60 VAC to 12V lead acid battery

Excellent!  Then, the 60V rating of the TPS54x60 IC is a great fit.

Whenever the input to the charger is lower than the battery voltage (plus the dropout of the charger), then no current will flow to the battery--the battery will not charge.  If you must charge at 8.5V in, then yes you need a SEPIC (or flyback or forward) converter.  The SEPIC would convert the input voltage to a DC output voltage that is larger than the maximum pack voltage (usually 13.8V) plus the dropout of the charger.  The dropout of the charger depends on which pass element topology you pick and this is explained in the bq24450 datasheet.

For the SEPIC controller, I recommend the TPS40210.  It's input voltage range only goes up to 52V, so you would need to add a resistor and zener diode on its Vdd to drop the input voltage to below 52V for the controller to operate from.

Designing a SEPIC with this wide of an input range may not be worth it.  You may be better off in terms of size, cost, complexity, design time, etc. just designing a buck front end and accepting not always being able to charge at the lower input voltages.

Yes, the TPS54x60 would step down the rectified voltage to a level that is high enough to charge the battery pack.  The bq24450 would then use that voltage to charge the pack properly and safely.  The bq24450 controls a pass element (transistor) in the active region to regulate the voltage and current applied to the pack.

What exactly do you mean by load management?  What 'load' are you referring to, the battery or something else?  The MSP430 could likely do what you need.  There is an entire forum dedicated to it.

For hysteresis, the TPS54x60 family has an adjustable UVLO feature.  You can optimize the start and stop voltage for the buck converter with any hysteresis level.  There is information in the datasheet on how to do this.

Chris,

I'm working with Jimit on this project.  First off, thank you for all of your help.  Hopefully, I can explain the load management portion to you.  Basically, we have a wind turbine that is connected to a permanent magnet generator. Our first goal was to convert the varying AC from the generator to a steady-fixed DC voltage to charge a battery.  Our second goal was to implement power management.  If the wind energy starts to decrease, the RPMs on the generator will decrease and the input voltage will decrease.  We don't want the generator RPM to drop too low otherwise we'll risk having the turbine blades coming to a standstill.  Then, it would take a lot of wind to get things moving again.  So, can we use a microcontroller to monitor the input voltage and stop charging the battery if the input voltage drops below a certain level?  On the flip side, we'd also like to cut off the input voltage from the rest of the circuit if it goes above 60 VAC.  We are trying to power a set of sensors (total power consumption = 15 Watts).  So, our plan was to connect the sensors to the battery.  If there is little or no wind, the sensors would draw current from the battery.  Then, if there was sufficient wind, the battery would be charged (as the sensors continued to draw current from it).  I'm not sure if this is the right approach?  Should we connect the sensors to the output of the buck converter AND the battery?  Then, with enough wind, our circuit would power the senors plus charge the battery.  With little or no wind, the microcontroller (MPS430) could be used to stop powering the sensors and stop charging the battery and just have the senors draw from the battery?  Sorry for the long post. 

Thanks again for your help.

-Kaleb

Yes, you can feed the input voltage to a uC and it can turn off the front end switcher, TPS54x60, through its EN pin.

You certainly need to protect against an input voltage that is too high.  Above 65V and the TPS54x60 could be damaged.  If the input voltage changes slowly enough, you could monitor it with the uC and turn off a 100V or 200V FET in series with the input supply before it gets too high.

How much current can the wind generator provide?  This is sounding more like greater than the 2.5A the TPS54260 is rated for.  A better power solution might be to use a flyback or forward converter that could be designed for any voltage and power level.  This could be designed to regulate an output voltage that is either higher or lower than the input voltage.  And it would still be able to regulate if the wind generator voltage goes too high.  You can post in the isolated controllers forum for the best parts for this, as I am not an expert in that area.

Hi Chris,

I was just checking the datasheet for  TPS54260EVM-597 and it says that the output is 3.3V. Now, is this fixed output or it's for the 12V input? If this is fixed, what is the best way to boost it up to 15V?

Thank you,

Jimit

The TPS54260 IC is an adjustable output voltage part.  The EVM is set to output 3.3V.  The datasheet explains how to change the output voltage by changing the 2 feedback resistors.  The IC will always attempt to regulate the output voltage to what it is set to and will do this as the input voltage is varied.

Hi Chris,

I realized that after I posted the question. Thank you for the information anyway. I was trying to find some kind of curve for output power/current vs. input voltage but couldn't find one in the datasheet. What is the amount of output current when the input voltage goes toward the maximum? Also, I need to charge the battery at 13.8V so how much current will be supplied at that amount of voltage?

I'm not sure that I understand the question.  The TPS54260 will regulate the output voltage.  It provides a DC bus from which the bq24450 and any other circuits can draw power.  The bq24450 will charge the batteries and it sets the charge current.  As long as the total current draw on the DC bus (the output of the TPS54260) does not exceed either the current limit of the TPS54260 or its thermal limit, then the bus voltage will remain constant and everything is ok.

So, the TPS54260 provides a voltage.  The bq24450 (or any lead acid charger) will provide a CC/CV (constant current or constant voltage) supply that is needed to charge the pack.

What is your fast charge current going to be?

Hi Chris,

Thank you once again for all the help. So we're trying to charge a 12V 7Ah lead acid battery with the following specs:

If your maximum fast charge current will be only 600 mA, then you can use the lower current TPS54160 instead of the TPS54260.

On paper, all that is required to modify the TPS54260 EVM to get 15V out is the feedback divider, which you correctly calculated.  However, the power stage and loop response should be at least checked based on the new level of Vout.  The applications section in the datasheet gives the equations to calculate the necessary parameters.  The EVM was only designed as a 12V in supply, thought it may work up to 60V.  The input caps and diode are sized well enough, but the output caps voltage rating needs to be increased to at least 25V and the inductor might need to change.  Of course, any changes you make can and should be implemented and tested on the EVM.

For the charger, if you want 14.4V out at 600 mA, which is what your boost charge specs are, then you will need more than a 15V input.  The output voltage of the TPS54260 will need to be increased.  I am assuming here that you will use the circuit in figure 9 of the datasheet--with Dext and a PNP pass element in the common emitter configuration.  As explained on page 13 of the datasheet, the different transistor configurations give different amounts of power lost in the driver IC (bq24450) and have different amounts of headroom, delatV.  This deltaV is the minimum drop across the transistor.  Other drops include the 250 mV drop across the sense resistor and the drop across Dext.  Also, you will have drops in the PCB traces and any cables/wires between the TPS54260 and bq24450 and the batteries.  The sum of all these drops added on top of your 14.4V is the absolute minimum voltage that the TPS54260 needs to output.  I would bump it up by at least 0.5V to give some margin.

hfe should be spec'd in most transistor datasheets.  It might be called 'gain' or something similar.  It will have no units and be around 100 typically.

Well, we are not too concerned with the boost voltage as for right now because our main goal is to present a demo for 15 mins to show that we are able to charge the battery at 13.8V at a good amount of output current. Ideally, I was thinking going with 800mA to 1A. But, I know that the TPS54260EVM is already equipped with parts for 600mA and hence didn't want to change that. 

Also, we are running out of time and only have one week left so we want to change the least amount of parts on both EVMs.

The TPS54260 EVM as shipped can do up to 2.5A load.  You need to at least increase the voltage rating of the output caps and change the output voltage to get it to work.  You may need to change other items, but you won't know until you try it or do some calculations as in the datasheet.

Depending on your bq24450 design, you may not be able to charge at all going from 15V to 13.8V.  There is an EVM for the bq24450 that charges at 450 mA.  I recommend at least 2V of headroom using the EVM as assembled.

So if I change my Vout from the TPS54260EVM to 16V that gives me enough headroom.

After calculating, I get R6 = 190Kohm. Also I will be changing the two 47uF cap to 25V rating. The only problem with that is the size. I can find 16V caps in the same size. Do you think that would be ok or we are cutting too close?

Ceramic capacitors suffer from a 'DC bias' effect.  Under a DC bias voltage, the actual capacitance of the capacitor will decrease.  With 16V on a 16V capacitor, you have very very little capacitance left--10% would not be unexpected.

You can use 4 x22 uF.  Or you can redesign the loop compensation for less output capacitance.  You probably don't need that much since this will be powering a battery charger.

How do you calculate the transient load response? Or do you just have to set it yourself? 

According to the datasheet, the transient load response is 3% change in Vout for a load step from 1.5A to 2.5A.

If i calculate the output capacitance using the same conditions, I get the following parameters for output capacitor. 

Lo=10uH

Vf=14.2554V 

Vi=13.8V (input capacitor voltage)

Ioh=2.5A

Iol=1.5A

Cout > 3.13uF

Can I do that? or do I need some other value for transient load response?

And if that is correct, the closest value I find with same size and 25V rating is 3.3uF cap.

The transient response is a systemparameter that you set.  Will you have the sensors running off of the output of the switcher?  If so, when they turn on--that is your load transient that you need to design for.  They will transition from some low load to their full load and their input voltage needs to remain within a certain level.  Use those specs for the datasheet equations.

If the only load on the switcher is the charger, then you don't have much of a load transient requirement.  I would recommend designing for a no load to your fast charge current load step and keeping the output within 5% of its initial value.

The capacitance that you calculate is the minimum required to give the desired transient response.  You would need to increase it as you did to find a standard value capacitor.  I recommend adding 50-100% additional capacitance to account for DC bias loss and to give some margin in the calculations.

Vi and Vf are the initial and final voltages of the output capacitor during a load dump--when the load is removed.  This would be your 16V output voltage for Vi plus your maximum allowed output voltage for Vf.  For your design with just a charger, you can ignore this spec as the charger can handle up to 40V on its input.  If you have the sensors running on the output, then this voltage needs to be less than what they can handle.

That makes a lot of sense.

The switcher will be connected to the charger only. For prototyping, we are not going to have any load on the charger. But if the sensors were to be attached, they would be attached to the output of the battery not the switcher or the charger.

So, I realized I had made a mistake in calculations. However, calculating the Cout with no load characteristics and keeping the switching frequency at about 300KHz, I get the value of Cout as 9.66uF. Giving 100% additional capacitance, the closest value to that with same size gives me a value of 22uF.

That sounds very reasonable.  You could also use 2 10uF's in parallel.

Hi Chris,

My circuit works great and charges the 12V battery perfectly. 

Thank you so much for all your help. 

- Jimit Shah