12
\$\begingroup\$

I have come across many mobile phone chargers/power supplies that produce a regulated 5 volt output but I don't see any voltage regulator or a Zener diode inside. I want to know how these power supplies produce the regulated output.

Moreover, why can't we simply provide a 4.1 volts input to batteries instead of using 5 volts power supply with an overcharge protection circuit?

\$\endgroup\$
13
  • 2
    \$\begingroup\$ For the second question; because you are connecting power to a phone, not directly to a battery. \$\endgroup\$ May 24 at 12:14
  • 11
    \$\begingroup\$ Re, "why 5V," Answers on the linked question explain why you need more than 3.7V to charge a nominally 3.7V battery, but they don't say why 5V (as opposed to 4.8 or 5.2 or 6V). The reason is historical: 5V was the supply voltage for digital logic circuits for several decades starting back in the 1960s. Then, when lower voltages were used for "core" logic, it usually would be converted down from existing 5V supplies. That practice still was common back when USB was invented. The infrastructure for 5V power was wide-spread and cheaply available. It was an obvious choice at the time. \$\endgroup\$ May 24 at 13:33
  • 5
    \$\begingroup\$ @SimSon: i've popped dozens of usb chargers, mostly cheap, and never seen what you describe. They always use SMPSs because the chip is (much) cheaper than a copper winding big enough to produce an amp or two. \$\endgroup\$
    – dandavis
    May 25 at 7:54
  • 3
    \$\begingroup\$ @SimSon, no they don't. Can you explain how you know so surely that "Many of those cheap "chargers" work with a simple transformer/rectifier and the actual charging circuit is in the phone itself" ? \$\endgroup\$
    – TonyM
    May 25 at 8:35
  • 2
    \$\begingroup\$ Iron transformers working at mains frequency are soooo 1980s... bulky, heavy, hot and sometimes smelly. 1990s were the transition time and in 2000 no one produced transformer-based power bricks. Fast forward 2010 - even welding machines are switching mode. \$\endgroup\$
    – fraxinus
    May 25 at 11:59
30
\$\begingroup\$

Phone chargers are indeed usually a 5 V regulated power supply.

Here's an example of a simple circuit that is commonly used:

enter image description here

Source.

This is a flyback converter circuit.

The output voltage is regulated even though it's not immediately obvious how that's done. But note the winding "Na" on the left, it is one of the transformer's windings.

This winding supplied power to the IC and at the same time the voltage across this winding is coupled (through the transformer) to the voltage at winding "Ns" which feeds the output.

The IC has a feedback loop where is measures the voltage at winding Na (voltage Vaux) and regulates the on/off switching of the transistor connected to the DRV output of the IC, such that the voltage Vaux has a certain value. When that is done, the output voltage will also be fairly constant (it does not need to be exactly 5 V, up to 5.5 V is allowed, usually, in a good design).

There is indeed no zener diode in this circuit but there usually will be a "bandgap circuit" inside the IC. A Bandgap circuit can provide a reference voltage that is almost constant over temperature. On-chip Zener diodes do exist so that could be used as well. This circuit (or the zener diode) will provide a constant voltage that is used as a reference voltage to regulate the output voltage.

why can't we simply provide a 4.1 volts input to batteries instead of using 5 volts

Because the battery charger circuit needs to:

  • know the temperature of the battery

  • know at what current the battery should be charged (note that this current varies with the battery's charging state, when the battery is almost empty or almost full, the charge current needs to be limited).

  • know the type (Chemistry) of the battery so that the "stop charging" voltage is known and correct.

Because of this the charge controller cannot be in the supply, it has to be in the device (phone). Also having the charge controller in the device means you can charge it from any 5 V USB output which is convenient.

\$\endgroup\$
8
  • 3
    \$\begingroup\$ Also people typically want their phone to stay powered on while connected to the charger. A fully external charging circuit would thus have to distinguish the current going into the battery from the power consumption of the device. There would thus likely have to be two sets of wires and two separate regulation circuits in the charger. \$\endgroup\$
    – TooTea
    May 24 at 12:57
  • 1
    \$\begingroup\$ I think 5.5V is above the standard (without looking it up). I have seen ICs designed to be powered by USB 5 V that (very stupidly) are rated at 5.3 V max. (One such was initially used in an "open cellphone" design.) \$\endgroup\$
    – Russell McMahon
    May 24 at 13:26
  • 1
    \$\begingroup\$ @RussellMcMahon The 5.5 V is also from my memory. I worked on DCDC converters years ago and there the requirement was even above 5.5 V to prevent a cheap (poorly regulated) supply outputting for example 6 V and blowing things up. So yes, 5.3 V is indeed on the very low side. You'd want to add some protection between the USB connector and that IC. \$\endgroup\$ May 24 at 15:51
  • 7
    \$\begingroup\$ USB +5V relates to TTL +5V. It was +/- 5%, so 4.75V .. 5.25V became USB standard as well. On the other hand, some chargers went as high as 5.7V when loaded in order to overcome the cable resistance and deliver higher charging current. They even used to boast about this ability under different 3- or 4-letter abbreviations. Phones and other similar devices seem to tolerate this. Of course, the proper solution (to use higher voltage) came later (as QC2.0/3.0) and was standardized even later (as USB PD). \$\endgroup\$
    – fraxinus
    May 24 at 17:24
  • 2
    \$\begingroup\$ @ChrisH indeed it is quite broad. Phone manufacturers face the dilemma of either making it broad or dealing with a RMA avalanche because of low quality chargers. And the phones that are QC/PD compatible simply accept anything in their QC/PD range (with added safety margins) no matter if the charger does any negotiation about the matter. I am sure I once charged a Samsung phone directly from a somewhat depleted car battery with no apparent ill effects. The voltage was somewhere between 11.5V and 12.5V. The phone happily said "charging rapidly" and charged at the usual "rapid" rate. \$\endgroup\$
    – fraxinus
    May 25 at 10:12
11
\$\begingroup\$

Complementary to main answers:

It is crucial to note that attempting to charge a Li-ion cell by simply applying 4.2 V (or whatever maximum voltage is required) is an extremely bad idea because:

  • One normal charging method is to use CCCV (constant current / constant voltage) charging where a constant current is applied while Vbat is under 4.2 V and then a constant voltage is applied until Icharge reduces to some predetermined value, when charging is stopped.

    If instead a constant 4.2 V is applied longer term and the cell is "floated", it will initially be damaged and in due course destroyed. In addition, lithium may "plate out" which is usually "a very bad idea indeed".

  • If the cell is below about 2.5 V unloaded normal practice is to apply a low "trickle current" until Vbat rises to say 3 V. Applying 4.2 V limited to even normal charging currents to a cell with Voc < 2.5 V can result in "vent with flames". This is invariably fatal for the cell and quite possibly for the device it is installed in.

  • If the cell is less than 2 V Voc it is usually deemed irrecoverably dead. Normal advice is that to recover a cell from below this voltage is impossible and may, as above, lead to "vent with flame". In reality SOME cells may SOMETIMES be recovered from below 2 V BUT they are liable to be damaged and are thereafter suspect safety wise.

  • There are various methods of fast charging. These usually DO apply a voltage and current limited supply to the cell. Icharge is usually greater than the usual specified charging current. A fully or partially discharged cell voltage ramps up towards 4.2 V. When 4.2 V is reached the voltage is removed and after a pause is reapplied. This is repeated until Icharge falls to some prespecified value. However, this is NOT the same as "just applying 4.2 V", requires suitable current control, and is (IMHO - not tested) liable to shorten battery lifetime compared to more usual CCCV methods.

\$\endgroup\$
2
  • \$\begingroup\$ Fast charging also typically requires monitoring the temperature of the cells to make the whole process at least halfway safe (especially so for lithium-ion chemistries). \$\endgroup\$
    – TooTea
    May 26 at 10:42
  • \$\begingroup\$ @Tootea True. As does CCCV charging. LiIon is a relatively easy chemistry to charge - but things do "sometimes go wrong". I've only ever had one "vent with flame" experience - but it was every bit as good as you hear of :-) \$\endgroup\$
    – Russell McMahon
    May 27 at 4:03
8
\$\begingroup\$

To answer one point ...

Moreover, why can't we simply provide a 4.1 volts input to batteries instead of using 5 volts power supply with an overcharge protection circuit?

Because to get current to flow you need a higher voltage than your target. With your approach the current would quickly fall to a very low value when the battery terminal voltage reached 4.1 V and it would take multiples of the desired charging time to complete the charge. Running from a higher voltage means you can feed a constant current over a wide range of battery voltages and therefore shorten the charge.

\$\endgroup\$
7
\$\begingroup\$

I don't see any voltage regulator or a zener diode inside.

In all of the phone chargers I have examined, there is a very small switching regulator chip. The chip often has an internal Zener diode to provide a bootstrap supply for the chip. The chip is a switchmode regulator and sends modulated width pulses through an inductor / flyback transformer to get a 5V output. That output is smoothed by (usually) two capacitors. It is then fed into USB connector.

\$\endgroup\$
2
\$\begingroup\$

Almost everyone dislikes 'proprietary' chargers

Why can't we simply provide a 4.1 volts input to batteries instead of using 5 volts power supply with an overcharge protection circuit?

You're talking about putting the battery charge-control circuitry in the wall-wart instead of the phone. That's the same idea, really, as putting the Electric Vehicle charger in the shore-side equipment instead of in the car.

Segue into this video, much of which is worth watching, but it goes to your point directly at 2:20.

Because nobody liked it when phones had proprietary wall-wart chargers, which would be necessary due to ever-changing battery tech (there are many kinds of lithium battery for instance).

So you're at the office and your phone is flat, so you email around "Does anybody have a Nokia 7-series phone? I need to charge". Remember that anyone? That was no fun.

Contemplate power tool batteries in the 2000s

Consumers wanted to buy a full suite of tools in the same battery system, with interchangeable packs. They were very distrustful/resentful of what they perceived as "planned obsolescence".

So manufacturers were locked into a standard built around the dominant 1990s battery, nickel-cadmium (1.2V nominal). The best they could do was nickel-metal-hydride (NiMH) which had the same nominal voltage, and similar enough charge curves that NiMH chargers could charge NiCd batteries without damage (still necessitated replacement of NiCd chargers). They did not have a viable upgrade path to lithium.

Because reaching it would require a customer-enraging reboot of their battery pack form-factors. This was directly caused by the blunder of putting the charge-control circuitry inside the charger -- if it had been inside the battery pack, the problem would be solved.

That's why you don't make that mistake.

Now, USB has settled in as the standard for phone and small device charging, and that is a wonderful thing.

\$\endgroup\$

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

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