Almost too afraid to ask: why not use a diode to lower voltage supply? - Page 1
Forum section: electronics / beginners

author: Watth
author_rank: Frequent Contributor
timestamp: January 01, 2022, 01:48:49 pm
content: It's such a simple solution that I'm afraid I missed something obvious. Can a diode be used to lower voltage ? E.g. lower from a battery or set of battery to power components that accept lower voltages ?

author: wraper
author_rank: Unknown
timestamp: January 01, 2022, 02:09:45 pm
content: Quote from: Watth on January 01, 2022, 01:48:49 pmIn some cases it can be done with reason, however in most cases it's a dumb endeavor adding a diode for such reason. You still won't get stabilized output voltage but also uselessly lose energy on that diode. Also voltage drop is dependent on current consumption and temperature, so even if input voltage is stabilized, output voltage after the diode won't be that stable. When diode is added for polarity protection, voltage drop is an undesired side effect.

author: madires
author_rank: Super Contributor
timestamp: January 01, 2022, 03:07:53 pm
content: I don't see the power loss as a large problem because you would have that too with a linear regulator. When Vcc may vary by a few 0.1V and just a low current is involved diodes are fine for dropping a small voltage. Above 2 or 3V voltage drop I'd use a jelly bean regulator.

author: wraper
author_rank: Unknown
timestamp: January 01, 2022, 03:15:55 pm
content: Quote from: madires on January 01, 2022, 03:07:53 pmA battery was mentioned. You would use LDO in a battery powered application if you need a stabilized voltage but don't want to bother with SMPS or current consumption is too low for SMPS. Using a diode in battery powered application means that you will have reduced operating voltage range since output voltage will follow declining battery voltage.

author: madires
author_rank: Super Contributor
timestamp: January 01, 2022, 03:40:44 pm
content: Yep, the output voltage follows the battery voltage in that case. However, the circuit's power requirements matter.

author: Wallace Gasiewicz
author_rank: Super Contributor
timestamp: January 01, 2022, 04:01:29 pm
content: Strings of series diodes can be used to decrease the voltage, however all the power goes thru each one. Each diode drops the volts. A germanium will drop by less than a Silicon diode.You can also use a Zener "in line", if your V+ is a 9 Volt battery and you put a 5 Volt zener "in line" you will get 4 volts available on the other side of the zener. Again all current goes thru the zener.I have seen these configurations used in older HP test equipment when low power circuits are involved.

author: BeBuLamar
author_rank: Super Contributor
timestamp: January 01, 2022, 04:34:53 pm
content: I do use a diode for one such application often. Most of our machinery have regulated 24VDC supply. From time to time I would need 10VDC to power a potentiometer to supply a voltage speed reference to a variable speed drive. I would use a 14V Zener diode in series with the potentiometer to have the 10VDC. 10VDC power supplies are actually rare. You can find a 12VDC easily but not 10VDC.

author: nfmax
author_rank: Super Contributor
timestamp: January 01, 2022, 04:40:39 pm
content: I used a single diode to drop the voltage from a stack of 4 AA cells, to power only one IC - CD4060 oscillator/divider. It turns out the power consumption of the oscillator increases rapidly with supply voltage, so the power reduction from a 0.6V lower power supply (with fresh batteries) more than outweighs the power 'wasted' in the diode. And the 4060's outputs still swing high enough to fully drive the inputs of the remaining 4000 series CMOS. It should make a significant improvement in battery life.

author: T3sl4co1l
author_rank: Super Contributor
timestamp: January 02, 2022, 02:05:19 am
content: You can. Do you want to, though?The point of regulation is, when you have a load that can't handle a wide range of supply voltage, but your source spans a wider range than that. Say you have an an MCU that needs 2.5-3.6V. But a battery that does 3.3-4.2V or something. Well, in this case, it might be adequate to simply drop the 0.6V or so.What if the battery is 3-6V (like 4 x AAAs)? You can't just drop some voltage off the top, the span is much too wide -- so the dropper must be variable, and so we need a regulator. Which, we could still set at 3.3V nominal, or just shy of 3.6V maximum, and allow its output to drop as the input fades below that (most LDOs, simply track the input when there's not enough overhead to regulate properly). Or we could set just above the 2.5V minimum, and have no change in performance or whatever, all the way down 'til the battery is empty. Which way you choose in this situation, doesn't really matter, at least to the extent that nothing being powered has any tighter limitations of course.The range can come into play at AC as well -- and usually with a narrower margin. A small wiggle in supply voltage can be coupled into, say, analog audio circuits, adding noise or hum or whatever. Maybe a particular load can only handle 100mV of ripple, or 10mV, or even less, in some frequency range. If that load also draws varying currents, say of 10mA amplitude, then to maintain its say 100mV voltage spec, we need a supply impedance below (100mV) / (10mA) = 10 ohms at those frequencies. If we simply wire a series resistor to that load, we can drop some voltage, but only up to this value -- and only if the DC current is stable to suit. We can get better stability with the logarithmic curve of a diode Vf, but this still has some resistance (roughly Ri = 26mV / If), and we will again fail spec if it's too much.We can however, help the AC impedance with a bypass capacitor, or other filtering. Actually, as it happens, most LDOs have a fairly high impedance at AC too, so need to be helped with a capacitor -- make sure you use the right type (usually ceramic or tantalum for low/modest ESR respectively), they can be unstable (prone to oscillation) otherwise.So, the most common use case for diode or other dropping, is where the load current is fairly stable (so the voltage drop is also stable, and the change in load current (AC) is modest), and where the source voltage is fairly stable. Maybe you need 2.5V for a few IO banks in a digital circuit, and already have 3.3V regulated; a diode might be a suitable choice here.BTW, germanium diodes are nonexistent today; schottky diodes have similar voltage drops however. The curve is a bit softer than PN diodes, so keep that in mind. Also, Vf depends on temperature, another source of variability.Tim

author: wizard69
author_rank: Super Contributor
timestamp: January 02, 2022, 04:03:49 am
content: I've actually have seen diodes in products that was apparently put there as voltage drop. That was a guess on my part but I've also seen diodes used as capacitors and that actually called out on a print.

author: Gyro
author_rank: Super Contributor
timestamp: January 02, 2022, 10:18:25 am
content: Satellite/Cable/Terrestrial set-top box designs transitioned from separate tuners and main PCB mounted QPSK/QAM/COFDM demodulator ICs, to all-in-one integrated single canned NIMs about 25 years ago. I remember that Samsung used the diode drop method to power the (now integrated) 2.5V demod ICs from the incoming 3V3 supply. I think that counts as a seriously high volume production use case.These days, with 1.8V Silicon, I guess LDOs are used.

author: Wallace Gasiewicz
author_rank: Super Contributor
timestamp: January 02, 2022, 12:21:05 pm
content: All diodes have capacitance. Some types ate designed for this function. They vary the capacitance in response to voltage across the diode. They are called varicaps and are used in tuning circuits for resonant frequencies in many radio functions.Much less expensive than mechanical variable capacitors.

author: Siwastaja
author_rank: Super Contributor
timestamp: January 02, 2022, 12:34:07 pm
content: Sometimes it just happens that all you need is approximate voltage drop of around half a volt. And it happens you already have a general purpose schottky or standard Si diode in your Bill Of Materials. In such case, there's nothing wrong using it. I definitely have done this once or twice.Just be aware that the voltage drop over diode is not a constant; it's a logarithm of current. If the load current varies over many orders of magnitude, drop over the diode is far from constant. So if you add a diode to prevent parts from failing because of overvoltage, then if the load goes into very low consumption, the drop over the diode can be just like 0.2V even if it's 0.7V during active period.

author: Capernicus
author_rank: Frequent Contributor
timestamp: January 02, 2022, 12:35:59 pm
content: I dont see a problem with it, using diodes might be easier at the time. If your starting from a definite voltage rating, that is quite constant, I dont think a regulator is mandatory.Another cheap fix is just a good ol' voltage divider. (with a couple of resistors.)pos+__res1_____________.................|.................\.................|..................|..............res2..........LOAD/OUTPUT.................|..................|-..________|__________/negthe voltage output is the relative ratio of the resistors.

author: MK14
author_rank: Super Contributor
timestamp: January 02, 2022, 02:22:25 pm
content: Quote from: Capernicus on January 02, 2022, 12:35:59 pmThat doesn't really work well in practice. Because the output voltage, will vary (drop), considerably, as the load current varies. Some kind of voltage regulator, would usually give much better results.

author: Gyro
author_rank: Super Contributor
timestamp: January 02, 2022, 02:54:17 pm
content: Quote from: Capernicus on January 02, 2022, 12:35:59 pmVoltage dividers are useless for deriving supplies (as the OP requests). They only provide the calculated ratio when they are sourcing zero current. For the upper resistor to be low enough in value to source any sensible supply current, the lower resistor will need to be correspondingly low value, wasting huge amounts of power relative to the load.The diode drop method will work in limited circumstances (where the required drop and poor regulation match up with one or two diodes), otherwise active voltage regulation is the answer. Resistive dividers are never a solution for a current drawing load.

author: Capernicus
author_rank: Frequent Contributor
timestamp: January 02, 2022, 03:04:06 pm
content: Quote from: MK14 on January 02, 2022, 02:22:25 pmIn some cases (Like for a robot for example) The load is constant, then it is ok in this case.And If it turns out there is a variable load in your machine, you can modify the design so it is constant.

author: MK14
author_rank: Super Contributor
timestamp: January 02, 2022, 03:16:49 pm
content: Quote from: Capernicus on January 02, 2022, 03:04:06 pmMy understanding is, that in the 'old' days of, valve (tube) electronics. When electricity was especially cheap and plentiful. Doing what you just said, and using dropping resistors (which, are effectively creating voltage dividers), for the big series connected (valve/tube) filament strings, was sometimes done. Also, cheap IC voltage regulators, didn't exist then, and the regulators of the day (such as in the 1950's), would have been big, somewhat expensive, regulating tubes, which were not especially impressive or accurate, compared to modern technology.In principal, you are right. If it is definitely a constant load, doesn't need especially precise voltages and other things, such as fault conditions, don't apply. Then it could be used. But I still suspect, proper voltage regulators, would still be a better approach, as it would give more accurate power usage, consistent results, be more fault tolerant and less dramatic when things go wrong.

author: Gyro
author_rank: Super Contributor
timestamp: January 02, 2022, 05:25:14 pm
content: Yes, if the load draws a constant current, or has a constant resistance, then you can just use a simple single series resistor. The load replaces the lower resistor, it is not in addition to it.

author: m k
author_rank: Super Contributor
timestamp: January 03, 2022, 03:07:06 pm
content: Another volume use was PC RTC circuit.

author: MK14
author_rank: Super Contributor
timestamp: January 03, 2022, 03:30:48 pm
content: Quote from: m k on January 03, 2022, 03:07:06 pmI'm guessing, without researching it. But, I suspect the Quiescent Current, of a voltage regulator. When powering a very low current load, such as a CMOS (presumably), real-time clock circuit, intended to keep going, for many years, maybe even 10 years, on a (usually) very low capacity battery (such as a CR2032 coin cell). Would discharge the battery, way, way too quickly, even if the Quiescent Current, was quite low, for most applications.So, using diode drop(s), to get the voltage into a suitable range and/or lower the voltage, to lower the current consumption (which applies in some cases). Makes perfect sense.

author: BrokenYugo
author_rank: Super Contributor
timestamp: January 03, 2022, 05:42:02 pm
content: Maybe PCs were doing something different but I thought the diodes in a RTC/SRAM battery circuit were only there to keep the Vcc rail from trying to charge the battery, and the battery from trying to power everything on the Vcc rail. IIRC Schottky is recommend to keep the drop low and maximize usable battery capacity.

author: wraper
author_rank: Unknown
timestamp: January 03, 2022, 05:48:51 pm
content: There is no need to drop battery voltage in PC RTC circuit. It's an unwanted side effect.

author: capt bullshot
author_rank: Super Contributor
timestamp: January 03, 2022, 05:52:02 pm
content: Quote from: Gyro on January 02, 2022, 10:18:25 amBack in the days, I've done a similar thing: 2.5V DSP core supply from 3.3V by a simple diode drop. Didn't work out well because the diode drop was too high at low ambient temperatures and the DSP went haywire. My fault, of course, didn't check all limits over all conditions thoroughly enough. Replaced by an LDO and everything went fine.

author: m k
author_rank: Super Contributor
timestamp: January 04, 2022, 04:36:33 pm
content: Quote from: MK14 on January 03, 2022, 03:30:48 pmI was thinking the time before.But maybe back then the whole PC thing wasn't really a volume yet.There were also a possible feature where clock stopped before setup memory were lost.

author: tkamiya
author_rank: Super Contributor
timestamp: January 05, 2022, 08:06:50 am
content: I've done this and it works just fine.In earlier days of my hobby, I didn't have a proper power supply. So I took four AA batteries to make 6 volts, and one serial silicon diode to drop it 0.7V. TTLs were perfectly safe and functional at 5.3V.

author: Siwastaja
author_rank: Super Contributor
timestamp: January 05, 2022, 09:40:06 am
content: Quote from: tkamiya on January 05, 2022, 08:06:50 amThis works because the TTL logic and any pullups always draw significant quiescent current, preventing the diode Vf going catastrophically low. Besides, the TTL ICs likely would have been fine at ~6.5V (4 * full 1.5V cell) to begin with. So that silicon diode actually dropped something between 0.4-0.7V, making the supply never exceed 6V at least by much.For example, picking any random TTL datasheet,http://ee.hawaii.edu/~sasaki/EE260/Labs/Datasheets/7400.pdfabsolute maximum rating for supply voltage is given as 7.0V.Modern CMOS parts tend to have much larger ratio between minimum and maximum currents. Minimum current can be so tiny that a normal Si diode that you expect to drop 0.7V only drops like 0.2-0.3V. OTOH, maximum peak current could be in amperes, making the drop highly variable between 0.2 - 1.2V. Adding extra bypass caps to account for this won't pay off, so just using a regulator tends to be simpler.But yes, it does have usable edge cases where it works out. Just be aware of the risks.

author: Canis Dirus Leidy
author_rank: Regular Contributor
timestamp: January 05, 2022, 02:40:07 pm
content: Such method (diode as voltage dropping resistor) was used in the external battery module for theSverdlovsk-4 light meter: three mercury batteries or NiCd accumulators were normally used there, while the module was intended for AA batteries.