Zener supply - Page 1
Forum section: electronics / projects

author: OM222O
author_rank: Frequent Contributor
timestamp: August 18, 2018, 12:54:34 pm
content: I wanted to design a voltage clamping circuit that takes an input voltage of 9-22 volts and clamps it to a maximum of 12v (i.e output range of 9-12v) to drive low power chips (total of 0.3 watts).Should I just use a simple zener + series resistor to power my chips or is it better to use the zener as a voltage reference and use a unity gain op amp to drive my load? the cost/space difference is not that important. what is the drawbacks of using an op amp if any?

author: ocset
author_rank: Super Contributor
timestamp: August 18, 2018, 02:17:34 pm
content: Ok so you say your output is 0.3W….so 9v and 0.3W means that you have 33mA . When your input is 22V, then the power dissipation in the regulator will be 433mW.If I were you I’d use something like a LT1934-1 (or similar) to give you the power.Or maybe if you want to do it linear then use a SOT89 (or bigger) package linear regulator.Or an R/ZENER/BJT regulator, with the BJT being SOT89 or bigger.

author: OM222O
author_rank: Frequent Contributor
timestamp: August 18, 2018, 02:50:49 pm
content: I don't really want to go below 9 volts so a linear regulator or another transistor is not really a good idea.

author: JS
author_rank: Frequent Contributor
timestamp: August 18, 2018, 03:08:54 pm
content: Linear regulators are a better choice than the zener, do the math for the zener, put a 1W zener maxed out, at 12V that's about 80mA plus the 0.3W load that's 25mA.So 105mA, for 12V output 22V input 10V diff, 95Ω.Now lets go at 9V input, I'd assume the load keeps the same current.95Ω x 25mA = 2.4V6.6V will be the output when the input is 9V using a zener blowing 1W and a resistor blowing another W for a 0.3W load. Witha linear regulator you can get the heat dissipation at about 0.5W and the output voltage at 9V input can be over 8V, like it could be 8.8V for some parts.JS

author: Zero999
author_rank: Super Contributor
timestamp: August 18, 2018, 03:38:11 pm
content: Quote from: OM222O on August 18, 2018, 12:54:34 pmA unity gain op-amp would be no different to using another transistor, because its output stage is just transistors.You don't want to go below 9V, yet the minimum input voltage is 9V, so a zener and resistor won't work, because the resistor will always drop some voltage, which would be unacceptable when the input is 9V.How about a low-dropout regulator?

author: OM222O
author_rank: Frequent Contributor
timestamp: August 18, 2018, 04:25:29 pm
content: The drop across the resistor is not gonna matter much (a few hundred mv worst case scenario) compared to 1.3 or 1.4v of an LDO ... also a unity gain op amp will compensate for any drops ... that's actually how you can build an "ideal diode" which has no drops across it. I'm just wondering if it has any negative effects like worse clamping (chips have a maximum of 14volts VCC but should run at 12 nominally), so I don't want to go too high, but they're performance falls horribly around 8.5volts and they stop working at 8 volts, so I want the absolute 9-12 volt range that I can get supplied.

author: JS
author_rank: Frequent Contributor
timestamp: August 18, 2018, 04:45:04 pm
content: I did the math and with 25mA load you wont get under 2.4V drop at the resistor. An LDO won't have over one volt, more like 200mV to 300mV. So you are thinking things the wrong way. The only way to work with an opamp if you use a P channel mosfet as output and an opamp diving it but stability might be an issue.JS

author: Siwastaja
author_rank: Super Contributor
timestamp: August 18, 2018, 05:45:23 pm
content: Quote from: OM222O on August 18, 2018, 04:25:29 pmThe market is full of LDOs with dropout way below 1.3 or 1.4V, down to around 0.1V. For example, see all listed below 0.5V max dropout:https://www.digikey.fi/products/en/integrated-circuits-ics/pmic-voltage-regulators-linear/699?k=&pkeyword=&pv2065=570&pv2065=573&pv2065=1280&pv2065=21&pv2065=933&pv2065=436&pv2065=29&pv2065=324&pv2065=333&pv2065=5&pv2065=439&pv2065=186&pv2065=337&pv2065=1034&pv2065=1047&pv2065=359&pv2065=244&pv2065=623&pv2065=109&pv2065=54&pv2065=527&pv2065=629&pv2065=298&pv2065=934&pv2065=1077&pv2065=390&pv2065=409&pv2065=93&pv2065=48&pv2065=737&pv2065=141&pv2065=1347&pv2065=851&pv2065=1033&pv2065=57&pv2065=114&pv2065=1049&pv2065=484&pv2065=602&pv2065=1048&pv2065=242&pv2065=497&pv2065=536&pv2065=596&pv2065=763&pv2065=329&pv2065=850&pv2065=299&pv2065=236&pv2065=34&pv2065=241&FV=8f40064%2C8f40069%2C8f40087%2C8f40088%2C8f40093%2C8f40010%2C8f400a8%2C8f400aa%2C8f400b1%2C8f400bb%2C8f400bc%2C8f40013%2C8f400c2%2C8f400db%2C8f400e4%2C8f40019%2C8f4001a%2C8f4001c%2C8f40118%2C8f4011d%2C8f40022%2C8f40023%2C8f40029%2C8f40034%2C8f4003c%2C8f40043%2C8f40044%2C8f4004d%2C8f40051%2C8f40053%2C8f40054%2C8f40059%2Cffe002bb%2C20440409%2C2044040a%2C20440414%2C20440417%2C20440418%2C20440419%2C20440435%2C2044006d%2C20440072%2C20440500%2C20440543%2C2044008d%2C204400ba%2C20440002%2C20440015%2C204400ec%2C204400f1%2C204400f2%2C204400f4%2C2044001d%2C2044012a%2C2044012b%2C20440144%2C20440149%2C2044014d%2C20440151%2C20440022%2C20440167%2C20440186%2C20440199%2C204401b4%2C204401b7%2C20440030%2C204401e4%2C204401f1%2C20440005%2C2044020f%2C20440218%2C20440036%2C20440039%2C2044023a%2C2044023d%2C20440254%2C2044025a%2C2044026f%2C20440275%2C204402e1%2C204402fb%2C20440352%2C20440353%2C2044005d%2C204403a5%2C204403a6&quantity=0&ColumnSort=0&page=1&stock=1&nstock=1&pageSize=25719 results, in stock.Also note that if you want to minimize the dropout in your proposed shunt regulator resistor, you need a massive shunt current, and massive power dissipation. Just do the math at minimum and maximum operating points and you'll find out this is not a good application for shunt regulator - just use a LDO series regulator.

author: Zero999
author_rank: Super Contributor
timestamp: August 18, 2018, 07:25:20 pm
content: Quote from: OM222O on August 18, 2018, 04:25:29 pmHow did you work that out? Show your calculations.You've not thought this through. When the input voltage is 9V, the zener diode will do nothing, so all you have is a resistor in series with the load, with a voltage equal to IR across it. Suppose you select the value, so there's only 200mV across the resistor. Before you said the power is 0.3W, which at 9V would be a current of 33mA. The resistor would have to be R = V/I = 0.2/0.0333 = 6R to drop 200mV at 33mA. Now suppose you select a 12V zener, to limit the voltage. The current through the resistor, when the input voltage is 22V, would be I = (VIN-VZENER)/R = (22-12)/6 = 10/6 = 1.67A, giving a total power dissipation of 36.67W.QuoteAs mentioned above, there are plenty of low dropout regulators, which will drop a couple of hundred mV at 33mA.QuoteHow are you going to power the op-amp?You obviously studied the output characteristics of op-amps.The maximum output voltage is always considerably less, than the power supply voltage, even for rail-to-rail types. The only way your idea would work is if the op-amp was powered by a higher supply voltage, than the output voltage. In short, a unity gain op-amp with a 9V zener reference at the input, will behave identically to a 9V linear regulator, with similar drop-out characteristics.

author: NiHaoMike
author_rank: Super Contributor
timestamp: August 18, 2018, 10:11:40 pm
content: A PMOS based LDO can have close to zero dropout voltage.

author: coppercone2
author_rank: Super Contributor
timestamp: August 18, 2018, 10:20:49 pm
content: I think his question is about the transient response characteristic of the simple zener diode circuit in comparison to the LDO and robustness for transients.The zener/resistor circuit is much simpler in terms of manufacturing process (in the factory) and the RF behavior should be more linear, since a typical LDO is quite complicated in comparison to a diode. I think the diode will have a worse tempco but lower noise, compared to the bandgap reference that is probobly used inside of a LDO reference voltage. It's voltage stability should be better with temperature cycling and stuff too since its a simpler structure. A regulator will have over 10 transistor in it.I would like to know myself. How does the frequency response look like?A beefy zener diode might be more robust then a LDO. I like the idea of a beefy ESD diode followed up by a zener diode circuit. It seems like a cool idea.https://www.onsemi.com/pub/Collateral/HBD854-D.PDFthis has some information about zener diode capacitance and frequency response.

author: Zero999
author_rank: Super Contributor
timestamp: August 18, 2018, 11:29:50 pm
content: The problem with a simple zener is the resistor would have to be very small to get a low dropout voltage, which would result in a huge current, at higher voltages.I've simulated all three ideas given here: a simple zener with a resistor, an op-amp and an LDO. V1 is the supply voltage which is swept from 9V to 22V and is shown on the horizontal axis. All circuits have a 33.33mA constant current load, which might be different to what your circuit draws and is just a guess.At first the zener appears to be good, but look at the current though the series resistor R1, it's over 1.6A, the voltage does rise above 12V, at higher input voltages and the zener diode is running well over its maximum rating of 1W: it's dissipating nearly 20WThe op-amp is poor because its output stage has a high dropout voltage. When the input is 9V, the output is only 7V and it doesn't reach 12V, until the input is 14V.The LDO is the best because it doesn't draw excessive current and hardly drops any voltage, when the input is low.

author: OM222O
author_rank: Frequent Contributor
timestamp: August 19, 2018, 03:23:54 pm
content: The op amps seem to have a high dropout ... I knew the simple zener + resistor couldn't really hold up to the task which is why I wanted to follow it with a unity gain op amp, but as others mentioned, it will have issues even if it's a rail to rail op amp. A low dropout regulator with less than 300mv drop seems to be the best option here. Thank you all.