It has occurred to me that almost no battery charger would be required if the battery pack was of the same voltage as grid supplied power.

I build electric bikes, and the voltage on these has been ever increasing for obvious reasons (smaller diameter cables and connectors vs high current, over volting motors for additional speed, lower current mosfets required in the speed controls Etc.).

90 volt systems have now become fairly common.

Now I was pondering, if we were to increase the battery pack up to, say, 32 LifePo4 cells (3.6 volts constant voltage charging) couldn't I just rectify the 117 volt grid power and charge directly from that? I know voltage will shift with the rectification, but I don't know the specifics. In either event the number of cells charging could be adjusted to the closest match of what was coming out of the wall. Yes the final constant current phase of charging would be skipped, but this is a negligible amount of power anyway. Am I missing something? Is there a reason this won't work?

  • \$\begingroup\$ Your phases are the other way round - it's constant current most of the way up. \$\endgroup\$ – pjc50 Nov 7 '13 at 22:14

117VAC is a sinewave that rises up to a peak of \$\sqrt2\times 117 = 165.5V\$ then back through zero to a negative peak of -165.5V then back through zero and repeats.

enter image description here

How does this allow you to charge a battery of 117V even if you rectified it - you'd still get peaks of about 164V falling to zero then back up to 164V.


First let's adjust the battery pack voltage to 165.6V or 46 cells, because the peak voltage of the AC signal is \$\sqrt { 2 } * 117 = 165.46V\$. Then rectify the voltage, so we aren't reverse biasing the battery pack:

AC to Battery Circuit

Now apply that AC voltage to the battery pack when it is fully discharged to 92V (2V per cell) and the only time a charge current will flow is when the AC line voltage exceeds the battery pack voltage.

In this graph, V(n001) is the AC voltage. V(n002) is the voltage on the battery's positive terminal. I(V2) is the current into the battery, assuming a 10\$\Omega\$ internal resistance, which is a value I randomly selected.

Plot of voltage and current in AC to battery system.

This has 2 potential problems:

  1. I don't think this is a major issue, as the current may be low and the effects negligible, but it's not good to have all the current in the system flow at only one point of the wave form. This is an example of non-linear/non-sinusoidal distortion power factor. (As opposed to displacement power factor, which is linear about out of phase.) This is similar to a basic switched mode power supply, which only draws power when the rectified AC voltage exceeds the storage capacitor voltage. From wikipedia:

    the input current of such basic switched mode power supplies has high harmonic content and relatively low power factor. This creates extra load on utility lines, increases heating of building wiring, the utility transformers, and standard AC electric motors, and may cause stability problems in some applications such as in emergency generator systems 3

    How does this cause heating of the building wiring and utility transformers? Here is one way to think about it: a circuit could draw a steady sinusoidal 5A at 60Hz, or it might draw 20A spikes at 60Hz. Heat created by current passing through a wire goes up with the square of the current and the decrease in time during which the current is flowing does not compensate. (\$P={ i }^{ 2 }Rt\$) I've had personal experience with this when trying to maximize efficiency with a bicycle hub dynamo. No current would flow until the voltage exceeded the storage capacitors, then the hub was practically shorted and a large portion of the power was lost to heating up the dynamo. (If you prefer printed books, another good discussion of this is in Electromagnetic Compatibility Engineering by Henry W. Ott, chapter 13 section 9. 5)

  2. This is essentially a constant voltage across the battery, which increases to the maximum battery voltage. This can cause cell balancing issues and excessive current, which can damage the LiFePO4 batteries.

Controlling the current through the batteries and limiting the power factor distortion can both be solved with, for example, a Buck-boost Converter 4 using a current controlled feedback loop or another switched converter topology and specifically designed charge controller IC.

At this point, the charger is just as complex as the original system that had a lower battery voltage. It may be that requiring a smaller voltage step down could improve system efficiency, but I believe this would depend on the overall design and component selection.

Another interesting thing to point out here, is that the current would be delivered in pulses. I have seen some comments that pulsed current charging is safe for LiFePO4 batteries, but I haven't conducted or read any research regarding this, and would recommend further investigation.

Schematic and graph were generated by LTSpice. 6

  • \$\begingroup\$ Can you justify your point 1 a little better, please? It appears to me that the current is always exactly in phase with the voltage, so all of the power is real power and there is no reactive power at all. To me, this seems like a PF of 100%. Exactly how does this result in "extra load on utility lines" and "increased heating of...wiring"? What's the sentence just before the start of your quote..."as a result" of what? \$\endgroup\$ – Joe Hass Jan 11 '14 at 13:09
  • \$\begingroup\$ The current is in phase, but non-linear, which results in harmonic distortions. A rapid change in current causes harmonic noise. These harmonics won't be in phase. There is a whole section on wikipedia that does a great job of explaining how harmonic distortion from a rectifier will decreases the average power transferred to the load. en.wikipedia.org/wiki/Power_factor_correction#Non-linear_loads </br> "As a result" of the preceding sentence: "which only draws power when the rectified AC voltage exceeds the storage capacitor voltage, 'as a result'" There rest is a quote from wikipedia. \$\endgroup\$ – Krh3o Jan 11 '14 at 19:34
  • \$\begingroup\$ Here is the whole quote: "Simple off-line switched mode power supplies incorporate a simple full-wave rectifier connected to a large energy storing capacitor. Such SMPSs draw current from the AC line in short pulses when the mains instantaneous voltage exceeds the voltage across this capacitor. During the remaining portion of the AC cycle the capacitor provides energy to the power supply. As a result, the input current of such basic switched mode power supplies has high harmonic content and relatively low power factor. This creates extra load on utility lines..." \$\endgroup\$ – Krh3o Jan 11 '14 at 19:47
  • \$\begingroup\$ I will edit my answer to provide some clarity on displacement power factor vs distortion power factor. Please let me know if there is anything else I should improve. Thanks! \$\endgroup\$ – Krh3o Jan 11 '14 at 19:56

To answer your last, and most important, question: the voltage across each cell needs to be kept within the correct range. The batteries will not be physically identical, and so different voltages will appear across each battery in the series chain. With lithium batteries, this will likely result in fire and/or explosions.

Proper battery chargers have means of balancing the voltage across the cells.


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.