loading 6 NiMH cells in series with max 0.25A

I have a source with 5V and 13W.  But I may only have from this source 5V and 3W for my battery charging.

So I must restrict the battery current to ca. 2,5W (Battery current: I = P/U   2,5W / 9,6V   =0,26A).

My idea is an             boost regulator from 5V to ca 12V        and then         the BQ2000 (he can restrict the current to 0,25A).

Attached is the overwiev from my idea.

What do you think, is this so possible ?

I would take the BQ2000T to terminate with time or T/t

Yes, I understand your system.  Let me know if you need recommendations for the POE controller/converter or the step-down converter.  Have you thought about using the high power POE standard to get 25W out of the port?

Since you can only guarantee a C/8 charge rate for your batteries, you cannot accurately detect termination, even if you use dT/dt.  In applications like yours that use current limited or power limited input sources, nickel batteries aren't the best choice.  Either lead acid or lithium, which both require a CC/CV algorithm, would be better options.

Ok,

but do you think this circuit with bq2000T does work anyway ?

(5V    ----->   boost to 12V       --------->     bq2000 (charging 6 x NimH) = 6 - 9,6V    ------------>   step down to 5V)

Yes, that topology will work.

But you should not charge nickel cells at a C/8 rate unless your pack manufacturer is ok with that.

If you are just charging nickel cells, then you can use either the bq24401 or bq2000T.

- The PIN 8 (MOD) is nowhere specified.  How shall I connect this pin ?

- Where will specified how large the inductor shall be ?

- Where shall be the load connected ?

Which part did you end up selecting?  Did you have a look at its datasheet?  There is also an EVM for the bq2000 that I suggest you look at.

The MOD pin is the output of the chip and drives a FET usually.  A gate drive circuit is usually required to buffer the output to make it strong enough to drive a FET.

The inductor should be sized as in any buck converter, based on the desired ripple current.  The switching frequency of the charger is a function of Rf and Cf, as explained in the datasheet.

By load, I assume you mean battery?  It should be connected as shown in figure 7 in this datasheet: http://focus.ti.com/lit/ds/symlink/bq24401.pdf 

Yes, I have looked to the datasheet.

I have the eavaluation board DV2000TS1 here.

Enclosed I have the schematic from the BQ2000

My questions:

- Pin 8  (MOD):   Can I directly drive the gate of a p-channel MOSFET ?    (please look to the enclosed schematic)

- Inductor L:        How large (Henry) should be the inductor ?                       (please look to the enclosed schematic)

No, you cannot drive your FET in that manner.  Please use the gate drive circuit on the EVM.  There are 2 reasons for this.  One is you need a level shift circuit to properly drive the FET.  The MOD pin outputs a high or low, Vcc (5V) or ground.  The gate of your PFET needs to be at 12V for it to be off and at ground (or some voltage below 12) to be on.

The other reason is that the gate drive level shift circuit buffers the MOD output.  MOD is only capable of sinking 20 mA, which is not much at all for a gate driver.  Thus, in the schematic in the datasheet, MOD drives a transistor which drives the main transistor.

Also in your schematic, the LED pull up voltage needs to be Vcc.  You will destroy the part if you apply greater than Vcc to any pin.  In addition, the diode needs to turn the other way, so that IC sinks the current through the diode.

You will also need an input capacitor to this circuit, just like you would for any buck converter.

Based on your approximate switching frequency of 110kHz, I recommend somewhere in the range of a 330 uH inductor for your 200 mAcharge rate.

OK, I have change this points:

- 2 Transistors with reference to Ground  instead 1 MOSFET

- LED to 5V and the right polarity  (the resistor will be chang to 560 R - it is not yet in the schematic)

- input capacitor

- 330 uH inductor

Enclosed the schematic.

- Will this so work ?

Since you have the EVM, you can modify it to your schematic and see if it works.  That is why we make these EVMs--for customers to play with and modify.

I would recommend the extra circuitry (R3, Q2, D3, C5) on the gate of the main transistor.  This helps turn it off quicker, which will increase your efficiency.  In addition, the resistors in the gate drive path may be too large for a quick turn on of the transistor.  You will want to experiment with their value.

For a load current of 0.2A   I need the  "current measuring resistance" of 0.24 Ohm.                    (datasheet: Imax = 0.05 / 0.24 Ohm)

Than,  the resistor has a power loss from 1.5 W at a discharge current of 2.5A                             (P = I2 * R).

- Does this resistor not become too hot ?

Yes, your math is correct for a charge current of 0.2A.

However, you don't need to connect the second buck converter across the battery and sense resistor--you can connect it just across the battery if you like.