I´m a new designer and i´m trying to implement the bq24450 in order to charge a VRLA battery 12V;7Ah.
I´ve two questions :
a) I think that the formula written at page 15 of datasheet regarding Rb calculation is not complete
Should be Vth = Vref x (( Ra+Rb + Rc// Rd ) / (Rb + Rc//Rd))
Making the calculation I got Rb = 16,7 K using data of the example not 36,85559 K of the example .
Am I wrong ?
b) How can I get the SOC ( State of charge ) of the battery using this IC BQ24450 ?
I thing that I can use a PIC in order to manage the SOC ( in % ). If so, what are the steps that I can follow in order to do it ?
You are correct. Rb is around 16.7k. We will correct the datasheet (and fix the formatting).
The bq24450 does not report any voltage or current information that would be necessary for gauging. I suggest posting a new post on the battery management solutions forum (the forum above this sub-forum) with a title of "gas gauge for VRLA" or something similar.
Let me ask another question.
As far I understand, the external pass transistor will run in the active mode ( not as a switch ). I am wrong?
Assuming that I want charge a battery 12V / 7Ah and the supply voltage will operate between 20V-30V, and the configuration of bq24450 shall be with pre-charge and a Dext device, the topologie used is the NPN Emitter Follower.
The results were as follows:
Is it ok ?
Correct, the BQ24450 is a linear charger. Your chosen transistor looks acceptable, but keep in mind your operating conditions. How long/often will you run at 30V in? At 750 mA charge current, that is over 10W lost in the transistor. You might look into stepping your input voltage down with a switcher, such as the TPS5410.
By my calculations, I get Rd=422k. The other values looked correct.
Yes, Rd=422K ; I got it, but I defined 1% range around it, so the values were 417K and 426K and I picked 412K. My mistake.
How do you get Pd = 10W; the formulat in datasheet is confused
Using Vinmax = 30V; Vout = 14,4V ;hfe = 200;Imax charg = 750 mA and Rp = 1200 ohm ,
It hardly will run at 30Vin , but can happen.
I'll see the TPS 5410 in order to reduce Vin.
Pd is the power lost in the IC. For the emitter follower topology, this power is equal to the drive current power minus the power lost in Rp. The second term in the equation should read: (Ichg/ hFE)^2 * Rp. Thus, it is simply an I^2R formula.
The power lost in the NPN transistor is the voltage across the transistor times the collector current. This is equal to the charge current (max) * the max differential across the transistor, which is Vin (max) - Vout (min). I suggest stepping the Vin voltage down.
Stepping Vin Voltage down using for instance TPS5420 and using the switcherPro Design the new output Voltage range is 18,186V - 17,175 V, which change the topology.
It´s not an issue, but in fact the initial NPN Emitter Follower will change to the another one Common Emiter PNP, as DV changed from 4,6 V to 1,8V.
Do you agree with it?
Yes, with the TPS5420's maximum duty cycle of 87%, the highest output voltage you can reliably achieve from a 20V input is 17.4V. This gives 3V of headroom for the drop across the transistor and sense resistor, which should be just enough for your 14.4V output. But a change of topology does give more headroom.
Can a MOSFET be used as the pass transistor to reduce loss?
If so, do you have any suggested components/configurations?
A MOSFET would not significantly reduce the power lost in the system. The main source of power loss is because the pass element (either BJT or FET) is being operated in the active region.
The output drive of the BQ24450 is single-ended. It can only source or sink current but not both. This is okay for BJTs, which are current driven devices, but is not okay for MOSFETs, which require charge to be added or removed from their gate. For example, if the BQ24450 is setup in the NPN emitter follower topology and uses an NFET, then there is no way to pull down the gate and remove charge. Possibly a resistor from DRVE to ground would provide the pull down, but that is extra loss.
TI now carries MOSFETs: http://focus.ti.com/paramsearch/docs/parametricsearch.tsp?family=analog&familyId=1647&uiTemplateId=NODE_STRY_PGE_T
The Pd formula in the datasheet is *very* confusing...it would be nice if you would use parentheses to indicate order of operations.
Can you reply with the Pd formula parenthesized (and yielding the same results as the application example on page 15 of the datasheet)?
Per the example,
Vinmax = 13v
Rd = 464000
How do you get 126mW from
Pd=(Vinmax-0v7) / hfe * Imaxchg - (Imaxchg^2) / (hfe^2) * Rd
In general if you could post a reply with the corrected formulas and values from the datasheet example it would be very helpful.
Designing to a 6v 4.5A battery charger with 12v input and target voltages of:
Vth = 5.25, Vfloat = 6.90, Vboost = 7.35
I calculated as follows:
Data sheet suggests an input bias current through Rd of 50uA (see page 15) so: Rc = 2.3 / 50uA = 46K Closest 1% resistor is 46.4K
When charging is completed, Vfloat maintains the battery charge at 6.9v. STAT1 (pin 10) turns off, to enter float mode removing resistorRd from the voltage divider so that the V_SENSE (Vfb) input is determined by Vbat * (Rc) / (Ra+Rb+Rc). The charger drives VSense to Vref (2v3), so we want Ra+Rb=Rc so that Vbat * Rc / 3Rc = V_sense = Vbat/3 (6.9/3=2.3). Since Rc=46.4K we know that Ra+Rb=92.8K.
When fast charging, STAT1 pin 10 goes low which adds Rd to theresistor divider. The V_sense input is determined by Vsense = Vbat * (Rc||Rd) / (Ra+Rb+(Rc||Rd))and the chip attempts to drive Vsense to 2v3 So we wantRc||Rd to change the value of Rc such that Vbat is 7.35v.Since Ra+Rb=91.8K we have: 2v3 = 7v35 * (46.4K||Rd) / (92.8K+(46.4K||Rd)) 474772 = Rd Closest 1% resistor = 475K So (Rc||Rd) = 42271K
Note: Rc||Rd denotes the parallel value of Rc and Rd
Vth defines the threshold for turning on Qext; the desiredthreshold 5v25. Vth is sensed at pin 12 (CE) and whenit is below Vref (2v3) fast charging is disabled.The voltage at pin 12 when power is first applied is defined by Ra / (Ra+Rb+(Rc||Rd))Vth should be kept below 2v3 until Vbat climbs above 5v25 Rc=46.4K and Ra+Rb=92.8K and Rc||Rd=42271 2.3 = (5.25 * Ra) / (Ra+Rb+(Rc||Rd)) Ra = 59174 Closest 1% resistor = 59.0K
Since Ra+Rb = 92.8K and Ra=59K Rb = 33.8K Closest 1% resistor is 34K
Final values: Ra=59K, Rb=34K, Rc=46.4K, Rd=475K
Final voltages: Vfloat=6.91, Vboost=7.36, Vthresh=5.27
Your reasoning to get Ra and Rb is correct, but this equation is a little off: 2.3 = (5.25 * Ra) / (Ra+Rb+(Rc||Rd)). Ra in the numerator should be Rb + Rc||Rd to get the usual voltage divider form. Thus, Rb would equal 16.9k and Ra would be 75k as noted earlier in this thread.
There is a typo in the Pd formula on page 15. Rd should be Rp, as the formula correctly contains on page 14. This gives 126 mW for Pd. We are currently updating the datasheet for clarity, but in the meantime here is the correct formula for Pd for the circuit on page 15:
Pd = [(Vin(max) - 0.7V) * Imax-chg] / hFE - [(Imax-chg / hFE) ^ 2 * Rp]
Using the David Albert´s example can anyone explain how to selec the external pass transistor ?
First, this is the corrected example, I goofed on one of the calculations, thanks to Chris for pointing that out.
My design parameters: Final discharge voltage (1.75v/cell) = 5.25v (Vth) Float voltage (2.25v/cell) = 6.75v (Vfloat) (2.25 instead of 2.30v/cell for longer life) Boost voltage (2.45v/cell) = 7.35v (Vboost)Data sheet suggests an input bias current through Rdof 50uA (see page 15) so: Rc = 2.3 / 50uA = 46K Closest 1% resistor is 46.4KWhen charging is completed, a float voltage is applied (Vfloat)to maintain the battery charge. The desired float charge is6.75v and is set when STAT1 (pin 10) turns off, removing resistorRd from the voltage divider so that the V_SENSE (Vfb) inputis determined by Vbat * (Rc) / (Ra+Rb+Rc). The charger attempts to drive VSense to Vref (2v3), so we want 2.3 = (6.75 * Rc) / (Ra+Rb+Rc) (Ra+Rb) = 1.9347826*Rc (Ra+Rb) = 89774When fast charging, pin 10 goes low which adds Rd to theresistor divider. Note: Rc||Rd denotes the parallel value of Rc and Rd Vsense = Vbat * (Rc||Rd) / (Ra+Rb+(Rc||Rd))and the chip attempts to drive Vsense to 2v3 So we wantRc||Rd to change the value of Rc such that Vbat is 7.35v.Since Ra+Rb=89774 and Rc=46.4K we have: 2v3 = 7v35 * (46.4K||Rd) / (89774+(46.4K||Rd)) 40891 = (46.4K || Rd) 344408 = Rd Closest 1% resistor = 348K Note: then (Rc||Rd) = 40941KVth defines the threshold for turning on the fast charger(external pass transistor) and should be above the batteryfinal discharge voltage to prevent outgassing due to fastcharging while the battery is severely depleted. The desiredthreshold is 5v25. Vth is sensed at pin 12 (Chg_En or CE) and when it is below Vref (2v3) prevents fast charging). The voltage at pin 12 when power is first applied is defined by the resistor divider (Rb+(Rc||Rd)) / (Ra+Rb+(Rc||Rd))Vth should be kept below 2v3 until Vbat climbs above 5v25 Rc=46.4K and Ra+Rb=89774K and Rc||Rd=40941 2.3 = (5.25 * Rb+(Rc||Rd)) / (Ra+Rb+(Rc||Rd)) 2.3 = (5.25 * (Rb+40941)) / (89774+40941) Rb = 16325 Closest 1% resistor = 16.2K
Since Ra+Rb = 89774K and Rb=16.2K Ra = 73574 Closest 1% resistor is 73.2KSo: Ra=73.2K, Rb=16.2K, Rc=46.4K, Rd=348K Vfloat = 6.74138v Vboost = 2.3 * 130715 / 40941 = 7.34336 Vthreshold = 2.3 * 130715 / 57266 = 5.25
Hopefully I got it right this time...
Afonso, the pass transistor selection is made based on input voltage, peak charging current, and hfe.
For example: Vcbo, Vceo = 80V, Icmax=3A, hfe min = 25 @ Ic=1A
The current through the pass transistor during bulk/boost charge is determined by Risns where 250mV/Risns = charge current
Chris, I couldn't find a TI part, but parts like a TIP32B or BD242B seem like a good fit and can dissipate a lot of heat with a proper sink.
What is your target bulk/fast-charge current and voltage?
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