I have extract this schematic from youtube (only charger parts) and create this on falstad because I want to create 3.7V battery charger. The original charger use 1N4007 diode with BC547 transistor. I have only 1N4001 diode and 2N3904 transistor, can I use this in place? Is this circuit good or bad? I read my specs for my battery and I read 4.2 V for charge and 2.75 V for discharge, 3.7 V 70 mAh, but, my first TP point is 5 V. But probably, the battery act as resistor and then voltage down. I learn about to simulate lipo for confirm that. For any advice, thank you, sorry for my english.

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    \$\begingroup\$ There is no CC, CV , Vref or 10%CC disconnect , so it is not a good Li-ion charger design. \$\endgroup\$ Jun 20, 2018 at 13:28
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    \$\begingroup\$ 1) I consider "falstad" for educational purposes only, it isn't really intended to design practical circuits. Especially not chargers for Lithium based cells. 2) Simulating this circuit is OK but no more than that. If you really need a working and safe Li-Ion cell charger, this is not it. Li-Ion based cells need to be charged quite carefully, you will need some chip for this, the TP4056 is a common type. On ebay you can buy ready made modules with this chip. These can charge a cell from a 5V supply. 3) your circuit is just too simple to be a proper Li-Ion charger. \$\endgroup\$ Jun 20, 2018 at 13:31
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    \$\begingroup\$ No, you can not use a voltage divider to create a safe charger for lithium batteries. You must be able to limit the charging voltage quite precisely, even as the charging current varies a lot. \$\endgroup\$ Jun 20, 2018 at 15:55
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    \$\begingroup\$ falstad simulates very well Sure, for educational purposes it does. It cannot simulate an IC. You don't have the schematic of the TP4056. Maybe the TP4056 can be simulated in LTSpice. You keep insisting to "design" a Li-Ion charger. There are several reasons why you should not do that: 1) it is too complicated for a beginner 2) Falstad isn't suitable for this. But sure, go ahead and ignore the advice of basically everyone else here. We're just trying to help you hitting a brick wall at some point and being disappointed in electronics and giving up like so many have done already. \$\endgroup\$ Jun 20, 2018 at 16:32
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    \$\begingroup\$ There's no need to be sorry for being a beginner, we were all beginners at some point. I know you're not questioning my opinions, I just want to steer you in the right direction to increase your chance of success. Electronics is fun but there are so many options, ways to learn today that it's easy to get lost and lose interest. And that would be a pity. \$\endgroup\$ Jun 20, 2018 at 18:42

1 Answer 1


Is this circuit good or bad?

Very bad.

1) For a "fully discharged" battery (2.75 V), it will only produce 22 mA of charge current. 5V -2.75 = 2.25 volts across 100 ohms.

2) At "fully charged" (4.2 V) it will continue to supply 8 mA of current into the battery.

So it will provide a VERY slow recharge, and will then do a sort of "trickle charge" after a very long time, which is not how you should charge Li batteries.

3) At no time will the "Full" and "Charging" LEDs change their brightness, since they are driven by the 5 volt supply, and that will not change with the current levels mentioned above.

4) The "Full" LED will never be on very brightly. Assuming that the "Full" LED is blue, it will have a turn-on voltage of about 3.5 volts. Then the 10k resistor ensures that the LED current will never get to more than about 0.08 mA ((5 - 3,5 - 0.7)/10000). This is hardly bright.

5) At the same time, this will produce 0.08 volts across the 1k resistor and the NPN base, which will not turn on the transistor. So the "Charging" LED will always be full on. Assuming a red LED with a 2 volt Vf, this will produce about 15 mA in the LED.

6) If you were to connect the 10k to the battery itself, that would at least make some sort of conceptual sense, but it still wouldn't work. The combination of 10k and diode drops would ensure that the LEDs would never change.

EDIT - Having watched part of the Youtube (I don't speak the language used - Hindi?) It's clear that the 5 volt supply must be current limited to the desired charge current. Also, the 10k shown in the OP is incorrect - it's a 10k variable resistor which is set for some lower resistance than 10k. Presumably the procedure for setting the pot is discussed in the video, but I am unable to evaluate it. Under these circumstances, the circuit might work, sort of.

An important point to consider is that the circuit has no automatic shutoff. The user must monitor the circuit, and when the "Full" LED turns off the user must disconnect the battery. The claims in the Youtube that the circuit is safe are only true if this occurs. Otherwise, the guy who made the video is simply displaying his incompetence.



I forgot to mention that the Youtube circuit has no resistors in series with the battery. I have no idea where those 50 ohm resistors came from, and I have no idea why anyone would want to put them there. At the very least, they will prevent drawing more than a few mA from the battery to the load. If the "Blink LED" is the only load, it will work well enough, but at the same time is unnecessary given the 1k limiting resistor.

So, yeah, this is a bad circuit.


  • \$\begingroup\$ Thank's for your reply, i have put 5V on input but in tutorial it's not specified. Does this change anything about the reliability of this scheme? I guess no, but ... \$\endgroup\$
    – f14284
    Jun 20, 2018 at 18:50
  • \$\begingroup\$ What I do not understand also is why this is on youtube and that a lot of people try this circuit, that the author of the video answers and says that it is functional, there is still a risk of explosion of the battery, it's dangerous no? \$\endgroup\$
    – f14284
    Jun 20, 2018 at 18:59
  • \$\begingroup\$ From the tutorial comments, I confirm 5V for input. I read from user : 'Is it safe to charge lithium battery using this charger?' And the author : 'Yes my dear friend....' . But you bring me the elements that make me think the opposite but that I can not confirm by my lack of knowledge so I do not know what to deduce. \$\endgroup\$
    – f14284
    Jun 20, 2018 at 19:16
  • \$\begingroup\$ Thank you very much for your time and your answer. About your first edit: Indeed I do not know the language of the author either .Hindi I think too. For variable resistance 10K, yes actually I knew it but I wanted to try like this to start. I also asked the question for the charging current and you just confirmed what I learned by reading on the internet. \$\endgroup\$
    – f14284
    Jun 21, 2018 at 19:27
  • \$\begingroup\$ For the monitoring of the battery until the full charge of it I suspected but nothing confirmed it to me since I am currently learning the charging process of lipo batteries. About your second edit: 50 ohm resistors are simply tests of a representation / simulation of a lipo battery. This is to ignore. All schemtic right part are tests. \$\endgroup\$
    – f14284
    Jun 21, 2018 at 19:29

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