# How much can I push the limits of a device's VDD?

On a PCB I'm designing, I'm using two chips with very similar voltage requirements. Among other voltages, one requires 1.2v, and the other requires 1.1v. in fact, looking at the datasheet, the allowable voltage ranges slightly overlap.

The USB7206 has a range of 1.09v to 1.21v (nominally 1.1v).

The CX3 has a range of 1.15v to 1.25v (nominally 1.2v).

Would it be crazy to pick a single voltage to run both of them? Could that be 1.2v, or would 1.175v be more sensible?

• Also note that there is quite some additional safety margin to keep the chip operational across all process corners and operating temperatures. If you can avoid going to the lowest and highest temperatures, chances are that the limits given by the datasheet are way too strict – michi7x7 Apr 6 at 10:58
• "Would it be crazy to pick a single voltage to run both of them?" No, but how stable is your PSU? Is it stable enough to make 1.2V really 1.2V and not 1.22? – Mast Apr 6 at 13:01
• @michi7x7 Don't post answers in the comment section, because now you can write weird things without getting downvotes. For example I see no "+/-10%" and the voltages are absolute maximum, not typical operating voltages. Your comments are now written before the correct answers. – pipe Apr 6 at 13:51
• @pipe oh, you're right. I only calculated the upper bound of the USB7206, +/- 10% is what I usually use for my designs. – michi7x7 Apr 6 at 14:47
• USB7206 seems to say "Operating Voltage 3.3V". Perhaps you are thinking of particular I/O pins and not VDD? – Ben Voigt Apr 6 at 21:53

Sure, you can use single supply for both, if you are confident that the supply voltage right at the chips will be withing operating conditions of both chips at all times.

This includes calculating the ripple+noise, load regulation, and any voltage drop over PCB track resistance and filtering ferrite beads that may exist on the supply path, under all input voltage and output current conditions your design will encounter.

• Worth noting that the USB7206 has 1.21 V listed under its absolute maximum rating, together with a dedicated note warning about the consequences such as "Some power supplies exhibit voltage spikes on their outputs when AC power is switched on or off". They seem worried, probably for good reasons. – pipe Apr 6 at 0:28

As long as you are operating both devices within their specified operating voltages you should be good to go.

• Agreed, I just wanted to mention that the selected voltage regulator will probably need an accuracy of +/- 2% instead of the less expensive +/- 5%. – Elliot Alderson Apr 5 at 15:09
• Thanks, I should add that I'm using a switching regulator. – Rocketmagnet Apr 5 at 15:21
• @Rocketmagnet - "I should add that I'm using a switching regulator." Irrelevant, unless you are suggesting that the regulator has a high ripple voltage. – WhatRoughBeast Apr 5 at 16:24
• Switching regulators have been known to allow fast transients under certain conditions, like power on/off. And those can erode IC circuits, leading to reduced MTBF. I encountered this issue once with a prototype design that kept failing after a few weeks on my bench. I doubled down on static control and new surge protectors on my bench outlets, but that didn't make a statistically significant difference in MTBF. Power cycling tests increased the failure rate, so we started watching for transients on all power and ground pins. Sen the failed chips back to the producer and they identified... – jwdonahue Apr 6 at 1:11

If you don't want to gamble and can only generate 1 voltage, produce a stable 1.2V and use an LDO to drop 0.1V from that level to produce a stable 1.1V. For example NCP136 can drop as little as 0.06V. As @pipe wrote in a comment, it could be cheaper (albeit less efficient) to generate 1.1V from 3.3V using dirt-cheap regulator which then will drop 2.2V

If you do want to gamble, consider making a prototype PCB where the LDO can be bypassed, which can be used to test how critical it is to supply both ICs with an intermediate voltage. And then calculate how much you will spend on testing, and how much you will save on not buying the LDOs, and only do actual testing if there's a benefit on the horizon.

PS. If anything, I would check whether I could do without a power plane, if that is your actual constraint. Putting large decoupling caps, (and perhaps even two LDOs on both sides on the chip) might be enough. In many cases, a voltage at the limit of the allowed range is worse than not having a power plane.

• The USB chip requires a 3.3 volt source as well, so it's probably better to use that as a source – pipe Apr 6 at 13:26
• @pipe Yes, that could work too. – Dmitry Grigoryev Apr 6 at 13:28

It looks like almost any design using both of these is also likely to have a 1.8V supply as well as these two Vcore supplies.

You might be able to get by just starting from the 1.8V supply, and putting a diode in series with each of these.

A diode has a forward drop of about 0.65 volts, which varies (ever so slightly) with the current being drawn. So, starting from 1.8V and a nominal 0.65 volt drop, each of them is at 1.15 volts--(just barely) within spec for both.

Since the 7206 draws more current, it'd see a slightly greater drop across the diode, bringing its supply voltage closer to its nominal 1.1 (while the CX3 drawing a bit less would probably maintain at least 1.15 all the time).

If you wanted to do this more accurately, you could use Schottky diodes, which are normally rated for more specific forward voltage drops. Getting one rated for a 0.6 V drop, and the other for a 0.7 V drop would put each of them at rated voltage (and still let you use one regulator instead of 2 or 3).

• This is a very interesting idea. The main problem I have is with the number of power planes on this design all fighting for space, and I was wondering if I could merge two of them. I'm not too bothered about the regulators themselves. But, thanks. I'll consider it. – Rocketmagnet Apr 6 at 9:24
• Series diodes are not voltage regulators, and should not be used as such. "A diode has a forward drop of about 0.65 volt" - there are literally hundreds of thousands of diodes which don't, and so said generic statement is wrong. "which varies (ever so slightly)" - it varies exponetially with current, so going from 1uA to 1mA would cause a very large change, not a slight change. – Tom Carpenter Apr 6 at 9:46
• Seriously, don't do that. As soon as the chips stops drawing power (for any reason) you will have 1.8 V on Vdd and blown gates in your IC. If you really wanted to do that, at least ensure a constant current through the diode by using a resistor parallel to the IC. – michi7x7 Apr 6 at 10:45
• But why use a diode instead of an LDO? Or two LDOs, one per chip? This suggestion makes little sense to me, unless OP is extremely constrained in space and/or budget a diode is a very bad idea. On top of that, this does not answer OP's question. – Vladimir Cravero Apr 6 at 12:12
• That "about 0.65V" of voltage drop is actually 0.5...0.8V across different physical diodes, current variations and temperatures. If you build 1000 devices, 10 will blow up on first use, another 5 will fail after several hours of use, and 5 more will fail when used outdoors. – Dmitry Grigoryev Apr 6 at 13:05