2
\$\begingroup\$

This Wikipedia entry specifically on the SMPS part, says:

Switched-mode power supplies are always regulated. To keep the output voltage constant, the power supply employs a feedback controller that monitors current drawn by the load.

However, I've found a range of mobile phone powersupplies, which claim to be switched-mode, but specify their output as being 4.7-6.5VDC at 900mA max, rated for 4W. Given that these were switch-mode and a range being specified, is what struck me as being contradictory. So, I am wondering if these are fakes ?

In fact, I went ahead and bought one of these, but at the moment, I am far from home and don't have my tools. It seems to be able to charge my Android phone, and is also able to power a RaspberryPi -- though I tried just the boot sequence and then switched-off, afraid of damaging the board. One thing in favor of their claim to being switched-mode, is that they are pretty light weight, and relatively small.

So, is the range, only because their manufacturing is not under tight quality control, and they actually operate at a fixed + regulated output voltage, but there could be variations within that range, from one piece to another ?

Edit: The power-supply is as such meant as generic / back-market replacement charger for Blackberry smartphones, that expect 5V supply.

My primary use for these supplies is to power a cache of RaspberryPi's (model-B), with minimal peripherals. The supplies available closer home, cost more than twice as much, so was wondering if this could be a good money saving bargain.

Edit2: (Feb 11, 2013) If the switch-mode power supply output is indeed unregulated, would a LDO as an intermediate step be a good way to regulate the final output to 5V ? Are there any inexpensive (<= $1) 5V LDO's which could do this job ?

\$\endgroup\$
3
  • 3
    \$\begingroup\$ This is just an aside to your question, but the Wikipedia statement is simply wrong. There are many commercial examples of unregulated switchmode power converters, in which the output voltage is simply proportional to the input voltage. \$\endgroup\$
    – Dave Tweed
    Commented Feb 11, 2013 at 4:51
  • \$\begingroup\$ The Wikipedia entry is now different unless my edit gets reverted ;) \$\endgroup\$
    – PeterJ
    Commented Feb 11, 2013 at 9:52
  • \$\begingroup\$ The no-load voltage of a poorly regulated power supply can be some way above the nominal output voltage, but putting a small load across the output - to draw say 10% of the rated maximum output current - can often bring the voltage into line. \$\endgroup\$
    – nekomatic
    Commented Apr 3, 2014 at 12:51

3 Answers 3

4
\$\begingroup\$

The 4.7-6.5V DC may well be within spec for charging a Blackberry (no idea) but on the Raspberry Pi the 5V side of the power supply has a SMBJ 5V transient voltage suppression (TVS) diode across the rail. That has a minimum breakdown voltage of 6.4V so at 6.5V it will be conducting to some degree and will possibly blow the 1A fuse which is a SMT part.

However if it doesn't blow the fuse the 5V rail also goes to the HDMI connector and the USB host connectors so it may cause problems with attached devices as well. I wouldn't recommend using that particular adapter and look for something that is regulated to within 5% of 5V as per the USB specifications, so 4.75V to 5.25V.

Also as Dave Tweed mentioned the Wikipedia article is incorrect (it mentions citation needed) and a switching power supply does not have to be regulated by definition. In this case however it may well be a regulated switch-mode supply, just one that is not regulated to a tight tolerance.

As per the updated question a typical linear regulator such as a 7805 has a drop-out voltage of around 2V, so that could end up being as low as 2.7V (if it works at all). Some low-dropout (LDO) regulators may be in the order of 200mV drop-out but as you may be starting below the specified voltage that wouldn't help getting things within spec the whole time.

The robust solution would be a buck–boost converter that can produce a constant voltage from a supply greater than or less than the output voltage. But in reality the cost / complexity would be greater than just purchasing a better supply.

\$\endgroup\$
3
  • \$\begingroup\$ Thanks @PeterJ. Edited my question slightly, to see if adding LDO can help. \$\endgroup\$
    – bdutta74
    Commented Feb 11, 2013 at 9:02
  • \$\begingroup\$ Just updated, but even a LDO with a really low drop-out wouldn't help because you're starting out with a voltage that may be below the spec. A buck-boost I've mentioned would likely cost more than $1 in large volumes, let alone as a one-off. \$\endgroup\$
    – PeterJ
    Commented Feb 11, 2013 at 9:07
  • \$\begingroup\$ Perfect answer, and makes complete sense. \$\endgroup\$
    – bdutta74
    Commented Feb 11, 2013 at 14:00
3
\$\begingroup\$

No, a switching power supply need not be regulated. There are entire families of devices known as isolated bus converters (for example) which operate at fixed duty cycle to convert DC to DC efficiently, and they're certainly switchers.

Even a regulated SMPS can have difficulty maintaining regulation at load extremes (usually very light or very heavy loads). This is normal, and usually some special techniques are needed to ensure regulation at absolutely zero load and at maximum load.

Adding an LDO can help. Bear in mind that the LDO headroom may sneak up on you when the DC/DC converter output is low, causing your output to go lower (don't use a 5V LDO on a 4.7V to 6.5V rail and expect 5V out all the time - the LDO will act almost like a diode once the headroom is breached, dropping a fixed amount of voltage).

\$\endgroup\$
1
\$\begingroup\$

wikipedia should say "SMPS are always regulated to keep output constant" (No full stop). And in general, any system (not only power supplies) need regulation to keep a variable constant, isn't that what control is for? Unregulated systems exist, including power supplies and SMPS. As they are not regulated, output is dependant on input, load, etc.

\$\endgroup\$

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service and acknowledge you have read our privacy policy.

Not the answer you're looking for? Browse other questions tagged or ask your own question.