5
\$\begingroup\$

I am reading the book Electrical Engineering 101.

When talking about NPN and PNP transistors, the book said this:

Use an NPN to switch a ground leg and a PNP to switch a Vcc leg. This might seem odd to you at first. After all, they are both like a switch, right? Well, they are like a switch, but the diode drop in the base causes an important difference, especially when you only have 0 to 5 V to deal with.

And below are the example pictures quoted from the book showing different designs with different robustness. But I cannot figure it out why.

Scenario 1 enter image description here

Scenario 2 enter image description here

Add 1

Thanks to Photon's reply, I found this reference: Active Mode of BJT.

There're 3 modes of BJT:

  • cut-off
  • active
  • saturated

If working in active mode, I think the voltage over Load is heavily affected by the Vin in ALL 4 pictures.

If working in saturated mode, I think the BJT are just switches and the voltage over Load are no different in each scenario respectively.

So how come the robustness difference?

\$\endgroup\$
  • \$\begingroup\$ How do you define "robustness"? Against which influence? \$\endgroup\$ – LvW Jun 18 '16 at 14:57
  • \$\begingroup\$ More or less robust in what context? \$\endgroup\$ – Peter Smith Jun 18 '16 at 14:58
  • \$\begingroup\$ What means robust / who says that? \$\endgroup\$ – KarlKarlsom Jun 18 '16 at 15:00
  • \$\begingroup\$ Sorry for missing some context. I just updated the question. \$\endgroup\$ – smwikipedia Jun 18 '16 at 15:02
  • 1
    \$\begingroup\$ Your edit does not change the question. (But your inferences about the different operating modes are mainly correct) \$\endgroup\$ – The Photon Jun 18 '16 at 15:24
3
\$\begingroup\$

The two circuits on the left are emitter followers. The BJTs operate in forward active mode (assuming \$V_{in}<V_{cc}\$). The two circuits on the right are common-emitter switching circuits. The BJTs may operate in saturated mode.

When operating in saturated mode, the power consumption of the BJT can be substantially lower than when in active mode. This leads to less self-heating of the BJT, and probably a longer operating life.

Note my use of "may" and "can". This all depends on the resistor values being chosen appropriately for the application.

\$\endgroup\$
1
\$\begingroup\$

The two circuits on the left are the emitter followers. As the name suggest their response actually much depends on what is happening on their emitters.

Imagine the two circuits on the left are used as the switch to control the load. As the load current getting higher and higher, there will be higher voltage drop at the emitter thus effectively reduce the base drive of the transistors and makes them less conductive. This is some sort of negative feedback or self current regulation.

The "less robust" statement about those two circuits probably describe the ability of the transistors to act as an ideal switch in those configuration.

Ideal switch should exhibit very low resistance under all load circumstances, obviously the two left circuit won't do that.

\$\endgroup\$
  • \$\begingroup\$ If I understand you correctly, I have a feeling that to make it more robust, we should try to make \$V_{BE}\$ un-affectable as possible. Though there's still a resistance between \$V_{in}\$ and \$B\$, but the small current \$I_{B}\$ make its effect trivial. \$\endgroup\$ – smwikipedia Jul 6 '16 at 8:00
1
\$\begingroup\$

In scenario 1, take the PNP on the left:

To fully turn the switch on, \$V_{in}\$ would need to go below ground to get at least 0.6V \$V_{BE}\$ which would require some source of negative voltage.

[Update]

In response to the comment for the PNP in scenario 1, it takes at least 0.6V to turn on a transistor base-emitter junction; with this scenario, the lowest the emitter (the output) could go to is about 0.6V for 0V in.

Circuit:

PNP non optimal switch

Response:

PNP response at 0V in

The NPN on the right would need perhaps a volt at \$V_{in}\$ (depends on resistor value) so it would be easy to operate as a switch in a 3.3V or 5V application.

For scenario 2, with the NPN on the left, \$V_{in}\$ would need to go above Vcc by at least 0.6V to apply the full value of Vcc to the load, necessitating a separate (or higher at least) supply.

On the right, the PNPn could take \$V_{in}\$ to ground to fully operate the switch (assuming that \$V_{in}\$ goes to Vcc for the off case).

So as switches, the solutions on the right are easier to fully implement, which could be called more robust perhaps.

\$\endgroup\$
  • \$\begingroup\$ Thanks. In your second paragraph about scenario 1 left pic, why \$V_{in}\$ has to go below ground? Why not just go below the \$V_{cc}\$ by 0.6V? I guess that will provide enough forward voltage drop between the \$EB\$ \$\endgroup\$ – smwikipedia Jun 18 '16 at 15:52
  • \$\begingroup\$ But in scenario 1 left, the emitter is at \$V_{cc}\$, not ground. So I think \$V_{in}\$ just needs to be \$V_{cc} - 0.6\$. Why have to go below ground? I think if we set \$V_{in}\$ = \$V_{cc}\$, the switch is off. Then if we turn the \$V_{in}\$ a bit lower, the switch will be on. \$\endgroup\$ – smwikipedia Jun 18 '16 at 16:07
0
\$\begingroup\$

The 2 circuits on the right are called common emitter. By adding a emitter resistor here as well you can increase the linearity and stability (while decreasing the gain).

The 2 circuits on the left are called common collector.

I made simple table for you to compare the properties of both circuits.

enter image description here

\$\endgroup\$

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.