Maker Pro
Maker Pro

Transistor Question

Muhasaresa

Jan 2, 2012
45
Joined
Jan 2, 2012
Messages
45
Hello everyone,

I have a question about transistors. When I first started learning about them in my free time, all the examples showed that a current has to flow through the base in order for the transistor to change states. Yay! I thought I understood....

...but now...when I was playing around on circuit wizard with a 555 circuit, pins 2 and 6 enter a voltage comparator ,which is made out of transistors, but there is only a voltage change but no current flowing through them. Does this mean a transistor changes states with changing voltage and no current?

Thanks a lot,
Muhasaresa
 

(*steve*)

¡sǝpodᴉʇuɐ ǝɥʇ ɹɐǝɥd
Moderator
Jan 21, 2010
25,510
Joined
Jan 21, 2010
Messages
25,510
Well, there are several types of transistors.

Bipolar junction transistors (BJT) come in 2 varieties NPN and PNP. These require a base current to switch on (not to change states)

Field effect transistors (FET) come in a plethora of varieties, but can be split in a similar way into N channel and P channel. These require a voltage to turn on (or turn off) and removing that voltage changes them back to the original state

I think it's possible that you're looking at a circuit using FETs.
 

KrisBlueNZ

Sadly passed away in 2015
Nov 28, 2011
8,393
Joined
Nov 28, 2011
Messages
8,393
Yes, bipolar junction transistors do require SOME base current, but how much current the base will draw depends on the circuit configuration.
The comparators in the 555 are made with bipolar transistors (at least, they were on the original devices), and they will draw a very small amount of current on pins 2 and 6. This current is specified as less than a microamp on each input, which is negligible in normal applications of the device, and is called the "input leakage current". Here is a full explanation.
In common-emitter configuration, which is what you're familiar with, the transistor is normally biased part-way into conduction (analogue circuits) or saturated when base current is present (digital circuits). In both cases, significant base current is involved, as you were taught.
The emitter follower (aka common collector) configuration is described at http://en.wikipedia.org/wiki/Emitter_follower but the text makes it sound much more complicated than it really is. But the first schematic diagram shows the general idea. I'll try to explain it now.
In the emitter follower configuration, the transistor's base is used for input, but the emitter is not grounded directly - it is grounded through a load resistor (or another part of the circuit). The collector is connected to a higher voltage.
Because of the transistor's behaviour, the emitter voltage "follows" the base voltage (with a voltage drop corresponding to Vbe), hence the name "emitter follower". A small amount of base current causes a larger amount of collector current to flow, and because the collector is connected to a higher voltage, the transistor "pulls" its emitter voltage up, so the emitter voltage "follows" the base voltage (with the Vbe drop).
If the emitter load resistor has a high value, the amount of collector current needed to pull the emitter up will be quite low, and the base current is much lower than that (the factor is the transistor's current gain, which is usually over 100 for small signal devices). This means that only a very small base current is needed. This is also described as "the circuit has a high input resistance".
The Darlington configuration is described at http://en.wikipedia.org/wiki/Darlington_transistor (quite a good description this time). It expands on this idea further. Two transistors are connected as cascaded emitter followers, and the combined current gain is typically 10,000 or more. This means that the base (input) current can be extremely low.
A look at the original 555 equivalent schematic (National and Motorola) shows that the comparators in the 555 use Darlington arrangements for their inputs. Pin 6's comparator uses NPN darlingtons, so it will sink a very tiny amount of current, and pin 2's comparator uses PNP darlingtons, so it will source a very tiny amount of current. These darlingtons are used as differential amplifiers ("long-tailed pairs", see http://en.wikipedia.org/wiki/Long-tailed_pair if you're keen) and this yields a very high input resistance and very low input current - specified as under a microamp for each input, even in the early devices.
I hope that was interesting :)
 

Muhasaresa

Jan 2, 2012
45
Joined
Jan 2, 2012
Messages
45
WORDS OF GOLD!!!! IT MUST BE THE TRANSISTOR ORACLE!!!! THANKS SO MUCH!!!

:D:D:D:D:D:D:D:D:D:D:D:D:D:D:D:D:D:D:D:D:D:D:D:D:D:D:D:D:D:D:D:D:D:D:D:D:D:D:D:D
 
Top