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I'm designing a Lead Acid battery charger with 3 stages. For the 2nd stage (constant voltage stage) as it is written in literature: the voltage applied on the battery must be dependent on temperature. So I thought of making a buck converter. The mosfet is controlled by IR2111 driver via a PIC Microcontroller PWM (Look at the circuit). Input = 18 V.

When I used a resistive load (R = 3 Ohm or 10 Ohm) the circuit worked very well: The driver worked, the output voltage is flat (no ripples) and up to 15.5V (with a duty cycle of 95%), the current is up to 5 amps with the 3 Ohm load (without overheating) ...

Now when I use the battery as a load to charge it (I have a discharged battery of 12V and its voltage is 9V) it doesn't work at all :( The output voltage is imposed by the battery so I find Vout = 9V. !!! Even if I change the duty cycle nothing will happen, the driver can not drive the mosfet and (Vgs is not square and equals 9V: seems that is related to battery voltage)

It seems that I must add something or .. I don't know .. I you can help me I will be very thankful :)

This is the circuit that I used:

enter image description here

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4 Answers 4

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Lead Acid cells have a very low internal resistance. So a LA Battery discharged to 9 V will behave as a 9 V voltage source with a low value resistor in series. So that is why you measure 9V at the output when you connect the discharged battery.

What happens in a "proper" LA battery charger is that it enters current limiting mode which means that is will charge the battery with a predetermined current for example 1 A untill the battery is sufficiently charged. Then the voltage limiting charging will occur.

I see no current limiting measures in your schematic. This is BAD I would never design a battery charger without current limiting. Especially for a switching converter where the inductor can saturate and currents can become exceptionally large.

It can be that the current limiting is in the 18 V supply. Possible but not so elegant. There are plenty of designs that integrate voltage and current limiting in one solution.

Oh that IR2111 is only a driver ! Geez, I expected it to be a proper SMPS chip. So you're just PWM controlling the MOSFET from a microcontroller ? Please consider implementing a proper switched more converter with voltage feedback and all. This is not a design that I would build.

Some other remarks:

Usually temperature controlled charging is not needed unless the LA battery is in an environment which can be very hot or cold. For normal room temperature a maximum charging voltage of 13.6 - 13.8 V is usually all you need.

Discharging a 12 V LA battery to 9 V is not good for the battery, the lower the voltage the more chance you have to permanently damage the battery (sulfite on the plates). I would not go below 11 V to keep the battery in good shape.

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  • \$\begingroup\$ For the current limiting I know that is not that elegant, but I have a power supply with a limiting current. Yes IR2111 is just a driver and I don't think that it is a problem as it works properly with a resistive load. Actually the voltage of the battery doesn't worry me, and I know that 12V battery discharged to 9V is bad. So how do you explain that if I use a power supply (13.5V) directly to the battery, the battery doesn't impose the voltage and the process of charge work ? So the problem is when I use the Buck :( \$\endgroup\$
    – Billben
    Commented Jun 2, 2016 at 11:28
  • \$\begingroup\$ Yes IR2111 is just a driver and I don't think that it is a problem as it works properly with a resistive load If you don't see the problem then you do not understand how a proper buck converter works. There's no voltage feedback in your circuit. There's no way to detect a saturated coil in your circuit. What you build is a "blind" PWM generator with an inductor and diode. I think it is no surprise that you don't get the voltage you expect. Sure it works with a resistor. But does a battery behave the same as a resistor ? \$\endgroup\$ Commented Jun 2, 2016 at 13:44
  • \$\begingroup\$ Okay maybe I don't understand how proper buck converter works ! Okay how can I solve this problem with feedback and ... so as to get the expected voltage in the output. Can you help with the modifications I should do ? Can you please give me a circuit diagram ? \$\endgroup\$
    – Billben
    Commented Jun 2, 2016 at 16:53
  • \$\begingroup\$ Is is not that simple ! What you want is to use a dedicated chip or even better, get a module from ebay like this: ebay.nl/itm/… It has two settings for voltage and current. It is cheap and will do the job. \$\endgroup\$ Commented Jun 2, 2016 at 17:15
  • \$\begingroup\$ While this answer has good points, it doesn't deal with the fact that the boost cap can't charge enough. \$\endgroup\$ Commented Mar 11, 2022 at 3:03
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Definitely something is missing in your circuit and that is charging path for the bootstrap circuit. Ideally capacitor C3 gets charged while the switching MOSFET is off. Here D3 cannot provide such charging. You need to provide a alternate path for such charging only then the boostrage voltage would be higher than your input voltage so as to switch ON the MOSFET M5.

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The buck regulator has 2 modes for inductor current. Ccm - continuos current mode and dcm - discontinues current mode. C3 needs sufficient time to charge up during the on time of diode D3; there is not much time for this in the dcm condition and swtiching M5 stops, even with a pulse input to the driver. I hope this helps.

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To regulate the voltage output of the buck converter just add a voltage divider to the output and have that read into the analog input pin of your micro-controller. Throttling the current can be done with a resistor in series with the battery, although as you say your power supply already throttles it before the buck conversion stage so you are all good! No you don't have to buy something off Ebay to make your setup work!

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