Bifet is a coined term meaning a combination of bipolar and FET, and a bifet amplifier is one that contains both types of transistors. These amplifiers arc designed to exploit the most desirable characteristics of each device, such as the very large input impedance of the FFT and ‘he large voltage gain of the 13JT. A good example is the cascodc amplifier, the hifct version of which is shown in Figure 11-27. Note that the input stage, QJ, is a JFET connected in a common-source configuruion and the second stage is a connected as common-basic amplifier.
The an ” cquiuulent circuit of The uc equivalent circuit or the bilct cascodc is shown in Figure 11-2X. The voltage gain or the first stage
Since the common-base input resistance 1′” is quite small.
While A”I is usually small (less than I). the large input resistance and small Miller capacitance of the JFET stage results in better high-frequency response than that found in the BJT version. As in the BJT cascade. the voltage gain of thc amplifier is achieved in the common-base stage
Since a B.lT umpliticr is generally capable’ or producing more voltage gain than its fo’ET counterpart. \V\: see that the bird amplifier uses the best features of both kinds of devices,
Besides blocking de. an important advantage or using a transformer for intcrstuge coupling is ihat the turns ratio can be designed to achieve impedance matching and therefore to maximize power transfer between stages. Recall that the turns ratio or a transformer defined by where N”. N, = number of turns in the primary al~d secondary windings
A transformer whose primary is driven by a signal source VI having resistance ‘I and whose secondary has a load resistance ‘I. connected across it. Assuming an ideal transformer (zero winding resistance and no power loss). it is easy to show that the resistance ri, looking into the primary winding is To achieve maximum power transfer from the source to the load in Figure 11-29, it is necessary (by the maximum power transfer theorem) that the resistance rIU seen hy l”H:s;~llal source he cquul to (he source resistance 1′” Thus, from equation.
In multistage transistor amplifiers, the signal source driving the primary winding is often a common-emitter stage having a large output resistance, and the load rcsisranee may be the much smaller value equal to the input resistance of the next stage. In that case. the ratio ””1. is greater than I and therefore, by equation 11-64, the turns ratio N,.N.”is greater than I. As a consequence, the coupling transformer is typically a st(‘p-t/OIVII transformer, meaning the secondary voltage is less than the primary voltage.
Commercially available transformers arc often specified in terms of their impedance ratios rather than their turns ratios. In these cases, the turns ratio can be determined using equation 11-64, i.e., by taking the square root of the specified ,Impedance ratio. For cxa-nplc, if the specifications on a transformer state that it has primary impedance 16 kH and secondary impedance Hi n (typical values for
matching a transistor output stage to a speaker).
A transformer used to match a 100-k!l source to a 250-!l load.
1. What should be the turns ratio of the transformer?
2. With the matching transformer in place, what is the rms voltage across the load?
3. What is the maximum power that can be delivered to the load?
l. From equation 11-64,
2. From equation I I -62,
We see that the resistance looking into the primary winding is J(X) kH, as it should be when load and source are matched by the transformer. Since the source resistance and r”, form a voltage divider across the X-V-rms signal source. the voltage across the primary winding is one-hut] the signal voltage:
3. Maximum power t ranslcr occurs when the source is matched to the load, and with the matching transformer in place. we have (When the source and load arc matched, the same amount of power is dissipated in the source as is delivered to the load: verify that fact in this cxample.)
The circuit in its SPICE formal. Note that we cannot isolute the circuit from the primary because SPICE request the lt there be a dc path to ground from every node. Thus, the secondary winding must have one side connected to node O.The statement Kl LP LS 1 makes the transformer ideal, since it specifies the coupling coefficient, k, to be 1. As is required to simulate an iron-core transformer in SPICE, the primary reactance at SOkHz (21T/Lp = 6.28 MO) is much greater than the 100-kO source impedance, and the secondary reactance (21T/ LJ = 1.57 kO) is much greater than the 2S0-0 load impedance. Dummy voltage sources VIP and VIS have been inserted to measure the primary and secondary currents. The results of the analysis(b). We see that “» = V(3) = 4 V and VJ = V(S) = 0.2 V, in agreement with Example11-1S. The primary current is i, = I(VIP) = 40 J1-A and the secondary current is is = I(VIS) = 0.8 mA. Note that.
Figure 11-32(a) shows II two-stage transformer-coupled amplifier. Note that there is no resistor in the collector circuit of either transistor, so the primary of each transformer is driven from a source (a collector) having relatively large resistance.
The secondary winding of the interstage transformer is connected directly to the base of 02. The 4.7-kO base biasing’ resistor is bypassed with a capacitor, so all of ·the ac voltage developed in the secondary winding appears from the base of 02 to ground. Note that both sides of the 1S-kO resistor are at ac ground. With the arrangement shown, the 02 bias resistors do not reduce the ac input resistance to the second stage, and the transformer does not short the de bias current to ground. The 1.2S-kO load is transformer-coupled to the output of Q2′ Note ti1at the load is shown completely Isolated from the rest of the circuit, in the sense that i. does not share a common ground reference with 01 and 02 (although it could, if desired). This is another advantage of transformer coupling: Individual amplifier stages and loads can be isolated from each other when desired or necessary. In fact, isolation transformers having 1 . 1 turns ratios are sometimes used specifically for that purpose.
The ac equivalent circuit of the transformer-coupled amplifier. The next example demonstrates that the transformer turns ratios were chosen to achieve impedance matching.
The transistors in Figure 11-32 have f31 = 105,r,,1 = 20 kH, f32 = 89. and r,,2 = 20 kil. Find the mid band voltage gain ut/us.
Solution 01 and Q2 arc biased identically. so the following computations for determining apply to both:
The ac input resistance at the base of Qz is ria f3zr t2 = (89)(9) = 801 n.Therefore, by equation t 1-66, the resistance looking into the primary of transformer TI is
When we compute the overall gain of the amplifier, we will take into account the fact that T, steps the voltage down by a factor of NJ N/. = 1/5, between the collector of QI and the ba ‘c of Q2.
We see that transformer 1’2 matches the output of Qz to the load. The ac load resistahce at the collector of Q2.
Taking into account the voltage division at the input of 01 and the step-down ratios of the transformers, we find the overall voltage gain.
The sign of the voltage gain, indicating the phase of the output with respect to the input, may be changed by the coupling transformers, depending on how each is connected In the circuit: Reversing primary or secondary connections will reverse the phase.