thermistor cold is high impedance and hot is low impedance?

[D.A.(Tony)Stewart]

So what does it do? it is there for what purpose?

Dan this getting tedious by repetition.

and is likely an inherent part of the epitaxial process of a darlington

 

[Audioguru]

So what does it do? it is there for what purpose?

Repeat, "It is probably just part of the manufacturing process", "is likely an inherent part of the epitaxial process of a darlington".
Then it has no purpose.

 

[PrescottDan]

inherent part of the epitaxial process of a darlington".
Then it has no purpose.

Ok, I was just confused about it, cause i have asked techs and they say they can't figure out what it does or why it's there.

So I guess it has no purpose

 

[mrinalmani]

The diode probably protects the Base-Emitter junction in case of reverse voltage. The B-E junction as a very low reverse breakdown voltage (around 10V) and can easily be destroyed if there's a reverse voltage across it.

 

[Disha Karnataki]

power transistors are always used in switching circuits.
Their switching capacity must be faster.
So let's say base voltage is removed the charges in conduction mode take time to go to their original positions(when no base voltage) this time taken is reverse recovery time. This reverse recovery time should not be more than time taken for the second gate pulse comes @ the base of the transistor.
Hence in order to give a easier path for the conducting electrons when the power transistor is switched off a diode is placed across the transistor
hope this helps..

 

[PrescottDan]

So the diode speeds up the switching on/off time or causes the switching on and off time to be faster of the transistors?

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The Diode changes the duty cycle on/off time of the transistor switching speed?

 

[Audioguru]

So the diode speeds up the switching on/off time or causes the switching on and off time to be faster of the transistors?

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The Diode changes the duty cycle on/off time of the transistor switching speed?

No.
The diode cannot protect the base-emitter junction from damage caused if its base goes more negative than about 5V than the emitter because the diode is not connected to the base.
The diode cannot speed up the switching time.
The diode cannot change the duty cycle.
Because the diode does nothing. The diode is part of the chip and has no spec's.

 

[PrescottDan]

Hence in order to give a easier path for the conducting electrons when the power transistor is switched off a diode is placed across the transistor

so what does this mean?

the diode is used to just bleed off, sink or shunt the voltage when the transistor is turned off?

 

[Audioguru]

so what does this mean?

the diode is used to just bleed off, sink or shunt the voltage when the transistor is turned off?

How can the diode "bleed off, sink or shunt the voltage when the transistor is turned off" when the diode is always reverse-biased so it never conducts and it never does anything?

I think Disha in India knows less about electronics than you do.

 

[PrescottDan]

The Diode is only turned on during the negative cycle of the output, the positive cycle of the output turns the diode off

If there was a Reverse DC voltage on the output of the transistors, the diode would sink the reverse voltage to protect the transistors

 

[Audioguru]

The Diode is only turned on during the negative cycle of the output, the positive cycle of the output turns the diode off

If there was a Reverse DC voltage on the output of the transistors, the diode would sink the reverse voltage to protect the transistors

What would cause the collector of an NPN transistor to go negative?
If the emitter is at 0V and an inductive load is between the collector and a positive supply voltage then the inductor will cause a positive voltage spike at the collector when the transistor turns off and the reverse-biased diode between the collector and emitter does nothing.

Darlington arrays also have this diode on each darlington. Mistsubishi says, "It is a parasitic diode and cannot be used".

 

[mrinalmani]

@ Audioguru
Yes you are right, the diode is not between B-E terminal therefore it does not appear to protect the B-E junction directly.
But when used in an inductive H-Bridge, the protective nature of the diode is apparent at once. Infact had it not been for the internal diodes, external ones would be required, thus adding to cost.

@ Disha
For speeding up the transistor, the diode must be connected between the C-B junction. And also the forward drop of the diode must be less than the forward drop of the transistor junction. You may want to re-check.

 

[PrescottDan]

But when used in an inductive H-Bridge, the protective nature of the diode is apparent at once.

why would u need a protecting diode when using transistors in an H-Bridge? so the positive cycle and negative cycle don't short out?

The Protecting diode protects the transistors during the "idle time/dead time" between the positive and negative cycles or when they are overlapping between positive and negative cycles in an H-bridge or complimentary power amps

What would cause the collector of an NPN transistor to go negative?

If the biasing resistors get shorted or open, which will have no DC offset voltage on the Base

The AC waveform on the input of the base will swing both positive and negative and the collector will also have a positive and negative swing or cycles with no biasing or DC offset

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For speeding up the transistor, the diode must be connected between the C-B junction.

Why does placing a diode between the base to the collector speed up the transistors switching?

 

[mrinalmani]

1. During the dead time (when all switches are OFF) the inductor current must find a path. And the diode provides this path.
2. Placing a schottky diode between the C_B junction provides a path for the stored charges in the junction to discharge externally. Unless the charges are removed, the transistor will not switch off.

 

[FvM]

First necessary comment, the thread discussion is running pretty much out of topic related to the original title. Rather unfriendly to future edaboard visitors seeking for specific help.

Although Audioguru has clarified that the parasitic darlington diode has no specification in any datasheet and is apparently even classified as unusable by manufacturers, I still hear idle speculations that it must serve a purpose or have been built-in intentionally. (A philospher would call this teleogical thinking...).

I agree however with mrinalmani that some bipolar output stages actually need reverse conducting diodes.

It's clearly the case for switching H-bridges with inductive load. And also (Audioguru should know) for audio power amplifiers. Otherwise in case of output overdrive, the energy of a reactive load (e.g. a speaker in resonance or a cross-over) would be absorbed by the reversed base-emitter diode and possibly kill it.

But apparently, you can't rely on the inbuilt darlington diodes for protection purposes and have to supply separate ones.

 

[Audioguru]

We are discussing a darlington NPN switch with its emitter connected to 0V. Its collector never goes negative.

An audio amplifier has its output transistors as emitter-followers so the emitter output of an NPN output transistor can swing above the collector supply voltage when it drives a resonant load which can damage the base-emitter junction without a clamp diode from the output to the positive supply.

In Google I looked at How to Speed Up Transistor Reverse Recovery Time and they showed how the Schottky diode from the collector to the base in a 74LSxx logic IC speeds it up: https://www.vk6fh.com/vk6fh/schottky diode.htm
Guess what the "S" stands for? https://en.wikipedia.org/wiki/7400_series