# how to think values of components???

Discussion in 'General Electronics Chat' started by indianhits, Aug 22, 2009.

1. ### indianhits Thread Starter Active Member

Jul 25, 2009
86
0
hello guys i want ti know how do you guys think of components values like resistors and capacitors value and all and how do you think or make your own circuit diagrams

and in CE amplifier circuit we use a resistor at the emitter to ground side what is the use if it?????????

Thank you!!!

2. ### beenthere Retired Moderator

Apr 20, 2004
15,808
294
The ability to design a circuit is the result of some time spent studying electronics. Some instruction in a classroom setting is good for a structured approach. With a solid grounding in the basics, it is possible to move on to more complex circuits, and learn how and why they operate as they do. With enough of that, one may start designing circuits.

Like everything else, it takes time and study to master the subject.

3. ### rjenkins AAC Fanatic!

Nov 6, 2005
1,015
69
If you try to put a linear bias on a transistor without an emitter resistor (or some form of feedback stabilisation), what voltage do you use?

The base-emitter junction starts to conduct somewhere round 0.6 - 0.7V and is dependent on temperature. You will not easily get a stable bias.

If you add an emitter resistor, you can bias the base rather higher (say 1V) and the excess 0.3 - 0.4V is dropped across the emitter resistor. For example say the resistor is 100 Ohm, the emitter current range is now 3 -4 mA and the collector current is about the same as it's effectively a series circuit.

Without a capacitor across the emitter resistor, the circuit gain is the ratio of the emitter resistor to the collector resistor. eg. with (100 Ohm emitter and) a 1K collector resistor, which would have 3 to 4 volts across it (assuming a high enough supply voltage), the amp would have a gain of 10.

The amp will also be linear over a larger range as the varying input voltage to the base (from whatever signal you are amplifying) can be larger amplitude without risk of either saturating or completely turning off the transistor.

Overall, it's more stable and more flexible to component and temperature variations.

4. ### rspuzio Active Member

Jan 19, 2009
77
0
In addition to what R. Jenkins pointed out, here is another
aspect of the issue:

With the emitter connected directly to ground, the the output of
the amplifier would vary as the exponential of the input voltage.
For most applications, this is not good --- we want the output of
the amplifier to be proportional to the input, not to the exponential
of the input. To be sure, decreasing the range of the input helps,
but it only goes so far.

Placing a resistor between the emitter and ground goes a long way
to solving this problem --- if you put a big enough resistor between
the emitter and ground, the relation between voltage of the base
and collector current is dominated by the resistor, which behaves
according to the linear Ohm's law with the non-linearity of the
transistor only having a small effect. For some applications, such
as an amplifier in a simple radio, this remaining non-linearity is
small enough that we can safely ignore it. If we are interested in
higher fidelity, then, since the deviation from linearity is small, we
can easily enough make it even smaller by some technique such as
canceling it against an opposite deviation (as in a push-pull design)
or by introducing a bit of negative feedback.

Since you asked about computing values, let's make a
little computation here. The exponential constant for a PN junction
is around 25 mV. Let's say that, as in the example R. Jenkins
gave, the emitter current varies between 3 and 4 mA. Then the
ratio of the currents is 4/3 = 1.33..., we have 25 ln 1.33 = 7.2,
which means that the base-emitter voltage drop only varies by
7.2 mV. By Ohm's law, the voltage across the 100 Ω resistor will
vary by 100 mV for the same range of output currents, so we see
that the resistor is responsible for 86% of the change in base
voltage over that range. I don't want to bore you with more
math, but if you do the algebra with the logarithms, you find
that the plot of current versus voltage differs from a straight
line by much less than a percent. Looking at a plot, you would be
hard pressed to distinguish it from a straight line; for not too
demanding purposes, this will do just fine.

5. ### indianhits Thread Starter Active Member

Jul 25, 2009
86
0
i am having hard time understanding electronics now i am completely dependent on this site cause our college madam is not teaching this stuff properly and i want to master this subject

Thanks guys.Thanks!!!