# Emitter bias

Discussion in 'Homework Help' started by Thevenin's Planet, Mar 22, 2009.

1. ### Thevenin's Planet Thread Starter Active Member

Nov 14, 2008
183
1
Hello ! GMT -7
This is about the Emitter bias circuit.Since it takes positive and neegative
power supply to bias the circuit,Why shouls I have to use two supplies rather then one,since the collector voltage must be +vcc and the emitter is grounded with the negative side of a power supply.What is I missing about the positve and negative potentials.

2. ### PRS Well-Known Member

Aug 24, 2008
989
35
You need to fully describe or draw the circuit.

3. ### Thevenin's Planet Thread Starter Active Member

Nov 14, 2008
183
1
Sorry for the delay,due to the fact that have no computer.
So the the circuit is a simple emitter bias circuit,that is, having a resistor attached to the collector terminal and the opppsite side connected to +vcc.The base terminal has a resistor attached to it and the opposite side of that resistor is also attached to +vcc.Which is shared by the collector resistor.The emitter terminal is connected by a resistor and its opposite end is connected to the Negative side of a second power supply and this second power supply positive side is grounded.
So why can I not use one supply since the Emitter is connected to a negative polarization. What's happenning? More current or what.
GMT -7

4. ### kakin New Member

Mar 24, 2009
9
0
if i understand you description correctly, the second supply is shorted to itself.. the other explanation would be that the second supply effectively lowers the bias voltage, for example if Vcc = 5v and Vs (second source) = 1v you have an effective 4v at the base.

I may be wrong, little rusty on analog.

5. ### PRS Well-Known Member

Aug 24, 2008
989
35
Your transistor is set up correctly and so is your supply. I'm not sure if this is a school project or your own experiment. If the former, your forced to use what is called a 'two-ended' supply because it's an assignment. If you're just experimenting, you can use a single ended supply.

To do this get rid of the 2nd supply altogether. Now ground is the negative lead on your supply and Vcc is the high side. Hook the collector resistor to Vcc, the emitter resistor to ground and your base resistor to Vcc -- just as before. But most amplifiers use two resistors at the base. These are in series between Vcc and ground. This forms a voltage divider with which to bias the base.

In short, yes, one supply works.

6. ### Thevenin's Planet Thread Starter Active Member

Nov 14, 2008
183
1
This is the circuit config.

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7. ### PRS Well-Known Member

Aug 24, 2008
989
35
Now you've got +Vcc and -Vcc and also ground. This circuit is very nice. I wish I had the facility to draw you a circuit with the same configuration but with only one source, but I do not. Perhaps someone else in this forum will do it for you.

8. ### Thevenin's Planet Thread Starter Active Member

Nov 14, 2008
183
1
Hello again !
The homework is attempting to understand what make electronic works. Regarding Kakin , I am assuming that you are thinking of a feed back circuit for controlling the bias point.PRS if I do what you suggested it seem to become
the common emitter bias.But the circuit suppose to have the intention to keep a constant current.Referring to the mysterious grounds or should I say the interchangeable concepts OF GROUNDS,but one of the main question is that the Ground on the above thread,is Rb grounded to +Vcc or -Vee. If its grounded to +Vcc then its what the frist post I made.If its grounded to the -Vee,then it seems to be a grounded base circuit ? My third intuition is that two batteries in series connection and the Rb is grounded between the connection where the two batteries positive and negative are connected leaving the two opposite end which is positive and the other negative polarity.Which having the positive end connected to the +Vcc and the other end of the series battery that is negative connected to Re end. Frist where is the ground connected ANYBODY.
P.S. kAKIN THAT bLOG OF YOUR GOING TO BREAK SOMEBODY MACHINE
GMT -7

9. ### PRS Well-Known Member

Aug 24, 2008
989
35
Like you said, you hook up your two batteries in series -- as you described -- and the connection between them is the ground. Connect Rb between the base and that point. The whole circuit sees this point as ground.

The circuit is a generic one as you can input a signal at the base and take the output off either the collector or the emitter. And if you put a capacitor from base to ground and input at the emitter it is also a CB amp. Without that it is either a CE or a CC amp.

CE or CC amps can be considered constant current sources, controlled by the voltage at the base. That's why the gain parameter gm is transconductance.

If you want to find the current do a KVL loop around Rb, Vbe and Re and -Vcc. If you need help on this let me know.

10. ### Thevenin's Planet Thread Starter Active Member

Nov 14, 2008
183
1
Hi !
I am assumming that a constant current has a specific amount of current that will not damper under a significant amount of current drain. A good analogy would be a large reservior.The water flow would not decrease or damper out because of the the large magnitude regrading a large draint of current . I am assumming again, that is why the battery is placed in the emitter,to get that magnitude.This is leading to the circuitry of the Long tail pair Differential amplifier. I become to know that is a very important circuit in the digital intergrated circuitry.So I figure I would start from the derivative of the circuit,the Emitter bias. Besides,it seems that it can be modified to be use as another important circuit,the Upper and lower comparator in the 555 timer.But there seems to still be a discretion of confusion, and is differial amplifier if splitted into halve the base will be connected to Vcc and the emitter to the second battery nagative potential. As I was describbing in the orginal post of this thread.I did can more understanding about the ground connection,Which seems to puzzle me alot when looking a schmatics of photo facts.

11. ### PRS Well-Known Member

Aug 24, 2008
989
35
Hello again, TP! How' spring in Portland? Here in Yakima it's still cold.

I'm not sure how to answer your question, and I guess it's because I'm not really understanding, though it could well be I just don't know the answers.

A constant current source is an ideal, but as long as it's not stressed by overloading it should remain fairly constant. With the BJT amp VB created a VE of VB-.6 volts, as you know. The voltage drop from VE across RE to -VEE sets the current through the transistor. The only way to increase or decrease the current, IC, is to change VB. This constitutes a voltage-controlled current source.

The battery at the emitter is a convience for biasing ICs. For example because of the ability to center the two supplies at 0 volts, an op amp will accept DC input and output a DC signal or some other signal that it centered on 0 volts. But two supplies are not a requirement of a constant current source. You can use one and still make the current constant due to the biasing.