Design problem from AoE - Page 1
Forum section: electronics / beginners

author: stoica adrian
author_rank: Contributor
timestamp: December 11, 2019, 08:20:05 pm
content: Hello, i work on this design problem from AoE and i need your help, can you please help me with some advice, on how to resolve this problem, Thanks.Design a +10V regulated supply for load currents from 0 to 100mA; the input voltage is +20 to +25V. Allow at least 10mA zener current under all(worst case) conditions.A. what power rating must the zener have?B. With the same specifications use a sener emitter follower. Calculate worst-case dissipation in transistor and zener.

author: iMo
author_rank: Super Contributor
timestamp: December 11, 2019, 08:44:45 pm
content: My advice to you would be to download the LTspice and draw the circuit in. You will learn how to design and simulate your AoE circuits.

author: todorp
author_rank: Regular Contributor
timestamp: December 12, 2019, 02:49:54 pm
content: Hi, a long time ago I found solutiuons to some of the exercises for the AoE 2nd edition. Look after page 17 (Bipolar transistors) in the attached pdf file. Some of the exercises in the 3rd edition are the same. Mind it is a good idea to first try to do the exercises as you have done and than use the solutions :-)

author: T3sl4co1l
author_rank: Super Contributor
timestamp: December 12, 2019, 04:25:32 pm
content: Incidentally, 100 ohms is not a "best fit" common value -- that will violate Iz(min). Note that Iz(min) / Iz(max) is a factor of 10%, and 100 ohms is larger than 90.9 by about 10%. Prefer smaller resistances (and adjust power ratings upward as needed). 91 and 82 ohms should be easy enough to find, otherwise 100 || 1k can be used.P_R is correct given the assumed resistance (100 ohms), but obviously it's a bit higher with the calculated resistance. By about 20%, or 2.7W. Because "10% more" (a ratio of 1.1) squared is about 1.2 ("20% more"); or exactly, 1.21.Pzmax is simply Izmax * Vz or 1.1W.For B, first, show the circuit. Consider the zener: what bias does it need*? Does it depend on the transistor characteristics? If so, which ones do we need to know to solve it? Should we begin selecting components immediately, or further consider the problem before looking up parts?*Note that Vz varies with Iz, and diodes are rated at a given Iz. So, if we're varying Iz, Vz will vary some as well. For sake of simplicity, let's just assume that we can select a diode which has the correct Vz at whatever Iz we're using in the circuit. We can worry about that later (when it comes to component selection).The transistor dissipation will be closely related to the voltage drop and load current; indeed for common components, it's fair to assume (because the error will be less than, say, 10%) that this is all of it, so we can immediately solve for the approximate power dissipation as simply(Vin,max - Vout) * Iout,maxor about 1.5W.Tim

author: iMo
author_rank: Super Contributor
timestamp: December 12, 2019, 08:09:02 pm
content: Your Rc calculation is not good, imho. The transistor has to get some Vce, say 3-4V.PS: also your calculation has to involve the transistor too. You need some Ibe to allow for some Ice.1) I_load = Ice + Ibe2) Vce = 3V (for example)3) Vload = (Ice + Ibe) * Rload = 10V (it will never be precise 10V with a zener)4) Ibe = Ice / beta (where the beta is an amplification coefficient of the transistor at specific Ice and Vce, 50 - 200 for example)5) I_Rb = Iz + Ibe6) Iz = (Vin - Vz) / Rb7) 0 = Vin - Ice*Rc - Vce - Vload8 ) Vload = Vz - Vbe = Vz - 0.7V (aprox)PPS: indexed the equations

author: T3sl4co1l
author_rank: Super Contributor
timestamp: December 12, 2019, 08:10:41 pm
content: Is that beta = 100 for the transistor? Then Ib,max = 100mA/100 = 1mA, so Iz,max = 10+1mA.Rc actually uses Vin - (Vout - Vce(sat)), not Vb -- Vce can saturate lower than Vbe. Practically speaking it's not only a small difference, but also one you would want to avoid with some margin (or alternately, Imax can be overestimated by a bit).Tim

author: hermitengineer
author_rank: Regular Contributor
timestamp: December 13, 2019, 10:46:38 pm
content: Quote from: stoica adrian on December 11, 2019, 08:20:05 pmHow much is "at least 10mA"? That condition seems open-ended. 500A is "at least 10mA". And your chosen max amperage will directly affect the required power rating of the zener. Seems like there are an infinite number of correct solutions for this question. Or am I missing something?

author: T3sl4co1l
author_rank: Super Contributor
timestamp: December 14, 2019, 02:50:49 am
content: Quote from: hermitengineer on December 13, 2019, 10:46:38 pmPedantically speaking, there should be a constraint calling for maximum efficiency / least overall current consumption / power dissipation, yes. That may've been understood elsewhere.Tim

author: hermitengineer
author_rank: Regular Contributor
timestamp: December 17, 2019, 05:52:16 pm
content: Oh yeah, the first condition is what I missed.Now, it appears that this question came before actual discussion of the emitter follower addition, so the author expects the answer to be a simple resistor and zener regulator.Now, with the first requirement: 0-100mA regulated 10V with input voltage between 20 and 25V, I was about to dismiss the second (at least 10mA under all conditions) as superfluous, but I think that means the author wants the zener to sink 10mA even with the 100mA load. So that means the resistor should be sized to limit current to 110mA instead of 100mA at 20V. For that requirement, the resistor should be 10V/110mA=91ohms.Any disagreement there?Anyway, that means the zener will take 165mA under the worst condition (25V input, no load). So its absolute bare minimum power requirement is 1.65W.