# Tag Info

2

The device is made to handle signals which normally operate between GND and V++. Any signal line transient that goes below GND is clamped by the lower diode to a voltage that is equal to the forward voltage drop of the lower diode. So you can see that clamping will be to less than about -1V in that case. (Clamping level really depends upon how much current ...

2

It's a simple matter of definitions. In either direction, there is a voltage above which the diode begins to conduct a large current for a small increase (or decrease in the reverse case) in voltage. The finer details of the current-voltage function in each direction are somewhat different, but as a first order approximation, above a minimum (reverse ...

1

Well, yes this could happen. But See, the Breakdown current is due to minority carriers which are very less than majority carriers. Now firstly if you apply a high voltage the current due to majority carriers will be high. Also, at breakdown majority carriers also along with minority carriers will be generated but the current due to majority carriers grew ...

0

There is no such phenomena as breakdown in forward bias conditions. Reverse breakdown is a sudden condition on reaching a certain reverse voltage. Forward voltage conduction happens continuously - there isn't a magic 0.7V where diodes start conducting - it's a gradual increase in conduction as soon as any positive voltage is applied - a lot of people say ...

1

It's difficult to apply a forward voltage of much more than 0.7V across a diode, because the diode conducts when forward biased! The graph below shows that the current through a diode increases very rapidly (in fact, exponentially) with increasing forward voltage. If the forward voltage across the diode were much more than 0.7V, the current flowing ...

2

I don't understand why people interpret this question as being about BJT, but with incorrect terminology. I see a perfectly well phrased MOSFET question, except that it lacks the term itself. NMOS connected in diode configuration: simulate this circuit – Schematic created using CircuitLab Since Gate and Drain are shorted, the following ...

1

D2 has 3V on its cathode and for it to conduct there has to be more than 3V on its anode. By simple inspection, the cathode voltage cannot be greater than 3V therefore D2 is not conducting. D1 has 1V on it's cathode and because D2 is not conducting we can forget about it as having any significant role in the circuit. Therefore the remaining supply voltage ...

0

I dont get this,why is V=1 V and not 3 Volt? Suppose V was 3 V. Then D1 would be strongly turned on and many amps of current would flow through it. But this current would have to come through the 2 kOhm resistor. Which would thus have a huge drop across it. Since the voltage at V is 3 V minus the drop across the resistor, this is a contradiction, so ...

2

Maybe, but probably not. A key feature of decoupling capacitors is that they are able to quickly supply pulsed currents as close as possible to the chip - without the inherent inductance of copper planes or traces that carry the average power from your power supply to the chip. When you are talking about picofarads decoupling capacitors, you're talking ...

1

I'm not sure if I understand your problem, but if you want to determine which of the four types of diodes you have you can look at the forward voltage (V_f) and reverse voltage when diode starts to conduct at reverse polarity (V_r) and maybe reverse leakage current (I_rl) before that point: Si-Diode: V_f ca. 0.6 V, large V_r, low Irl Ge-Diode: V_f ca. ...

0

It's digital logic, 3 being High and 1 being Low, anything in between won't have much effect. The left side is an OR gate. So its kinda like High or low = 1 or 0 = High High being 3 volts, and low being 2 or 1 volt relatively speaking. In the latter, right side, is like an AND gate, So it's kinda like High and Low = 1 and 0 = 0 Again, High being 3 volts ...

2

If there is no current drawn from $V_{out}$, then by Kirchoff's Current Law, the same current must flow in both the resistor and the Zener diode, and you want that current to be not more than the Zener's maximum rated current, to prevent damage to the Zener diode. If some current is taken from $V_{out}$, then the current through the resistor will indeed ...

1

AFAIK the statement that energy levels actually bend is wrong. We are just extrapolating diagrammatically to show how the two conduction bands and valence bands are linked with a space charge region in between. Lets just consider the case of a pn-junction. First, when the two parts, p and n are not forming a junction, then, you will agree that the fermi ...

2

The MCPH3 package nomenclature is an OnSemi name for the SC-70 package standard: The package specification document provides the above diagram.

1

Given that every reference to MCPH3 as a semiconductor package is in reference to an ON Semiconductor device, I am going to say that the designation is not controlled by any standards body and is instead only used by ON Semiconductor.

Top 50 recent answers are included