  24 February 2009
Author: Giorgos Lazaridis
BJT Transistor theory

As said, this setup achieves maximum voltage amplification ratio. Let's see how we can design such an amplifier. This amplifier will drive a high impedance load (10 Kohms). The DC biasing voltage is 12 Volts and the input signal is 10 mVolts p-p. The internal resistor of the signal generator is not significant so we leave it out of our calculations for now. The frequency is about 1KHz. The transistor is a BC338 with a typical hFE=250 (DC current gain). We will use the same value for hfe as well (AC Current gain) . First, let's design the circuit schematic: We can start the design by estimating the RC resistor. We will select one collector resistor with an educative guess, taking into account the load resistor RL. The higher the collector resistor, the higher the voltage amplification of the amplifier. The point is that we cannot have large collector resistor if the load resistor is low. Since the load resistor is high, we can start by selecting a medium to high collector resistor, like for example 1000 Ohms. As said this is an educative guess, so experience helps to choose. Let's draw the DC equivalent and do the DC analysis: We want the Q point to be in the middle of the load line, therefore we set the VCE to VCC / 2, which is 6 volts:

IC = VRC / RC = (VCC - VCE) / RC = (12 - 6) / 1000 = 6 mA

Now we can use the hFE parameter to calculate the base current. This is where most errors take place, since the hFE parameter is not very stable nor easy to measure. The typical hFE of the BC338 is 250:

IB = IC / hFE = 6 / 250 => IB = 24 uA

The base current must be 24 uA for the transistor to operate in the middle of the load line. We set the base current with the base resistor:

RB = VCC / IB = 12 / 24 => RB = 500 Kohm

It is quite large value, but that is normal for this amplifier. At more extreme example you may have results in the scale of Meg-Ohms. I will use a 470 Kohms resistor as a closer value. Since we've used different resistor from the calculated value, it is good practice to re-calculate the IC value for this resistor following the opposite way:

IB = VCC / RB = 12 / 470 = 25,5 uA
IC = IB x hFE = 25,5 x 250 = 6,3 mA

The difference is not very big, but if you want to be academically correct then you have to do this recalculation. Now we can proceed with the AC analysis of the circuit to estimate the characteristics of the amplifier. To do so, we make the AC equivalent: It is important to calculate the internal emitter resistance of the transistor and the base impedance before we start:

r'e = 25mV / IC = 25mV / 6,3mA = 3,96 Ohms
Zin(base) = Î² x r'e = 250 x 3,96 = 990 Ohms

The first characteristic that we can easily calculate is the input impedance of the amplifier. Since the base resistor r'e is parallel to the base impedance Zin(base), we use the Ohm's law to calculate the amplifier's input impedance:

Zin = RB // Zin = (RB x Zin(base)) / (RB + Zin(base)) = (470 K x 990)/( 470 K + 990) = 987,9 Ohm

This value can be used to calculate the AC voltage drop from the AC source on the internal generator resistor, but in our case we said that this resistor is zero, so there is theoretically no voltage drop. The next value we can calculate is the AC current in the base:

iB = uB / Zin(base) = 10 mV / 990 = 10,1 uA

Using the hfe parameter, we can calculate the ac current on the collector:

iC = Î² x iB = 250 x 10,1 = 2,5 mA

Now that we know the collector current, we can use the ohm's law to calculate the voltage drop that this current will generate onto the output resistor RC // RL. But first we calculate the equivalent resistor Rext:

Rext = RC // RL = (RC x RL) / (RC + RL) = 909,1 Ohms
uRc = iC x Rext = 2,5 x 909,1 = 2272,7 mV
Av = uRc / uB = 2272,7 / 10 = 227

So this amplifier will amplify the input voltage by 227 times. It is obvious that the voltage amplification is very big. One thing to determine is if this amplifier works as a small signal amplifier. Let's remember the rule once more:

The AC peak to peak current of the emitter must be smaller than 10% of the DC current of the emitter.

The AC current is 2,5 mA and the DC current is 6,3 mA. For the amplifier to operate as small signal, the AC current should be smaller than 10% of the 6.3 mA, this means that is should be smaller than 630 uA. The circuit does not operate as a small signal amplifier and this means that there may be some sort of distortion at the output. Let's make this circuit on a breadboard and see the results: The yellow channel of the oscilloscope is connected to the base of the transistor with the yellow wire to measure the input voltage. The green channel comes from the green wire which measures the output: First, let's examine the output amplitude. We calculated that the output had to be 2.2 Volts. With the oscilloscope we measure the output and it is 1.95 Volts. Although the numbers are not identical, we can still consider that the circuit operates normally, because the error is very small.

Before we calculated that this amplifier does not operate as small signal amplifier. Nevertheless, the output does not seem to be distorted whatsoever. The waveform is a nice smooth sine wave with no clipped or rounded edges. How comes that? Well, the answer is in the base-emitter characteristic. It is obvious that any distortion would start if the collector current was higher than 50mA in any temperature. Since our collector current is limited to 6 mA, the transistor works in the linear area of the characteristic and that is why we do not see any distortion.

Power Gain Calculation

As we saw in the Power Calculation and Efficiency page, we can calculate the power gain of this amplifier simply by using the voltage gain Av and the hfe parameter:

Ap = Av x hfe = 227 x 250 => Ap = 56750

So the power of the input signal is amplified by 56750 times.

Efficiency Calculation

This amplifier works as an A-class amplifier, which means that the efficiency will not be very good. To calculate the efficiency we need first to calculate the DC power that the transistor draws (PS):

PS = VCC x IS

As we saw in the previous pages, the DC current IS for this sort of transistor bias is equal to the ICQ current. Therefore:

PS = VCC x ICQ = 12 V x 6.3 mA => PS = 75.6 mWatt

Now we must calculate the output power that is provided to the load by the circuit:

Pout = u2out_p-p / 8 x RL = 5,15 / 8 x 10000 = 0.064 mWatts

So, the efficiency is:

Î· = Pout / PS x 100% = 0.064 / 75.6 x 100% => Î· = 0.084 %

Transistor Power Dissipation Calculation

Finally, we can calculate the power that will be dissipated on the transistor. The maximum dissipation happens when no signal is applied, so we will calculate this value using the DC voltage and current quiescence values:

PD = VCEQ x ICQ = 6 x 0.0063 => PD = 37.8 mWatts

Under Construction...

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