This tutorial actually meant to explain the practical importance of the well known transistor theories, characteristics plots and make use of them to design a high gain amplifier circuit using a single transistor and minimum number of other components.

We will analyze certain transistor characteristics and theories, summaries them, and by the end of this tutorial we will be designing a high gain amplifier by our own.

**Common emitter configuration:**

Let us consider a common emitter, fixed biased, class A amplifier circuit

*Fig. 1: Circuit Diagram of Class A Amplifier*

The hybrid equivalent of the same circuit is given below

*Fig. 2: Hybrid Equivalent Circuit Diagram of Class A Amplifier*

Here hi is the impedance found by any device connected at the input, hoe is the admittance at the output.

( hfe * ib ) forms a current dependent current source where, hfe is the forward current transfer ratio, and ib is the base current. ( hre * Vce ) forms a voltage dependent voltage source, the value is too low that we can ignore it. hre is the reverse voltage ratio.

Hence we can reduce the equivalent circuit as the following:

*Fig. 3: Equivalent Circuit Diagram of class A Amplifier*

From the figure 2 and 3, we can observe that BJT is a current driven device.

Or, info 1.

**Change in input current causes a corresponding change in the output current**

We can also observe from the above figure that, the output current ( Collector current ) is directly proportional to the input current ( base current ).

Or we can explain the relation using the equation

Ic = hfe * ib â€¦â€¦â€¦â€¦â€¦â€¦â€¦â€¦â€¦â€¦â€¦â€¦â€¦â€¦â€¦â€¦â€¦â€¦â€¦â€¦â€¦â€¦â€¦â€¦â€¦.â€¦.. eq 1.

## Transistor Characteristics

**Transistor characteristics**:

Now if we plot the output current ( ic ) Vs output voltage ( Vce ), for the different values of input current ( ib )

__hfe = 100__

*Fig. 4: Graph showing Characterstic Curve of a Transistor at Hfe = 100*

__hfe = 200__

*Fig. 5: Graph showing Characterstic Curve of a Transistor at Hfe = 200*

**Saturation region:**

Saturation region is the region in which the output current increases non-linearly with the output voltage for a particular input current.

**Cut-off region:**

Cut-off region shows the output current for zero input current.

**Active region:**

Active region is the region in which the output current varies almost linearly with the output voltage for all values of input current.

From the above figures ( 4 & 5 ), it is clear that, as the hfe increase twice from figure 4 to figure 5, the same set of input current yields twice the output current than the previous one.

We can see the effect more clearly from the following figure.

*Fig. 6: Graph showing Active Region of Transistor*

Thus we have the following information,

For a large variation in output current corresponding to the typically small variation in the input current, a transistor with high hfe is required.

**info 2****As the hfe increases, variation in output current corresponding to variation in input current also increases.**

So far we have seen the plots with variation in output current corresponding change in input current. Now let us see, how the change in input current affects the output voltage.

For the above purpose we make use of Load line.

**Load line is nothing but a straight line connecting maximum possible output current in Y axis to the maximum possible output voltage in the X axis.**

**From the load line we can find the output voltage corresponding to a particular input current.**

*Fig. 7: Graph between Output Voltage and Input Current of BJT*

From the above figure we have the following information:

__info 3.__

**Maximum value of output current is VCC/Rc and minimum value is almost 0**

__info 4.__

**Maximum value of output voltage is VCC and minimum value is almost 0**

__info 5.__

**As the input current increases output voltage decreases**

Now let us apply the load line concept to the previous transistor plots in figure 6.

*Fig. 8: Graph between Output Voltage and Input Current of BJT*

From the above figure, we can observe that

__info 6__.

**As the hfe increases, the variation in output voltage corresponding to the variations in input current increases.**

Now what happens if we increase the slope of the load line?

*Fig. 9: Graph between Output Voltage and Input Current of BJT*

It can be observed that the same amount of change in the input current yields even more change in output voltage than the previous plots, as we increases the slope of load line.

The slope of the load line can be increased by increasing the value of Rc.

So, we have found another key information.

__info 7.__

**Rc increases, the variation in output voltage corresponding to the variations in input current increases.**

Although the output swing increase with the increase in slope, the input range found to be decreased significantly.

So we should keep in mind that,

If we increase the slope of load line as large as possible we must keep the initial input current ( input current when there is no input signal applied ) as small as possible to keep the device in linear region for the complete input swing, to get corresponding distortion less output.

The input current can be reduced by increasing the Rb. From the transistor characteristics it is clear that the small valued input current allows wide range of output voltage i.e., from almost zero to around VCC.

Thus we can say that,

__info 8__.

**Rb should be kept extremely high so as to increase the linear range of output voltage swing corresponding to the typically small input signal current**

*Fig. 10: Graph between Output Voltage and Input Current of BJT*

If we are using a device at the input which can increase or decrease the input current from a mean value, for example a sinusoidal input. In such cases, to get a full output swing corresponding to the input swing, we should keep the initial output current and initial output voltage at a mean position called Q point, or quiescent point.

**Q point is the operating point in which we keep the amplifier, while there is no input applied, to get a complete, distortion less swing in output corresponding to an input swing, whenever an input signal within the expected range is applied.**

From the figure 10, Vce max is almost equals VCC, and Vce min almost equals 0. Thus we can assume that the output can swings between 0 and VCC, and hence it is appropriate to keep the output voltage initially half of the VCC. Similarly ib min is almost equals 0. We can assume that the input current swings between ib min and ib max, and hence it is appropriate to keep the input current initially at ib max/2.

While there is no input present,

VceQ should be VCC/2, ibQ should be ib max/2.

Or,

__info 9.__

**The value of Rc should be set such that Vceq = VCC/2 to get maximum output voltage swing**

__info 10__.

**The value of Rb should be set such that ibQ = ibmax/2 to allow maximum input swing**

**Summary:**

For a high gain amplifier,

1) Pick a transistor with hfe as high as possible to get large variations in output current corresponding to the variations in input current.

2) Rb should be kept extremely high so as to increase the linear range of output voltage swing corresponding to the typically small input signal current, also the value of Rb should be selected so as to keep the Q point exactly in the middle of expected input current swing.

3) Rc should be high valued so as to get large variations in output voltage corresponding to variations in output current, caused due to the variations in input current, which in turn caused due to the variations in the applied input signal voltage.

## Designing a high voltage gain amplifier

**Designing a high voltage gain amplifier:**

So let us start designing for the circuit shown in figure 1.

We pick the transistor BC109, as it is having hfe around 300.

We assume Vbe as 0.7 V for a typical forward biased junction

Hence VRb = 5 â€“ 0.7V = 4.3 V

Let us fix the operating current at 1uA, for that we have to calculate the value of Rb as,

Rb = 4.3 / ( 1 * 10-6 ) = 4.7 Mohms

Now ic = hfe * ib

ic = 300 * ( 1 * 10-6 )

ic = 0.3 mA

Now we can calculate the value of Rc as,

Rc = 2.5 / ( 3 * 10-4 ) = 8.2 Kohms.

If we need only the variations in input to be appear amplified at the output, we can use coupling capacitors at both ends. The value of coupling capacitors need to be calculated based on the frequency of signals that we use.

For audio amplifications, values below 100mfd will give reasonable performance.

This circuit can be used as a single stage audio amplifier in which the input is fed from a microphone and on the output side we can use a common 8ohm loud speaker. We can use a resistance value of 1k to 20K for pulling up the microphone depending on its type.

*Fig. 11: Circuit Diagram of High Voltage Gain Amplifier*

Filed Under: Tutorials

## Questions related to this article?

đź‘‰Ask and discuss on EDAboard.com and Electro-Tech-Online.com forums.

Tell Us What You Think!!

You must be logged in to post a comment.