I work the below circuit.

enter image description here

LDO is 3.2V fixed , Two diode is connected one at output and one at ground for compensate the temperature drift occur.

Op is connected to a micro controller VDD-BUS which takes generally 10mA current.

D1(SD103ATW-7-F) top diode which i mention in datasheet ( 0.32 at 5mA If),

So the output voltage (OP) should be a) 3.2V with respect to ground ( since both diode cancel each drop)

b) 3.4V with respect to ground ( since diode at ground will have very low current Iq of LDO)

Any adivse.

  • 1
    \$\begingroup\$ The error because the current has to go through the diode will be much, much bigger then any temperature influence. The diodes do NOT cancel each other drop as the microcontroller current is an average! \$\endgroup\$ – Oldfart Aug 16 '18 at 17:24
  • \$\begingroup\$ Let me start again: Why do you think your micro controller needs a temperature compensated supply? Many work between 2.5 and 5V. Or maybe 3v3 +/- 10%. \$\endgroup\$ – Oldfart Aug 16 '18 at 17:32
  • \$\begingroup\$ reason is OP also share the voltage coming form a power back up circuit , which feed power of 3.3V when we lost 5V, diode used at output to stop back feeding supply to LDO. \$\endgroup\$ – Bharav Aug 16 '18 at 17:34

If you add a single resistor to drive a few mA through the 6-1 diode then you'll be a lot closer.

The problem is that the voltage across the diode will be much less at 25~100uA than at 5mA. At various temperatures, it looks like the typical difference will be in the 100-150mV range (Fig. 1). As well, I'm not 100% confident that those numbers are not a bit optimistic. You might want to grab the SPICE model and do some simulations, but of course that will only tell you the typical behavior not worst-case.

The MCU current drain is likely also highly variable.

There are probably better ways of doing what you have in mind, but if your application will tolerate a few hundred mV difference maybe this is a usable solution.


From the comments it appears clear that the main purpose of D2 (?) is to compensate for D1 drop, while D1 is there as a part of a voltage maximizer.

If you cannot accept the voltage drop on D1 the easiest option is probably to use a schottky diode. They have a voltage drop as low as a few hundred mV, if you want to do better than that you will need a mosfet and a circuit to drive it.

  • \$\begingroup\$ Both diode comes in single pack , concern is there should be no drop if both diode is same construction, since i assume both drop will cancel out each other and output voltage from LDO will be un affected, correct me if i am wrong here. \$\endgroup\$ – Bharav Aug 16 '18 at 18:20
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    \$\begingroup\$ What makes you think that the ground pin of the LDO sources enough current to polarize the diode? The drop on the diodes is the same if they are on the same die, same temperature and same current. Or at least enough current. \$\endgroup\$ – Vladimir Cravero Aug 16 '18 at 18:59
  • \$\begingroup\$ they are on same die see PN SD103ATW-7-F \$\endgroup\$ – Bharav Aug 16 '18 at 19:04
  • \$\begingroup\$ Again, what makes you think that there is some current flowing from the GND pin of the regulator? \$\endgroup\$ – Vladimir Cravero Aug 16 '18 at 19:44

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