Hi everyone,

I have some questions regarding transistor as I came across this in a book. Here it goes:

All transistors have maximum collector-current ratings (​

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IC,max), maximum collector-
to-base (BVCBO), collector-to-emitter (BVCEO), and emitter-to-base (VEBO) breakdown
voltages, and maximum collector power dissipation (PD) ratings. If these rating
are exceeded, the transistor may get zapped. One method to safeguard against BVEB​

is to place a diode from the emitter to the base, as shown in Fig. 4.38​

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a. The diode prevents
emitter-to-base conduction whenever the emitter becomes more positive than
the base (e.g., input at base swings negative while emitter is grounded). To avoid
exceeding BVCBO, a diode placed in series with the collector (Fig. 4.38b) can be used to
prevent collector-base conduction from occurring when the base voltage becomes
excessively larger than the collector voltage. To prevent exceeding BVCEO, which may​

be an issue if the collector holds an inductive load, a diode placed in parallel with the

load (see Fig. 4.38​

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c) will go into conduction before a collector-voltage spike, created​

by the inductive load, reaches the breakdown voltage.

My questions are:
1) In the figure 4.38a, how the diode would prevent the conduction from emitter to base whenever emitter becomes more positive? The diode would be in forward bias condition!

2) In Figure 4.38b, how would the diode protect BVCBo since it's in forward bias? (almost same as the case above)

3) How did the diode protect transistor against an induction load (for eg: relay) and when does the spike occur? Appreciate if anyone could explain this by drawing a diagram.

Thanks..

 

I guess you composed your post in a text editor. If you would, please use NotePad or other editor that does not embed font and size codes, because it makes it difficult to read, and very difficult to use quoting in replying.

My questions are:
1) In the figure 4.38a, how the diode would prevent the conduction from emitter to base whenever emitter becomes more positive? The diode would be in forward bias condition!

That's why the diode is there. It prevents Vebo from rising above the Vf of the diode, which is usually much less than the Vebo.

2) In Figure 4.38b, how would the diode protect BVCBo since it's in forward bias? (almost same as the case above)

Got me there.

3) How did the diode protect transistor against an induction load (for eg: relay) and when does the spike occur? Appreciate if anyone could explain this by drawing a diagram.

The spike occurs when current has been flowing through the inductor and collector-emitter junction of the transistor for a period of time, and then suddenly the transistor is switched off. Without the diode, the inductor current would have nowhere to go!

As the magnetic field around the inductor collapses (quite rapidly) the current keeps trying to flow, and the polarity of the voltage across the inductor inverts. When current was flowing through the transistor, it was more positive on top of the inductor, and more negative at the bottom of the inductor.

The voltage at the top is held steady by the power supply, so instead of the top of the inductor becoming very negative, the bottom of the inductor goes positive, very quickly. Without the diode present, it is highly likely that BVceo will be exceeded. Instead, the diode becomes forward biased and provides a current path back through the inductor itself.

This is why they are sometimes referred to as "flywheel" diodes or reverse-EMF diodes; the current gets recirculated around the inductor until power is dissipated in the diode and the inductor itself.

 

Here's a simulation of what I'm talking about; two circuits that are nearly identical. Click on the attached to bring it up. You might have to click on it again to see it full-size in your web browser.

Q1 and L1 - no diode. Q2 and L2 - and a flywheel diode.

Look at the plots on the bottom.
V(Q2C) doesn't even reach 11v, because the current (red plot) gets recirculated through the diode.
V(Q1C) hits around 800V. Had that been a real-world 2N2222, it would've been vaporized.

 

Hi SgtWookie,

Thanks for replying my questions in the effort of providing simulations. Ya, I would definitely composed my post in a text editor if I copy the text directly from pdf file. Pardon my mistake for making it difficult to read.

The spike occurs when current has been flowing through the inductor and collector-emitter junction of the transistor for a period of time, and then suddenly the transistor is switched off. Without the diode, the inductor current would have nowhere to go!

Why would the transistor suddenly switch off? For normal usage, the transistor would not off right? It's just for assumption right?

As the magnetic field around the inductor collapses (quite rapidly) the current keeps trying to flow, and the polarity of the voltage across the inductor inverts. When current was flowing through the transistor, it was more positive on top of the inductor, and more negative at the bottom of the inductor.

What do you mean magnetic field around the inductor collapses? How do this scenario occur?

The voltage at the top is held steady by the power supply, so instead of the top of the inductor becoming very negative, the bottom of the inductor goes positive, very quickly. Without the diode present, it is highly likely that BVceo will be exceeded. Instead, the diode becomes forward biased and provides a current path back through the inductor itself.

Why would the bottom of the inductor goes positive since at the top of the inductor is positive power supply?

Sorry for so many questions from me...

 

Generally when you're driving a relay the transistor itself is being used as a low power switch, which switches the much higher power relay.

Transistors are often used in their digital modes, linear has its uses, but is generally hot (as in wattage).

 

V(Q2C) doesn't even reach 11v, because the current (red plot) gets recirculated through the diode.
V(Q1C) hits around 800V. Had that been a real-world 2N2222, it would've been vaporized.

How would V(Q2C) reaches 11V when the Vcc is only 10V? And why would V(Q1C) hits around 800V out of the blue? Can you please provide some in-depth explanations? Thank you

 

Generally when you're driving a relay the transistor itself is being used as a low power switch, which switches the much higher power relay.

Transistors are often used in their digital modes, linear has its uses, but is generally hot (as in wattage).

What are the consequences of a low power switch driving a high power relay??? Any side effect?

 

BVcbo means collector-to-base breakdown (collector more positive than base) when the emitter is open (unconnected). A diode in series with the collector (as in your schematic) will not protect the transistor in this situation, because the diode will be forward biased.
As you suggested, a diode in series with the collector will protect against excessive base-collector current in the case where the base voltage exceeds the collector voltage by more than 0.7V (this is not BVcbo). This is an unusual situation.

 

Why would the transistor suddenly switch off? For normal usage, the transistor would not off right? It's just for assumption right?

When driving things like relays, solenoids and other such loads, a transistor is generally used as a saturated switch; IE: either ON or OFF. In the simulation, SigIn is a 5v square wave, which is typically what a TTL or other logic device like a microcontroller would output.

What do you mean magnetic field around the inductor collapses? How do this scenario occur?

When current starts flowing through an inductor, a magnetic field builds up around it; when the current source is discontinued, the magnetic field around the inductor collapses.
Review our E-book: http://www.allaboutcircuits.com/vol_1/chpt_15/index.html

Why would the bottom of the inductor goes positive since at the top of the inductor is positive power supply?

It's because of the direction of current flow in the inductor. The collector goes positive due to the direction of current flow. In the simulation on the right (Q2) the collector only goes about 1v more positive than the supply due to the forward voltage of the diode.

 

How would V(Q2C) reaches 11V when the Vcc is only 10V?

The current path through Q2 is turned off, but current is still flowing through the inductor. The only path for current is through the flywheel diode. V(Q2C) increases as a result of the Vf (forward voltage) of the diode when current passes through it.

why would V(Q1C) hits around 800V out of the blue? Can you please provide some in-depth explanations? Thank you

It's due to the current flow through the inductor, and the lack of a path for the current. Had this been a real circuit, the 2N2222 transistor has collector breakdown voltage of around 60v; once the voltage on the collector exceeded that, the transistor would be destroyed.

 

Hold a relay's coil terminals in one hand.
Apply power to the coil and you won't feel anything.
Then disconnect the power to the relay coil and feel the 750V!

You will never forget to connect a protection diode to a relay coil.