6.301 Solid-State CircuitsRecitation 23: Zener Diode References (continued)Prof. Joel L. DawsonWe’ve talked about this before, but it bears repeating: True DC performance is every bit as difficultto achieve as high-frequency performance. By “DC” performance, we’re talking low DC offsets, inthe case of amplifiers, and low drift performance. Usually with drift, temperature changes are theenemy.With Zener references, we’ve already talked about using self-biasing to set the current through thediode. Why go to such pains? It turns out that with regards to temperature stability, Zener diodestend to have a “sweet spot”: run them at a particular current, and their temperature stability isespecially good.In the last class we looked at a particular self-biased cell. Let’s look at it again.CLASS EXERCISE: Argue that the output impedance of this circuit must be negative.VZ−+RVOUT6.301 Solid-State CircuitsRecitation 23: Zener Diode References (continued)Prof. Joel L. DawsonPage 2Another interesting tidbit about this circuit is that it is actually an example of positive feedback:After loop is broken, raising node (A) causes (B) to rise! We don’t get into trouble because the gainaround the loop is small (<1).Again, why go through the trouble of a self-biased stage? Let’s consider the alternative.A good Zener diode will have a small incremental resistance, rz. If we write out the response of theoutput voltage to changes in the supply, we get:BRABreakloop hereVZVCC−VZRVZΔVZΔIZ→ small!IZIZ6.301 Solid-State CircuitsRecitation 23: Zener Diode References (continued)Prof. Joel L. DawsonPage 3ΔVzΔVCL=rzrz+ RTime for some numbers. Let VCC= 15V, rz= 10Ω, VZ= 6.2V. Further, suppose that the “sweetspot” for IZ is 7.5mA, at which point the temperature coefficient is a mere 5ppm/°C. The value ofR is determined by the sweet spot:R =VCC− VZ7.5mA=15V − 6.2V7.5mA= 1.17kΩNow, what happens when there is a 1% change in VCC?ΔVz=Rzrz+ R⎛⎝⎜⎞⎠⎟ΔVCC=10r1.18kΩ⎛⎝⎜⎞⎠⎟150mV = 1.25mVThis, in the rarefied air of good voltage references, is huge. We can appreciate that by consideringwhat temperature swing would correspond to this kind of change.ΔVzVz=1.25mV6.2V= 200 ppm = 40°C( )5 ppm / °C( )For those of us who work in Fahrenheit, it would take a temperature swing of 72°F to cause thismuch change! The point is that the care you took to temperature stabilize the Zener will beundermined by fluctuations or drift in the power supply.6.301 Solid-State CircuitsRecitation 23: Zener Diode References (continued)Prof. Joel L. DawsonPage 4One way to improve things is to use pre-regulation:Otherwise, we go to any number of self-biasing techniques. The class exercise is a good example.Another Self-Biasing TechniqueAs usual, our old friend the op-amp comes in handy here.Notice that the op-amp is just connected as a non-inverting amplifier of gain k. For purposes of theDC bias calculation, notice that R and C have no impact.R1VCCR2VZRZ−+RV0CRARB6.301 Solid-State CircuitsRecitation 23: Zener Diode References (continued)Prof. Joel L. DawsonPage 5Simplifying, we can draw this asAnd now Iz is straightforward to calculate:Iz=kVz− VzRz=k − 1Rz⎛⎝⎜⎞⎠⎟VzThe nice thing about this circuit is that it let’s us get stable voltage references at values other than Vz.For example, suppose that we need an output voltage of 10V, and our Zener diode has a Vz of 6.2Vwith a “sweet spot” current of 7.5mA. We select k according tokVz= 10V ⇒ k =10V6.2V= 1.61Then, we choose Rz to get the Iz that we want.7.5mA =1.61 −1Rz⎛⎝⎜⎞⎠⎟6.2V( )Rz=0.617.5mA⎛⎝⎜⎞⎠⎟6.2V( )= 504ΩVZ−+RZ+kkVZV06.301 Solid-State CircuitsRecitation 23: Zener Diode References (continued)Prof. Joel L. DawsonPage 6This circuit has the drawback of demanding a relatively large amount of current from the output ofthe op-amp. We can improve on things by adding a pull-up resistor as follows:Rp can be chosen to provide much of the Zener current. In addition, it helps with startup for