# Common-Emitter Amplifier (Collector Voltage)

• roam
In summary, the conversation discusses a common-emitter transistor amplifier and the confusion over the voltage at the collector in the given circuit. The book states that the transistor has a β value of 99 and the voltage at the base is 6 volts, leading to a voltage at the collector of 7 volts. However, the assumption of a 0.7 volt drop at the base to collector junction leads to a calculated collector voltage of 5 volts. The conversation concludes that the book's statement is correct and there was a mistake in the calculation.
roam
Hi,

My question is regarding an example of a common-emitter transistor amplifier shown in my electronics book. Here is the diagram of this particular amp operating from a DC power supply:

My book says: "the transistor has β = 99, with no signal applied the voltage at the base of the transistor is measured to be 6 volts, so the dc voltages at the emitter and collector of the transistor are Ve=5.3 V, Vc=7 V respectively."

So I don't understand how you can have 7 volts at the collector in this particular circuit.

My book always makes the assumption that b-c voltage drop is 0.7. So we have

##I_e = \frac{V_e=6-0.7}{5.3 \ k\Omega} = 1 \ mA##​

And ##I_e \approx I_c##, therefore ##V_c=I_cR_c = 5 \ V##! So is the book wrong, or is there something wrong with my calculation?

We know that ##V_{cb} = 12-6 =6 V##. Hence Vc can't be greater than 6. I'm very confused and I think it might be a typo. Any help would be greatly appreciated.

Last edited by a moderator:
If IC = 1mA then the voltage drop across RC is 5V. Your book is correct.

1 person
roam said:
Hence Vc can't be greater than 6.

Why do you think that Vc can't be greater than the base voltage?

I agree with The Electrician, you are wrong to state that Vc-b =12-6=6V

My book always makes the assumption that b-c voltage drop is 0.7.
What polarity?

double check that statement. i'd have expected them to say that about voltage b-e not b-c.

jim hardy said:
What polarity?double check that statement. i'd have expected them to say that about voltage b-e not b-c.

My mistake, I meant to say b-e voltage drop is 0.7 volts. That is why we had ##V_e =6-0.7= 5.3 \ V##.

Last edited:

## What is a common-emitter amplifier?

A common-emitter amplifier is an electronic circuit that amplifies a signal using a transistor with the emitter terminal as the input, the collector terminal as the output, and the base terminal as the control.

## How does a common-emitter amplifier work?

The common-emitter amplifier works by using a transistor in the common-emitter configuration, where the input signal is applied to the emitter terminal and the output signal is taken from the collector terminal. The transistor acts as an amplifier by controlling the flow of current between the collector and emitter terminals, thus amplifying the input signal.

## What is the role of the collector voltage in a common-emitter amplifier?

The collector voltage, also known as VCE, is the voltage between the collector and emitter terminals of the transistor. In a common-emitter amplifier, it determines the amount of voltage gain and current gain of the circuit, and also affects the stability and linearity of the output signal.

## How is the collector voltage calculated in a common-emitter amplifier?

The collector voltage can be calculated using Ohm's law, where VCE = IC * RC, where IC is the collector current and RC is the resistance of the collector circuit. It can also be calculated using the voltage divider rule, where VCE = VCC * (RC / (RC + RE)).

## What are the advantages of using a common-emitter amplifier?

Some advantages of using a common-emitter amplifier include high voltage gain, high current gain, and low input impedance. It also offers good frequency response and is relatively easy to design and construct. Additionally, it can be used in a variety of applications, such as audio amplifiers, radio receivers, and signal processing circuits.

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