Sizing a zener and resistor for a regulator. - Page 1
Forum section: electronics / beginners

author: Hextejas
author_rank: Frequent Contributor
timestamp: January 09, 2018, 10:08:27 pm
content: I have a 13.7v DC supply that I want to regulate and drop down to about 10v. The load will be about 200ma.I have a schematic from the Forrest Mims book on learning electronics that has an exercise to do something similar.His exercise had Vin=12, Vout=6, load=.05maHis calculations led to a resistor of 60 ohms., and from memory it looked something like this.R = (Vin - Vz) ÷ load. I could be wrong on the formula as I am doing it from memory. Anyhow my questions are this.How do you start with the value of the zener. I guessed and picked a BZX559V1. For some reason I picked 9v as that was close to the 10 that I wanted.Next I calculated R to be 24 ohms but that led to a Bout of about 11.5. Too much. I doubled R to 47 and that got me closer.So, surely there must be a better way.Thanks

author: danadak
author_rank: Super Contributor
timestamp: January 10, 2018, 12:03:25 am
content: Design info here -https://www.onsemi.com/pub/Collateral/HBD854-D.PDFRegards, Dana.

author: Zero999
author_rank: Super Contributor
timestamp: January 10, 2018, 09:04:31 am
content: There is a better way. Scrap the zener idea altogether. Use an adjustable regulator, such as the LM317.

author: Paul Moir
author_rank: Frequent Contributor
timestamp: January 10, 2018, 12:08:25 pm
content: Buy 10 LM317, you'll use them.The point of a voltage regulator is to maintain a particular voltage no matter what the circuit connected demands. In real life this is up to a point, and this point is 200mA in your example. Also in your example you want to maintain 9.1V (from the zener) with up to a 13.7 input voltage. In this case R is simple to calculate V=IR, I = 200mA, V = 13.7-9.1, R = 23 ohms. No getting around that if you want 200mA into your circuit. If no current happens to be consumed by the connected circuit, then 200mA is going to have to be run though the zener. Since P=IV, that means 200mA is going to have to loose 9.1V through that zener, or 1.8W. The absolute maximum dissipation of the BZX559V1 is .5W with a 25 degree environment, so you're going to need a bigger zener.Hopping over to Digikey, I put in that I want a single zener, qty 1, in stock, voltage from 9-10.5V, and a wattage of something like 1.5W-5W. Sorting by price I have two options: A $.52 (CAD) 1SMA5925BT3G 1.5W 10V SMD zener or a $.53 1N5925BRLG which is thru-hole. Looking at the thru-hole one, you see it can totally dissipate 1W at 50C ambient. But we need to do 1.8W. Hmmm. Looking at the SMD one, we see it can totally do 1.5W when it's leads are 75C, but can only really do 0.5W when the environment is 25C. Another hmmm.Maybe we can solve this for more money. Get a 1N5346 for $.64. This can dissipate 5W according to the datasheet if we magically keep it's leads at 25C. But really since the junction to lead thermal resistance is 25C per watt and it would be pretty optimistic to get another 25C per watt out of the leads themselves, it's going to be pretty hot. We are running now around 9.5V with all that current going through the zener so that'll help. But maybe this can work. The resistor will need to be (P=V^2/R) (13.7-9.5)^2/23 ohms = 0.78W. Fortunately resistors don't care too much how hot they are so a 1W or 1.5W one will probably do. Check the datasheet of course!Compare all this to a TO-220 package LM317 which from Digikey costs $.84 for a genuine ST one. It can handle this all day long, and has a convenient screw hole which you can use to affix it to a heat sink if you need to help get rid of the heat. Also you can set it for exactly the voltage you want. It'll even shut down if it gets too hot.

author: Hextejas
author_rank: Frequent Contributor
timestamp: January 10, 2018, 12:44:40 pm
content: Thank you so much Paul, that is exactly what I needed. And including what to look for from the data sheet.The reason I used that particular zener is because I had a whole box of them.And I guess that explains why I burned my finger.😣Invaluable example of the work flow.As you might have guessed, this is all new to me and I still think it's a black art. 😁

author: Zero999
author_rank: Super Contributor
timestamp: January 10, 2018, 02:29:12 pm
content: Here's an LM317 calculator:http://www.reuk.co.uk/wordpress/electric-circuit/lm317-voltage-calculator/For 10V:R1 = 130RR2 = 910ROr:R1 = 120RR2 = 820R in parallel with 4k7.Or:R1 = 470RR2 = 3k3For the latter resistor values of 470R & 3k3, you'll need a minimum load of around 7mA to ensure proper regulation.Here's another, more generic potential dividers, which is better because it selects standard resistor values. It can be used for the LM317 by setting V1 to the desired output voltage and V2 to the LM317's reference voltage of 1.25V. The values generated for R1 and R2 will have to be transposed, for use in the above schematic.http://sim.okawa-denshi.jp/en/teikokeisan.htm

author: davy peleman
author_rank: Regular Contributor
timestamp: January 10, 2018, 03:45:15 pm
content: lm317 all the way

author: Hextejas
author_rank: Frequent Contributor
timestamp: January 10, 2018, 07:39:04 pm
content: Lm317 are on order with 2 day shipping so I will continue to use this over worked zener until it gets here.

author: james_s
author_rank: Super Contributor
timestamp: January 11, 2018, 01:57:50 am
content: The LM317 should do what you want. If you did want to build a regulator with a zener for such a high current what you need to do is add a pass transistor. The zener and resistor create a voltage reference which then controls the transistor which carries most of the current to the load. Using just the zener and resistor is only reasonable up to perhaps a few tens of mA.