ECEE 352 Analog Electronics DC Power Supply Winter 2016 This Document Produced By: Leo Filippini lf458@drexel.edu Instructor: Prof. Basavaiah basu@coe.drexel.edu TA: Zhihuan Wang zw78@drexel.edu The goal of this project is to design, simulate, and build a voltage regulator given a set of specifications (each group will be assigned a different output voltage to target). The following is the project outline: Understand the design flow and the function of each part of the circuit Size the components to meet the specifications Simulate and get confident about the design Build the circuit Remember to always work one stage at a time. When you re sure everything is working put them together. In this way it is easier to find and solve problems. Introduction If we are interested in the performance of a voltage regulator we can ask the following questions: 1. How much does change when changes? 2. How much does change when I L changes? 3. What is the power efficiency? The first two questions deal with the line regulation and the load regulation, respectively. The former can be written as Line regulation = (1) while the latter as Load regulation = I L (2) Let s now analyze the performance of the voltage regulator of figure 1a. From its equivalent circuit of figure 1b we can write R = V Z0 + r Z r Z R I L (3) R + r Z R + r Z R + r Z where I L depends of course on the voltage. The reason why I left it there is because we are interested in the load regulation which has the term I L in it. From equation 3 is evident that the line regulation is = r Z (4) R + r Z 1
R R I R I Z I L V Z0 r Z (a) Simple regulator (b) Equivalent circuit Figure 1: A simple voltage regulator using a Zener diode and its equivalent circuit while the load regulation is where if R r Z we can safely write I L = r ZR R + r Z = R r Z (5) I L r Z (6) Equations 4 and 6 are the results found in exercise 4.7 on page 192 of the book. So far we have addressed the first two questions and what is left to talk about is the power efficiency of our beloved voltage regulator of figure 1a. The easiest way to see this circuit s limitation is to take a look at the two extreme cases: loaded and unloaded. Let s assume, for the sake of simplicity, that our Zener diode is ideal, thus r Z = 0. In this case the current I R is constant no matter what the load is and we have I R = ( V Z0 ) /R. The power consumption is then ( V P = I R = + ) V Z0 (7) R which, as you can see, does not depend on the load. This means that the voltage regulator will always dissipate the maximum power it was designed for, even when there is no load. The goal of this project is to address these three limitations of the voltage regulator of figure 1a. Strategy In order to tackle the critical points of the simple voltage regulator we will do what engineers do: divide et impera 1. Instead of trying to solve everything we will take care of each issue one at a time then put together what we have. Figure 2 shows the approach we will take in designing the voltage regulator. We will have a first circuit addressing line regulation and a second one dealing with load regulation and power efficiency. In the next few sections we will analyze these blocks and eventually put them together to form the final voltage regulator that you will design, simulate, and build. 1 The famous latin idiom meaning divide and conquer 2
Improve line regulation Improve load regulation and power efficiency Figure 2: The building blocks of the voltage regulator for this project. Each block will take care of some of the issues we talked about, so that the overall system will have none. V B I B I Z I R I B V B R 1 (a) (b) (c) Figure 3: An improved version of the circuit of figure 1a, where the use of a current source lowers the line regulation of the circuit (remember that when we compute the line regulation we assume ). A single PNP can be used as a current source when in active region and V EB is constant. In order to guarantee immunity from variations, a circuit using a Zener diode may be used to provide a constant voltage V EB = V B = V Z0. Improve line regulation Equation 4 tells us that in order to reduce the line regulation we can Reduce r Z Increase R since we, as circuit designers, don t have control over devices parameters like r Z 2, we have to try and increase R as much as possible. Ideally we d want R so that the line regulation goes to zero. Taking this approach has a major drawback, that is it limits the ability to deliver high currents to the load. A solution is to use some component that has an infinite small-signal resistance: a current source. Figure 3a shows such a circuit, and it s easy to see that the line regulation in this case is zero (since no matter what is the current through the Zener diode is constant). 2 We could of course pick a different device, but even if we had the Zener diode with the smallest available r Z, it wouldn t solve the problem. What if we wanted an even lower line regulation? We have to find a way of improving the design without relying on better devices. 3
D Z1 R 2 Q 1 I Z2 I L R 1 D Z2 Figure 4: The final version of the circuit of figure 3a. The highlighted part is the current source of figure 3, where we added R 2 for improved performance. The next step is to build a current source to use in our circuit. There are many different ways of doing that, but the simplest current source is a PNP transistor in active region (see figure 3b). As long as V EB = ( V B ) V γ and V EC > V ECsat the transistor will be in active region and behave as a current source. As you know, in active region the current of a PNP can be expressed as I C = I S e V EB V T (8) where I S is the saturation current and V T the thermal voltage. For the circuit of figure 3b it is I C = I S e ( VB) V T (9) so the current depends on the line voltage, which is the opposite of what we wanted. How do we get rid of this dependence? We ll have to guarantee that V B changes with such as V B = V constant (10) so that the voltage V EB will always be constant 3. A way of doing so is to use the circuit of figure 3c. In order to simplify its analysis we ll assume that the I Z I R, thus the current I B coming from the transistor is negligible. Replacing the Zener equivalent model in this circuit we can write V B = ( V Z0 ) R 1 R 1 + r Z (11) which becomes equation 10 only when R 1 /(R 1 + r Z ) = 1. In order to do so we ll have to pick R 1 as large as possible. 3 Since we d have V EB = V B = ( V constant ) = Vconstant 4
V B I Z I B V B I L I Z I B I L (a) Single BJT (b) Darlington configuration Figure 5: The circuit of figure 3a with an improved load regulation and power efficiency Figure 4 shows the final circuit of the simple voltage regulator with improved line regulation. There is one difference with respect to what we just said, and that is the addition of R 2 on the emitter of the BJT. Can you come up with good reasons why that resistor is there? It s important to note that this circuit only improves the line regulation and that the load regulation is still given by equation 6. Improve load regulation and power efficiency From the current equations at the output node in figure 3a is clear that there is a direct relationship between the variations of I L and I Z (e.g. if the load current increases by 1 ma the current through the Zener diode will decrease by 1 ma). Putting this down mathematically we have I Z = I L (12) What if we could weaken this proportionality? In order to do so we can exploit the current gain characteristic of a BJT, as shown in figure 5a. Here we have I Z = I B I L β (13) where we used the approximation I E βi B for a BJT in active region when β 1. Clearly we want β as big as possible in order to reduce the impact that a variation on I L has on I Z. For this reason we can use a Darlington transistor (which is just two BJT connected as shown in figure 5b) in order to boost the current gain β and decrease the load regulation even more. You are encouraged to reason on why this approach also improves the power efficiency of the voltage regulator. What is the output voltage of the circuit of figure 5a? Using the Kirchhoff s voltage law = V Z V BE (14) which tells us that there is a dependence from the current I C through the term V BE, since we know that ( ) IC V BE = V T ln (15) 5 I S
R 3 Q 3 D Z2 R 4 Figure 6: The feedback version of the circuit of figure 5b. Note that the Darlington transistor is labeled because it can be seen as a single transistor with a large β and usually comes in a three-pin package. where V T and I S are the threshold voltage and the saturation current of the BJT, respectively. Let s rewrite the output voltage with this information and we get ( ) IL = V Z V T ln I S (16) where we assumed β 1 so that I C I L. From the last equation is clear that even if the Zener diode were ideal (constant V Z ) we d have a nonzero load regulation. There are three major drawbacks of the circuit of figure 5a: Doesn t work without load From equation 16 we know that when the transistor is off and = V Z, while when the BJT is on we have roughly = V Z V γ Can t finely pick Again, in the general case = V Z V γ but what if we want to pick an output voltage for which we don t have the right Zener diode? Load regulation is improvable From equation 16 we see that an increase in I L would lower by some amount, because the BJT would need a bigger V BE to deliver the increased current. We will need to counteract this phenomena by increasing V B so that = V B V BE stays constant In order to address these issues we will introduce a feedback, as shown in figure 6. Here the only differences from the circuit of figure 5b are transistor Q 3 and resistors R 3 and R 4 : the components that realize the feedback. Intuitively, a decrease in is read by Q 3 through the voltage divider R 3 and R 4. This is what happens: I L V BE2 V B3 V B2 (17) and this improves the load regulation since now we have V B2 that goes in the opposite direction with respect to I L. Also, note that even without load the output voltage will be constant since now resistors R 3 and R 4 guarantee that some current always flow through. 6
D Z1 R 2 D Z1 R 2 Q 1 R 3 Q 1 Q 3 R 1 R 3 R 1 D Z2 R 4 Q 3 D Z2 R 4 (a) All the previous blocks put together (b) Classic arrangement of the circuit on the left Figure 7: The final topology for this project. These two circuits are exactly the same, just in different arrangements. They are reported because the one on the left flows from top to bottom and it s easier to visualize, while the one on the right is usually found in literature and datasheets. Finally, the output voltage can be written as = (V Z2 + V BE3 ) ( 1 + R ) 3 R 4 (18) where we assumed that the current I B3 is negligible with respect to the current going through R 3 (so we can use the voltage divider equation). Equation 18 shows that the output voltage can be tuned by modifying the ratio R 3 /R 4, hence giving us a new degree of freedom when designing the voltage regulator. Putting it all together The final circuit is shown in figure 7 in two equivalent forms. Try to identify for each arrangement the building blocks we discussed in the previous pages. Remember: the designer must consider that the single blocks of figure 2 eventually are merged to form the circuit of figure 7. Therefore when picking the components for a block, one has to make sure that everything will work as expected when the pieces come together, too. For example, you must guarantee that each BJT will be in active region for every operating condition: ripple on, loaded and unloaded output, et cetera. 7
Specifications Your circuit must meet the following requirements Output voltage = 9 12 V (your TA will assign one to each group) Output ripple,ripple 50 mv Output power P O 1 W Load regulation 4 0.2 % Line regulation 50 mv/v Part list These are the parts that you will find in the lab and that you re encouraged to use. You can, of course, use any other parts you d like but, since some may not be available in the laboratory, you might have to buy them. Component Name Type Specifications 1N4007 Diode 1N4728 Zener Diode 3.3 V, 1 W 1N4734A Zener Diode 5.6 V, 1 W 1N4736A Zener Diode 6.8 V, 1 W 2N3904 NPN transistor 350 mw 2N3906 PNP transistor 350 mw TIP120 Darlington NPN 65 W Heat Sink for TIP120 Capacitor up to 1 mf Resistor any Table 1: The components that will be available to you for this project Please remember: do not use the components outside their power or voltage range! For example, electrolytic capacitors have a polarity and a voltage range. Not respecting these specifications may cause them to explode, so be careful. 4 Here we are using a slightly different definition of load regulation, namely (,Unloaded,Loaded ) /VO,Unloaded 8
I + D Z1 R 2 R 3 D 1 D 4 Q 1 C P D 3 D 2 Q 3 R 1 D Z2 R 4 Figure 8: The final circuit, from the transformer to the output voltage Sizing The purpose of this section is to guide you through the sizing of each element in the circuit of figure 8, given our specifications. First of all we want to know what the average will be, and the ripple on that node (which of course will be different from the ripple on the output node). Since our regulator will provide a constant voltage, the current drawn from it is at least I + = / which we can also write as I + = P O / (since we have a specification on the power). To be sure that our circuit will work also in the worst case scenario 5 let s double that current and write I + = 2 P O (19) where I + is the current drawn by the regulator, as shown in figure 8, and is the output voltage of our specifications. With this value we can now compute the ripple on the node, which is ripple = I+ 2C P f (20) where f is the frequency of the input voltage (60 Hz). Since we want this ripple to be as small as possible, we will pick a capacitor as large as possible (remember, the limit is 1 mf). Once we know the ripple that the node will have in the worst case (when there is a load at the output node) we can calculate and also avg = peak 2V γ 1 2 ripple (21) min = peak 2V γ ripple (22) where peak is the peak value of the line and V γ the voltage drop across a diode of the rectifier. These values, V avg + and min, are very important, since we will have to guarantee that our regulator works when is in this range. 5 To be more precise we would have to budget some power for the circuitry of the voltage regulator. In other words we d need to take into account that some power is wasted to make the circuit functioning. 9
Let s start by sizing R 1 and picking D Z1 : as we saw in equation 11 we want R 1 to be as big as possible, hence we will use the Zener diode with the smallest V Znom. Also, we ll have to make sure that I Z1 I ZT 1 (where I ZT 1 is the test current for the Zener diode D Z1, check the datasheet) so then we have a boundary for R 1 R 1 = min V Z1nom I Z1 min V Z1nom I ZT 1 (23) where V Z1nom is the Zener voltage of D Z1 and we assumed I Z1 I B1. How do we pick R 2? It depends on the current I C1 R 2 = V Z1 V EB1 I C1 V Z1nom 0.7 I C1 (24) where I C1 should be large enough to provide for I Z2 and I B2, so assuming I Z2 I B2 we can simply say I C1 I ZT 2 (25) and pick R 2 consequently. Let s now turn our attention to D Z2, R 3, and R 4. When picking the diode we have to guarantee that the transistor Q 3 will be in active region, so so V CE3 = V BE2 V Z2 1.4 V V Z2nom > V CE3sat 0.2 V (26) V Z2nom < 1.6 V (27) Now from equation 18 you can pick R 3 and R 4 ratio. Finally, for these resistors to form a voltage divider we have to make sure that I R4 I B3, thus R 4 = V Z2 + V BE3 I R4 from which we also get the value for R 3. V Z2nom + 0.7 V I R4 V Z2nom + 0.7 V I B3 (28) 10