New answers tagged diodes
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I think you are going to get a little disappointed with the BAS16 especially as the leakage value will slightly change with the measured voltage at the junction of R1 and R2 and temperature. I'd be more inclined to choose something like the BAS716. It has a typical leakage current of below 2 nA at 125 degC and a guaranteed max value of 50 nA at 125 degC: -
...
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I think it’s very confusing, since all the current are flowing from
high potential to low potential.
The simplest way to imagine why this is the case is to add parasitic junction capacitance to a diode.
When a diode is forward biased this capacitance is charged to 0.7V.
But when you apply reversed biased voltage the discharge capacitor current will ...
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It looks like your graph "is wrong" because, with a forward biased voltage on the diode, the current cannot possibly reverse. This graph is not alone; I see plenty on the internet with this confusing approach. I'm calling it a confusing approach because I'm giving them the benefit of the doubt that some words associated with those graphs may shed light on ...
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First you need to find a relay that rated 24 V/80 A (or higher current) and then wiring should be like this:
simulate this circuit – Schematic created using CircuitLab
When V1 is connected it will energize the relay and your "equipment" get its power from V1 and when disconnected relay will switch to V2.
You probably have some capacitors at the ...
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I guess you are using that classic yellow chinese multimeter. The diode setting is used for testing circuit continuity, i.e., to check whether there are any breaks in the circuit. They are also used to test whether leds are working or not and display the resistance of the closed circuit. This is all. But I need you to edit your question and post pictures of ...
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It is a switch mode regulator. It first turns on to allow current to flow into the coil, and when it turns off, current must keep flowing to output to release the energy to output, so the diode allows it. The less voltage drop over diode, the more efficient the buck converter is.
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Figure 1. Buck converter.
Inductors don't like you to change the current suddenly. When you try large voltages are induced.
In the on-state current flows through the inductor to the load and the filter capacitor.
In the off state there is no current supplied by the PSU. The inductor tries to keep the current flowing and since the right side of the inductor ...
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D1 is not best described as a 'catch' diode. That's a better description of the diode across a relay coil, which 'catches' the high voltage transient when its driver turns off.
In a buck converter, best to call it a 'freewheel' diode.
During the on phase of the IC, the output is taken up to the input voltage. Current flows from left to right in L1, ...
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how would i achieve 80,000 amps
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The simplistic approach that you show will not work for almost any distance between the devices. You will need to apply some more technology to have any chance of getting anything to work similar to what you envision.
For close proximity where you align the transmitter physically close to the receiver you may be able to get this to work by adding an ...
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Additional to the points added by @jusaca already:
The power up sequence of the supplies can be different. one can rise faster than the other. The slower power supply will load the faster power supply
The diodes will also serve the purpose of reverse voltage protection
even if the cases are ideal, the wiring resistance, the internal series resistance of ...
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It would work fine without the diodes in an ideal world, where both voltage sources would really have exactly the same voltage. But in the real world this is not the case - both voltages will differ slightly from each other (e.g. one has 23.99V and the other one 24.01V).
This voltage difference between the sources will result in a compensating current ...
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HP used to sell its 5082_2935 schottky (and others) with 100 picosecond time constants
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It depends on the type of diode used, for example;
The frequency rating of the diode
The material the diode is made of
the type of diode you want to use
rectifier diodes= low freq
shottky diodes= medium freq
RF diodes = high freq
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[The datasheet] does not really say the maximum input voltage the diode can handle.
It really does.
Figure 1. The reverse voltage is specified at 60 V maximum.
So does it mean that theoretically I can use any voltage as long as it above the forward voltage and is within current limits?
If it is only ever biased forward then yes, but then why would you ...
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There can be several good choices. As of now I think the selection you have done fair enough and should work.
The new chosen diode had maximum Reverse voltage of 50 V Rating and it should be okay for a DC motor. If you have a chance, you can choose a diode with higher reverse Vorlage Rating.
The \$V_{GS}\$ of chosen new MOSFET is even better than the ...
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D1 protects the regulator if the input voltage is shorted or there is a heavy load on the input when it is turned off. Otherwise the output capacitor(s) will discharge through the parasitic diode and transistors in the regulator and the current might be high enough to damage the regulator. It does not take the place of a flyback diode across an inductive ...
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Is the “feedback” diode acting as a “fly back” diode if there’s an
inductive load on the output?
No it isn't. See this note from TI in their LM317 regulator data sheet: -
See this from ST: -
Is D2 acting as protection to the rest of the circuit in the case of
reverse voltage coming in at the output?
It would do this.
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Is the “feedback” diode acting as a “fly back” diode if there’s an inductive load on the output? If so does this voltage just dissipate across the diode? Wouldn’t it also go back into the input supply and cause damage?
No
The quote you included explains it well;
“D1 connected across the input to output terminals is for protection and prevents the ...
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I believe the writer there was conflating two issues. D2 would be the likely saviour of the part due to inductive loads, since the current would continue to flow in the same direction and would pull the positive rail below ground. D1 is there to prevent the part from being reverse biased as a result of a sudden loss of input voltage.
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The diode adds a path for reverse current in case when applied voltage to the output is higher than regulator voltage. The reverse current would flow also without that diode, but the LM317 isn't designed for that, so this reverse current would overheat the silicon and burn the LM317, so that's why the reverse diode is added.
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It is there to prevent the input from becoming too much lower of a voltage than the output. If the input gets too low vs the output, the internals can become damaged.
Other regulator types may not be susceptible to being damaged when reverse driven and thus not need a diode.
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I think you will need a higher voltage to supply the LED.
Depending on the LED you are using, the forward voltage is between 4 to 6 volts at 20ma.
This device has a much higher forward voltage drop than a standard LED.
Look on the spec sheet for the Forward Voltage Vf for the exact LED you are using.
Note on the second page that the Forward Voltage vs ...
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In the lower than rated current range, the Rs is inversely proportional to forward current until bulk series Rs dominates .
This really means the Rs is more logarithmic below rated current and more constant above rated current.
answered Feb 19 at 23:37
Tony Stewart Sunnyskyguy EE75
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It is probably expecting you to use a more detailed zener model of the constant voltage model.
This is the model in Zener model in Sedra Smith. The slope is pretty linear so it basically models the Zener as a constant voltage in series with a resistance Rz.
You have the knee current, and the zener voltage and current deeper into the curve. But You seem to ...
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What are the resistor values, and what voltage is applied across the circuits? You can determine the current being used.
(Vs - Vf)/R = I
So say the signals being monitored are 5V, the Vf you measure on a good LED is 2.2V, and the resistor is 51 ohm (5,1 x 10^0), then (5V - 2.2V)/51 ohm = 55mA which seems like a lot.
Digikey has a boatload of 0805 LEDs, ...
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The diodes D1 and D2 will each have a very similar voltage drop across them. Therefore they will cancel each other out, and as D1 is connected to 0V, there will also be 0V below D2.
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That circuit uses a JFET as a very low leakage diode, presumably, but there's no clear requirement for such a function as a similar diode is built into the CMOS gate, and 100K is not a particularly high resistance.
For example this application note shows a diode-connected JFET used where fA matter.
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According to the picture, the 6.8 V is across the series combination of both Zener voltage and the resistor.
With out further context, I can only go ahead with the image. As per the image, the equations derived are right.
It is the representation of Zener diode as a whole. Hence the statement, the Zener diode has 6.8 V across it. When you split Zener ...
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Based on the logo I'd guess it's most probably from a Chinese semiconductor fab called Hanking Electronics.
Their main logo is slightly different, but the H part is a pretty good match. I tried to find some examples of smaller package products from them, but yeaaah. Good luck finding a supplier, I can't. Seems like these days they're all about the MEMS.
...
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Searching for "2020R Diode" shows several options, such as On-Semi MUR2020R in the correct TO-220 package.
This may not be the exact manufacturer, but it will likely be a compatible alternative. There are many series numbers that are made by multiple manufacturers, including unnamed fab houses.
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You don't mention the inductance of your load, but if it's not a drive motor for an aircraft carrier it probably won't store enough power to heat up your zener. Most likely, since you're driving it push-pull, the body diodes will provide flyback protection on their own, but the zener won't hurt anything. In any case, current from your 3.3-27V supply will ...
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You do not need an LED with the exact same current rating. An over-rated LED will work fine.
Furthermore, indicator LEDs are not usually driven with the highest possible current, so it is unlikely that you even need to consider the drive current at all, and almost any LED of the correct color and package will do.
The exact brightness, color, and beam ...
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In practice, a transistor is generally considered "saturated" when increasing or decreasing the base current/gate voltage incrementally will not proportionately affect the collector/drain current. The C-E voltage of a bipolar will usually be about 0.2V under these conditions, or the D-S voltage will be small compared to its limits.
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Yes, that equation assumes Ib=0. We all know that can't be true IRL, but if the transistors aren't in saturation, it's likely that Ib is far lower than the bias currents used in the resistor/diode network. So the estimate isn't a perfect solution in the real world, but it's close enough...and engineering is never about perfection, it's all about being ...
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While researching something for a project of mine, I got the idea to reverse engineer the digispark USB connection for clues, and the D3 diode had me stumped for days, days! But I think I finally figured it out:
What is the purpose of D3?:
D3 is to protect your laptop from the things you're likely to be doing with a crazy cheap prototyping board kind of ...
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This is more of an intuitive answer rather than a very technical one.
Let's look at the following circuit. In layman's terms, here a power source of 5V(connected at the top) is trying to establish a current in the circuit. But a battery connected with opposite polarity(with respect to the source voltage) is trying to oppose that current.
Now consider this ...
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The function of R2 is to regulate the led current, which is very important. However, with 3.8V over the led and min.0.7 V over Q1 only leaves 0.5 V over R2. Not enough to really regulate the led current. When the led heats up, the led voltage will get a little lower, the voltage over R2 the same amount higher and the current quite a lot higher. Thermal ...
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if you have identical diodes, at identical temperatures in identical circuits , which is your circuit, the forward voltage will be 18 millivolts lower than for just one diode --- for silicon, at room temperature.
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For both diodes to be in parallel they must be connected having two common points.
You have different values for current because current differs with each component whereas voltage drop remains the same.
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I wouldn't attempt the diode solution. The commenters are correct in that wiring the two power supplies together like that is likely to cause problems; I've myself seen a PC that won't shut down properly if back-fed 5V through USB from a bad hub.
What I would suggest is connect both the grounds together, but connect only one of the power supplies. That ...
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The problem here is that OP thinks the voltage drop across the diode is VF (about 0.7 V); so, including a voltage source with the same voltage VF, the result should be zero. However, this is valid if sufficient current flows through the diode (e.g., it is a basic requirement for a Zener voltage stabilizer)... and this requires higher supply voltage, which in ...
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Well since no schemantic is provided i will assume you are talking about low power signal to high power signal. And you are talking about amplifier. Basically, you use transistor as a switch. low power signal controlls the switch which allows higher power signal to be conducted.
in case of the picture above transistor will not conduct any electricity.. If ...
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You can also connect a red LED instead of the diode, in the same direction as that of diode. It will give the user a quick visual indication also that he has connected something in reverse.
The forward voltage of red LED is about 2 V. And this will not harm the optocoupler. The heat generated due to the reverse leakage current and the lower forward ...
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The answer is ... #1. It protects the LED in the optoisolator from negative voltages that could applied by the user. From the datasheet:
The Zener diode will, of course, conduct as a diode in the forward direction if a negative voltage is applied, leaving the remainder of the input voltage across the LED until it starts to break down in reverse.
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There is a voltage vs current chart for the diode. Presuming a specific voltage and current level on the chart it can be determined that an infinitesimal change in voltage and current represents an equivalent resistance, therefore in series with the current limiting resistor. You can also guesstimate the led voltage, presume voltage current and power across ...
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I think there is a much simpler "Electronics 1" answer to your question.
The voltage across the diode and the voltage across the resistor must sum to that of the voltage source. Thus if the voltage drop across the diode is its forward voltage drop (which would be the case in a simple ideal diode model - and that's the simplification I'm suggesting you are ...
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Sample Schematic
So please find for your entertainment, analysis of the following circuit:
simulate this circuit – Schematic created using CircuitLab
(Most of the material that follows here can be readily found at this Wikipedia site: diode modeling. I will, however, take a different approach to their closed solution answer.)
Shockley Diode ...
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If you have a graph of current versus voltage for a diode you can draw "load lines" on them to solve your question. Here's one I had created for LEDs running from a 5 V supply. The voltages are higher than for a regular diode but the principle is the same.
Figure 1. The simple circuit.
Figure 2. Current versus forward voltage for a range of LEDs of ...
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The voltage drop on a (real) diode is not fixed, but varies with current, temperature, and perhaps other conditions.
While in your circuit you slowly raise the source voltage, the voltage drop on the diode will also raise: it will be never greater than the source (otherwise you have a generator, not a diode). So a current flows, the resistor drops some ...
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