# ADP2291 Battery Charger Circuit is not working as expected

#### turgut

Nov 8, 2022
15
Hello everyone!
Im designing a battery charger circuit with ADP2291. You can see the datasheet in this link.
Normally this circuit should work like this. When you choose RS=100mohm you have to get Imax=1.5A.(Changing between Vin=4.5-6V)

Here are my inputs,
-Vin=5V
RS=100mohm
-My battery voltage between 3-4.2V. Not less than 3V and not equal or higher than 4.2V. (I tested is my battery working normal. So its working normal.)

My outputs like this,
-Vout=4.2V (as expected)
-Vrs=70mV (it should 150mV)
-Imax= 210-240mA and Ibat=200-230mA (it should 1.5A max)

I exactly set the figure.22 and figure.23 from datasheet. I also tried lots of different way. But I couldnt find the answer. Can anyone help me about ADP2291 or why its not working as expected?

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#### Harald Kapp

##### Moderator
Moderator
Nov 17, 2011
12,722
The application circuit states 750 mA (using a 200 mΩ sense resistor):

You try to increase the current to 2 × 750 mA = 1.5 A by changing RS to 100 mΩ.
The transistor used for Q1 is a FZT549. Acc. to the datasheet this transistor is designed for a continuous collector current of 1 A only. Using it for a 1.5 A application will drastically shorten its lifetime.
Assuming it works: At 1.5 A ICE the voltage drop across Q1 is VCE ~ 1 V. Your battery voltage may be high enough such that the battery voltage + the voltage rise due to charging reach 4 V (5 Vinput - 1 Vdrop and thus charging current will be severely limited.
Try with e.g. a 6 V power source...

Next step ist to use a transistor for Q1 that matches the requirement for 1.5 A continuous ICE. A heat sink may be required to avoid overheating of Q1.

#### turgut

Nov 8, 2022
15
The application circuit states 750 mA (using a 200 mΩ sense resistor):
View attachment 56849
You try to increase the current to 2 × 750 mA = 1.5 A by changing RS to 100 mΩ.
The transistor used for Q1 is a FZT549. Acc. to the datasheet this transistor is designed for a continuous collector current of 1 A only. Using it for a 1.5 A application will drastically shorten its lifetime.
Assuming it works: At 1.5 A ICE the voltage drop across Q1 is VCE ~ 1 V. Your battery voltage may be high enough such that the battery voltage + the voltage rise due to charging reach 4 V (5 Vinput - 1 Vdrop and thus charging current will be severely limited.
Try with e.g. a 6 V power source...

Next step ist to use a transistor for Q1 that matches the requirement for 1.5 A continuous ICE. A heat sink may be required to avoid overheating of Q1.
I changed FZT549 (1A) with FZT749 (3A) and I also changed BAT1000 with 1N5822 but still didnt work.

#### Harald Kapp

##### Moderator
Moderator
Nov 17, 2011
12,722
The Schottky diode adds another 300 mV to 400 mV voltage drop.
Have yout ried a higher input voltage?

#### turgut

Nov 8, 2022
15
The Schottky diode adds another 300 mV to 400 mV voltage drop.
Have yout ried a higher input voltage?
i tried till 11V. Im really tried lots of different ways.

#### Harald Kapp

##### Moderator
Moderator
Nov 17, 2011
12,722
Collector and Emitter of Q1 are correctly connected? When you swap C-E, the transistor will still work but with a drastically reduced gain which may be insufficient to drive the collector current.

Another option to try: remove the FET 2N7002, leaving the ADJ pin open:

Thus you can exclude any interference from the shutdown circuit.

Make sure R5 is 500 kΩ, not accidentally much lower. This may possibly prevent the /CHG signal to pull down the gate of the MOSFET. Measure the gate-source voltage of the MOSFET to ensure it is below the threshold voltage (<< 1 V).
Measure the voltage at the ADJ pin, it should be ~ 3 V (see the table above).

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#### turgut

Nov 8, 2022
15
I have better and new results. First I changed FZT549 with a new one. Also Im not using 2N7002 as MOSFET now. I changed 2N7002 with IRFZ44N for testing. I set figure.24 (charge termination) again with this components. New results like that,
-Vin=5V Ibat=130mA
-Vin=6V Ibat=260mA
-Vin=7V Ibat=395mA
-Vin=8V Ibat= 516mA
Now I can getting higher currents. But FZT549 is over heating when voltage increase 6V. Also its not charging battery as it should be. 12-13mV charging per 10 minutes.

#### turgut

Nov 8, 2022
15
Collector and Emitter of Q1 are correctly connected? When you swap C-E, the transistor will still work but with a drastically reduced gain which may be insufficient to drive the collector current.

Another option to try: remove the FET 2N7002, leaving the ADJ pin open:
View attachment 56852
Thus you can exclude any interference from the shutdown circuit.

Make sure R5 is 500 kΩ, not accidentally much lower. This may possibly prevent the /CHG signal to pull down the gate of the MOSFET. Measure the gate-source voltage of the MOSFET to ensure it is below the threshold voltage (<< 1 V).
Measure the voltage at the ADJ pin, it should be ~ 3 V (see the table above).
I think I find the problem. Still i want to ask am i thinking right? FZT549 and FZT749 both of max power is 2W. Im getting 400mA when Im giving 5v. That means 2W-max value for the transistors. They cant get higher miliAmps because they are in maximum power.
My idea is if i change the transistors with higher watts it should work. I need 750mA-800mA when Vin=5V so if i choose 3-4W transistor my circuit will work as expected.

#### Harald Kapp

##### Moderator
Moderator
Nov 17, 2011
12,722
The power dissipation of the transistor is given (mainly) by P = ICE × VCE.
With an input voltage of 8 V battery voltage of 4 V and a current of 0.4 A this means P = (8 V - 4 V) × 0.4 A = 1.6 W, well within the limits of the transistr, if well cooled.
At an input voltage of 5 V the power is accordingly less.
Cooling (heatsink) is a good idea anyway.
You haven't repsonded on my question about correct connection of collector and emitter (post #6).

#### turgut

Nov 8, 2022
15
The power dissipation of the transistor is given (mainly) by P = ICE × VCE.
With an input voltage of 8 V battery voltage of 4 V and a current of 0.4 A this means P = (8 V - 4 V) × 0.4 A = 1.6 W, well within the limits of the transistr, if well cooled.
At an input voltage of 5 V the power is accordingly less.
Cooling (heatsink) is a good idea anyway.
You haven't repsonded on my question about correct connection of collector and emitter (post #6).
All connections are true. I checked lots of times. In addition My Vin=5V and Vbat=3.7~3.8V. Im getting 400mA current in circuit. (i see 400mA on DC Power Supply screen.) But I need ~750mA to charging the battery. When i increasing the voltage till 6V, current tracing the increase voltage till Vin=6V. But not as expected which datasheet telling you can get 750mA when your Vin=6V. I have not a heatsink if i have i would tested it. Also FZT549 is heating when voltage is equal or higher than 6V and not giving the current that i want.

#### turgut

Nov 8, 2022
15
I said I set completely the figure-22 and the figure-23 in the datasheet. The circuits are not working as expected. This part is important "The circuits are working but not as expected."

In addition i find a list in a web site which i added the link. This list is telling that which products' examples are false and which products' examples are true. According to the list the examples are false which has ADP2291. So i will add link and photos about ADP2291. Im waiting different ideas, examples or true components which i can use. Also i wonder is this component working healty or the component has a problem?

If i choose different transistors such as ZXT790AKTC, BDP954H6327XTSA1 and DXTP3C60PS-13 circuit can work? Because FZT549-FZT749 are really heating even if has heatsink.
Also im going to change resistors with different values such as 50, 100, 200, 300 milliOhms or maybe little higher.
Are there useful or not? Its my first project and its getting complicated.
Im completely open any different suggestions.

A photo is from my circuit and a photo from the list.

#### Harald Kapp

##### Moderator
Moderator
Nov 17, 2011
12,722
This list is telling that which products' examples are false and which products' examples are true.
I think you misinterpret the list. "Samples" doe not refer to the circuit examples in the datasheet but whether sample components are available or not.

I set completely the figure-22 and the figure-23
Break down the problem. Try to realize the circuit from fig. 22 without the shutdown mechanism first. Get it working for 1.5 A as required.
Only then add the shutdown logic.

FZT549-FZT749 are really heating even if has heatsink.
Of course they get warm. They dissipate a lot of power, see post #9. Use a true power transistor in e.g. in a TO-220 or a TO-247 case which incidentally makes it easy to attach a suitably large heat sink.

A few questions:
- Is your 5 V regulated power supply capable of delivering the 1.5 A?
- Is C6 placed as near as possible to U2? Using short wires? Try an additional 100 nF ceramic capacitor directly from U2 pin 6 to U2 pin 2.
- Do you have a solid ground (-) system? Thick, short wires? Ideally all connected to GND of U2 (Pin 2) in a star connection.
- Do you have access to an oscilloscope to measure the input voltage and the output voltage?

The chip may oscillate due to insufficient bypassing (C6) or a bad ground system. Your multimeter will show the DC component of the oscillating current as a mean value.
Lacking an oscilloscope, you can measure the voltage across R6 using the AC-range of your multimeter. If the current oscillates, you will see a noticeable AC component of the voltage (and hence the current through R6) instead of a negligible AC component when there are no oscillations.

#### turgut

Nov 8, 2022
15
I think you misinterpret the list. "Samples" doe not refer to the circuit examples in the datasheet but whether sample components are available or not.
I understand that part OK. Thank you.

- Is your 5 V regulated power supply capable of delivering the 1.5 A?
2 or 3 times pin 4 touched the pin 3 then i saw 1.3A or higher so i think yes delivering.

Is C6 placed as near as possible to U2? Using short wires? Try an additional 100 nF ceramic capacitor directly from U2 pin 6 to U2 pin 2.
Do you have a solid ground (-) system? Thick, short wires? Ideally all connected to GND of U2 (Pin 2) in a star connection.
It is not near as possible because circuit is on the breadborad now with the soldered wires by hands. ( Every one wire near 10cm and have solder ) Im just doing lab tests if i can saw it works i will get as pcb.

Do you have access to an oscilloscope to measure the input voltage and the output voltage?
I will work on it.

#### turgut

Nov 8, 2022
15
Do you have access to an oscilloscope to measure the input voltage and the output voltage?
When i gave Vin=6Vdc from power supply, I saw Vin=5.4V and Vout=4.14V on the oscilloscope screen.

#### Harald Kapp

##### Moderator
Moderator
Nov 17, 2011
12,722
2 or 3 times pin 4 touched the pin 3 then i saw 1.3A or higher so i think yes delivering.
That is not proof. The power supply needs to be able to supply 1.5 A while still delivering the set output voltage. The specifications of the power supply should give that information. Lacking a spec sheet, you can measure the output voltage while drawing 1.5 A to check the power supply's capabilities.

When i gave Vin=6Vdc from power supply, I saw Vin=5.4V and Vout=4.14V on the oscilloscope screen.
The oscilloscope is meant to see excess AC ripple, not to measure DC.
Anyway, assuming Von is the voltage at U2, pin 6 (by the way: it is good practice to label the voltages, currents etc. in the schematic so we are sure to have a common understanding - that's why I "assume") and Vout is at connector J2, Pin 1, then the voltage drop across R6 and Q1 is 1.26 V. Please supply a measurement of both the voltages across R6 and across Q1. With this information we can check whether it is an issue with the resistor or with the transistor.

Just for testing the transistor: disconnect the base of Q1 from U2, pin 1. Instead connect the base by a 110 Ω resistor to ground (-). This should drive the base with ~40 mA. What is the output current in this case? What is the VCE in this case? (Also note at which input voltage you measure.)

#### turgut

Nov 8, 2022
15
Just for testing the transistor: disconnect the base of Q1 from U2, pin 1. Instead connect the base by a 110 Ω resistor to ground (-). This should drive the base with ~40 mA. What is the output current in this case? What is the VCE in this case? (Also note at which input voltage you measure.)
I did and the results like this.
Vin=5V Ib=32mA I=300mA Vce=-70mV
Vin=6V Ib=36.3mA I=620mA Vce=-140mV
Vin=7V Ib=39.4mA I=1A

Also i tried IRF4905 instead of FZT549.
With FZT549 DRV Pin voltage is 3.16V (Vin=5V) and 4.35V (Vin=6V)
With IRF4905 DRV Pin voltage is 0.2V (Vin=5V) and 1.25V (Vin=6V)

#### Harald Kapp

##### Moderator
Moderator
Nov 17, 2011
12,722
Your measurements show that the transistor is saturated (VCE is only a few millivolts.
Looking at the measurements with 5 V input voltage:
IC/IB = 10. The transistor's typical gain is much higher, therefore the gain is not the current limiting factor.
R6 with 100 mΩ is also not a limiting factor as the voltage drop across R6 is only 0.03 V (100 mΩ × 0.3 A).
The only explanation (apart from a wiring error) is that the battery itself limits the current because it is fully charged. In that case the battery voltage rises when trying to charge the battery and thus reduces the voltage difference between input voltage and battery voltage so much that the current is limited.
Use a battery that is definitely discharged to avoid this effect. Or use a 2.7 Ω / 6 W resistor instead of the battery for testing. YOu may not have such a power resistor. You can also use 12 × 33 Ω / 0.5 W resistors in parallel.

Also i tried IRF4905 instead of FZT549.
The ADP2291 is meant to be used with a bipolar (PNP) transistor. The IRF4905 is a MOSFET. Using a MOSFET instead of a bipolar transistor is not covered by the datasheet.

#### turgut

Nov 8, 2022
15
Your measurements show that the transistor is saturated (VCE is only a few millivolts.
It was my bypass measurement. But this measurements which i add are not bypass. Im trying lots of thing but anything is not working.

#### Harald Kapp

##### Moderator
Moderator
Nov 17, 2011
12,722
Assuming RS = R6, your measurements make no sense.

Let's look at BCX53-16 at Vin = 4.5 V:
With VS = 80 mV and RS = 100 mΩ The current through RS is IS = 80 mV / 0.1Ω = 800 mA.
With an output current of Iout = 30 mA this means that the base current out of Q1 is 770 mA! This is simply not plausible because at this base current Q1 will be fully saturated and Vout = Vin- VCEsat = 4.5 V - 100 mV = 4.4 V, not 4.19 V.

#### turgut

Nov 8, 2022
15
i used 2 different test;
1) Two parallel power resistors which 5w & 10ohm
2) A power resistor which 5w 2.2ohm
Results are same. They are not working. I = 0 when i use power resistors as battery. But when im using battery that battery charge near 3.7-3.8V. Current is like in the table. (100-200mA)
Also i did a different thing to understand is my circuit cutting current when battery voltage is higher. So connected a different battery which battery charge is 4.1V. This 2 battery is same brand. They are completely same except their charge. The results like that,
-Battery charge=3.7v Vin=5V I=130mA
-Battery charge=4.1v Vin=5V I=60mA
So current cutting proterty is working. But still im not seeing higher currents like 500-750mA.

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