Happened across this E-book page:
http://www.allaboutcircuits.com/vol_3/chpt_4/2.html

The first Figure (c) is incorrect. The PNP transistor will act as a voltage follower (common collector) rather than a switch. The load (lamp) needs to be between the source negative side and the transistor's collector.

Re-use of figure letters on the same page is not a good idea, as it can lead to confusion.

The 2nd figure (a) and (b) are not correct. The transistor will not saturate; Vce will not go below around 0.63v, or the Vf of the BE junction.

 

This was brought up a long while back. I tend to agree with Wookie on this one.

(c) needs the transistor and the battery flipped.

 

(c) needs the transistor and the battery flipped.

No. No. It won't make any difference.

(c) needs the transistor and lamp flipped, as below.

 

It won't?

I disagree, and it matches the illustration better.

 

I disagree, and it matches the illustration better.

It take me a while to find out what exactly you disagree.

You would like your version of the diagram better, which is just the same version as I posted, albeit drawn differently.

 

This was brought up a long while back. I tend to agree with Wookie on this one.

(c) needs the transistor and the battery flipped.

No. No. It won't make any difference.

You stated this was incorrect. In what way is it wrong?

I redrew it verbatum to the statement you disagreed with.

 

I'm sorry Bill, my mistake.

Its my poor understanding of the word "filpped" that has caused all the trouble.

I had taken it to mean "swapped" but then I realise there is difference between flipped and swapped.

See, I have made the same mistake again above in the reply: quoting myself:

(c) needs the transistor and lamp flipped, as below.

 

No problem, writing text doesn't let nuances that speech does come through.

I think we are all agreed, problem verified.

 

I have updated the image at ibiblio.org by flippping both the pnp and the battery.

 

OK, addressing on the following page:
http://www.allaboutcircuits.com/vol_3/chpt_4/2.html

The 2nd figure (a) and (b) are not correct. The transistor will not saturate; Vce will not go below around 0.63v, or the Vf of the BE junction.

Existing image:

Proposed change:

At that particular stage, it may simply confuse a new reader to introduce the concept of Ib=(Vcc-Vbe)/(Ic/10) for forced beta. However, it's rather incorrect to call a transistor saturated, when the original circuit's Vbe could be reduced by increasing Ib.

 

Completely agree with SgtWookie on this one. The existing circuit never saturates and the transistor ends up wasting far too much power and probably overheating.
It came up in this thread:
http://forum.allaboutcircuits.com/showthread.php?t=36650

 

I think if you look again that problem has been fixed.

Compare the illustrations shown at the beginning of this thread to the current ones.

 

Sheesh, I missed post #10. I think Sarge has nailed it again. Should have been in a different thread, but it is a problem.

 

I just sent a note to Dcrunkilton to ask him to review and update the page.

Hopefully, he will update the schematics and at the same time correct the figure numbers, so that there are not duplicate references to (a) and (b) on the same page.

 

Existing image:

Proposed change:

At that particular stage, it may simply confuse a new reader to introduce the concept of Ib=(Vcc-Vbe)/(Ic/10) for forced beta. However, it's rather incorrect to call a transistor saturated, when the original circuit's Vbe could be reduced by increasing Ib.

Thanks SgtWookie,

I have changed the image at ibiblio (03078.png 03078.eps)

I don't think that changed images at ibiblio get detected (like changed text) . Therefore, we may need to inform Dave or Jjrap of the changed image.

Dennis Crunkilton