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Diode to offset a voltage regulator?

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I have a fixed regulator and need the output voltage to be exactly one forward drop over a diode higher than it is.


Is there any drawback apart from the obvious "all real men use adjustable vregs" in simply putting a diode between the ground on the vreg and the electrical ground?


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It works. Put the filter caps to ground, not the the ref/ground pin of the regulator. As you have it, the filter caps can not source more current than the regulator quiescent current (due to the diode), and when the caps are sourcing, your diode drop will be very unstable, as the diaode current will vary.


You may also want a resistor from the output to the anode of the diode, if you use a more modern regulator, or need to tweak the diode current. With the '140 as is, you get about 6mA, and for a small sig diode (like the 1N914) you are in the zone. If needed, select the resistor to provide the appropriate bias current for the drop you need.

THis can also help stabilize the diode current, and drop, against the transient response of the regulator. I would also put a 0.1uF cap across the diode.


See the data sheep for inre info about using the '140 as an adjustable reg.


Edit: Also note that the diode does not see load current. It sees only the regulator quiescent current (about 6mA). A small sig diode is sufficient. DIssipation should be approx 4mW

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Sorry, the regs I have are not LM140 but LE33CZ (yes that is supposed to be 3.3v). Despite my recklessness with the schematics (partly because of a bit to much of an "how big of a difference can it be" attitude on my part, sorry about that) your advice is nonetheless solid @@enl


@@roadrunner84 Yeah, It will have to be able to take the full load. It should however just have to dissipate 0.7*I Watts (since the forward drop over the diode is ~0.7V). With a peak load in this particular application of about 100mA that shouldn't be much of an issue though.

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If you are trying to raise the output by using a diode on the ground terminal of the regulator, the diode will NOT see load current. Load current is returned to ground independantly of the regulator. The diode will see the quiescent current of the regulator.


The LE33C has an ill-specified quiescent current of something less than 0.5mA to 3mA, depending on load (ignoring the inhibit state, where the output is, for practical purposes, floating).


This is a pretty wide range (factor of 6) giving a variation of about 50mV for the diode drop (conservatively). The current is low enough that any diode that can handle 5mA is fine (pretty much any small signal diode). Power dissipation is about 2mW max.


If you need more stable output than approx 50mV, use a 1.5K resistor from the output to the ground lead, so you have an ADDITIONAL 2mA current through the diode. This will restrict the current variation to less than a factor of 2, giving you about 15mV max variation. Presumably, a given device will be more stable, but you can't count on it. The enemy of good design is ignoring the min and max values in the spec, and unspecified values (like the min quiescent current) really play hob with non-linear relationships like the voltage-current relation for a diode.


See crude MSpaint diagram. Might need to up the cap across the diode to 1uF. The data sheet isn't complete enough to know it there will be stability issues with this one. fc is about 100kHz with 0.1uF, which is well within the spec of the regulator for stability, but raising the cap to 1uF will bring the frequency down to about 10Khz.


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  • 1 month later...

I studied about the LE33CZ voltage regulator that it has a very low drop out voltage and moreover it has a very low quiescent current so what I think it does not create any problem if you will use the diode to offset the output voltage of this voltage regulator but just put a 2.2uF capacitor for the stability of the voltage regulator.


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