The Biasing Of Bjt

1. Bipolar Transistor Biasing
2. The Biasing Of Bjt Property
3. Bjt Biasing Circuits

DC Biasing & AC Performance Analysis of BJT and FET Differential Amplifier Sub-circuits with Active LoadsAny op-amp worth its salt has a differential amplifier at its front end, and you’re nobody if you can’t design one yourself. So, this article presents a general method for biasing and analyzing the performance characteristics of single-stage BJT and MOSFET differential amplifier circuits. The following images show the general schematic for both kinds of differential amplifiers, often referred to as a differential input stage when used in designing op-amps. Notice that these types of differential amplifiers use to achieve wide swing and high gain. BJT and MOSFET differential amplifiers with active loadsDue to design processes and the nature of the devices involved, BJT circuits are “simpler” to analyze than their FET counterparts, whose circuits require a few extra steps when calculating performance parameters.

For this reason, this tutorial will begin by biasing and analyzing a BJT differential amplifier circuit, and then will move on to do the same for a FET differential amplifier. But it should be noted that the procedures to analyze these types of differential amplifiers are virtually the same.

BJT Differential AmplifierThe first thing needed is to configure the DC biasing. To accomplish this, a practical implementation of must be developed. A very popular method is to use a. A simple current mirror is shown below: BJT Current Mirror Figure 2. BJT Current MirrorIt is easy to understand how a current mirror works. Observe the equation governing the amount of collector current in a BJT, denoted:where:.

Base Bias of a BJT Transistor. Below is a typical BJT receiving base bias: V BB is the base supply voltage, which is used to give the transistor sufficient current to turn the transistor on. R B is a resistance value that is used to provide the desired value of base current I B. V CC is the collector supply voltage.

Bipolar Transistor Biasing

is the collector current. is the. is the DC voltage across the base-emitter junction. is the thermal voltage, typically 25 mV. is the quality factor, typically between 1- 2 and is frequently assumed to be 1. is the voltage across the collector-base junction.

is theNote: This equation may look intimidating at first, but what is important to understand is that the point of designing “by hand” is to get close. One should aim simply to get a good estimation of such parameters as necessary bias current, gain, input impedance, etc. In this way, computer simulations can analyze the hand-designed circuit in much closer detail, which greatly aids in the process of designing a real-life differential amplifier. Knowing this, the equations to be used in this tutorial will be rough estimates, but are still invaluable when it comes to designing these types of circuits. By assuming a very large equivalent resistance, one can estimate that the collector current through any BJT can be described by:What can be noticed here is that the only controllable variable in that equation is. All the other terms in the equation are constants that depend on either the environment or the actual physical size of the device. This means that for any two same-sized transistors, the currents through their collectors will be the same as long as the voltage across their base-emitter junctions is the same.

By tying their bases and emitters together, we can mirror the currents between them! In order to implement a successful current mirror, one transistor (here, ) must have a current induced in it to mirror it to the differential amplifier’s current source (here, ). After adding this current mirror to our BJT differential amplifier, the resulting schematic is: Figure 3. BJT differential amp with current mirror biasingIn order to properly bias this circuit, it is necessary to include.

The Biasing Of Bjt Property

Two things are accomplished by including in our circuit. One of them is that we can induce the current in, and thus, the current in. The other important thing this resistor does is drop a majority of the available voltage across itself, so that doesn’t have the entire voltage difference between the supplies across it!

To bias this circuit, the first thing one must do is determine what the desired magnitude of the current source will be. This parameter depends on how you want the circuit to operate, and is usually a known value. In this tutorial, we will assume we want an of 1mA. In order to determine the necessary size of, we analyze the loop that consists of:(KVL) around this loop reveals:These kinds of circuits are typically supplied rails of to. So, this tutorial will assume.For a given technology, all of the BJT transistors are designed to have the same turn-on voltage. This tutorial will assume.7 V for each BJT.

That being the case, and rearranging the above equation, results in:By introducing a resistor of to the above schematic, the bias current is now established at 1 mA. Due to symmetry, the currents through transistors and are each half of the bias current, described by:Now that we know the collector currents through and, characterizing the performance of this differential amplifier is a breeze. Since the parameters we are interested in (gain, etc) are small-signal parameters, the of this circuit is needed. To obtain this, a nice trick is to “cut the amplifier in half” (lengthwise, such that you only analyze the output side of the amplifier) to obtain: Figure 4. Small-signal model for above differential ampliferNote: even though the output signal is single-ended here, the output is still a result of the entire input signal, and not just half of it. This is because the small-signal changes in the currents flowing through are impeded from traveling down the branches controlled by current sources.

Also note that the connections between and the voltage-controlled current source (VCCS) indicate that the voltage that controls the VCCS is the voltage across. This is because the resistance in the emitter of these transistors has been omitted, due to its typically small value (10 to 25 ).

In addition to this, is assumed to be a small signal (AC) open-circuit. The frequency response has also been omitted, and the amplifier is assumed to be. Differential Mode GainIt is simple to see that (the small-signal output voltage) is equal to the current across the parallel combination of the resistors and multiplied by the size of the same parallel combination. Since we know the value of the current through this combination is equal to the input voltage multiplied by (the transconductance parameter):The transconductance parameter is a ratio of output current to input voltage. It is described mathematically as:and can be solved for thusly:In this example, is.5 mA and is 25 mV.

With these values, we compute:Now that the transconductance parameter is known, the only other values needed to compute the differential mode gain are and., so they will not have the same small-signal resistance, but the procedure to find these two values are nearly identical. The following equation describes the small-signal output resistance of any BJT:The parameter is typically given, and in this tutorial:Which would result in:andNow that the small-signal resistances are known, along with the transconductance parameter, the differential mode gain ( ) may be calculated:or, in decibels (dB): Differential Input ImpedanceThe differential input impedance of a differential amplifier is the impedance a “seen” by any “differential” signal. A “differential signal” is any and all signals that aren’t shared.

For instance, if:andthen the common mode signal and differential mode signals are:andTo find the differential input impedance, begin by following the loop consisting of:, as illustrated below: Figure 5. Loop analyzed in order to determine Rin(DM). We see that, in the differential signal mode, the path to ground only consists of of each input transistor. Since this is the case, the differential mode input impedance of any BJT diff-amp may be expressed as ( omitting emitter resistance and assuming matched):where:(current gain factor)A typical value for is 100, and knowing allows one to compute:So, for the BJT differential amplifier in this tutorial, the differential mode input impedance is: Common Mode GainThe CM gain ( ) is the “gain” that common mode signals “see,” or rather, is the applied to signals present on both differential inputs. A good op amp attempts to eliminate all common mode signals, but this is obviously not possible in the real world. However, one may compute the common mode gain by “cutting the amplifier in half” by observing one of the loops in the following diagram.

Bjt Biasing Circuits

The path differs from that of differential signals because common mode signals make it so that the two signal sources don’t “see” each other. Notice:Figure 6. Loop(s) analyzed to determine common mode voltage gain and input impedanceWe choose a loop and draw the small-signal model to obtain: Figure 7. Small-signal model for common mode input signalsSimilar to the output voltage of the differential mode small signal model, we can see that is the voltage across. We also know the current running through this resistance, and may equate the output voltage to:This time, though, isn’t distributed entirely over the resistances at the base. Instead, a fraction of the input common mode input signal is across the base-emitter junction.