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Reverse recovery time quote from Wiki: -
Following the end of forward conduction in a p–n type diode, a reverse
current can flow for a short time. The device does not attain its
blocking capability until the mobile charge in the junction is
depleted.
This is why some diodes are useless for switching circuits or power supplies or for radio frequency ...
5
Yes. Bidirectional ESD/TVS are agnostic to mounting position because either way presents exactly the same electrical characteristics - both terminals are ANODES
4
Bz1 is a self-oscillating warning buzzer, and the circuit around R1, T1 & Bz1 is an over-current warning.
The current from the power supply drawn by the load you connect to it ends up returning through R1. When the current increases sufficiently to cause enough of a voltage-drop across R1, T1 begins to turn on which in turn causes Bz1 to buzz at you.
...
4
In the datasheet of the old 1N4007 we see:
This maximum rating says that these diodes can handle a peak current of up to 30 A during a time of 8.3 ms.
Non-repetitive means that the diode has to "recover" after such a peak. If you repeat such a surge current too soon then the diode may be damaged. What time period is "too soon" and what is "OK" (the diode ...
4
The parallel Schottky diodes are there to allow a higher forward current through the load. Usually done to prevent scenarios where the load current exceeds the maximum current rating of a single diode.
https://www.daenotes.com/electronics/basic-electronics/diode-in-parallel
EDIT: You can refer to the datasheet, page 23 for a similar design example. The ...
3
You can't solve these kinds of problems with a simple equation because the diodes are non-linear.
You need to consider (analyze) four separate cases: both diodes conducting, neither diode conducting, only D1 conducting, and only D2 conducting. Find the currents through the diodes and the voltages across them in all four cases.
If you assumed that a diode ...
3
There are several things wrong with the diagram for this question.
The fact they define a Vz suggests that those are zener diodes, not ordinary diodes. They should be drawn as zeners.
The fact that the correct answer is +/- 12.2v suggests that the zeners should be drawn in anti-series, so that whichever way the output swings, they will start conducting at ...
2
As Bimpelrekkie explained, whether or not the voltage drop of the diodes is significant or not depends on what voltage you are working with.
Since I've been playing with these things for the last few days, I'll post some of the diagrams and measurements I've made.
I think the voltage traces explain fairly what happens to the output when the AC is not at ...
2
Depending on how much semiconductor devices you have studied you may or may not be able to track my description below. But here it is:
In forward bias, the N side injects electrons in the P and the P side injects holes into the N and as a result there is a certain distribution/profile of electrons and holes in the P and N region of the diode respectively. ...
2
You can use MOSFETs, but you must gang two of them source-to-source in each power path so that the body diodes prevent current backflow down the 'inactive' path.
One such example (by no means unique) is the LTC4418, a dual channel prioritized power-path controller. A sample application is shown below:
(Image courtesy of Analog Devices)
The voltage ...
2
Before you drive a current through it, a semiconductor’s population of holes and electrons is (roughly) in balance between thermal excitation that supplies them, and recombination that removes them.
When you drive currents through the semiconductor, those processes are still working. Thermal excitation will continue to create holes, which the current can ...
2
I would not use both a zener and a TVS diode. Seems a little pointless. Just one or the other.
One thing you can do is stick a regular diode in series with the zener. The regular diode should have a much smaller capacitance. With two capacitor in series, the smaller one will dominate similar to two resistors in parallel. The more you stick in series the ...
1
The LM7805 is basically an NPN Emitter Follower with feedback so removing the input makes the regulator high impedance unless you wish to block the ~4mA bias current .
You may verify my assumptions.
1
Obviously you can't ignore the Is temperature dependence or the diode temperature coefficient would be of the opposite sign (let alone the magnitude) of what we know it to be (~-2mV/K).
But basically you have the answer.
Vp = \$ \frac{n KT}{q} (\ln{(\frac {Id1}{Is}}) - \ln{(}{\frac{Id2}{Is}})) = \frac{n KT}{q} \ln{(\frac {Id1}{Id2}}) \$
dVp/dT = \$\...
1
C) 1k, 9.3mA
To get this answer we must assume that the 9V 1mA output and the diode forward current are occurring with opposite input polarities (ie. not at the same time). This is the only way the question makes sense because the circuit cannot put out 9V when the polarity is positive, while the diode can't draw current when it is negative.
First ...
1
Apart from protecting the Base from excessive negative voltage, the diode is necessary to provide equal charge and discharge paths for the coupling capacitor. Without the diode the capacitor will charge up as Base current flows into the transistor when the signal goes high, but will have no discharge path (apart from the Vbe breakdown at ~-7.5V) when it goes ...
1
Depending on duty cycle of the incoming "hard sync", you may need the diode to provide DC restoration.
1
From the look of it D2 is to protect Q1's base emitter junction from excessive reverse voltages. The presence of C2 means the Sync input signal is AC coupled and it eliminates any DC bias, but now the Sync signal will attempt to drive base of Q1 equally positive and negative. In the positive direction the b-e junction will conduct in normal transistor ...
1
D2 protects Q1 from reverse bias on its B-E junction.
A typical small-signal transistor can only withstand a few volts of reverse bias before it breaks down, and an antiparallel diode limits it to about 0.7 V.
R31 limits the current into the transistor when the input is positive, and it also limits the current through the diode when the input is ...
1
To make sense the circuit should have a connection from the top of the capacitor to the input of the Schmitt trigger. To protect the input on power-down from the charge held in the cap you need a diode that connects from the input to Vcc. But you already have a diode there, its the top one of the 2 in series.
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As exciting as these are, @Bimpelrekkie is right. They are useless without the support electronics, and that is radically different for each module.
Just buy a new laser module with known specs and connections.
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If I am not seeing wrong, you used 10k resistors instead of 1k. Not a bad idea, but then R2 and R4 should be 20k. The breadboard is a mix of the 1st and 2nd schematic. I'm sure this is the problem, and also explans why the negative half has slightly less amplitude than the positive one.
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The Vf drop of the diode does have a significant, cumulative effect. Each stage reduces the output voltage by one Vf drop.
In the example below, there are two stages. With a 10Vp-p input, the output voltage is 18.66V instead of 20. The difference is two Vf diode drops of 0.66V each (1.33V total.)
simulate this circuit – Schematic created using ...
1
I do not understand if the voltage drop across the diodes in conduction is negligible or not
It is not a case of "negligible or not", as with many things in electronics: it depends.
The diodes have a certain forward voltage, typically between 0.5 V to 1.0 V depending on how much current flows though the diode (more current => more voltage drop).
If I ...
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