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16th September 2011, 06:27 PM  #1 
diyAudio Member

Jan's quizz in the February Newsletter
Thanks to all that took the trouble to answer my little quizz at the end of my shunt reg column in the September Newsletter.
The question was: given a 6.8V zener in the attached circuit, size R1 & R2 for 15VCD output. The insight needed was that the adj terminal of the 317 is always 1.25V below the output terminal. With the output terminal at 6.8V that puts the adj terminal at nominal 5.55V. So there will be 5.55V across R2 and 155.55= 9.45V across R1. According to ohms law, if we know the current through the resistors, we can calculate the R value, of course. Actually we can just pick a current we fancy it seems, but there's two other considerations. There's a minute current coming out of the adjust terminal, flowing through R2 which may upset the correct voltages so we chose the current high enough that the adj current is insignificant. The first answer I got was from Dick Middelkoop and he followed this exact reasoning, chosing 5mA current and getting R1 = 1.8k and R1 = 1.057 k, say 1k. Fred Dieckmann came up with a very similar reasoning but 3 days later. So, I nominate Dick for the Grand Prize, an F4 or F5 board set from the diyaudio store! Congrats! jan
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An engineer designing a feedback amp is a mixed emotions creature  Hendrik Bode Check out Linear Audio! Last edited by jan.didden; 16th September 2011 at 06:47 PM. 
16th September 2011, 06:31 PM  #2 
diyAudio Member

Several other replies:
A VERY complete and accurate reply came from Jean Herdies from Brussels. He actually calculate the values taking care of the adj current, so he needs much less current through the resistors. This would be a good solution for minimal power draw for instance in battery powered equipment. His solution is a model of accurate work, see below. Vo = 15V Vz=6,8V Vref= 1,25V Iadj= 50µA (from specs) Equations: (I1+I2) . R2 = Vz+Vref= 6,81,25=5,55V (a) (I1.R1)+((I1+I2).R2)=Vo=15V (b) => (Substituing (a) in (b)): I1.R1+5,55=15 => I1.R1=155.55=9,45V => I1 = 9,45/R1 (c) => substituing ( c) in (a) : (9,45/R1+50.106).R2=5,55 => R2 = 5,55/((9,45/R1+50.106) R2 depends solely on R1's value Let's assume R1 = 10 kOhms => R2 = 5,55/(9,45/10000+50.106)= 5,55/0,000995=5577,889 Ohms => R2 ~ 6k2 // 56k (5582 Ohms) Verification => substituing R2 and I2 in (a) : I1=(5,55/R2)I2 => I1~944,27µA => substituing I1, I2, R1 and R2 in (b): Vo=(I1.R1)+((I1+I2).R2)~14,99268V The real value depend on the precision of the components (Zener, Vref of the regulator, resistors) Value of R12 depends on the current drawn by the load, the value of the input voltage, the admissible power of the Zener the admissible power of the 317 and the load regulation (those values were not provided).
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An engineer designing a feedback amp is a mixed emotions creature  Hendrik Bode Check out Linear Audio! Last edited by jan.didden; 16th September 2011 at 06:47 PM. 
16th September 2011, 06:45 PM  #3 
diyAudio Member

Other solutions:
Kenneth: R1=1.7*R2 which is correct; John also calculated the ratio of R1 and R2 correctly; Koen mentioned correctly that the zener 'softens' the regulation and proposes to use a lowerimpedance reference (but Koen, Vadj = 6.81.25 not 6.8+1.25); Paul took the same route as Dick but selected 10mA standing current which leads to half the resistor values of course. That is the secong consideration I mentioned: setting the current too high leads to needless dissipation. Personally I would go to 3 or 5mA although 10mA isn't wrong of course. Omar also used a relatively large current: "Let R12 = 125R, R1 = 4K7 is ok for our purpose here. As long as you are using a 6.8V zener here, the original LM317 calculation Vout = 1.25 x (1 + R2 / R1), Vadj will become 6.81.25 = 5.55V rather than 1.25V. Hence the formula becomes Vout = 5.55 x (1 + R2 / R1). R2 is 8K." Sasmit took the opportunity to size the series R, R12: "The resistor (R12) sizing would depend on the supply voltage and the current draw. Given a 6.8 v zener and a 15V output, we would need at least 21.8V at the input. Assuming a current draw of 1A and supply voltage of 25V the resistor would need to drop around 3 volts. 3 ohms * 1 A , I think a 2.7 ohms 5 W resistor would work." Finally, Csaszar gave a correct solution but didn't explain how he got there. Thanks again guys! jan
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An engineer designing a feedback amp is a mixed emotions creature  Hendrik Bode Check out Linear Audio! 
16th September 2011, 09:47 PM  #4 
diyAudio Member
Join Date: Nov 2008
Location: Oakmont PA

Jan,
One issue about shunt regulators is that the bandwidth is better if you use a PNP on a positive rail as the shunt element. That way the final stage is a follower. (Follows for FETS, reverse polarity etc.) ES 
17th September 2011, 04:18 PM  #5  
diyAudio Member

Quote:
jan
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An engineer designing a feedback amp is a mixed emotions creature  Hendrik Bode Check out Linear Audio! 

17th September 2011, 05:17 PM  #6  
diyAudio Member
Join Date: Nov 2008
Location: Oakmont PA

Quote:


19th September 2011, 03:43 PM  #7 
Banned
Join Date: Aug 2009

For something complety different.
Why not put a 9 volt battery in series with the adjustment terminal. The very small adjustment current will act as trickle charge for the battery, and the battery adds practically no noise, and is free of dynamic impedance influence in this circuit location. You might have add some protection for power down conditions and I would suggest this would be best for an always on circuit which I belive all audio circuits should be except power amps. I think putting active voltage references between the out terminal of the LM317 and ground be looking for some real stability issues as well.
http://www.national.com/an/AN/AN181.pdf Better late than clever, Fred Dieckmann P.S. Doesn't that qualify for at least a Tshirt (Xtra large) 
19th September 2011, 04:09 PM  #8  
diyAudio Member

Quote:
You bring up several good points. The battery stuff will surely work but I personally don't like to use them as they always seem to die on me at the pinnacle of a great demo ;) I don't agree that the impedance in the output leg, a zener or other ref, worsens stability. In my experience, the larger the dynamic impedance, the *better* the stability. Which is logical; it decreases the loop gain, so less stability issues (but also less good regulation). Edit: No Tshirt I'm afraid; this is all for the honorary mentioning. ;) jan
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An engineer designing a feedback amp is a mixed emotions creature  Hendrik Bode Check out Linear Audio! 

19th September 2011, 04:23 PM  #9 
diyAudio Member
Join Date: Jan 2003
Location: ancient Batsch , behind Iron Curtain

I didn't even tried to participate , from 3 reasons :
1. you didn't provide BOM , and without BOM .... I'm  just as every proper DiyA  helpless ..... 2.there is more fun fiddling with really poor LM337 ; you used slightly better cousin , so there is no fun . 3. I like it shiny ... so I'll return it with LED String instead of zener ..... 4. I'm dumb ; and lazy ; fact . so  I'll return it with either R1 or R2 bypassed with small cap , depending which variant show less trouble on scope
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my Papa is smarter than your Nelson ! clean thread; Cook Book;PSM LS Cook Book;Baby DiyA ;Mighty ZM's Bloggg;Papatreasure;Papa...© by Mighty ZM 
19th September 2011, 04:29 PM  #10 
diyAudio Member

Yes a LED is a good solution. I don't know the Z of a LED but maybe it is lower than the 6.8V zener.
About being lazy; that makes two of us ;) BTW Looking forward to see you at BAF! jan
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An engineer designing a feedback amp is a mixed emotions creature  Hendrik Bode Check out Linear Audio! 
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