Diode in GND pin of voltage regulator - Page 1
Forum section: electronics / beginners

author: tester43
author_rank: Regular Contributor
timestamp: July 03, 2018, 09:49:24 am
content: Let's use an example: LM7805;Input voltage 12V - output 5V (let's ignore if it makes sense for the moment);I made an experiment: supplied 2V to GND pin of LM7805;As expected output vs GND from LM7805 jumped to 7V in reference to gnd - because I gave him a reference of 2V and left 10V as voltage difference so LM7805 has got its space to operate and produce it's 5V output to it's given reference (5V) which is 7V when compared to global gnd.I found that it's possible to put just a diode in GND to obtain similar result. How does it work with diode between LM7805 and GND?

author: wraper
author_rank: Unknown
timestamp: July 03, 2018, 10:13:17 am
content: 78xx have relatively high bias current of a few mA flowing through GND pin. Thus it creates forward voltage drop across the diode. That's why they suck as adjustable voltage regulator and LM317 with resistive divider should be used instead.

author: not1xor1
author_rank: Frequent Contributor
timestamp: July 03, 2018, 04:11:20 pm
content: Quote from: wraper on July 03, 2018, 10:13:17 amThere is not much difference. The quiescent current of 78xx is around 4.3mA (typical - ST datasheet) while LM117 is granted to be stable with a load of 5mA and LM317 with 10mAs (although the few I checked work even with as little as 3-4mAs).If somebody needs an higher voltage and is not concerned by the voltage drift can just put a LED in series with the ground pin to get an increase of few Volts (depending on the forward voltage of the led).

author: ArthurDent
author_rank: Super Contributor
timestamp: July 03, 2018, 05:00:51 pm
content: As not1xor1 said, LEDs work as low voltage zeners. I used one with a 1.7 VDC drop in series with a 5 VDC supply to get 3.3 VDC at low current for a small circuit I was working with. As a bonus, I also got a visual indication that the circuit was drawing current because the LED was on. The drop isn't as consistant as a real zener but it isn't too bad.

author: tester43
author_rank: Regular Contributor
timestamp: July 03, 2018, 06:00:41 pm
content: Hi, does it mean that on reference pin of lm7805 potential is higher than 0,7V when not grounded (so diode can conduct when it is connected to reference and ground)?

author: wraper
author_rank: Unknown
timestamp: July 03, 2018, 09:31:31 pm
content: Quote from: not1xor1 on July 03, 2018, 04:11:20 pmYou mix minimum load with bias current through GND (common)/ADJ pin. For LM317 ADJUST terminal current typ 50μA, max 100 μA. LM78xx passes bias current through the common pin but LM317 through the load.

author: Audioguru
author_rank: Super Contributor
timestamp: July 03, 2018, 09:58:07 pm
content: The voltage across a diode changes when its current changes and when its temperature changes.

author: innkeeper
author_rank: Unknown
timestamp: July 03, 2018, 10:03:29 pm
content: i've used diodes to offset voltages on regulators and zeners, and other places for a quick known voltage drop without using a zener...I think this is a fairly common trick.the caveats is that VF (forward voltage) changes with temperature and current. so keep that in mind if you're looking for any kind of stability and accuracy.

author: innkeeper
author_rank: Unknown
timestamp: July 03, 2018, 10:04:41 pm
content: Quote from: Audioguru on July 03, 2018, 09:58:07 pmlol you beat me to the comment...

author: KL27x
author_rank: Super Contributor
timestamp: July 04, 2018, 09:51:19 am
content: QuoteThere's no need to dig around for the right combinations of diodes to get approximately the voltage you want and the resulting instability. You can dial in w/e output voltage you want, very precisely. Put one feedback resistor between the OUT pin and the GND pin. Let's call it R1. And you put one resistor between the GND pin and ground. Call that R2. Your output voltage is now Vref * (1+R2/R1). Vref is the nominal output voltage of the regulator, so for an LM7805 it is 5V. If that equation looks familiar, it's because an LM317 operates pretty much exactly the same way as an LM78XX, except the GND pin is called an ADJ pin and its Vref (nominal output voltage) is 1.25V. (And on the LM317, the heatsink goes on the output pin rather than the GND/ADJ pin).This is just Kirchoff's Law. If the voltage between OUT pin and GND pin is 5V, you can look at the resistor ladder to figure out the voltage between GND pin and the actual ground. And add those 2 numbers together to get the voltage bewteen OUT pin and ground.Using this "trick", you have to consider that using resistors of too high value will cause instability/malfunction. But it doesn't take much. R1 1K or lower for 7805 will probably work in most applications to get output voltage up to 20V.

author: Zero999
author_rank: Super Contributor
timestamp: July 04, 2018, 10:12:30 am
content: Quote from: KL27x on July 04, 2018, 09:51:19 amThat will work, but you need to consider the bias current, i.e. the current through the LM7805's ground pin, even for modest value resistors, so the calculation is a little bit more involved than that.VOUT= VREF(1 + R2 / R1) + (IBIAS× R2)

author: wraper
author_rank: Unknown
timestamp: July 04, 2018, 10:38:07 am
content: Quote from: Hero999 on July 04, 2018, 10:12:30 amBias current will also very significantly depend on particular IC, temperature, input voltage and output current. So using 78xx like this should be avoided.

author: KL27x
author_rank: Super Contributor
timestamp: July 04, 2018, 11:08:32 am
content: You can choose your own compromise with the feedback resistor method. You can squash the effect and offset of the bias current by choosing resistors that are low enough in value, at the cost of higher quiescent current. W/e your application, you can figure out for yourself if the regulation and the quiescent current are acceptable.In practice, this works perfectly fine. Of course neither of these things would ever find their way into a PCB unless you happen to have a ton of 78XX in your inventory which you want to get rid of. For prototyping and learning, they are both quite useful.The bias current is included in the equation in datasheet for virtually all LM317. And 99% of EE will ignore it for 99% of their applications. Just because you can doesn't mean have to. The end result of your circuit is going to be what it is, and if precision and temp stability is that important, tweaking is going to be virtually inevitable. And you are probably not going to be using an LM78XX to begin with.

author: Zero999
author_rank: Super Contributor
timestamp: July 04, 2018, 11:18:11 am
content: Quote from: KL27x on July 04, 2018, 11:08:32 amBut you have to go very low, to make the bias current negligible. For example, you want 10V, so use 470R resistors with the LM7805, for R1 & R2, the output voltage will be nearly 12V, an error of 20%, because the typical bias current is 4.3mA, according to the data sheet, but it could be as high as 8mA, giving just over 13V out.The bias current needs to be factored in, when calculating the resistor values and even then there's the change due to the input voltage, output current and temperature, which will ruin the performance slightly.QuoteBecause with the LM317, it's typically 50μA, nearly two orders of a magnitude lower than the LM7805 and the percentage variation due to temperture, input voltage and load current, is much less.In short, using the LM7805 as a variable voltage regulator is a hack. Use the LM317 in the first place.

author: KL27x
author_rank: Super Contributor
timestamp: July 04, 2018, 11:25:12 am
content: QuoteThis equation might work for LM317, but I don't think the bias current of 8mA will make that effect on the LM7805. It's been awhile since I did this, but I have a feeling even 5%erroroffset(which is fixable) with 470R R1 at double Vref is wayyyy too much. I think this is case of applying wrong equation correctly? Using values like this, I seem to recall ignoring bias current and getting output I wanted. Maybe I am wrong.anyhow, I agree fundamentally with what you said Hero, I just don't think it's significant as you think.I shall have to break out the breadboard and try it. If I have any 78xx.* edit: Well, color me wrong. Your application of the equation seems to be spot on. Using 500R resistors for R1 and R2, I got output voltage of 13V, lol. With this large of an offset contributed by the bias current, I suspect the diode/zener trick could be more temp stable when increasing output voltage by more than a couple volts.For small bump in voltage within a few diode drops, any effect would be proportionally less, of course. So I'm still conflicted if this isn't still better than diode trick for small increase in output voltage.I understand what Wraper is saying, now. Yeah, 78XX are technically not as good as LM317 in any way, except when they're good enough. 2 extra resistors is more cost of manufacture. Or more complex BOM and inventory if you are the manufacturer.

author: Zero999
author_rank: Super Contributor
timestamp: July 04, 2018, 12:42:23 pm
content: I think the diode hack will be slightly better than two resistors, because the voltage drop will be more stable, for changes in the bias current.Of course this isn't something which should be designed in. The purpose of such hacks are to get one out of trouble, when they've make a mistake and ordered the LM7805, rather than the LM7812 and there's no time to get the correct part.

author: not1xor1
author_rank: Frequent Contributor
timestamp: July 04, 2018, 05:09:47 pm
content: Quote from: wraper on July 03, 2018, 09:31:31 pmYou are right... and wrong... as I wrote "quiescent current" and "minimum load current"... so no mix-up.And in any case, regarding the question of the original poster, an LM78xx is perfectly equivalent to an LM317 with built-in voltage divider.

author: not1xor1
author_rank: Frequent Contributor
timestamp: July 04, 2018, 05:27:12 pm
content: Quote from: Hero999 on July 04, 2018, 12:42:23 pmOr use an opamp as a buffer to provide the required bias current while programming any voltage (within reasonable parameters) you need.

author: C
author_rank: Super Contributor
timestamp: July 04, 2018, 08:45:35 pm
content: Step one could be putting better labels on the chips.For a 7805 you have V+, 5 Vdc & 0 VdcYou could label a 317 as V+, 1.25 Vdc & 0 VdcIf current does not change at the 0 Vdc pin then a simple resistor will work. If it does change then you need more voltage regulation.The 317 uses a voltage divider, You can also do this with a 7805.You could use a zener, diode or LED to get less voltage change due to current change.A bypass resistor will allow a smaller current zener, diode or LED and adjusting where on current knee it will operate.This is all just working with current change and there are many ways to do the same, some better then others.Extreme is just a voltage regulator to raise 0 volt pin.C

author: Zero999
author_rank: Super Contributor
timestamp: July 05, 2018, 08:48:14 am
content: Quote from: not1xor1 on July 04, 2018, 05:27:12 pmWhat do you mean?That sounds silly, when the LM317 or the appropriated LM78xx could be used in the first place, to give the correct voltage. The resistor/diode hack is only useful for correcting a mistake: unsolder the regulator's ground pin and add a diode or series resistor and another resistor from the output to the ground pin.

author: not1xor1
author_rank: Frequent Contributor
timestamp: July 06, 2018, 04:16:36 am
content: Quote from: Hero999 on July 05, 2018, 08:48:14 amI'm referring to one of the circuit applications published in the ST datasheet (and quite likely in the original one of NSC).You might even program a different voltage by using a couple of high value resistors as divider and a series of NPN + PNP emitter follower to roughly compensate the Vbe drop.I like to think about various possible solutions... just for fun.

author: Relayer
author_rank: Contributor
timestamp: July 06, 2018, 10:23:19 pm
content: Hello tester43,Here's just a simple reason you would insert diode/s to ground:If say you only had a 12V regulator (i.e. 7812), but you wanted13.4V on the output, you just insert 2 diodes in series and it willgive you the voltage you need...Regards,Relayer

author: wraper
author_rank: Unknown
timestamp: July 06, 2018, 10:33:28 pm
content: Quote from: not1xor1 on July 04, 2018, 05:09:47 pmIt's not an equivalent at all.78xxLM317

author: not1xor1
author_rank: Frequent Contributor
timestamp: July 08, 2018, 06:33:36 am
content: Quote from: wraper on July 06, 2018, 10:33:28 pmequivalent adjectiveus ​ /ɪˈkwɪv·ə·lənt/equal to orhaving the same effect as something else:https://dictionary.cambridge.org/dictionary/english/equivalent#dataset-american-englishBTW the original (i.e. Latin) meaning is just "works like, has the same value as", not "it is equal to".Even if you buy the same IC (e.g. LM358) from different manufacturers, you get something that is equivalent but which is not an exact copy.

author: wraper
author_rank: Unknown
timestamp: July 08, 2018, 07:29:15 am
content: Quote from: not1xor1 on July 08, 2018, 06:33:36 amYep, it does not work like LM317 with resistive divider. If we add resistor divider to LM317 and assume it's now became a common pin, then bias current through this pin won't depend on load, temperature, input voltage. It's not equivalent because there is a huge part of the circuit attached to common pin of 78xx besides internal resistive divider (yes it's there). It's basically will be a stable current source, therefore it will be different what you can do with it. Say you could just attach another resistor on GND side to increase the voltage, and it will be stable. If you do this with 78xx, it won't be stable.BTW don't forget what you said:Quote from: not1xor1 on July 04, 2018, 05:09:47 pm