# Tag Info

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Another approach is to note that the IV characteristics of the combined diodes is the same as the IV characteristic of $D2$. Hence $v=-15V$, so $D2$ is on and $D1$ is off. Note that the IV characteristic of two components in parallel is given by $(v,i_1+i_2)$ where $(v,i_k)$ is characteristic of the individual components.

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Still, I don't see how it affects the temperature rise since, e.g. in a buck-boost converter, the diode and transistor in the module will never conduct at the same time. Just because two circuit elements are electrically connected in series or parallel doesn't mean their thermal paths are connected the same way. In fact, normally we'd treat the thermal ...

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Here's another bad water analogy. simulate this circuit – Schematic created using CircuitLab Figure 1. I said this was bad. Water is being fed from a +15 m head through a pipe with some resistance to flow. There are two outlets - one at +10 m and one at -15 m. To what level do you think the tank will fill?

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I figured it out with adding diodes and selected another regulator which has enable pin. The circuit is below.

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Diode D3 is wrongly represented in this Digispark schematic diagram and does not comply with the actual hw configuration. In fact the Digispark hardware includes it with reversed polarity, so that the USB is able to provide +5V to the device, while an external power supply will not feed current to the USB (in order to protect external host devices connected ...

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It's a unipolar TVS (Transient Voltage Suppression) diode. The package size and markings may be able to lead you to a replacement. It may be a 3.0SMCJ24A There is a non-zero chance that whatever killed the protective device (generically called a TVS) has killed more devices. If the proprietary microcontroller (MCU) U4 is fried, you're probably going to be ...

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I started this answer on the second question you asked on this subject, and moved it here when the other question was closed. I've tried to make it address both questions. The machines you are looking at use the bremsstrahlung effect to generate short wavelength electromagnetic waves. The X-rays and gamma rays are generated inside the field emission diode ...

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No, because the load current also flows through R2, and that happens all the time — even when VZ is not conducting.

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From this question you can see the typical schematic of a two-speed inexpensive hair dryer. The heater forms a voltage divider to feed the low voltage DC motor (via a bridge rectifier), and you can switch an additional rectifier in series for 'low'. The fuse (FU = 熔断器) and thermostat (ST = 恆溫器) are for obvious safety reasons. If the air flow is obstructed ...

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The drift current is limited by the available charge carriers arriving at the reverse-biased junction. As a higher field draws more carriers, carrier concentration decreases. The resulting dependence of the current on the height of the potential barrier is weakened by the limited availability of charge carriers.

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The resistor is there to control the gate charge current and through that the MOSFET off-to-on time. It's important for the whole bunch of reasons. My personal favorite is that MOSFET going on too fast creates current through the miller capacitance (D to G) and makes itself oscillating (sometimes kills the driver and other circuits, if totally out of control)...

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The heat sink doesn't just dissipate heat. When you connect both devices to the same heat sink, the heat sink will quite effectively carry heat from one device to the other. So, even when one of the devices isn't dissipating much (if any) power, if the other device is dissipating power, it's going to heat up both devices. For an extreme example, consider two ...

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Surely it must depend upon the details of the heatsink. I can think of en example where the volume is identical. Envisage a prismatic heat-sink with one device in each half. The heat-sink would require a certain volume and the devices would attain attain a certain junction temperature. Now split the heat-sink between the two-devices with an infinitesimal cut ...

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A couple of other people have mentioned Schottky diodes, but could I please make one point: NEVER REPLACE A ZENER DIODE BY ONE OF HIGHER VOLTAGE. And yes, I've seen it done in the field.

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Assuming Vf of the diode is neglected when On for simplicity since the type is unspecified, the solution is logical and simple. ￼ For 0< Vin< Vb. R1 and R2 conduct as a simple voltage divider Vout/Vin=R2/(R1+R2) until Vin=Vb. when Vin>Vb the battery and R(D1)+R(Vb) values are low (0) which shunts R1 then. Vout=Vin-Vb and R2 may reduce not ...

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But how can I know which of 2 diodes will turn on first as V_in increases from -inf? Well there are tools that can help such as a simulator but I guess you might not be that keen on simulation so, ask yourself, what maximum positive voltage can be at the input that doesn't cause current to flow through D1. No current in D1 means you can analyse stuff ...

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The compressor will draw 10x it’s rated current on startup, so your supply may be over stressed.,If not add a 10% rating dummy load such as a bulb then it will be more stable. But the duration will be much less than a single 50,60Hz cycle if rated for 400A single surge current on a 15A diode. So Answer=No problem Your problem is lack of specs on power of ...

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For silicon diodes there no downside generally to buying a higher PIV rated device within reason, up to 400V anyway. Schottky diodes tend to have a higher Vf if they are rated for more voltage so there is a downside (they run hotter and you lose some voltage drop). Can it pass 10A continuously? I would say you need to have a fairly good heat sink on those ...

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Yes, you can substitute a higher spec part for a lower spec one for a case like this. The second question, however, is a bit trickier. The data sheet you referenced has this graph: So as you can see, the part's ability to pass the rated forward current depends on the ambient air temperature.

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Old question but I wanted to close this with an answer. I wound up placing a relay in series with the driver output. It was acceptable to wait for a bit in my application. I therefore turned off the LED driver and opened the relay contacts. I then proceeded to measure the voltage on the output of the LED driver and waited until it discharged down to a ...

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In your first picture, you have two resistors, Rs and Rp. These are their reference designators. Their values are, by default, an undefined R, which needs to be changed into something that will evaluate to a numeric value. In your case, it looks like you defined some values, named after the reference designators, Rp and Rs, but you have not assigned them. ...

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simulate this circuit – Schematic created using CircuitLab Figure 1. Two options. The diodes sketched in your diagram will work. Figure 1a shows them drawn a little more clearly. Whichever way the motor is running the current will be diverted to the power supply - if it can take it. Figure 1b uses two Zener diodes back-to-back to allow for forward and ...

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There is no need for a zener diode, a standard general purpose diode is good enough. A flyback diode (https://en.wikipedia.org/wiki/Flyback_diode) that short the back EMF (voltage) that can harm your driver electronic circuit.

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The single coil latching relay circuit should be as shown. Interlocking the push button switches would prevent dual-supply short-circuit, should both switches be actuated simultaneously. An RC snubber across the relay coil would suppress voltage transients during turn-off.

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This is not a diode, but a tantalum capacitor. When powered up, there should be about 3.3 volts across it. To build off of what I replied to one of your earlier posts, try removing the capacitor and powering the device up. Decoupling capacitors might not be required for it to function. Replace it with any moderate capacitance cap if it still doesn't work. ...

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6.5 digits are enough to see some otherwise subtle effects: Unintentional thermocouples between different metals in your setup. You do touch them by fingers, don't you? You fingers are hotter than the environment and the couples gradually cool down for a while. A semiconductor voltage references drifting away because of self-heating and even aging. Load ...

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It is called a voltage multiplier, or more specifically a voltage doubler.

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This is basically a "voltage doubler" circuit. There are many variations of this but as you can see, that's basically what you get across Vc1 to Vc2. It's 2X V1's peak.

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The first half-cycle charges C1 so Vc1 goes positive while C2 is still uncharged. The second half-cycle charges C2 so that Vc2 goes negative. Since Vc1 is still positive the difference between the two is doubled. The circuit is called "a voltage doubler". simulate this circuit – Schematic created using CircuitLab Figure 1. A more usual way ...

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It's a voltage doubler circuit. The output voltage (ignoring diode drops) is double the peak voltage of the AC input. Since the capacitors charge on alternate half-cycles, the total voltage goes up in a somewhat lumpy fashion. The output capacitor C3 causes the output to rise more slowly and in lumps than if you had two larger caps in series. At the first ...

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Note 9 on page 12 of the manual states: Specifications are for the voltage measured at the input terminals. The 1 mA test current is typical. Variation in the current source will create some variation in the voltage drop across a diode junction. Figure 1. The meters in question. Image taken from the manual. I think you're seeing two issues: While the ...

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The maximum load current is when then zener no longer has sufficient current to regulate. Since we don't have any information about the Zener diode itself, we can assume it to be perfect so it stops regulating when the zener current drops to zero. That's actually a reasonable assumption for an 8V zener. The worst-case conditions on the low side are minimum ...

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It's a precision half wave rectifier. It can be considerably improved by connecting another diode with cathode to the op-amp output and anode to the inverting input. This stops the output slewing towards the negative rail when the input is positive. Note that the output impedance is not symmetrical. It's low for positive going output but 10k otherwise. This ...

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An explicit model to be used for the diode is missing in the original question, however the diode parameters provided ($V_d$ and $R_d$) point towards using a piece-wise linear model for the diode. Therefore, your circuit should be redrawn as follows. simulate this circuit – Schematic created using CircuitLab Assuming $V_{out} > V_d$, the ...

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This is a full-wave precision rectifier. When a negative voltage is applied to the input it acts as an inverting amplifier, so Vout = -Vin. When a positive voltage is applied, the op-amp rails near the negative supply rail and the positive voltage finds it's way to the output, Vout = +Vin (if loading is minimal on the output). The op-amp output has to slew ...

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The first equation is correct. The 0.7V comes from the 2nd zener diode which is forwards biased so becomes a standard diode. The other diode is in reverse zener breakdown (becaues Vi >> Vz) so the total voltage across the two diodes is Vz + 0.7. Now that you have the voltage on both sides of the resistor, the current through it is that voltage ...

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I suppose it could be simulated but where’s the fun in that? I think this isn’t so much a clamper as a level shifter. Since there’s no explicit ground in this circuit let’s arbitrarily call the bottom node ground, or at least our zero voltage reference point. At t(0) the input is zero, and Vout is -2.7V (diode plus battery). C1 is now holding a 2.7V charge. *...

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The way I would do this is to observe that there is 100$\Omega$ to -1V and 100$\Omega$ to +1V, which means the Thevenin equivalent at the Vo point is 50$\Omega$ to GND. There is 1A flowing through D2 because of the current source so the voltage at Vo will be -50V and the voltage at the current source will be -50.7V. Anything (except an open circuit) in ...

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Your intuition is correct, however you assumed that the voltage drop on the current source I1 is zero, which is incorrect. It is like to assume that a current flowing through voltage souce is zero. On the contrary - you can imagine that a current source generates whatever voltage is needed in order to cause the current to reach the specified value. Lets ...

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because diode requires a minimum of 0.7V to conduct Not true. diodes conduct current all the way down to virtually zero volts applied across them: - Picture from this very good article I think that should explain all your questions.

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Why is the positive charge not distributed in the entire N type region (as there are free electrons for conduction in the N type region)? It is, but its normally cancelled out by the presence of an equal number of electrons. In a semiconductor there are 4 main sources of charge. Electrons and holes are mobile charge carriers, and ionized donors and ...

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The base-collector junction will act as a forward biased diode, so if the zener voltage is less than 5V (the unloaded potential of the resistive divider), the base will be at the zener voltage, and the collector about 0.7V less. Thus, for a very light load, and a Zener below 5V, there will actually be some voltage regulation at the output of this utterly ...

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To give you an idea of the iterative method let's assume a value of 4K ohms for R1 And let's make an initial guess of 0.3V across each diode totalling 0.6V Given 1V - 0.6V = 0.4V drop across 4K ohms will provide a current of 0.1mA check using the Diode model the corresponding forward voltage (x2 for two diodes) and get a new estimate for the voltage across ...

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No, you're asked to find the output voltage within a tolerance of 15mV. See above. Initial assumption doesn't matter too much, it saves iterations if you are closer. If you assume a current (it will be fairly low in this case, maybe uA for a silicon diode), then calculate Vo, then you'll have a new current based on (1V-Vo)/R1. Calculate Vo again, repeat ...

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For relay controls, you generally associate ORs with parallel wiring and ANDS with series wiring. So can you simply wire the GPIO and the USB power supply in parallel to the relay coil? Probably not. They are separate voltage sources and mingling them will have unpredictable unintended results. With relay controls, a simple answer is to add interposing ...

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It is there to protect IC's that require the 12V at all times that the 5V is present (possibly older memory ICs). If the 12V drops faster than the 5V supply on power down, you could have 5V still when the 12V bias is gone. This insures that if the 12V drops faster, it will not 'over-run' the 5V as they approach 0V. Some chips can be destroyed if the 12V is ...

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there's probably a part in there there that will be broken if the 12V input is significantly less than the 5V input. the diode prevents this.

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Those are all the same* diode, just in different package. B330 in SMC B330A in SMA B330B in SMB B330C in SMC * The only difference I've noticed so far (in Diodes Inc's datasheets) is a lower peak forward surge current (I_FSM) of 80A for the B330A,B and B330C compared to the 100A rating of the B330 (nothing). I wonder why there is a difference in the ...

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Depending on the accuracy needed, you can assume the forward voltage drop across a diode is 0.7 V and in reverse bias it is an open circuit. Increased accuracy would require iterating with the diode equation for a more exact solution.

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VCC = 5 V, RL = 100 Ω, IL = 5 mA Read what it says; load current is 5 mA. This means that the load voltage is 0.5 volts and not 5 volts. 20 mA is not present in this formula. If you want high speed you sacrifice output voltage level because the load resistance has to be naturally smaller for high-speed: -

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