Multiple 1N400x diodes to drop voltage before linear V regulator? - Page 1
Forum section: electronics / projects

author: george graves
author_rank: Super Contributor
timestamp: August 07, 2014, 02:49:26 am
content: Linear voltage regulator....Input is 14vOutput is 5vCurrent is 300mASo need to get rid of (14v-5v)*0.330mA=~3watts------------My idea is this....input 14 volts --->1n4007--->1n4007--->............---->1n4007---->1n4007---->5 volt VReg------->OUTPUT(330mAout)So the idea i had with diodes is basically, trying to drop some of the voltage before the linear reg so that a huge heat sink isn't needed (switching V Reg isn't an option).Is it wrong to use diodes to drop the input voltage? I'm using one diode already as a reverse polairity protection. Why not a few more? (must be something wrong with that idea, since I haven't seen anyone do it!)I know using a power resistor/zener is another option(as well as a second v-reg), but I wanted to know if/why diodes are a bad idea.(currently rewatching Dave Electronics Thermal Heatsink Design Tutorial #105 to calculate heat sink size - cause I have no clue what size 3 watts needs)Thanks!

author: c4757p
author_rank: Super Contributor
timestamp: August 07, 2014, 02:54:34 am
content: Diodes will work, resistors are just typically cheaper. You could do all that with just one big power resistor.However, 3W isn't a huge amount. That's easily done with a moderately sized heat sink, and with the exception of "buck converters are more efficient", there's not really much of a reason not to just dump it into one.

author: mariush
author_rank: Super Contributor
timestamp: August 07, 2014, 03:13:31 am
content: The voltage drop on diodes varies with current and temperature. At 300mA, you may have only 0.5v drop, at 600mA you may have 0.7-0.8A drop. Diodes usually are also bad at transferring heat... you have Power = Vdrop x I so 0.15w in each diode which is ok, below the typical 250mW or higher, but still a waste of space on pcb.As c4... says, at 0.3A you can make it with just a reasonably sized heatsink but if you want something cheaper, you could use a 7809 to drop the 14v down to 9v and then use the 7805... this way each regulator will generate about 1.5w of heat, so a very tiny heatsink on each will be enough to keep them at ok temperatures (hot, but below the maximum temperature).

author: SirNick
author_rank: Frequent Contributor
timestamp: August 07, 2014, 04:04:33 am
content: Why isn't it done more often? Well... honestly, because it's an ugly hack.All you're effectively doing is shedding heat over more surface area (dissipating in the diodes as well as the reg), and although diodes will work for this, two or more cascaded regulators also work, while potentially also coping with a wider input range. Or just use a big honkin' heat sink and deal with the heat in a brute-force manner. The difference is mostly academic though.For the majority of cases, by the time you're uncomfortable with the amount of heat dropped by a (single) linear reg, it's just time to move to a switching reg.That said, you actuallydosee cascaded diodes used to provide a more or less predictable voltage difference between two points in a circuit -- like between two legs of a transistor. (IIRC, there were a couple such circuits shown in the mic preamplifier video.) After all, diodes are cheap and readily available. But this is usually low-current stuff, and in a place where a proper regulator wouldn't be a logical choice.

author: void_error
author_rank: Frequent Contributor
timestamp: August 07, 2014, 07:26:03 am
content: At 330mA, assuming your regulator is not a LDO type, you'll have to drop about 3V (or maybe you can get away with 2V) across it to maintain regulation so you end up dissipating 0.33*3=0.99W on your regulator so you'll have to use a small heatsink anyway for TO220 devices (unless you want your reg to run really hot).Here's an example:7805 5V linear regulator (data is from the fairchild semi datasheet)junction to ambient thermal resistance (Rth J-A): 65C/Wfor 1W and say, 40C (TA) ambient temperature your regulator's junction temperature (TJ) will be:1W*Rth J-A+TA=1*65+40=105C, pretty hot I'd sayEven if you use diodes/resistor to drop the input voltage you'd still have to use a small heatsink for your regulator to get 330mA out of it.

author: Kjelt
author_rank: Super Contributor
timestamp: August 07, 2014, 07:39:20 am
content: You can do it (taking into effect the diode characteristics mentioned in posts above) but the reason you should not in this century is that you waste more energy then you use and that nowadays you can cheaply and easily save it by using a Switch mode regulator like you can buy on a ready made pcb at ebay for $1,50 (look for LM2596).Unless it is an ultra low noise supply design for audio or RF (then you probably won't use the 7805 anyway) I don't see the point to consume 4,5W's while only needing 1,5W.So you have the option of using the SMreg in front of the 7805 then set it at 3V higher so 8V and you still loose 1W in the 7805 + 0,5W in the SMreg instead of 3Wor you can use the SMreg directly to drop the 14V to 5V and then you have about 75-80% efficiency thus only loosing 0,7W.

author: richard.cs
author_rank: Super Contributor
timestamp: August 07, 2014, 08:40:37 am
content: I would be inclined to do this with resistors. If your supply never drops below 14V and your 300 mA is the maximum current you draw then to allow 2V across the regulator at full load you need to drop 7V. 7/0.3=23.3, using a 22R would drop 22*0.3=6.6V and dissipate 2W.Dissipating the excess power in resistors has advantages over doing it with diodes or the regulator because 1) they're cheap per Watt, 2) The junction to case thermal resistance of cheap diodes is fairly poor, 3) resistors can safely run hot so you can dissipate the power from a smaller surface area than you could with semiconductors. Moving some of the dissipation from a semiconductor to a piece of carbon-coated ceramic is a reasonable thing to do so long as you know that all you're doing is increasing the surface area to shed heat from, the same amount of power is lost.I wouldn't consider it an ugly hack, I often include it in designs at work where it's considered a reasonable design approach.

author: Psi
author_rank: Super Contributor
timestamp: August 07, 2014, 08:48:09 am
content: i've been known to put a few diodes in series when powering something with a voltage a bit close to, or outside, the specified voltage range.The LCD screen for my mini quad is rated at 7-15VDC, which is good for running from a 3S lipo (10.5 - 12.6V) but no good for a 4S lipo (13.2 - 16.8V).With 3 diodes added inside the screen i can drop a 4S lipo down to within range while still being in-range for a 3S. Being able to run from ether battery is really useful and diodes are cheap and easy to wire up.

author: Richard Head
author_rank: Frequent Contributor
timestamp: August 07, 2014, 10:47:56 am
content: There's no advantage in using series diodes over a series resistor. There's no point in using two series regulators unless you already have a large heatsink available.You need to dissipate a certain amount of power so use the best device for that, a power resistor.The advantages of a resistor over a semiconductor is that they can operate at a higher temperature, are more reliable and are lower cost.Just remember, a 5W resistor can't actually dissipate 5W under realistic conditions.

author: dannyf
author_rank: Super Contributor
timestamp: August 07, 2014, 10:50:33 am
content: QuoteIt works, if the current is sufficiently stable or sufficiently high.Otherwise, the voltage varies greatly with current.

author: eneuro
author_rank: Super Contributor
timestamp: August 07, 2014, 11:34:22 am
content: Quote from: Richard Head on August 07, 2014, 10:47:56 amWhat do you think about using Zener diode before linear V regulator with power resistor?Idea behind this is to protect additionaly linear regulator against too high imput voltage which could be higher than 27V or even 36V while sometimes I've 24-26VAC secondary transformer output, so close to 37Vmax when additional capacitors are used.Of course it depends on current needs for regulated voltage, but while I'm usually below 100mA @ 5V (about 0.5W regulated power), it looks like it makes sense, while capacitors before linear regulator are charged to lets say 9V, so lower current flows to them at higher voltage and I've additional some kind of overvoltage protection which keeps my LM7805 in safe input voltage range, and not needed power is disipated in power resistor and zener diode when input voltage is above expected voltage level?So, maybe a few stages Zener diodes with power resistors could disipate better than series diodes with bonus overvoltage protection?

author: dannyf
author_rank: Super Contributor
timestamp: August 07, 2014, 11:38:35 am
content: QuoteAt 300ma of load current? I don't think such zeners exist - you can, however, combine power transistors with zeners to form "power" zeners.

author: Richard Head
author_rank: Frequent Contributor
timestamp: August 07, 2014, 01:39:30 pm
content: My suggestion was based upon an input voltage of 14V maximum, as per your original post. If you want to be able to handle a 36V maximum continuous input then you will require a Zener to clamp the input to about 30V maximum to be safe. Most 78xx series linear regulators have a max input voltage of 36V if I remember correctly. The series resistor must now dissipate the additional power due to the Zener, but that shouldn't be too great. It sounds like this sort of voltage is really worst case and not continuous.

author: wagon
author_rank: Frequent Contributor
timestamp: August 07, 2014, 01:56:00 pm
content: I was installing a box of electronics I built onto a machine (a mulcher, with a 350hp diesel motor). The design used an LM2917. On the bench it worked perfectly, but it wouldn't run on the machine (using the tacho signal from the variable reluctance sensor on the flywheel). I worked out that it was putting out too much signal. It's a 2 hour round trip to get stuff from my workshop. All I had was a bandoleer of 1n5404 diodes. I made a long string of the diodes and stuffed it across the output of the sensor - like a big, bodgy zener. It worked and stayed that way!

author: David Hess
author_rank: Super Contributor
timestamp: August 07, 2014, 02:20:11 pm
content: Quote from: Richard Head on August 07, 2014, 10:47:56 amThe series diodes will have a lower impedance so the voltage at the regulator input will be more stable over the range of operating currents.Quote from: dannyf on August 07, 2014, 11:38:35 amZener diodes with a rating of 5 watts are readily available.

author: David Hess
author_rank: Super Contributor
timestamp: August 07, 2014, 02:27:15 pm
content: Quote from: eneuro on August 07, 2014, 11:34:22 amThe way I have done this in the past is to use a single transistor cascode in front of the regulator with a zener diode between the transistor base and regulator output and a resistor from the collector to the base to supply bias power. The zener holds a constant input to output voltage differential across the regulator and excess power is dissipated in the transistor. The regulator's current limit protects the transistor and the transistor's high Vce protects the regulator from high input voltages.A better design replaces the resistor with a wide voltage range current source and adds over voltage shutdown to protect against excessive power dissipation in the transistor.With the above, I have designed low voltage power supplies which could withstand input voltages of 400 volts or higher indefinitely during fault conditions.

author: digsys
author_rank: Unknown
timestamp: August 07, 2014, 02:32:33 pm
content: The LM317HVT goes to 60V IP ! Just add an appropriate heatsink (and the 2 set resistors of course).I still don't see why you wouldn't just use of of the many cheap S/Mode drop-in replacements. They have very low ripple, which is easilyreduced even further.

author: dannyf
author_rank: Super Contributor
timestamp: August 07, 2014, 03:10:41 pm
content: QuoteWhat about picking actual parts for this assignment and then maybe we can see if you can find that zener?QuoteThat's the concept of pre-regulator. LEDs work in place of the zener and double as indicators too.QuoteOnly to the extent that the input voltage isn't too high. Or the dissipation on the pre-regulator can be significant.QuoteIt is unlikely to get your input voltage that high, with moderate current output (1amp or so for example).

author: David Hess
author_rank: Super Contributor
timestamp: August 07, 2014, 03:46:12 pm
content: Quote from: dannyf on August 07, 2014, 03:10:41 pmI did not make the suggestion without having checked pricing and availability:http://www.mouser.com/Semiconductors/Discrete-Semiconductors/Diodes-Rectifiers/Zener-Diodes/_/N-ax1mhZscv7?P=1z0w7s7Z1z0w89lZ1z0w7thZ1z0w749Much higher power zener diodes in metal packages used to be commonly available but are not something I would design in anymore unless there was no other choice.QuoteDidn't I mention something about that in my post?A better design replaces the resistor with a wide voltage range current source and adds over voltage shutdown to protect against excessive power dissipation in the transistor.QuoteThis was in the 100 milliamp range but high voltage surges were part of the specifications that the design had to meet. I suspect they were thinking in terms of load dumps, lightning strikes, and actual battle damage.

author: dannyf
author_rank: Super Contributor
timestamp: August 07, 2014, 05:13:07 pm
content: QuoteMaybe a schematic with values / parts would help.Quote100ma at ~400v drop can create lots of heat.and if your goal is to protect against surges, there are better ways.

author: David Hess
author_rank: Super Contributor
timestamp: August 07, 2014, 06:18:25 pm
content: Quote from: dannyf on August 07, 2014, 05:13:07 pmThe issue was the feasibility of dropping the input voltage to a linear regulator with a series string of diodes or in this case a zener to distribute the heat dissipated. A schematic is hardly needed to replace a bunch of series connected diodes with a zener diode and looking up the part number of an inexpensive and available 5 watt zener diode is trivial.QuoteIt sure can but it was not required to withstand that level of overload indefinitely. It ended up being several orders of magnitude tougher than required.It looked a little odd to have a big surface mount power MOSFET as a cascode for a small SOT-223 regulator. My boss was doubtful as well so I made of demonstration of plugging it into the nearest wall socket. While not required, the design was proof against negative surges also.He was similarly doubtful about meeting the low temperature operating requirements until I pointed out that I was doing destructive testing at that point and the environment chamber was dribbling liquid nitrogen out the front and the thermometer was pegged at -299.9F. Early component selection had ruled out liquid electrolytic capacitors and high dielectric constant ceramic capacitors among other things.QuoteShunt protection like a clamp at the input was forbidden and overload recovery had to be fast and without damage. The device could stop working but not fail after normal power was restored.This was a project where we had problems because of a requirement for hermetically sealed packages where feasible. The glass diodes used for ESD protection had excessive leakage do to photo-generation. We ended up painting them black after assembly.

author: SeanB
author_rank: Super Contributor
timestamp: August 07, 2014, 06:45:16 pm
content: I think I have enough zener diodes though. I use them as small signal diodes if the peak voltage is less than 20V. With a controlled forward current they glow brightly and you can stretch them out a little to make a glass fibre. Another box has 12V diodes on bandoliers. Got them at auction years ago for under $2 the lot.

author: David Hess
author_rank: Super Contributor
timestamp: August 07, 2014, 06:57:44 pm
content: I picked up a roll of 3000+ 1N270 germanium diodes at a swap meet many years ago. I finally had to repackage them because the glue holding them into the strips was drying up allowing them to escape.They are very handy for low current switched capacitor charge pumps, baker clamps, low current switching regulators, and low bias RF detectors.

author: SeanB
author_rank: Super Contributor
timestamp: August 07, 2014, 07:01:44 pm
content: Same here, they do tend to get all over the place from the bandoliers. I could do with a couple of dozen of those diodes. Want to swap some of each, I can see just how many fit in a padded envelope?

author: eneuro
author_rank: Super Contributor
timestamp: August 07, 2014, 08:30:40 pm
content: Quote from: David Hess on August 07, 2014, 02:27:15 pmI've made TL431I overvoltage detection circuit which can protect power supply (and itself) against longer overvoltages, but sometimes in simple designs when small controll current is needed and no issues with a few wats lost I put before LDO LM2940CT-5 zener diodes with bigger resistors to ensure this regulator will not switch off when unexpected higher input voltage hits this circuit despite it has transients protection to 60V for given amount of time as I remember.Than powered circuit has a chance to "fight" and switch off something in the case of overvoltage-not just only let switch off regulator and force circuit to restart if BOD enabled in my AVR MCUs...

author: Alex Eisenhut
author_rank: Super Contributor
timestamp: August 07, 2014, 11:24:49 pm
content: What's supplying the 14V? One trick I did a long time ago was to reduce the filter capacitor after the bridge to increase the ripple voltage. As long as you keep within the limits, this reduces the heat dissipated by the regulator, and the regulator still performs per spec. Especially since they are tuned for PSRR near 100ish hertz.

author: David Hess
author_rank: Super Contributor
timestamp: August 08, 2014, 06:24:00 am
content: Quote from: Alex Eisenhut on August 07, 2014, 11:24:49 pmBe careful that the input capacitor is not too small. Sudden load changes will create voltage spikes from the transformer's leakage inductance which may be large enough to damage the regulator. I think Bob Pease mentioned this in one of his articles.

author: komet
author_rank: Unknown
timestamp: August 08, 2014, 08:37:36 am
content: Quote from: SeanB on August 07, 2014, 06:45:16 pmYou should make a video of this. I for one would watch it.

author: eneuro
author_rank: Super Contributor
timestamp: August 08, 2014, 10:40:53 am
content: Quote from: David Hess on August 08, 2014, 06:24:00 amThat is rightI've found it sometime ago when 2xIRFZ44N 0.018 RDSON each in pararell started to warm up too much at very small current, while those voltage spikes from switching off transformer secondary (21Vmax in normal conditions) created around 65VDSmax spikes and those Mosfets started work as 55V zener diodes in avalanche in my first version of desulfator... It was simply buck converter topology step down, but... without inductorThan I've run simple circuit simulator and it was clear that adding a few big capacitors in pararell on transformers secondary after full bridge let me charge old car battery in desulfator mode (without inductor) without problems and those Mosfets now are only warm at 5A-10A . Hopefully I had LDO LM2940CT-5 voltage regulator and quite big capacitors on its input and additionaly 24V zener diode too, so nothingso far.

author: electrophiliate
author_rank: Regular Contributor
timestamp: August 08, 2014, 11:18:07 am
content: Yes it should work, as will the other methods discussed on this thread. IIRC, the heatsink required to dissipate 3 watts shouldn't be big anyway. But if you're really pressed for space and/or efficiency, the best choice given the modest 14V in and 5V @ 300mA out, is a small switch-mode dropin replacement for linear regulators that someone already suggested earlier. I recently used the R-78E5.0-0.5 for a very similar situation and it was about $4:http://www.recom-power.com/pdf/Innoline/R-78Exx-0.5.pdf

author: dannyf
author_rank: Super Contributor
timestamp: August 08, 2014, 11:38:19 am
content: The diode approach will not work for a dynamic load (like a digital circuit), due to the voltage dropped by the diode string being dependent on the current drawn by the load.Here is an example.14v DC dropped by 5 diodes to about 10-11v. The load is a sinus current load (to introduce variability in the load). With a capacitor to help smoothing out, the voltage on the load varies about 1v. Without the capacitor (which is the design we are talking about here), the voltage varies about 3v, with lots of high frequency content.

author: dannyf
author_rank: Super Contributor
timestamp: August 08, 2014, 11:49:59 am
content: A single zener can hardly provide sufficient dissipation.However, you can build one with the help of a power device.Here is an example of how it can be done with a zener + nmos. The voltage drop is approximately the zener's voltage + the mosfet's Vgs. As you will notice, the power zener is a 2-terminal device and can easily replace a regular zener.

author: SirNick
author_rank: Frequent Contributor
timestamp: August 08, 2014, 07:12:01 pm
content: By that point, you've created a regulator, so why not justusea regulator?This is getting academic, way beyond the point of practicality. I'm sure the OP just had a lightbulb moment, then wondered why no one had ever thought of it before. The answer being, they have -- but there are almost always better ways to do it.