# What is voltage gain of a transistor amplifier and how is it affected by frequency?

I'm actually concerned about the frequency response of a bipolar transistor.

I searched online for an answer to this question but there was not a simple answer for a high school student to understand. So I would like a simple answer. • Is this related to a specific conceptual question? Right now this is too broad. There are too many different transistor types and speeds and how they work at high frequencies is pretty complicated.
– Brandon Enright
Nov 19, 2013 at 15:20
• This question appears to be off-topic because it is about engineering.
– user27730
Nov 19, 2013 at 15:40
• @Qmechanic, without a schematic of a specific amplifier configuration, ee.se would be likely to close this as too broad. Nov 19, 2013 at 17:09
• That is the scariest schematic I have seen in my life. I recommend this one from hyperphysics: hyperphysics.phy-astr.gsu.edu/hbase/electronic/ietron/npnce.gif The answer is it comes as a result of modeled capacitance. The full answer is one of the most satisfying and interesting things I have ever heard. The Streetman & Banerjee book probably talks about it. Nov 19, 2013 at 17:54
• @HL-SDK, they're pretty much the same schematic. Differences come from (a) PNP vs. NPN and (b) showing input and output ports. Mar 1, 2014 at 10:22

(Added much later) Zeroth, a bipolar transistor is actually a very complex device, when it comes to the actual device physics. Voltage across the emitter-base junction (base bias) determines the current into the emitter, which all goes into the base region. The device construction determines how much of that current recombines in the base (and becomes base current) and how much continues on into the collector. (The current "splits".) The ratio of collector current to emitter current is called $\alpha$ (alpha). $\alpha$ generally ranges somewhere between 0.95 and 0.995.

Usually, ratio of base current to collector current is more useful, as that's the DC current gain ($\beta$) (beta) of the transistor.

A bit of algebra will show that

$\beta = \frac{\alpha}{(1 - \alpha)}$

or, alternatively,

$\alpha= \frac{\beta}{(\beta + 1)}$

For example, $\alpha = 0.95$ gives $\beta = 19$, and $\alpha = 0.99$ gives $\beta = 99$.

First, a transistor is usually modeled as a current-controlled current source. Voltage gain is a function of $\beta$ and the circuit values. Current gain does vary with operating parameters.

The venerable 2N2222A specifies a minimum $\beta$ of anywhere between 35 and 100. At Vce = 10V and Ic = 150 mA, the device advertises a minimum $\beta$ of 100, and a maximum of 300.

Second, to a first approximation, the transistor may be regarded as a fixed-gain device at frequencies from DC up to what might be called the "corner frequency". The corner frequency is the gain-bandwidth product $f_T$ divided by transistor's $\beta$.

The venerable 2N2222A advertises a minimum $f_T$ of 300 MHz, measured at Vce = 20V, Ic = 20 mA, and f = 100 MHz. With a minimum DC current gain of 100, the corner frequency will be at least (300 MHz)/100 = 3 MHz. Below 3 MHz (or above, depending on the device and the operating conditions), the device has a fixed current gain. For frequencies above 3 MHz, the gain rolls off. At 30 MHz, you would expect a minimum current gain of 300 MHz / 30 MHz = 10.

Your mileage WILL vary. You design conservatively, and you test your circuit, ideally with several transistors.

This gets you started.

TheHybrid Pi Model might be your best bet at understanding the frequency response of a BJT. But the most significant effect on the response will be the base emitter capacitance(C-pi), which will interact with r-pi to create a low pass filter which will cause the gain to decrease as frequency goes up. Also, there is also other effects to take into account but I think those would be out of the scope for a high school student.

From a very general perspective, the bandwidth limitation comes from the fact that any transistor will have a small capacitance between any of its terminals. These capacitances can be either parasitic or intrinsic to the physics of how it works in the first place, so having some capacitance is unavoidable.

It turns out that capacitance looks like very small resistors at sufficiently high frequencies. This means that the terminals will be shorted with each other at high frequencies, undermining the amplification properties of any given configuration.

As you are a high school student ,Voltage gain of an amplifier is ratio of amplified o/p voltage and given i/p voltage. But there are many other parameters dependent on the voltage gain.

The bipolar transistors are current amplifiers. The reason is that the base-emitter act exactly like a diode. No matter if you use a low power transistor such as the 2N2222A or test with a large TO-220, you will find +0.7 volt on the base when the emitter is 0.0 volt (for NPN).

The gain of transistor in datasheets is always given as a ratio of the input/output current. The input current flow from the base to the emitter. The output current flow from the collector to the emitter. It is fascinating to use a 1.5 volt battery in serie with a resistor (from 1 ohm to 1 K ohm) connected on the base and be able to dim a 120 Watt light bulb connected on the collector of the transistor (and the other wire of the light bulb connected to a high positive voltage). You can find powerful transistors already mounted on good heat-sink inside any CRT tv - easy to find on garbage collecting day - everybody is replacing these CRT by flat screen tv now. You get also a strong enough high voltage supply (often +30 Volt and +120 Volt) in these same discarded CRT tv.

I presume that the battery called "Vbb" in your schematic was representing the equivalent 0.7 volt source that is inside the transistor. It is good engineering to represent a diode like if it was made of a switch (opening or closing depending on the polarity presented at the diode), a resistance (very small... less than 1 ohm) and a voltage source of 0.7 volt.

To avoid the nightmare of testing so many transistors until finding one with the correct gain, the standard approach consist in adding a small resistor on the emitter of the transistor to decrease the gain to a known value. For example, suppose the transistor gain range from 35 to 100 like the 2N2222A mentioned above, you may choose the resistance value to get a gain of 10. With that resistor, the gain will always be 10 for every circuit, no matter if the particular 2N2222A is really capable of 100 or only 35.

To compensate for the gain decreasing with frequency, the engineers often add a capacitor in parallel with that resistor which is added on the emitter pin of the transistor. This high pass filter effectively acting as a short circuit when the frequency is high enough

For a transistor amplifier,the frequency response curve indicates, that the voltage gain is low at high and low frequencies and high in the middle frequency range...