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I'm looking for a diode with a specific forward voltage (about 1V) when current is around 150mA (though the max current should be around 250mA or higher).

The closest I've come so far is the 1N270 (1V forward voltage around 200mA), and 1N4001 (about 0.7V forward voltage around 150mA). Does anyone know of anything closer to what I want, or know of a way to look them up? DigiKey is overwhelming when I just need something similar to the 1N270.

Thanks for the help!

EDIT

Apologies for the lack of background. I'm essentially looking for a component that can take a 3.7V nominal LiPo (4.2V when full, 3V when "empty") down to about 2.8-3.6V. I'm trying to power a device that typically runs on 2 AAs in series (3V). It has a bit of tolerance on the voltage. The device has a warning light that comes on at about 2.4V. The device operates at about 150mA on average (though it may range from about 100mA - 250mA).

I would like to avoid the warning light coming on too close to the LiPo's cutoff voltage (3V). I can use a 1N4001, which gives about 0.7V drop on average for my current. I'd like to have slightly more drop (closer to 1V) to give more time for the warning light to show before the battery cuts out (maybe at 3.2V - 3.3V on the LiPo). I know the discharge curves of LiPos get pretty steep towards the end, and I also don't want to push the LiPo too hard by having the user run it down all the way.

To sum up:

  • Source (LiPo) is 3.7V nominal, 4.2V when "full", 3.0V when "empty"
  • Device runs nominally at 150mA
  • Device warning light at about 2.4V (at device)
  • Would like 2.4V on device to map to about 3.2V-3.3V on LiPo
  • This leads to about 0.8V-0.9V desired voltage drop from LiPo to device

I understand that I can use a resistor, though I wanted something that gave a slightly less current-dependent voltage drop, and the forward voltage curve of the 1N270 and 1N4001 fit the bill fairly closely.

I also understand I can use a 3V regulator, and I've found a good one with low quiescent current. Though that reduces the effectiveness of the device's warning light.

I hope that explanation helps illuminate the purpose.

Big question: There's got to be a better way. What component(s) can I use instead to give a more predictable/consistent voltage drop?

EDIT 2

I was originally going to use the 3V MCP1700 LDO regulator(it's got a nice low quiescent current), with 0.18V typical dropout. The problem is that the LiPo will cut off at 3.0V, far above the 2.4V warning voltage... Using the MCP1700 would provide unregulated voltage only for about the 3.0V - 3.18V range. Any ideas?

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  • \$\begingroup\$ The voltage you see will either be typical or maximum (at some temperature or perhaps across temperature); there is no guarantee of the specific voltage at any particular forward current. \$\endgroup\$ – Peter Smith May 3 '18 at 7:07
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    \$\begingroup\$ This looks like an XY problem: OP needs to explain what (s)he is trying to achieve. Using a diode as a voltage source is not a good idea. \$\endgroup\$ – Vladimir Cravero May 3 '18 at 7:27
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    \$\begingroup\$ It's a bit like asking where you can find a sports car with a top speed of 50 mph. \$\endgroup\$ – Andy aka May 3 '18 at 7:31
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    \$\begingroup\$ You want a Low-Drop-Out (LDO) regulator. Some have the option of just following the input voltage when it drops below the target output voltage, thus making the warning light useful again. \$\endgroup\$ – loudnoises May 3 '18 at 14:17
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    \$\begingroup\$ Note that if you wanted a diode with a particular voltage drop, you want a Zener diode and you want to install it backwards so you're using its reverse voltage. Zener diodes are a type of diode designed for a particular reverse voltage drop, such as 1V. \$\endgroup\$ – immibis May 4 '18 at 5:07
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A specific voltage drop for a diode is an incidental, poorly controlled specification. If you're looking for it, then you're probably going about the design in the wrong way. 'Software smell' is a well-known phrase in software engineering. I would like to coin the term 'hardware smell' for this sort of quest.

Diodes are designed to minimise the forward drop for any given technology. The voltage drop remains affected by current and temperature according to the Shockley equation, and then by inherent resistance, which can increase the drop quite significantly at high currents, and is not very repeatable from device to device.

You may want a voltage reference, in which case there are many two and three terminal devices which work better than diodes. Or you may want a voltage regulator, in which case the same applies.

You may want a specific voltage drop in a circuit, in which case a fed-back system with a FET or BJT, opamp and reference voltage would be far superior to a diode.

Or you may simply want to drop a (say) 5v supply down to 4v more cheaply than using a regulator. In which case, check the tolerance of your supply, and the tolerance of your load, and do the worst case sums. You may find that even a perfect 1v drop is not adequate, or that you can tolerate a silicon diode in series with a schottky diode (nominally 1v ;-)) on worst case tolerances.

As an alternative, a commonly used programmable voltage dropper is an 'amplified diode'. This is often used as a bias component in audio amplifier output stages, to tune the output devices' standing current.

schematic

simulate this circuit – Schematic created using CircuitLab

This has two inaccuracies, transistor beta variation and Vbe versus temperature (though in the audio amplifier application, the second is a positive benefit).

In this case, the resistor ratio is set for a nominal 1v drop. Adjust the resistor ratio slightly to adjust the drop. You could replace R1 and R2 by a potentiometer. You could even replace R1 with a schottky diode. This combination could well be superior in a few spec points to the two diode 'solution', though still temperature sensitive.

Using low value resistors will tend to swamp beta variation, however the lower their value, the higher minimum current will be required to get the voltage drop. This may or may not be an issue in your application. Once your load switches off, the current and so the voltage drop will fall, and if the LiPo is still going, the load will see a higher voltage, which may cause hunting. This may still be the case if the LiPo switches off. This will still be the case for the diode 'solution', but that will have much lower minimum currents.

The correct (non-smell) way to do it is to sense the LiPo voltage, deliver your battery warning from that, and LDO to the load. However, if the battery and the load are already separately sensed, then what you're trying to do is perhaps the cheapest ugliest way to get to your goal.

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  • \$\begingroup\$ Could I also potentially use two diodes that have smaller voltage drops in series, to sum to my desired 1V? Could this be viable as long as I can tolerate the variances introduced by slight changes in current and temperature? \$\endgroup\$ – menehune23 May 3 '18 at 22:13
  • \$\begingroup\$ Are you using a protected or unprotected LiPo? In other words, is the '3v cutoff voltage' the voltage at which your protected LiPo switches off, or the voltage you don't want your unprotected LiPo to drop below as you'll damage it? As for two diodes, you mean like the 0.7v silicon diode and the 0.3v sckottky that I suggested in the last paragraph of my answer? \$\endgroup\$ – Neil_UK May 4 '18 at 3:55
  • \$\begingroup\$ The LiPo has a protection circuit built in. As for the series diodes question, I re-read your answer and saw that right after I posted my comment. Do you have some part numbers for two you'd recommend? \$\endgroup\$ – menehune23 May 4 '18 at 4:38
  • \$\begingroup\$ There's somewhere between 1 million and an infinite number of diode types (perhaps I exaggerate). However, I'd just go with a 1N400x for the silicon, and a random 'several amp' rectifier for the schottky. The reason for using 'large' diodes compared to the required current is that the residual resistance will be small, and you're more likely to get close to the nominal voltage drop. I'll post an alternative voltage dropper. \$\endgroup\$ – Neil_UK May 4 '18 at 4:45

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