I am having a little trouble understanding a common emitter circuit.

In a common emitter circuit, if the emitter of the bipolar transistor is directly connected to ground, how would there be voltage gain? Wouldn't the voltage just be the forward-biased diode voltage + the reverse biased diode voltage? I attached an image of the type of circuit I am thinking of.

Thank you for your help!

 

There is a slight amount of resistance in the emitter that is dependent on the emitter current in bipolar transistors.
The circuit gain would then be Rc/Re which would be high but unstable due to temperature and is why it is not used as an amplifier but as a switch.
SG

 

There is a slight amount of resistance in the emitter that is dependent on the emitter current in bipolar transistors.
The circuit gain would then be Rc/Re which would be high but unstable due to temperature and is why it is not used as an amplifier but as a switch.
SG

In most common emitter circuits the input is too the base and the output is from the collector. So the voltage on the emitter is not the output.
An emitter follower is a whole different tale. In that case the emitter voltage subtracts from the base voltage, a type of negative feedback. Are you really asking about COMMON emitter, or about emitter FOLLOWER circuits? They are quite different.

 

In most common emitter circuits the input is too the base and the output is from the collector. So the voltage on the emitter is not the output.
An emitter follower is a whole different tale. In that case the emitter voltage subtracts from the base voltage, a type of negative feedback. Are you really asking about COMMON emitter, or about emitter FOLLOWER circuits? They are quite different.

I am talking about a common emitter. In terms of the gain, I mean the voltage above the collector of the resistor compared to the voltage supplied to the base. The circuit I am thinking of is the circuit in the picture, and the only addition would be a resistor before the input signal at the base. Sorry for any confusion!

 

You must be thinking of the diode model of a BJT.



This is a very crude simplification. A BJT is much more that two diodes placed back-to-back. The impurity doping concentrations between the base and collector is much lower than that between the base and emitter.

When base-emitter current flows, it causes collector-emitter charges to flow.

The collector current = beta x base current.
beta = current gain is typically 100 to 300.

β = α / (1 - α)

 

Okay, that is helpful. I guess I am still confused on how there is voltage gain. Does having the current being amplified by the transistor also amplify the voltage through the diodes? Because I thought that the diodes would still asymptote at a certain critical voltage value, so they wouldn't be able to go above a certain amount...

Thank you!

 

You must be thinking of the diode model of a BJT.

View attachment 171160

This is a very crude simplification. A BJT is much more that two diodes placed back-to-back. The impurity doping concentrations between the base and collector is much lower than that between the base and emitter.

When base-emitter current flows, it causes collector-emitter charges to flow.

The collector current = beta x base current.
beta = current gain is typically 100 to 300.

β = α / (1 - α)

Okay, that is helpful. I guess I am still confused on how there is voltage gain. Does having the current being amplified by the transistor also amplify the voltage through the diodes? Because I thought that the diodes would still asymptote at a certain critical voltage value, so they wouldn't be able to go above a certain amount...

Thank you!

 

I thought that the diodes would still asymptote at a certain critical voltage value, so they wouldn't be able to go above a certain amount.

Yes, the voltage tends to asymptote at high currents, but that's well below the typical base current of a transistor.
A diode current increases logarithmically with voltage without theoretical limit.

Below is the LTspice simulation of the circuit to show how it works.
To simply the simulation, I used a current source for the base, biased at 1μAdc offset with a 0.2μApp AC signal (yellow trace).
As you can see, this cause a base-emitter input voltage of about 623mVdc with a 5.2mVpp AC signal (blue trace).
The collector current through R1 from this input is about 303μA (indicating a transistor β current gain of 303) with an AC current of about 59μApp (red trace).
This gives a AC collector signal of 590mVpp (green trace).
The AC voltage gain is thus about 590mV/5.2mv = 113.

Make sense?

 

Yes, the voltage tends to asymptote at high currents, but that's well below the typical base current of a transistor.
A diode current increases logarithmically with voltage without theoretical limit.

Below is the LTspice simulation of the circuit to show how it works.
To simply the simulation, I used a current source for the base, biased at 1μAdc offset with a 0.2μApp AC signal (yellow trace).
As you can see, this cause a base-emitter input voltage of about 623mVdc with a 5.2mVpp AC signal (blue trace).
The collector current through R1 from this input is about 303μA (indicating a transistor β current gain of 303) with an AC current of about 59μApp (red trace).
This gives a AC collector signal of 590mVpp (green trace).
The AC voltage gain is thus about 590mV/5.2mv = 113.

Make sense?

View attachment 171161

Thank you so much! This helped me understand the voltage gain a lot better!

 

The best way to see how amplification is achieved is by drawing the load line on top of the BJT current-voltage characteristic curves.

The blue lines show how the collector current IC changes with collector-emitter voltage VCE for a fixed base current IB.
The red line is called the load line and represents the collector load resistor RL.
This is drawn by connecting two points A and B on the chart.
B is the supply voltage VCC.
A is the maximum current IC(max) = VCC / RL

In this example VCC = 12V
RL = 1.2kΩ
Hence IC(max) ought to be 10mA.

The transistor is biased for a Class-A amplifier at the Q-point where IB = 46μA.
Observe what happens when a sine wave signal is impressed on the base such that IB min and max are 20μA and 80μA.
The transistor's operating points will traverse from a low point at N to a high point at M.
Follow what this does to the collector current and the collector voltage.

From this graphical analysis, you can determine the AC current gain, about 5mA/60μA = 83.

Note that the output voltage is out of phase from the input base current (inverted).
The collector currrent is in phase with the base current.
The voltage VL across the load resistor is amplified and in phase with the base current.
The collector voltage is out of phase because the collector voltage is Vcc - VL.

 

Okay, that is helpful. I guess I am still confused on how there is voltage gain. Does having the current being amplified by the transistor also amplify the voltage through the diodes? Because I thought that the diodes would still asymptote at a certain critical voltage value, so they wouldn't be able to go above a certain amount...

Thank you!

The voltage is not amplified across the transistor. The current through the transistor is amplified. The voltage is amplified across the load resistor since the voltage across the resistor is simply obeying Ohm's Law where V = I x R.

 

I am having a little trouble understanding a common emitter circuit.

In a common emitter circuit, if the emitter of the bipolar transistor is directly connected to ground, how would there be voltage gain? Wouldn't the voltage just be the forward-biased diode voltage + the reverse biased diode voltage? I attached an image of the type of circuit I am thinking of.

Thank you for your help!

Yes, the Forward Voltage across the B-to-E junction is FIXED, at nearly 0.7 volts
No, the Reverse Voltage across the B-to-C is NOT fixed, and it will vary, it must vary.

The current in the Base is multiplied by the GAIN and this HIGHER current flows through the Collector.
The voltage at the Collector varies as the Collector Current varies because
the Voltage Drop ( OUTPUT SIGNAL ) at the Collector = Source - ( Collector Current x Rc )
Substituting ...
The Voltage Drop ( OUTPUT SIGNAL ) at the Collector = Source - ( Base Current x Gain x Rc )

Therefore the small Base Current does control the large Output Signal.

 

Yes, the Forward Voltage across the B-to-E junction is FIXED

Not exactly.
That voltage does vary slightly with a change in the base current, as you can see in my simulation.

 

Therefore the small Base Current does control the large Output Signal.

No - from the physical point of view, this is not the case.

Not exactly.
That voltage does vary slightly with a change in the base current, as you can see in my simulation.

In fact - it is the voltage that causes a currrent variation.
Any input signal causes the voltage across the B-E junction to vary slightly. And this voltage variation causes the collector current to vary. This relationship is expressed by the well-known exponential Shockley equation. As a secondary effect (unwanted, but not to avoid) there is a base current which determines the input resistance at the base. But this current has nothing to do with the amplification properties of the BJT.
The bipolar tansistor acts as a (non-ideal) voltage controlled current source - and voltage amplification results because the collector resistor converts the output current at the collector node into a corresponding output voltage.
This can be verified by evaluating the voltage gain formula for a common-emitter stage (without feedback):

d(Vout)= - d(Ic)*Rc with d(Ic)=gm*d(Vbe)
Gain A=d(Vout)/d(Vbe)= - gm*Rc.

The parameter gm is called "transconductance" iand is defined as gm=dIc/d(Vbe). This parameter gm is identical to the slope of the mentioned Shockley equation Ic=f(Vbe).

 

Not exactly.
That voltage does vary slightly with a change in the base current, as you can see in my simulation.

Irrelevant to this discussion

 

Also note that grounded emitter (collector, base) is notnecessarily the same as common emitter (collector, base). It is the common electrode which determines the general characteristics of the circuit.

 

No - from the physical point of view, this is not the case.

In fact - it is the voltage that causes a currrent variation.
Any input signal causes the voltage across the B-E junction to vary slightly. And this voltage variation causes the collector current to vary. This relationship is expressed by the well-known exponential Shockley equation. As a secondary effect (unwanted, but not to avoid) there is a base current which determines the input resistance at the base. But this current has nothing to do with the amplification properties of the BJT.
The bipolar tansistor acts as a (non-ideal) voltage controlled current source - and voltage amplification results because the collector resistor converts the output current at the collector node into a corresponding output voltage.
This can be verified by evaluating the voltage gain formula for a common-emitter stage (without feedback):

d(Vout)= - d(Ic)*Rc with d(Ic)=gm*d(Vbe)
Gain A=d(Vout)/d(Vbe)= - gm*Rc.

The parameter gm is called "transconductance" iand is defined as gm=dIc/d(Vbe). This parameter gm is identical to the slope of the mentioned Shockley equation Ic=f(Vbe).

The BJT is classified as a Current Controlled Device.
The keyword here is "Current".

The GAIN ( Bf ) of the BJT is defined as ...
Bf = Ic / Ib
the ratio between the larger CURRENT flow in the Collector path vs the smaller CURRENT flow in the Base

How you get the current to flow into the Base is irrelevant to the BJT transistor.

If you want to classify the BJT as a " (non-ideal) voltage controlled current source " , that is your prerogative.
Your "AC Small Signal Analysis" is not relevant here, for this High Level discussion
He just wanted know how the Collector Voltage changes at a "High Level".
You guys get way too wrapped up in your, irrelevant esoteric details.

 

I am having a little trouble understanding a common emitter circuit.

In a common emitter circuit, if the emitter of the bipolar transistor is directly connected to ground, how would there be voltage gain? Wouldn't the voltage just be the forward-biased diode voltage + the reverse biased diode voltage? I attached an image of the type of circuit I am thinking of.

Thank you for your help!

Your circuit is a little more complicated than just a 'common emitter'. Common always means 'grounded'. However, your circuit is also an 'open collector'. That means this BJT is actually being used to pull the collector to ground, zeroing the output, when a positive voltage is applied to the base. Electron flow is from ground to positive- that is why in a BJT, the emitter is called the emitter, and the collector the collector. The emitter emits electrons into the junction that are captured by the collector.

In the case of your circuit, the positive pull-up resistor on the collector is what holds the voltage high, when the BJT is not conducting (base is negative). When base is positive, the ground overcomes the pull-up resistor on the collector, and holds the CE junction at the ground level, bringing output low on the collector.

 

Sometimes the threads do seem to wander into the more theoretical realms. Enough to say that the base current controls the collector current, and the collector resistor results in there being a change in the voltage. Likewise, the base voltage drives the base current, thus the voltage applied ultimately controls the collector voltage by changing the collector current. That is the mechanism of a common emitter circuit provides voltage gain. Sort of complicated, I suppose.

 

Common always means 'grounded'

Here's a 'for instance'.
Which electrode is grounded?
Which electrode is common to both input and output?