Is there a secret over-engineering prize here for figuring a way to power something that was originally packaged and powered with a 59¢ watch battery?
Thank-you gotpasswords, I was beginning to think I was the only engineer not working in gold-plated aerospace heaven…
BTW, 5 diodes in series would be a rubbish regulator unless the current drain was constant, as the forward voltage drop across a diode is proportional to the current through it. When the current is zero, the forward voltage drop is zero also. That’s why I biased the 2-diode shunt circuit I proposed at 5mA - (safely) assuming a current requirement of <1mA, they’ll drop about 0.6 to 0.7V each, depending on type.
I design stuff for bespoke excellence and quality, and I design stuff for production in the millions of units, and it’s useful to know the difference between the two. If the application is low-current, then a simple shunt regulator is the most reliable option. Zeners will fail (s/c usually) if they’re overheated or subject to high energy spikes, but if you can avoid this they’re great.
There’s plenty of lovely shiny ways to get 1.5V regulated from 5V, but I challenge anyone to find one simpler, cheaper and more reliable than a diode shunt regulator. They’re used in billions of products worldwide, and the quantities involved are so huge that any inherent unreliability in the technique would produce an unacceptable level of returned faulty units.
This is incorrect. Up to a certain current level, the forward voltage drop does increase nonlinearly with current, however, after that point, the voltage drop is nearly constant. See the diode conduction curve here. Note the sharp knee at the bottom of the curve. When the voltage has exceeded the bandgap voltage (about 0.7 V for a typical silicon diode, as shown here), large increases in current result in only a small increase in voltage drop. Only within the very narrow range of current inside the “knee” of the curve is the voltage drop approximately proportional to current flow.
It only looks like a knee on a linear current axis. On a log current axis, it’s a straight line. All the way from 0,0 to the upper operatin limits. For a silicon diode at least; the law is more e^whatever for germanium, and I’m not-sure-what for LEDs, though the latter have the sharpest ‘knee’ of all when viewed on a linear scale. If they were less noisy they’d make a great voltage reference in their own right.
Well, duh. What do you think “proportional” means?
My sloppy definition, QED. Indeed the word “proportional” does imply a linear relationship, unless specifed otherwise, e.g. “inversely-proportional”. I merely meant to imply that the forward voltage drop of a diode rises with forward current. It never truly flattens out, even on a linear scale. The actual relationship is generally logarithmic or has an exponential component. For example, a voltage drop across a typical 1N4148 silicon signal diode will be 0.5V at 0.1mA, and 0.7V at 5mA. That’s pretty poor regulation for such a small current range. You could stabilise the forward voltage drop by swamping the load current with a second, larger shunt load, but then you might as well use a shunt regulator and save some diodes.
Quite true, but I was assuming the 5V was coming from a PC or some kind of logic board, in which case there’ll be 5mA to spare easily. Even on a laptop. But it’s kind of fun to design micropower stuff - it certainly has its applications.
There are no 5V batteries. None of the usual elements for battery electrodes produce such a useful voltage.
Oh for crying-out-loud. (I am spending way too much time on this.)
Let’s assume the load pulls 10 mA ±30%, which means it pulls anywhere from 7 mA to 13 mA.
If I use the “5 silicon diode” regulator, and assume the following:
Ideality Factor = 1.4
T = 295 K
I[sub]o[/sub] = 3 X 10[sup]-11[/sup] A
Then we have the following for one diode:
V[sub]1D[/sub] (@ 7 mA) = 0.6855 V
V[sub]1D[/sub] (@ 10 mA) = 0.6982 V
V[sub]1D[/sub] (@ 13 mA) = 0.7075 V
The forward voltage drop for 5 diodes would be:
V[sub]5D[/sub] (@ 7 mA) = 3.4275 V
V[sub]5D[/sub] (@ 10 mA) = 3.491 V
V[sub]5D[/sub] (@ 13 mA) = 3.5375 V
Assuming a supply voltage of 5 V, the load would receive:
V[sub]L[/sub] (@ 7 mA) = 1.573 V
V[sub]L[/sub] (@ 10 mA) = 1.509 V
V[sub]L[/sub] (@ 13 mA) = 1.463 V
So when the load current varies by ±30%, the load voltage only varies by ±4%. Not great. But for some applications, it’s good enough.
And yes, I know there are factors I’m not taking into consideration, such as temperature variations, variations in the ideality factors, variations in I[sub]o[/sub], variations in the supply voltage, etc… But if the regulation is good enough for the load, then… it’s good enough!
Now having said that, would I use 5 diodes in series? No. I would use an LM317. Would I use a Zener as a shunt regulator? Nope, for two reasons: 1) It’s not efficient (only important if it’s battery powered). 2) There’s no built-in protection circuitry.
Hah! I was going to say the exact same thing, but with more swearing. C_M, a 7mA to 13mA draw range isn’t likely, and if you’re going to use any form of diode drop you need to account for a minimum current, because when the circuit is only pulling a few microamps quiescient, there’s going to be nearly 5V on your 1.5V rail.
The built-in protection for the zener circuit is the resistor. Sometimes you’ll see a mains-powered zener shunt circuit that uses a capacitive dropper to minimise heat dissipation, and these can indeed pass transient spikes straight onto the zener, turning it into a not-so-useful 0V reference. The way around this is to employ a small resistor in series with the capacitor just to limit the surge current. It works a treat, especially if you use a slightly inductive wirewound resistor. The Philips AC01/2/3 series WW resistors are great as they can be used safely as impromptu fuses in cases of massive transients. Metal film resistors are not recommended as they pop open circuit when overloaded, and definitely don’t overload a carbon film resistor or it will burst into flames and/or eject a red hot steel slug across the workshop.
Hey antechinus, put us out of our misery and build one of the bastards, would you? Any one, I don’t care.
Settle down fellas. In the interests of implementing the most complicated solution I have decided to use the LM317. Heck, it is much more efficient too - considering the application, I look forward to saving 2 cents per annum on power.
I would have used the zeners, but I feel guilty after the effort that Crafter Man has put in. Buying the battery was not an option.
I now have in my hot little hands the LM317, an IN4004 (dont have a 4002), a 820ohm and 150 ohm resistor. I even have a 10uF electrolytic to put across the output.
Might take 10 minutes to put together, but at least I will get out of doing some housework.
Thanks for your help.
PS the little chip plays the University of Wisconsin ‘theme song’.
After last Saturday you went to all this trouble or On Wisconsin?
Even I feel cheated and I didn’t contribute. 
Ahhahh. OnWisconsin is what it is called…I did not know that dude.
At least it is not badger, badger, badger, badger, badger, badger, badger, badger, badger, etc…
For the University of Wisconsin that’s the name and it’s now the state song also. In addition, high schools all over the US have adopted it including mine, Cherokee, IA, even though you must slightly mispronounce Cherokee in singing it.
Onward Cherokee, onward Cherokee,
Rush right through that line,
Push the ball right through the center,
Touchdown every time.
Onward Cherokee, Onward Cherokee,
Fight on for your fame,
Fight fellows, fight, fight, fight,
We’ll win this game.