# Buck converter charge battery control output voltage

I have a buck converter charging 8 series of Li-Ion batteries ( 8 x 4.2 voltage of charging) with a source of 36 V. I am using this schematic (from here):

I know that for a resistive load, the output is controlled as $$V_o=D V_{in}$$

But, here the load is not resistive. How to control the output voltage? can we control it as above without feedback?

Does duty-cycle D = 8 x 4.2 / 36 works here?

• Why don't you want to use feedback? That's both the standard and the best way to do it. And, without feedback you can't limit your charge current, so you're likely to blow up your batteries. – Nick Johnson Jun 22 '15 at 9:52
• @NickJohnson The batteries are protected. In the first step I just need something that works. First I should solve voltage control, then I will control the current. Do you see the duty-cycle formulation still valid for this problem too? – barej Jun 22 '15 at 10:03
• Protected against what? Most LiPos have circuits to protect against overdischarge; that doesn't mean you can just hook them up to an arbitrary voltage source and expect them to charge fine. – Nick Johnson Jun 22 '15 at 10:29
• And you don't need to solve this yourself - there are plenty of charging solutions out there already. – Nick Johnson Jun 22 '15 at 10:29
• @NickJohnson They are protected against overcurrent too. So what is your suggestion? could you introduce me some of those solutions? – barej Jun 22 '15 at 10:36

## 1 Answer

If you used a synchronous converter, its duty cycle x input voltage is pretty much the output voltage when averaged. If you are intent on doing this without feedback you should strongly consider a synchronous solution - it has a high side and a low side MOSFET. In effect it is a half bridge solution.

Having said all that, you need to modify duty cycle on the fly to cater for changes in input voltage levels so this requires some analogue signal processing.

• Could you please mention the formula between Vin, Vout and the duty-cycle? – barej Jun 22 '15 at 11:07
• It's simple. Vout = Vin x duty cycle. If the duty cycle is 50% then Vout = Vin x 0.5 BUT remember this is only for a synchronous (two MOSFET) buck converter. Think of it like a changeover switch where the switch pole is the output - the switch can only connect to incoming 0V and Vin. If you toggle that switch at a high rate with 50:50 duty cycle, the average voltage has to be half x Vin. – Andy aka Jun 22 '15 at 11:35
• Except for voltage drop at switch and diode, in my case is vout=vin x duty-cycle? – barej Jun 22 '15 at 12:15
• No, you need to have an active MOSFET where the diode is to get this relationship. – Andy aka Jun 22 '15 at 13:46
• what voltage do i get if i keep using diode? – barej Jun 22 '15 at 13:54