Current Amplification Help (page 2)
Because a small change in I B results in a large I C variation when the bias is right, a transistor can operate as a current amplifier . The extent of such amplification can be expressed in terms of what happens with either static (steady) or dynamic (varying) input signal current.
Static Current Amplification
The maximum obtainable current amplification factor of a bipolar transistor is known as the beta of the transistor. Depending on the way in which a transistor is manufactured, the beta can range from a factor of a few times up to hundreds of times. One method of expressing the beta of a transistor is as the static forward current transfer ratio , symbolized H FE . This is the ratio of the collector current to the base current:
H FE = I C / I B
For example, if a base current I B of 1 mA produces a collector current I C of 35 mA, then H FE = 35/1 = 35. If I B = 0.5 mA and I C = 35 mA, then H FE = 35/0.5 = 70.
Dynamic Current Amplification
Another way of specifying current amplification is as the ratio of a difference in I C to a small incremental difference in I B that produces it. This is the dynamic current amplification , also known as current gain . It is customary to abbreviate the words the difference in by the uppercase Greek letter delta (Δ) in mathematical expressions. Then, according to this definition,
Current gain = Δ I C /Δ I B
The ratio Δ I C /Δ I B is greatest where the slope of the characteristic curve is steepest. Geometrically, Δ I C /Δ I B at any given point on the curve is the slope of a line tangent to the curve at that point.
When the operating point of a transistor is on the steep part of the characteristic curve, the device has the largest possible current gain, as long as the input signal is small. This value is close to H FE . Because the characteristic curve is a straight line in this region, the transistor can serve as a linear amplifier if the input signal is not too strong. This means that the output signal waveform is a faithful reproduction of the input signal waveform, except that the output amplitude is greater than the input amplitude.
As the operating point is shifted into the part of the characteristic curve where the graph is not straight, the current gain decreases, and the amplifier becomes nonlinear. The same thing can happen if the input signal is strong enough to drive the transistor into the nonlinear part of the curve during any portion of the signal cycle.
Gain Versus Frequency
In any particular bipolar transistor, the gain decreases as the signal frequency increases. There are two expressions for gain-versus-frequency behavior.
The gain bandwidth product , abbreviated f T , is the frequency at which the current gain becomes equal to unity (1) with the emitter connected to ground. This means, in effect, that the transistor has no current gain; the output current amplitude is the same as the input current amplitude, even under ideal operating conditions. The alpha cutoff is the frequency at which the current gain becomes 0.707 times (that is, 70.7 percent of) its value at exactly 1 kHz (1,000 Hz). Most transistors can work as current amplifiers at frequencies above the alpha cutoff, but no transistor can work as a current amplifier at frequencies higher than its gain bandwidth product.
Current Amplification Practice Problems
A bipolar transistor has a current gain, under ideal conditions, of 23.5 at an operating frequency of 1,000 Hz. The alpha cutoff is specified as 900 kHz. What is the maximum possible current gain that the device can have at 900 kHz?
Multiply 23.5 by 0.707 to obtain 16.6. This is the maximum possible current gain that the transistor can produce at 900 kHz.
Suppose that the peak-to-peak (pk-pk) signal input current in the aforementioned transistor is 2.00 mA at a frequency of 1,000 Hz. Further suppose that the operating conditions are ideal and that the transistor is not driven into the nonlinear part of the characteristic curve during any part of the input signal cycle. If the frequency is changed to 900 kHz, what will be the pk-pk signal output current?
First, note that the current gain of the transistor is 23.5 at a frequency of 1,000 Hz. This means that the pk-pk output signal current at 1,000 Hz is 2.00 μA × 23.5 = 47.0 μA. At 900 kHz, the pk-pk output signal current is thus 0.707 × 47.0 μA = 33.2 μA.
Practice problems of these concepts can be found at: Semiconductors Practice Test
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