I am using a li-ion battery charger IC that regulates an output to 4.2V. I have a schottky diode in series with the output that then connects to the battery. How do I determine the voltage drop?

If the charger puts out 100mA do I go by the VI curve of the diode at 100mA? Or if the battery voltage is at 3.5V and the output is 4.2V then the drop has to b 0.7V regardless of current? What if the VI curve says at 0.7V the diode is at 200mA, then this case is impossible as the charger can only do 100mA. I'm confused to figuring out the voltage drop of the diode.

enter image description here

  • \$\begingroup\$ A key question is WHY? do you have a series diode. The result depends on your charger IC which you should have told us) and the circuit (which you should have shown us) but usually this means that the battery will be very substantially low on deliverable capacity (and long on cycle life). It also MAY mean that the charger IC will not change from CC to CV mode ever and will keep trying to charge the battery indefinitely. Unless there is a superbly good reason to do this and you or the person who designed the circuit know exactly what they are doing then it is a "bad" [tm] thing to do. ... \$\endgroup\$
    – Russell McMahon
    Jun 30, 2014 at 22:36
  • \$\begingroup\$ ... If the charger IC has a timeout timer it may prevent strange happenings. Or may not. \$\endgroup\$
    – Russell McMahon
    Jun 30, 2014 at 22:42
  • \$\begingroup\$ @RussellMcMahon, what makes a series diode a particularly bad idea? \$\endgroup\$
    – sherrellbc
    Jul 1, 2014 at 1:22
  • \$\begingroup\$ @sherrellbc - see additions at the end of my answer. The last nite hopefully clears up the "X sets Y" confusion and the one above that addresses the series diode. \$\endgroup\$
    – Russell McMahon
    Jul 2, 2014 at 1:08

5 Answers 5


The diode doesn't know about the voltage appplied to the circuit - it only knows the current passing through it, so you should use that current (the acutal current flowing through the diode, not the rated current of the power supply) to look on the V/I curve to find the voltage drop in the diode.

  • \$\begingroup\$ Okay, so what am I assuming wrong in this scenario. Charger can do 100mA output max. Battery is at 3.5V. Charger outputs 4.2V. The VI curve says at 100mA there is a 0.3V drop. What happens in my circuit. Battery can't get any charge? Or the output is forced to 3.8V? \$\endgroup\$
    – ACD
    Jun 30, 2014 at 16:58
  • 1
    \$\begingroup\$ @ACD, the charger will adjust the output voltage lower to whatever it must be such that the charge current is limited to 100mA. If, at 100mA charge current, the battery voltage is 3.5V and the diode voltage is 0.3V, the charger voltage must be 3.8V. \$\endgroup\$ Jun 30, 2014 at 17:15
  • \$\begingroup\$ @Peter, this doesn't answer the question of how to analytically calculate the current that will flow. \$\endgroup\$
    – Brad
    Jun 30, 2014 at 22:19

The voltage across and current through the diode are not independent but are related by the diode IV curve.

And, since the diode is in series with the charger, the diode current will be limited to the maximum current the charger can supply.

So, for example, if the charger can provide a maximum of \$100\mathrm{mA}\$ of charging current, then the maximum voltage across the diode will be the given by the point on the IV curve for \$I = 100\mathrm{mA}\$.


The datasheet specifically states this controller eliminates the need for external diodes, but I can understand what was going on here, that someone wanted to charge their Li-Ion cell less to greatly extend its service life by dropping termination charge voltage by maybe 0.1 to 0.2V, BUT the diode is not necessary for this.

Per the schematic you have a basic two resistor voltage divider in the feedback loop and can simply change the 274K resistor to a little higher value to pull the voltage a little less towards ground, to lower it. For example instead of 4.2V the datasheet gives a value of 332K ohm for 4.1V output.

The important factor is where the diode is, that since it is outside of that resistor feedback loop it won't effect the output voltage at the IC but will at any point outside of the regulator loop like at the battery.

This brings to mind another possibility for the diode, that its purpose is to prevent battery discharge through the voltage divider resistors when the circuit is not at a voltage higher than batt voltage, for example when it's not sunny out since it's solar powered.

As the battery voltage rises the current through it will drop below 100mA because it's a voltage regulating circuit. As the current continues to drop, the diode Vf will continue to drop until it passes no current at all.

On the diode datasheet where you see the curve for Vf vs current, you should follow the curve all the way to the bottom (left), which is the bare minimum forward drop for that species of diode. Below that current value all bets are off and you have to experimentally determine the addt'l minor forward diode drop but it won't be much.

For something like a 1A 1N5817, it would be about 0.2Vf so the battery would charge to about 4.0V, but as mentioned above you could just change the feedback resistor to accomplish this instead of adding the diode, OR to readjust the output to compensate for the 0.2V drop of the diode.


I think it depends on the V-I characteristics of your charger IC too. If it's a bench supply that's limited to 100mA, then Alfred is correct. But if it's simply a 4.2V charger that's "rated" for 100mA, it may "fold-back" and actually deliver less current at lower voltages. Or it may deliver more current, overheat, and fail.

Check the datasheet for the IC. You are looking to find a current such that the diode's voltage (for that current), plus the battery voltage (whatever it is at a given time) equals the IC's output (for that current).


The diode voltage drop will be determined entirely and only by the current flowing in it.
The current flowing will be set by the overall circuit and may be affected by the diode voltage drop.
So the two will interact even though the actual diode voltage is solely diode current dependent.

However: A key question is WHY? do you have a series diode.
The result of including one depends on what charger IC you are using (which you should have told us) and the circuit (which you should have shown us) but usually this means that the battery will be very substantially low on deliverable capacity (and long on cycle life). It also MAY mean that the charger IC will not change from CC to CV mode ever and will keep trying to charge the battery indefinitely.

Unless there is a superbly good reason to do this and you or the person who designed the circuit know exactly what they are doing then it is a "bad" [tm] thing to do.

Worst case - and this depends on the circuit - the charger IC may not swap from CC to CV mode, as the battery voltage that it sees appears too low, the battery will approach 4.2 V with diminishing current but never reach it, so charging will continue as the battery slowly asymptotes to 4.2V (as current falls Vdiode drops) and if you leave the battery connected it will be damaged and its lifetime greatly shortened.

If the charger IC has a timeout timer it may prevent strange happenings.

What is the charger IC?
Show us the circuit diagram.
Why are you using a series diode?

A series diode will prevent (almost) back feeding of battery into the charger IC when the power to the charger IC is removed. Any IC worth its salt will be designed to handle this and will typically draw around 1 uA (maybe 5 uA worst case for good ICs). In most scenarios this is acceptable. 1 uA = 0.168 mAh/week. 10 uA = 1.68 mAh/week.
10 uA = 88 mAh/year, so if you want very long periods between charging a 10 uA drain may be untenable.

A series diode will protect against reverse battery connection but is not a good solution due to the problems discussed above.

If back feeding prevention and/or battery polarity protection is desired a MOSFET can be configured to do both jobs. Ask if relevant.


In summary, you are trying to achieve a relatively minor circuit feature which could be obtained in some other manner by playing fast and loose with the LiIon charging algorithms in a way that no circuit ever used commercially attempts to do. The results can be expected to be bad.

You need to say WHY you wish to use PGOOD in this manner and wish to avoid "doing it properly". Reasons may include lack of space, lowest cost or not enough knowledge of hos to do this or ... . Until people know why and what you want solutions are unlikely to be 100% good.

The added circuit diagram shows the battery supplied by the series output diode and the IC's battery voltage sense feedback divider connected prior to the battery. this means that the battery voltage will be a diode drop lower than what the IC "sees" - OR the IC will see a voltage a diode drop higher than the battery actually is. This is bad.
An alternative is to connect the feedback divider at the battery, outside the diode.
This is also bad.
The reasons are covered more or less in the notes I've written previously.
LiIon batteries are reasonably easy to charge if done correctly. But they are quite easy to damage if charged incorrectly. A crucial part of part of charging is to know when to swap from CC (constant current) to CV charging. This occurs at Vbattery = 4.2V - with possible adjustment for temperature. Adding a diode as shown will cause change to CV mode when the battery is at about 3.9V. This is well before completion of the CC phase and will lead to an early slow down of charging. As battery voltage slowly increases Icharge will drop and lower current leads to lower Vdiode so as Ichg drops to near zero I diode will drop to near zero and ultimately the battery will get to almost 4.2V.This will take MUCH longer and it's possible that the IC will not change to CV mode at all (as perceived Vbat is too low). This means the battery will be 'floated" at somewhere in the 4.0 - 4.15 range. It will have significantly lower capacity AND may be damaged.

Also, without wading carefully through the data sheet and/or playing with the IC itself I can't be sure, it MAY be that the IC not seeing a battery will cause it to start up in the trickle start "dead battery" mode.

I mentioned the "divider outside diode case in my prior notes. It has its own issues and, as the IC, sees the actual battery voltage it probably fails your PGOOD requirement.

Diode confusions:

The confusion as to whether
diode current sets diode voltage or
diode voltage sets diode current
is caused by the fact that neither does either in a real-world circuit.
Diode current and voltage are set by a known relationship.
Ignoring second order effects (eg temperature), then
If you know V then you know I.
If you know I then you know V.

IF you connect the diode across an ideal voltage source then V does control I.
If you connect the diode across an ideal current source then I does control V.
BUT if you connect the diode in a real-world circuit I and V vary as the circuit adapts to what it "sees" until a steady state is reached (or it oscillates in some cases).
The diode I & V conform to the devices I & V curve.

  • \$\begingroup\$ This topic has always been perfectly unclear to me. From theory, we study that the PN junctions require a particular bias voltage in order for current to flow - yet other sources state the voltage developed across the diode is a derivative of the current through them! \$\endgroup\$
    – sherrellbc
    Jul 1, 2014 at 1:24
  • \$\begingroup\$ From the outside looking in, it would seem that the (constant current) IC charger have an output voltage that is only sufficiently forward biases the series diode such that 100mA of current would flow. As the battery voltage rises, the output voltage would have to rise in proportion to compensate for the Vf required to maintain constant current. Eventually, the IC would reach its max output voltage and .. the current would eventually trickle to a stop as the diode becomes reverse biased? \$\endgroup\$
    – sherrellbc
    Jul 1, 2014 at 1:32
  • \$\begingroup\$ I am using the LTC3105 charger. I need a diode in series with the output so I can use the PGOOD pin to notify me when the battery is actually being charged. Without the series diode the battery powers the PGOOD pin even if no charging is occurring. As long as the battery is not dead PGOOD will be high. The series diode means PGOOD is only high when the battery is charging. Here is a modified picture showing my circuit: imgur.com/40MQOBg I appreciate the reply. \$\endgroup\$
    – ACD
    Jul 1, 2014 at 14:16
  • 1
    \$\begingroup\$ Well I added the diode because I need an indication of when the battery is actually being charged by the solar cell, not an indication that the battery is not dead. Is there a better way to do this without a diode? \$\endgroup\$
    – ACD
    Jul 2, 2014 at 13:19

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.