Electronics 102 - Lesson 5 - AnswersQuestion 1I changed the value of R1 to 1.13 k and obtained very close to 9.0 V at the output. I could have changed the value of R2 as well (to a higher value) and obtained similar result.However, since the voltage across R2 is constant (it is equal to the Zener voltage + 0.7 V), by keeping the value of R2 constant, we keep the current through the R1-R2 chain also constant.
Question 2Resistor R5 is in parallel with R1-R2 that are in series. At the current 9 V output voltage, R5 draws a little less than 1 mA, where R1-R2 draws about 1.5 mA. 1.5 mA should be sufficient as a pre-load, so we should be able to remove R5. Run a DC Sweep simulation while sweeping the load current from 0 to 0.2 A, with R5 at its current value of 10 k, and after removing R5 altogether and compare the results. I have only included one plot because there is no appreciable difference, even at low current where the effect of R5 would be the most visible.
Question 3The ripple rejection of the circuit in lesson 4 was 36 dB at 1 kHz. This circuit only has 26 dB, a significantly poorer performance. Yet, this circuit has much better load regulation. How is that possible? Let's think about how ripple gets into the output in the first place. Ripple is present at the source and is fed into the collector of Q2 and into R3. The voltage modulation on the source modulates the current through R3 and therefore modulates the voltage on the collector of Q1 and the base of Q2. Bipolar transistors present a high impedance on their collectors, and the base of an emitter follower stage also is a high impedance point. This makes it easy for the ripple current through R3 to modulate the base voltage of Q2 and the ripple appears on the output. The closed loop operation tends to reduce the ripple, but it cannot eliminate it completely once it is allowed in the loop. In the circuit of lesson 4, the ripple current through R1 was fed into the Zener diode, which is a low impedance point, hence the better ripple rejection. How can we improve the ripple rejection of the circuit of lesson 5? We must find a way to reduce the ripple current before it gets to the base of Q2. This can be done as follows: We have split the R3 resistor into two resistors in series of half the value, and placed a decoupling capacitor in between. From a DC standpoint, the total resistance is the same, but from an AC standpoint, we have made a low pass filter. Definition: A decoupling capacitor is a capacitor placed between a point on the circuit and ground for the purpose of filtering (removing) excessive AC voltages. See how the ripple rejection is now better than 40 dB at frequencies below approximately 800 kHz.Done!!! |