# Tag Info

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You are asked to find the operating point such that the small-signal resistance is 5 ohms, i.e. $$\frac{dv_d}{di_d} = 5\,[\Omega]$$ You can obtain this by both sides of the given equation relating voltage and current, then rearranging as needed. The most straightforward way is to rearrange the equation above to a transconductance equation: \begin{align}\...

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It sounds like you are half-wave rectifying the mains when you should be full-wave rectifying the mains. There are thousands of articles on this topic on the Internet. Figure 1. Half-wave rectification. Image source: Electronics-Tutorials. It also sounds as though you shouldn't be doing this work on 240 V circuits with such poor understanding of the ...

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The diode "forward voltage" (or forward voltage with ohmic resistance) is a simplified heuristic model that assumes that the diode's voltage is nearly constant with forward current, but this model breaks down (is not accurate) when dealing with interactions near that voltage threshold. The schockley diode equation is a better model of the diode's V-...

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This is not a correct schematic for ideal diode rectifier. The feedback must go to the negative input of opamp. The negative feedback will eliminate the voltage drop of the diode D2. This is an "ideal diode" single phase rectifier. I've added a test point at the opamp out and smaller input voltage for better understanding.

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It is a gas discharge tube (GDT) in a glass package. In some power supplies they are shunting one of the common mode choke coils, so this is not an unique design. It makes somewhat sense that it seems to shunt high voltage transients so that they bypass the common mode choke neutral coil.

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My speculative answer: You wouldn't put a unidirectional device across an input CM winding for clamp suppression. It would conduct every half-cycle, after all. A bidirectional TVS or a gas-tube would work for clamping - in both cases, should sufficient voltage be present to cause breakdown, the TVS or gas tube would clamp the voltage. (This sort of event is ...

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The name of such circuit is a diode clamp or diode limiter. Assuming the dioded are ideal, they do nothing until they get forward biased, and they only get forward biased if input voltage exceeds the range of power supplies, i.e. goes above positive supply, or below negative supply. It limits the voltages at the op-amp input to be within supply voltages to ...

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The curve is much more useful in the reverse breakdown mode. Especially for avalanche diodes, which every "Zener" diode over above 5V actually is. simulate this circuit – Schematic created using CircuitLab Here I have swept the current from 100uA to 100mA on each type of "regulator" diode. Compare the regulation of each: ...

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all answers to your 3 questions are the same: That's up to the model you're studying here, not universally fixed. We can't tell you how the capacitance depends on the frequency in your model, because we don't know much about that model. To be exact, we only know that this is some model you've found, it applies to a forward-biased diode with a small AC signal ...

3

Consider this diode (about 0.7V Vf) in parallel with an LED (about 2V Vf). Of course the actual voltage across each will be the same. Vf, as we are talking about it here, is not the real forward voltage except in very particular circumstances, it's an approximate voltage that you would measure when some particular (sensible in the context) current flows ...

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The Shockley equation with minor modifications is what is used in SPICE simulations for forward current in silicon, Schottky, and LED diodes. The parameter list is given as (from here): The main difference between the different types (other than the number such as Is which is quite different for an LED) is the temperature dependence of the saturation ...

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For ideal components, the circuit is described by ill defined mathematical equations. The ideal diode when reverse biased will present infinite resistance. The ideal voltmeter also has infinite resistance. So the voltage division formula will be $V_{\text{voltmeter}} = \frac{\infty}{\infty + \infty}$. For realistic components, the volt meter would read ...

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The Shockley diode equation is a model for p-n junctions. One can derive such equation without ever mentioning the name 'Silicon', because most of the semiconductor physics equations used are true for both direct band gap semicondutors and indirect ones. As a model, it works as long as you keep the model within the assumptions you make. The equation you ...

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It appears to be a 1SMB51AT3 TVS diode rated for 51V with a current rating of 7.3 amperes. The housing size and shape match, as well as the position of the codes as seen in the linked datasheet.

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Assuming used pulls or something like that, if you put your multimeter on diode range and test the outputs and inputs relative to ground and you see diode junctions there, it's "probably" good. If they are brand new, they're probably good, period. A more thorough test, even a simple functional test, is going to require building up a circuit similar ...

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None of those values make sense. Red is usually around 1.8V, green is a little over 2 V. You can't measure the forward voltage of an LED with just a voltmeter. You have to connect your LED to a current source and measure the voltage across the diode when current is flowing through it. Many multimeters have a diode test function that does exactly what I've ...

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It is an automatic switch to turn the LED off during charging (and allow most of the charge current to flow into the battery). When the mains is connected and the cathodes of the 1N4007s go negative, the base of the BJT is pulled to approximately 0V so the LED turns off, and most of the current flows into the battery, except for the current through the 520\\$\...

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Diodes will not work and can damage your fan. What you need is 3x SPST relays with 240VAC coils and rated to switch your fan motor load. The relays have contacts isolated from the coils so they prevent "backflow". In the schematic below, SW1..3 represent the individual pyroelectric sensor units, each switching a single cubicle lamp. One relay coil ...

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If you want to understand diodes in forward conduction look at the current that flows through them with different forward voltages applied. Such as the 1N4148: - Above picture taken from here. They are quite nonlinear because with 0.74 volts applied you get a conduction of 10 mA. With 0.62 volts applied they only pass 1 mA. That's roughly a 10:1 drop in ...

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Are there two components of diffusion current (valence band AND conduction band) contributing to the generation of the PN junction voltage? Yes If so, do those components stay exclusively within their respective bands across the junction and throughout? Electrons and holes that diffuse initially stay within their respective bands. Electrons in the ...

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You can do that, but it stops being a RC low-pass filter when you do: When you say "R1 and C1 form an RC low-pass filter", that means you assume the output signal is the voltage over C1, which means that whatever "sees" that voltage has a high input impedance and doesn't actually draw significant current from the RC circuit. That means in ...

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You should put one in though it is not as critical in this case since the a manual switch is a lot hardier to voltage spikes than a transistor which is often the first thing to blow from the spike. Even so, having one will prevent the switch from arcing and wearing it out. You should put one in anyways just so the spike doesn't affect other things connected ...

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It is a 36V +/-5% 500mW Zener diode.

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As has been mentioned, the schematic should be changed so that the feedback is to the negative terminal of the op amp rather than the positive terminal. Here is my updated circuit: SPICE simulation of the circuit gives data for the following plot: I used the following model for a mostly ideal op amp: * Mostly ideal op amp * in+ in- vcc vdd out .subckt ...

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You can simplify the circuit and get better resilience to spikes with just one resistor in series with the LEDs that are ant-parallel. No diodes required. A bicolor two-terminal LED will work directly as they have the LEDs in anti-parallel. A capacitor across the diodes would reduce the effect of spikes even further. simulate this circuit – Schematic ...

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This would be a better setup: simulate this circuit – Schematic created using CircuitLab None of the diodes have to block anything over a few volts (LEDs are normally specified at -5V max, though they can typically withstand quite a bit more, at least for a while, though there is long-term damage possible). C1 is optional, but a 0.01uF/50V ceramic ...

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You use the 1 mA column if you are intending to run your zener at a current of around 1 mA. You use the 5 mA column if you are intending to run your zener at around 5 mA. For the 12 V BZX84 diode in the link you supplied, the typical values are 50 Ω at 1 mA and 10 Ω at 5 mA. That's a huge performance improvement, at the cost of power consumption, if you run ...

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Marcus has given a very general answer, which is completely correct. But I think we an actually give some useful answers for the specific model you've presented. How is capacitance of the diode dependent on the frequency of source? Ideally, it shouldn't. Practically, the model might fail at very high frequencies due to the inductance that Andy mentioned in ...

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In addition to Andy's answer: See notes (2) and (3), you may need to protect the inputs by adding a series resistors (RC filter) to limit the input current below 5mA.

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