What are the disadvantages of using a diode to drop Vcc by a fixed amount?

Many circuits nowadays accept only 3.3 V as input voltage, maximum 3.6 V, but they can work often down to 2.7 V.

Since Lithium batteries (LiPo and LiIon) provide 4.2-3.3 V when charged-discharged (not completely, but substantially), a diode could be used to drop the supply by a relatively constant 0.55-0.7 V (depending on current rating of the diode and current circulating) to obtain (for 0.6 V) 3.6-2.7 V.

A capacitor has to be connected after the diode to reduce noise and current peaks.

Provided that the final device accepts 3.6-2.7 V, what are the disadvantages of this solution?

The advantages are minor cost, simpler connections, no quiescent current of a (linear) regulator.

In my specific case I'm thinking about a ESP8266 powered by a 18650 Li-Ion battery: in deep sleep the quiescent current of the regulator (2-3 uA) is significant compared to the chip itself (20-30 uA). A more compact setup is also a good advantage.

• The obvious one - power wasting. – Eugene Sh. Sep 27 '17 at 15:41
• A linear regulator would have the same, after all some sort of voltage drop is needed. – FarO Sep 27 '17 at 15:46
• Because it doesn't drop VCC by a fixed amount : the drop depends on load current and temperature. It can sometimes be good enough but is it worth the analysis and risk of problems? – Brian Drummond Sep 27 '17 at 15:49
• @FarO It would. It's a general disadvantage, not comparing to linear regulator. – Eugene Sh. Sep 27 '17 at 15:51

First, the diode forward voltage may vary over temperature. For example, the jelly bean 1N4004 is likely to change its forward voltage by 100 mV as it heats up from 25 C to 100 C.

Second, if the load circuit has a very low power sleep mode (as many battery powered circuits do) the diode drop could be well below the nominal 0.55 to 0.7 V drop we normally assume.

Here's the I-V curve over temperature for another common part, 1N4148:

If this were used to drop voltage in a device with a 50 mA operating mode and 0.1 mA sleep mode, with operating temperature range from 0 to 85 C, the drop across this diode could range from about 0.375 to 0.9 V, which is a wider range than the allowable Vdd range for many chips.

Many circuits wouldn't be adversely affected by this power voltage uncertainty. But many others would. And in any case it is much simpler to design with the guaranteed 1% accuracy available from linear regulators than to worry about 100 mV or more of uncertainty given by the diode solution (but also watch out for linear regulators not dealing well with micro-power loads).

• I thought 0.55 V is the bandgap (or similar) and therefore also the value for an infinitely small current. Clearly I was wrong. – FarO Sep 27 '17 at 16:08
• @FarO, no, think about the Shockley diode equation. V goes to 0 when I goes to 0. – The Photon Sep 27 '17 at 16:09

For a WiFi chip, "function" and "performance" over a range of supply voltages means 2 different things. There is always performance tradeoffs.

As well as low drop being a requirement, so is stability, during Tx mode drawing 170 mA or low idle consumption of 0.9mA during a light sleep. These tradeoffs come with experience and thorough understanding of test spec conditions.

When the input supply range can be 33% of the 3.3V min Vbat at 10% SoC, this means a wide range of tradeoffs. Although the IC also has a 33% range from minimum Vdd it is not matched the battery so that adding an offset of a diode drop only increases the dynamic range.

Therefore the only solution I would consider is a 0.15V max drop at 170mA 3.6V LDO, something like this $0.15 200mA LDO It looks more like you need 0.3V Drop to 3.3V uC and 3.6V to WiFi chip. I would consider this Buck-Boost solution. • 200 mA may not be enough and surely 25 uA quiescent is too much, but I get the idea. MCP17xx are the option I had as alternative, specifically a 3.0 V version. – FarO Sep 27 '17 at 16:11 • what is your spec? – Tony Stewart Sunnyskyguy EE75 Sep 27 '17 at 16:12 • You may be better off with a buck boost regulator – Tony Stewart Sunnyskyguy EE75 Sep 27 '17 at 16:15 • 250 mA, 3.0-3.3 V output, 3.3-4.2 V input, quiescent current not comparable to the consumption of the device in sleep (30 uA). The uC and the WiFi chip are the same unit, the external sensors also work down to 3.0 V. MCP170x seems to work well. – FarO Sep 27 '17 at 16:36 • But like I said voltage stability affects performance on the WiFi chip for the Rx and Tx specifications, which are not always shown other than 1 voltage. So a buck boost regulator with good filters will perform better if designed properly and operating it on the minimum voltage. When the added BOM cost is$2 who cares how many parts are used, unless you plan to make a million of them. – Tony Stewart Sunnyskyguy EE75 Sep 28 '17 at 1:24