iC = βiB and iE = (β + 1)iB,
and the emitter is separated from the base by a diode. In order for this diode to conduct current, it must be forward biased with 0.65V.
iB ≈ 0 and iC ≈ iE.
These simple rules are similar to the rules we use with operational amplifiers. The analysis approach usually follows these steps:
1. Calculate the transistor base potential vB by assuming that no current enters the base (i.e., iB ≈ 0).
2. Calculate the potential vE at the emitter of the transistor using vB. For an npn transistor,
vE = vB − 0.65V,
and for a pnp transistor,
vE = vB + 0.65V.
3. Calculate the emitter current iE using the emitter voltage vE and the rest of the circuit.
4. Assume that iC ≈ iE and analyze the rest of the circuit.
• Because we know vE, we usually know iE as well. So our iE dictates what iC should be.
However, keep these notes in mind.
• For an npn transistor, active mode requires vC − vE > 0.2V. For a pnp transistor, active mode requires vE − vC > 0.2V. If this condition is violated, the transistor is saturated, and the analysis cannot continue using these simple rules.
In design problems,
change parameters (e.g., resistors, supply rails, etc.) to prevent saturation.
• Sometimes it’s easier to find vE first and use it to calculate vB.
• How “small” iB must be to neglect its effect depends on the circuit. In particular, iB × RB must be very small, where RB is the the Th´evenin equivalent resistance looking out of the transistor base.
• When the collector–emitter potential in an npn transistor drops to below 0.2V, the transistor saturates and leaves active mode.
• When the emitter–collector potential in a pnp transistor drops to below 0.2V, the transistor saturates and leaves active mode.
The pnp current mirror is relatively simpler than other current sources and has high compliance.
1. The left pnp transistor is “diode-connected,” which means that its collector and base are shorted together. Because the base is 0.65V below the emitter, the collector is held at 0.65V below the emitter as well. Therefore, the transistor
looks like a diode.
2. Potential acrossresistor RC is 11.35V, and so the collector current on the left transistor is (11.35V)/RC.
3. The left and right pnp transistors share the same base and emitter nodes. If we assume that they are perfectly matched transistors, they should have the same collector current. Therefore,
4. In order to keep vCE,right > 0.2V, vout < 11.8V. Therefore, the device has a compliance of 11.8V.
Additional transistors can be added in the same way as the right transistor in order to have multiple outputs all with the same current. Other current mirrors are (more) immune to these problems, but they usually require more complex configurations and/or tightly matched BJTs. IF YOU USE THIS CURRENT MIRROR IN THE LAB, BE SURE TO MINIMIZE THE TIME THE CIRCUIT IS POWERED ON.
where β ≥ 100 for small-signal BJTs. That is, small-signal input impedance is approximately β(REkL). A low RE increases transistor power dissipation and input current required, but it is required for clipping immunity, as shown below. High-β transistors reduce input current demand even for low RE.
Clipping on negative swing: Consider the simplified circuit in Figure 6.1(b), which replaces the transistor with a “dynamic variable resistor” that “pulls up” the output toward 6V. Increasing the size of this resistor moves the output lower. The lowest output occurs at transistor cutoff, as the pull-up resistor is infinite (i.e., an open circuit). Hence, the lowest output is (−6V) × L/(RE + L). A small RE reduces this distortion but also increases power dissipation. Not surprisingly, there is a power–distortion tradeoff.
Unable to “sink” current: The output cannot swing lower than (−6V) × L/(RE + L) because ground current from the load can only travel through the RE resistor. The npn emitter diode cannot sink current, so it all must go through RE. If RE were replaced with another “dynamic resistor,” more negative output swings could be produced. In other words, RE should be replaced with another transistor, as in section 7.
Negative swing caveat: Pulling the input signal down very far will cause the transistor’s base–emitter diode to breakdown. Care should be taken to prevent this from happening. Prevent vin from swinging more than 6V lower than the emitter. For protection, add a reverse-biased diode from the base to the emitter.
Pushing, Pulling, Sourcing, and Sinking: It is a “push–pull” amplifier because the upper npn transistor “pushes” (i.e., sources) current for positive input and the lower pnp transistor “pulls” (i.e., sinks) current for negative input. It is a “class-B” amplifier because only one transistor is active at a time. That is, for sufficiently positive signals, the upper transistor will be active and the lower transistor will be in cutoff mode (and vice versa). In Figure 7.1(b), this case corresponds to the PNP resistor being ∞ and the NPN resistor being some finite resistance that puts the desired vout between 0V (from ground) and 6V. For the class-B amplifier, only one “dynamic resistor” has a finite resistance at a time.
Distortion and Power: The class-B amplifier does not “waste” power because transistors only dissipate power (i.e., activate) when absolutely necessary to produce output. This property is attractive in low-power applications, but it comes with a power–distortion tradeoff. For signals between −0.65V and 0.65V, both “dynamic resistors” are infinite (i.e., “off”), and so the output is 0V. Larger signals that swing through this “dead zone” suffer significant distortion. We show in section 8 that some of this crossover distortion can be removed at the cost of additional power dissipation; we use extra biasing power so that the “dynamic resistors” turn “on” more easily.
Practical Applications: Circuits like this one (and the one described in section 8) are often used as output stages of buffering devices, like operational amplifiers. As the combination of two emitter followers, it provides very high power gain and is still be able to handle positive and negative signals (i.e., both sourcing and sinking load current). Because of the distortion it adds, this simple push–pull stage will only be used in low-power applications that sacrifice distortion for reduced power consumption.
Biasing diodes: The diodes move some of the “dead zone” from the middle of the input range to the ends, which reduces distortion while decreasing maximum input swing. Ideally, the diodes would have the same potential drop as the base–emitter diode in each of the transistors, which is usually true for a prefabricated transistor/diode array (i.e., inside an integrated circuit). In our case, our discrete components are not matched, so the output still has crossover distortion (e.g., consider signals between −0.05V and 0.05V). Additionally, as vin swings and ibase varies, vD changes slightly, so there is a new source of distortion.
Class-B push–pull amplifier: The amplifier is still a class-B amplifier because only one transistor is active at a time. However, the two diodes make the transistors more sensitive to small signals, and so they activate more readily than in the case in section 7. The “dead zone” that causes output distortion is smaller.
Practical Applications: Circuits like this one (and the one described in section 8) are often used as output stages of buffering devices, like operational amplifiers. As the combination of two emitter followers, it provides very high power gain and is still be able to handle positive and negative signals (i.e., both sourcing and sinking load current). Because of the distortion it adds, this simple push–pull stage will only be used in low-power applications that sacrifice distortion for reduced power consumption.
Class-AB push–pull amplifier: The amplifier bridges on being a “class-AB amplifier.” If additional biasing diodes and emitter resistance are added, both transistors will be active simultaneously for all small signals, which is like a class-A amplifier (i.e., both transistors are active even though only one is “needed”). However, if these signals swing out of a small region (determined by load and how many diodes are added), the amplifier behaves like a class-B amplifier (i.e., only one transistor is active). The class-A operation “wastes” power for less distortion, and the class-B operation introduces distortion to save power.
Power and distortion: The circuit “wastes” power in its biasing diodes in order to reduce output distortion. Increasing the biasing current will further reduce output distortion as the vD drop will depend less on the load. There is always a tradeoff between distortion and power. Clean amplifiers use extra power for low distortion of small signals; the extra power is not simply for production of large (e.g., high volume) signals.
Depending on your component choices and signal source, the circuit in Figure 3.1 may load the source so that the input signal is noticeably attenuated when connected to the circuit. That is, at signal frequencies, the input impedance of the circuit may be low compared to the output impedance of the signal source, and so dissipation in the signal source causes attenuation of the signal entering the circuit. To ensure the current into the base of the transistor is negligible, the biasing network must have a relatively low equivalent resistance at DC when looking out of the base. However, there is a clever method we can use to raise the impedance of the network at signal frequencies when looking out of the capacitor. By bootstrapping some of the transistor’s output signal back into the input, we can make the input impedance (at signal frequencies) very large (i.e., approximately βRE, the input impedance of the transistor). Consider the modified circuit in Figure A.1.
Instead of Equation (3.3) and Equation (3.4), assume that
Otherwise, components can be chosen exactly as before. The bootstrapping capacitor CB must be very large so that it looks like a short circuit to signal frequencies. Theoretically, the resistor RB can be chosen arbitrarily. As long as Equation (A.1) can be met, a high choice of RB (e.g., RB > 1kΩ) is a good idea. The signal at the transistor’s emitter follows the signal at its base. At signal frequencies, CB acts like a short circuit, and so both ends of RB see the same potential. Hence, RB carries no current at signal frequencies. Thus, the R1–R2 divider cannot load the input source because no current from the source makes its way across RB (i.e., RB ≈ ∞ at signal frequencies). The current through R1–R2 that would normally come from the source comes from the output instead. This method is called bootstrapping because we use the circuit’s own output to reduce current required from the input.
