2013-04-22 07:43 AM
I have built my own evaluation circuit as below. It seem to work ok when I hook it up to my bench PSU with 2.6v input and the output connected to a standard 5v LED. Output measures 4.9v as expected. The circuit is only populated with C1,C2,C4,L1 & U1. Capacitors are all MLCC's.
However when I link the /SHDN input to ground the LED dims but still remains on. My DVM says the /SHDN is definatly 0v but the output is still giving 2.0v and the LED current is 3mA. The datasheet says :''Shutdown
In shutdown mode (SHDN pulled low) all internal circuitries are turned off, minimizing the
current provided by the battery (I
SHDN
< 100 nA, in typical case).'' So why does not L6920 not fully shutdown ? It seems to be still conducting between Vbatt(7) and Vout(8) when it is in shutdown. So how can this be correct or useful in a battery application ? Its just going to run the battery flat. #l6920db2013-04-24 01:58 AM
Hi,
due to the topology, you will have the Vin-VF
of bulk diode of high side driver (see block diagram on DS) and for sure you will not have a true shutdown... Best Regards Vincenzo