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MCP73831

Hi

I am playing with the MCP73831 Li ion charger ic.
I have used the example circuit as per the datasheet and it is built on a professionally made PCB, however i am having a couple of problems that im a little stumped on.

The charger itself works but im not sure that its correct.

Source voltage is regulated 5v power supply. However when connected to the charger it drops to 4.85v

The output voltage from the charger is 4.05v where it should be 4.2v

When the battery charge voltage reaches about 4v the status led flickers. I cannot find any info relating to this error.

Has anyone used this chip that may have any suggestions as to the problem.
 

Harald Kapp

Moderator
Moderator
The symptoms you describe let me assume that your regulated voltage source does not deliver enough current for the charger. Try a smaller charge currrent or a power supply with more power.

The flickering LED indicates that the charger IC changes between different operating states (datasheet table 5-1).

Do you have both bypass capacitors in place?
 
Hi

Thanks for your reply.

I also thought this may be due to current draw requirements.
I am using a 5v 1amp regulated supply and i have a 2.49k resistor on the prog pin. This sets charge current to 400ma.

Both 4.7uF caps are in place.
 
Not sure if this helps.

If i remove the battery my supply voltage returns to 5.05v however the output voltage is still only 4.05v

From the datasheet it says these are fixed output, maybe a faulty ic??
 

Harald Kapp

Moderator
Moderator
The datasheet also states on page 3: VDD=Vreg+1V
This means that the input voltage needs to be at least 1V higher than the output voltage. For 4.2V output voltage you need at least 5.2V input voltage.
 
Ohh, i missed that part. I wrongly assumed this ic would run from a USB port.

There appear to be quite a few schematics online indicating this also.

I have a 6v power supply that i can try. Hopefully this will solve the issue.
 
I just looked at that part in the datasheet you reffered to. It says VDD = Vreg +0.3 and also VDD = Vreg +1.0.

Is it not a bit odd to state both??
 

Harald Kapp

Moderator
Moderator
Electrical Specifications: Unless otherwise indicated, all limits apply for VDD= [VREG(typical) + 0.3V] to 6V,

The important words are:
- unless otherwise indicated
- typical

Vdd=Vreg+1.0V falls under "unless otherwise indicated", because it does indicate otherhwise.
It also falls under "typical" because it is never guaranteed that typical data are achieved by a component. Always watch out for worst case (max. or min.) conditions and/or parameters.
 
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