I'm planning on building a small constant current power supply for charging lithium-ion batteries on the go. While they normally use constant current AND constant voltage, I've elected not to include the latter for design concerns - it's not necessary since I won't need to fully charge the battery, and constant-current will suffice for about 70% capacity depending on charge rate.

Looking at normal CC-CV chargers (such as one for the LTC1541, seen here: http://cds.linear.com/docs/en/design-note/dn188f.pdf), it seems this may be achievable with a Schmitt trigger combined with a current-sense resistor and voltage reference, as well as an inductor for filtering and MOSFET for switching. I envisage a system whereby the current sense (resistance would be <0.1ohm) is connected as close to ground as possible, and the voltage drop across the current sense compared with a voltage reference; the hysteretic output should yield a small current ripple but hopefully should keep it relatively constant.

I imagine a voltage drop across the current sense of about 0.05V, driven by a current of 2A. Since the ripple needs to be fairly low - say, plus/minus 0.02A, which corresponds to a very low fluctuation in voltage - would the comparator-driven schmitt trigger be able to detect this change and switch the MOSFET effectively?

While I understand the circuit theory I'm afraid I don't have much experience building these sorts of circuits (I normally buy micro power supplies), so I'm not sure what sort of values to use, nor the feasibility of something like this. I'm aiming for high-efficiency of the circuit, so a linear regulator is out of the question.

  • \$\begingroup\$ Do you have a problem using the conventional switching method such as the LTC1541? \$\endgroup\$
    – Andy aka
    Jun 4, 2014 at 13:53
  • \$\begingroup\$ @Andyaka - I think you may not have looked at [his LTC1541 reference ](cds.linear.com/docs/en/design-note/dn188f.pdf) - it's about as non standarda a LiIon charger as you'd hope to find, The IC is somewhat time honoured and here is pressganged into an unusual role. \$\endgroup\$
    – Russell McMahon
    Jun 4, 2014 at 16:06
  • \$\begingroup\$ @RussellMcMahon No I never looked at it - I was really trying to get the OP to justify why (s)he felt it wasn't justified. \$\endgroup\$
    – Andy aka
    Jun 4, 2014 at 16:28
  • \$\begingroup\$ The LTC1541 operates in a linear fashion. Excess power is dissipated as heat. This is too inefficient for me. \$\endgroup\$ Jun 4, 2014 at 16:46
  • \$\begingroup\$ @Andyaka - Reason given was " ... I'm aiming for high-efficiency of the circuit, so a linear regulator is out of the question." How that applied was not necessarily obvious from the text. \$\endgroup\$
    – Russell McMahon
    Jun 4, 2014 at 23:48

1 Answer 1


What you describe is workable - it's essentially a buck converter - but your exact circuit is not specified and there are many ways to go from a somewhat correct general concept to an incorrect circuit.

The design note 188 that you reference is about 15 years old and its time has passed. While it is a marvel of improvisation, you can get better results for less cost using modern ICs.
I'm aware that you only cited it as an example, and that you wanted a switch mode version, but using a dedicated switch mode converter IC would be a good idea at minimum, and using a LiIon charger IC would be an even better one.

An LTC1541 - datasheet here costs about $3 in 1's. It's quite a useful IC if you need its features. With a 5uA typical supply current it could not fight its way out of a wet paper bag, and a slew rate of 8 mV/uS means that you have to be clever to use it and the characteristics of the device can greatly affect circuit performance.

AN188 loosely uses the term "float voltage". A Lithium Ion battery must not be "floated" in the sense the term is usually used. They intend it to mean "hold at CV while charge current ramps down due to battery chemistry action" - but the applied voltage should be removed once current has fallen to some preselected percentage of Imax.

Once you have terminated charging there will usually not be any need to restart charging until there has been some load current drawn. This is because an unloaded LiIon cell will maintain a close to full state of charge over considerable periods. If it is going to stand unloaded for weeks or months then testing battery voltage for possible restart of charging may be in order - but the load from the voltage measuring circuit can easily exceed the self discharge rate.

If you are trying to produce a low cost design with a high element of DIY in it you could try the venerable MC340-63 switching regulator IC. These are about as cheap as you can buy a SMPS IC for and can implement almost any topology. As you have noted, to make a constant current driver you compare the drop across the current sense resistor with a reference voltage. The MC34063 has an unfortunately high reference voltage (about 1.2V) so you can use an opamp or comparator. This adds cost and complexity, but using an eg LM358 dual opamp is almost as cheap as the MC34063. Scale current sense resistor with opamp up to reference voltage. Figure 6. in this applicationnote give the general idea - in this case opamp inverts and it's a CUK converter, but the principle is the same.

  • \$\begingroup\$ The MC340-63 sounds interesting. However, you mentioned that the reference voltage is very high. Surely a simple voltage divider - much like the LTC1541 - would allow you to adjust it as you see fit? \$\endgroup\$ Jun 4, 2014 at 21:04
  • \$\begingroup\$ @AlexFreeman - See example I cited. If the reference is higher than the source signal you usually* divide the reference down OR multiply the sense signal up until they are equal at the design current and then compare them. So you need a comparator. If the reference is suitably low you can set a sense resistor to a value si=uch that Vsense = Vfer at the desired current. When Vref = 1.2V you do not want to drop 1.2V across your sense resistor [!] so you must divide the reference down OR multiply the sense voltage up. Which is what the opamp is for.[*There are "clever" 'dangerou' alternatives.] \$\endgroup\$
    – Russell McMahon
    Jun 4, 2014 at 23:54

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