# Tag Info

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Thanks for all the info. My Multi meter is a Greenlee DM-810A so it will read both diode and ohms. And yes I have checked others and they do show OL in one direction and a value in the other. Also, when I put the meter into the Ohms setting this diode shows 000.8 either way. Sounds like a dead diode.

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That is the correct formulae for an abrupt junction, just be sure to use the right temperature for the thermal voltage $V_T=\dfrac{KT}{q}$ and the right intrinsic carrier concentration $n_i$ at that temperature.

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As noted - diode test ranges usually inject a small current. Readings are usually mV drop but no guarantee. 1N5819 is 1A Schottky. should read around 300 mV forwards and open circuit in revere direction. If you have another diode see if it shows a differential reading when swapped. If meter has a low Ohm range (200 Ohm max often) the you can usually ...

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Usually, the multimeter will give you a reading in only one direction: the forward biased direction. Some expensive meters can do reverse biased voltages, i.e. zener diodes, but most multimeters will only do one direction. So, you'd have to connect the positive lead to the anode and the negative lead to the cathode. Then the meter will give you the ...

1

What is show depends on your meter and your diode. For a silicon diode like 1N.... I would expect the meter to read about 0.300V (for Schottky) to 0.700V for a regular type in forward biased direction and OL (overload, infinity) in the reverse biased direction. Sounds like the diode is bad, you should check the other ones too.

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We can't know what your multimeter does when set to "diode". Unless you explain exactly what the meter does on that mode, there is no way to know what 0000 means or what the reading should be for each direction. However, you should certainly get some assymetry when measuring a diode. On the Ohms scale, the resistance reading should be significantly higher ...

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No, the two circuits are not the same, even after ignoring the dead short of the input in the second circuit. Both circuits will produce about the same output for a valid input voltage. The difference is the load presented to whatever is driving the input. In the top circuit, the transistor is used in what is called the emitter follower configuration. ...

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LED Strip Basics As you might be aware, these LED strips come as parallel groups or 3 series LEDs with one series resistor. Connecting 12V to the main connectors is all it takes to light them up. They can be cut apart, but only in groups of three at the appropriate markings on the strips. The embedded resistor value varies for different types of strips (LED ...

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The circuit idea is fine and so is keeping the ground/power plane away from the protection device. If you are able to define a component shape/footprint/decal in eagle then, define an SMD component with two pins that are connected by a solid lump of copper. The copper is part of the component footprint. Thus when you route your connections, there will be ...

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KVL rules here. The zener branch (R1 and D1) will be at the same voltage as the capacitor branch (D2, R2, and C1), which will be 9V. If you want the zener to limit the cap voltage, you need to move D2 to the midpoint junction of R1 and D1. simulate this circuit – Schematic created using CircuitLab Don't forget that D2 will eat a few hundred ...

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Putting the 2.2V zener in parallel with the capacitor does not cut into the charging efficiency at all — the zener doesn't conduct at all until the capacitor is fully charged. But once it does conduct, it needs to absorb all of the energy coming from the dynamo, so make sure it's suitably rated.

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I would do this: simulate this circuit – Schematic created using CircuitLab Or this: simulate this circuit By putting two capacitors in series you double your cost, energy capacity, and maximum voltage. You could use a reverse-biased zener for the diodes also, but using a forward-biased LED also gives you a visual indication of "fully ...

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You can use a buck regulator to convert higher voltages to regulated lower voltages quite efficiently but I question the need. It's a wind up torch and I don't think it's a big-deal to waste a few joules of energy by using an inefficient mechanism. However, if you can't find a solution that adequately (albeit inefficiently) protects and charges your 10F ...

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The external 5V source via R3 produces about 0.6V on D1's anode. Ignore the input signal for now. The 0.6V level on D1 is transferred, via R1 to the anode of D2. Because the cathode of D2 is connected to 0V via the 10k resistor, D2 is on the verge of conducting - this is where you need it to be for a semi-decent precision half wave rectification of a ...

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The $R_1$, $R_3$ and $D_1$ circuit basically creates a 0.6V bias on the other side of the capacitor, so that a positive swing in the signal does not have to overcome a 0.6V hurdle. $D_1$ and $R_3$ form a shunt voltage regulator. The 0.6V voltage is conveyed to $D_2$ which is on the verge of conducting, as a result. So only a small positive ...

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You need something that can handle 5A at least, if it's a 55W bulb! Also, there will be about a volt of drop across those diodes. Also, you need at least 50V rating to handle load dumps when the starter is involved! Digi-key seems to indicate 10A1CT is the cheapest through-hole that matches the specs: ...

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CircuitLab solves the circuit because it doesn't simulate effects like junction temperatures reaching beyond the limit, so that semiconductors melt. A diode is not a fixed voltage drop. Current through a diode is related to voltage by an exponential equation. That exponential equation goes on forever: for any imaginable voltage, you can find a current. ...

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The circuit as shown is not viable - or you could analyze it in two phases, if you must: Phase 1: Each 1n4148 diode is rated for 200 mA continuous, 450 mA peak repetitive current. When wired as indicated, each diode will drop approximately 1 to 1.5 Volts (Fig.3 in datasheet) before the current exceeds absolute maximum rating As the supply voltage is 5 ...

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Use a p-channel MOSFET for reverse protection. It is much more efficient, does not drain a lot of heat, and less power is wasted. Here is a YouTube video talking about it's efficiency, and following is a picture of a schematic. Diodes aren't horrible though,and they are pretty efficient compared to the old selenium rectifiers or vacuum tubes or crystal ...

0

Perhaps I don't fully understand; but couldn't you simply place a single diode in series with the battery? The graph below is the voltage vs. current relationship of a diode. Diodes don't "Breakdown" until the 70-100 volt range. So even if you were to accidentally reverse the voltage with 6 volts, the current through the diode would be very close to zero. ...

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It's never recommended to parallel diodes unless they're on the same die (i.e. a double diode inside a single TO-220 or TO-247 package) since you can never guarantee that they'll share the current. A better option is a MOSFET with an 'ideal diode' OR-ing controller. These devices detect body diode conduction, then turn the MOSFET on so that you don't burn ...

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"Shorting out" something is layperson's description of a large number of electrical/electronics failures. It rarely is accurate. What more likely happened is that you did not have a current limiting resistor or there was one but its value was too small and as a result, too much current went through the diode and turned it into a very short-lived LED.

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I came up with the following circuit that does what you ask: 5 volts from the "SWITCH" source will turn on M1 and M2, connecting the 3 volt rail to the load, and pulling current out of the right hand transistor of the differential pair Q3 and Q4. This turns off Q2 and shuts down the 5 volt rail. When Q1 turns off by applying 0 volts to "SWITCH", Q4 is ...

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Measure currents around the 3V MOSFET M2 when the switch is closed. I have a feeling that 5V on its drain is forward biassing the body diode and limiting VOUT to 3.6V (or some value dictated by M6's ON resistance. You may need a pair of MOSFETs there, in series, (usually with their sources connected together) to provide isolation in both directions. (Edit: ...

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