# Buck converter voltage breaks down with load A4403

I am currently working on a Buck Converter that should charge my battery pack 4S2P. The problem is that by increasing the load the voltage is dropping, and with 450 mA (depending on the value of the inductor) the voltage will completely break down.

1. Allegro A4403, Vin= 20VDC, Vout=16.8V, Iout,max= 1.2A
2. R1 = 688k first, then I changed it to 984k.
3. L1 = 10uH...86uH, higher value less load is possible
4. R3=R4=110mOhm; R5=30k,R6=1k5,
5. I tried different diodes, all Schottky but none worked out

Problem:

The voltage decreases as the load current increases. This is because the Toff time increases, which means the inductor cannot build up enough energy to maintain the target output voltage of 16.8V along with the increasing load current.

As the Toff time increases, the frequency decreases by the same amount, tone remains constant. Whether R1 (determines the tone time) is 1Meg or 500kOhm does not affect the maximum possible load current, but only the applied output voltage, at different load currents. Thus, the frequency is not the trigger, of the problem. With increasing inductance, the max. possible load current falls. At 10uH 450mA can flow for a short time, at 82uH less.

The voltage, which is applied to the current measuring resistor is deultlich too high. It is -10V for 250mA. With a current of 250mA and a current measuring resistor of 0,11Ohm only 27,5mV should flow. This is much too high and is outside the max. rating for this pin of -1.0V...0.5V. In case of doubt the device is destroyed.

Since a too high current is "measured", which should not be there, I wonder from where this comes. Some diodes were tried, which did not help. The last one was the 10MQ060N from Vishay. I have also changed the ICs which has not resulted in any change. Also I bypassed the current sense resistor and connected the diode and sense pin directly to MAsse. In doing so, I was able to measure the same waveform at the sense pin (although direser was jumpered directly to ground).

My assumption so far is that wrong diodes were installed, which passed the high voltage directly to the iSense pin, whereupon a pledge (for toff) was formed, which passed the signal directly to Isen. The feedback voltage is for the set 16.8V at 0.8V it is for the case that a very small load flows at >0.8V, because in this case the output voltage is also above 16.8V (18.9V). Again, the IC does not regulate down and there is clearly too much voltage flowing. Why it does not reduce the voltage, although the feedback voltage feeds back the correct value is a mystery to me.

• So are you trying to connect a buck regulator with constant voltage output directly to batteries? If so, stop immediately. A4403 is not a battery charger, but a voltage regulator. Batteries are not charged with constant voltage, but according to their chemistry. Most likely the regulator is hitting overcurrent protection. Sep 10 '21 at 8:49
• Hello @Justme thank you for your response. For now i am running it on an variable load for testing purpose learn.sparkfun.com/tutorials/… In the end i want to charge the li-ion with constant voltage and then adjust the voltage by the DIS-Pin due to PWM. With the variable Load connected, when i start to draw current the voltage is dropping. What i dont understand is, that the output voltage is 18.9V with no load/50mA, but it should be 16.8V as set by Feedback resistors. Sep 10 '21 at 8:59
• You can't charge li-ions with constant voltage, and you can't adjust the voltage by sending PWM to DIS pin either. How is your PCB layout? Why there is no filter on Isen pin? Sep 10 '21 at 9:27
• I am using the EVAL-Kit from Allegro. digikey.com/en/products/detail/allegro-microsystems/… I changed the resistor values and the inductor. It should work with the programmable load no? I mean 16.8V and 1.2A should be possible. In this test setup there are no batteries. Sep 10 '21 at 9:31
• Why does your schematic show a 3.3 volt output when your words are saying you set it for 16.8 volts? Sep 10 '21 at 9:54

There are two things in the data sheet that suggest to me you have chosen the wrong device for your application.

1. The schematic you have posted is directly from the data sheet and it specifies a minimum input voltage of 9 volts to achieve 3.3 volts regulated at the output. That's a whopping 5.7 volts of headroom that is potentially needed to make this chip deliver.

Your requirement is for 16.8 volts from a 20 volt supply. Headroom is only 3.2 volts and that concerns me.

1. The second thing is this table: -

And that is telling me that the device chosen is for low voltage outputs i.e. much lower than 16.8 volts. With a maximum output voltage of 5 volts at 3 amps, that's an output power of 15 watts. You appear to be expecting 16.8 volts at 1.2 amps i.e. an output power of 20.16 watts.

Nothing in the data sheet would convince me to use this chip in your application.

I also have serious doubts about the choice of your inductor value. You appear to have too much inductance (quote L1 = 10uH...86uH) and that certainly won't help.

• Thank you Andy. I also had this doubt. Is there any way to verify if i can use it for higher voltage applications? Due to the bad situation at the market i couldnt find any good alternative. I think I'll have to start looking again for a chip that is available and not too expensive. Do you have any recommendations? Sep 10 '21 at 10:19
• No recommendations given that if you are not in the UK then anything I might find that's available may not be available to you. If you are in the UK or can access Farnell, RS, Digikey and Mouser then use their search engines and go for TI or Linear Technology parts. Sep 10 '21 at 10:42
• I am in germany and using all of the named sources. Thank you for your advice. Sep 10 '21 at 11:01