductors (with resistance) Insulators aka dielectrics (with leakage R) and capacitance between conductors. Capacitors will also conduct according the rate of change in voltage applied. Ic=CdV/dt THe path of least resistance will conduct the most, but not all of the current. That's more accurate. Charged ...


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Current doesn't all go down the one path of least resistance, it divides among all paths, inversely proportionally to the resistance of each path. Therefore resistors in parallel will have lower resistance than the lowest value parallel resistor. This goes for resistors in parallel with a wire as well. A wire is just a resistor with a very small resistance....


2

Hints Try redrawing it like this: - This would be the first stage reduction of the problem by redrawing and shifting things around. The next stage would involve combing the remaining 50 volt source with the new 20 volt source - again this is a significant reduction in order to make the problem easier to solve. However, if you can't follow what I've done ...


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As a general rule of thumb: there is always a transition between "on" and "off". For your solar cell that means: no light -> no voltage , a lot of light -> maximum voltage, maximum current. You can try this yourself, if you equip yourself with a cheap multimeter and use this as a replacement to the motor in both: current mode and ...


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"Partially cloudy sky" and especially "Indoor light" are not a good recipe of testing a PV module. It may sound somewhat surprizing, but human eyes are not very good at estimating a total amount of light even to an order of magnitude. In a sunny day one may get as much as 1 kW of light per sq. meter. Clouds may lower it to as low as 10W ...


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