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# electronic Transistor bias current, what is?

Hello everyone,
I'm studying my note on electronic transistor and op amp. I've doubt and very very unclear of
1- what is bias current?
2- What is the power dissipate in a transistor?

This is not a homework but i dont quite clear of these two at all.
Thank you :)
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sokhodom
4 Solutions

Commented:
A bias current is one that prefers one type of voltage over another.  This is not electrically correct, but it happens a lot, especially with older circuits.
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Commented:
Everyone's a comedian.

In a typical amplifier circuit you'd like it to amplify AC voltages, such as speech or music.  But transistors are single-polarity devices, they don't like being fed AC.    So you have to "bias" the transistor.  for example let's say you have a transistor that likes to work with between +5 and +10 volts, and you want it to amplifiy a 2 volt peak-to-peak AC signal.

You choose up some resistors to bias the transistor, so when no signal is coming in, the resistors feed the transistor let's say halfway between its limits-- say 7.5 volts.   Then we add in the AC signal, so the transistor is working between 6.5 and 8.5 volts.   That keeps the transistor in its happy zone.

That's a very short description of biasing-- ti's more involved that that as you have to supply biasing voltages and currents to both the input and output terminals of the transistor.  Also you have to ensure the bias tracks any temperature or gain changes in the transistor, which can be a big hassle.

The power dissipation is the difference betweenthe power going into the transistor and the power coming out.  The difference gets dissipated as heat.  For example, let's say the transistor above was drawing 1 Ampere with the 7.5 volts across it.  With no signal in and no signal out, all that power (7.5 watts) goes into heating the transistor.  You usually have to draw the heat away with a metal heat-sink and/or fan if the power goes up more than half a watt or so.

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Author Commented:
grg99, that's a great and detail explanation in layman term, Thank you.
I would like to clarify here:
-According to what you said: bias is actually a DC current, am i right?
-why do we need a resistor? Is it becoz we use voltage supply and hence need a resistor so that we can get current (from Ohm's law?)
-You said: "But transistors are single-polarity devices". What does this mean? Single polarity?
-So the power dissipation is according to the input resistance and output resistance of the transistor itself, right?

dastrw, thank you for trying to explain but i dont quite understand. Hope you could provide a more layman and beginner type of answer. But thank you.

Cheers, :)
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Commented:
My \$0.02 =)

---According to what you said: bias is actually a DC current, am i right?
Correcy. The ac power output is actually a result of a transfer of energy from the applied DC source. Biasing means establishing a fixed level of current and voltage - which establishes an operating point on the transistors "characteristic currve". The point at which you bias the transistor depends on how you are using it; As a switch, for amplification (ac or dc)...etc.

To add some more on the "happy zone" - think about the dual supply on the op-amp; +/- 15 volts. This allows the amplified AC output voltatage to fluctuate in both the pos/neg region (From -15v to +15v). Now, if you were to disconnect the neg supply the output cannot swing in the negatative portion anymore - unless you bias it a 7.5 volts, now it can swing from 0v to 15v, and it will distort of course if you are trying to amplify it beyond what the source can offer.

-You said: "But transistors are single-polarity devices". What does this mean? Single polarity?
A dc voltage can only be one polarity - pos or neg. But an ac voltage is alternating, between pos and neg. The transistor needs to be biased to allow for the range where the amplified wave can alternate.

---why do we need a resistor? Is it becoz we use voltage supply and hence need a resistor so that we can get current (from Ohm's law?)
Thats correct, as the small internal reesistance is not sufficient to create a bias current. For the simpliest bias circuit, a fixed bias, you just connect a resistor to the base and a resistor to the collector from the supply. This will create a base current => Ib = (Vsource - 0.7) / Rb and a collector current equal to => Ic = BIb (where B-The transistor's beta value). The voltages can then be found using voltage loop methods...

---So the power dissipation is according to the input resistance and output resistance of the transistor itself, right?
Sort of. The efficiency is the output power/input power. Usually the max dissipation level is a function of the collector-emittor voltage and collector current. Pmax = Vce(Ic) So, yes it is the resistances that setup the collector current and the voltages. Normally the spec sheet will list the maximum operating range and power dissipated.
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Commented:
>> That keeps the transistor in its happy zone.

A bias current or voltage is a current or voltage need to keep the device in an operational state. In junction devices like transistors the bias current floods a region with electrons enabling conduction. A normal silicon NP junction requires a bias voltage of 0.7 volts before it can conduct.

Quite often devices have non-linear characteristics particularly at low currents or voltages. Consequently circuitry is added to enable a bias current to flow bringing the device into the better operational area.
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Commented:
Hi sokhodom,
I will try to explain it from beginning...

I am assuming a NPN Bipolar junction transistor.

Collector (C)
/
Base (B)/
-MWMW--|
\
V
Emitter (E)

Now you may be knowing that the BJT can be put to work in 3 modes.
1) Cutoff region -- when the C & E are nearly open.
2) Saturation region -- when the C & E are nearly short.
3) Active region -- when the C & E are in the 'in between' state.

1 & 2 are used when BJT is used as a switch in relays, power converters etc.
3 is used when BJT is to be used as an amplifier etc.

Now let us understand biasing.
|Â¯Â¯Â¯Â¯|
| N  | <-- Collector
|    |
|Â¯Â¯Â¯Â¯|
| P  | <-- Base
|    |
|Â¯Â¯Â¯Â¯|
| N  | <-- Emitter
|____|

Now when you put a voltage across C & E, you will generally give +ve to C and -ve to E
top N is +ve and bottom N is -ve. Since P is not given anything, BC is reverse biased and BE is forward biased, exactly like a Diode. Now you have to forward bias the BC junction to 'short' C & E.

Now consider a small current to B, which will push holes (I am considering convensional current) through P to both the N regions. As a result, BC is slightly forward biased. Now a small current flows through CBE. If you increase the base current, BC will be more forward biased and you get more current through CBE.

Since the base current is used to forward bias the BC junction, it is called "Bias current".

If you have any doubts, I will try to clear :)

Bye
---
Harish
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Commented:
I wanted to give the transistor :-/

This may be right :)

Collector (C)
/
Base (B)    /
|
-MWMW--|
|
\
V
Emitter (E)

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Commented:
|Â¯Â¯Â¯Â¯|
| N    | <-- Collector
|       |
|Â¯Â¯Â¯Â¯|
| P    | <-- Base
|       |
|Â¯Â¯Â¯Â¯|
| N    | <-- Emitter
|____|
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Commented:
oh ! that nice !

5 years here, never saw this technology known well here.

yw,
tal
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Commented:
I wanted to show the resistor as well...

Collector (C)
/
Base (B)     /
|
_/\/\/\/\__|
|
\
V
Emitter (E)

sokhodom, you have also asked the purpose of the resistor...

You know when the transistor is working in Active or Saturation region, B & E are shorted (have a very small resistance of about 5 ohms).
Typically you give 10 or 15 V as base supply. Now, ignoring the 0.7 volt drop,

I = V / R = 15 / 5 = 3A

It may seem to be a small current. But in case the junction is shorted, the current will be dangerously high (not for the operator, but for the device itself). So the resistance is required.

But a more important reason is that... BJT is a current operated device. So you should have a current source for the base. As you know current source is nothing but a voltage source with very high resistance.

Suppose you have a supply of 10 V and base resistance of 10 kiloohms, then it is equivalent to 10/10k = 1mA currnet source with 10k parallel resistance (By Norton's Theorem or source transformation). This means that the source provides nearly constant current. That is the purpose of high resistance included with the voltage source
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Author Commented:
Great Information,
Thank you all :)
Cheers
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