DCR010505 question

Haroldo Amaral

Other Parts Discussed in Thread: DCR010505

I'm testing the DCR010505 and came across the following problem. I tried the basic circuit and without a load the output is 5.02V and the ripple is around 5mV (measured with the myDAQ).

When I turn the load there is a drop and the output is around 4.27V and the error pin goes to the low state.

Trying to solve this problem I put a ceramic capacitor of 100nF between the 0V pin and the isolated ground (this pin was direct connected to the isolated ground). And this solved the problem. With the load the Vout is around 4.99V and the ripple around 15mV.

But why? (I'm using a breadboard to the test).

over 10 years ago

0 Tom Guerin over 10 years ago

TI__Mastermind 38230 points

Haroldo:

The output circuit Vrecitified to Vout(0V) must have a 1uF ceramic capacitor for required filtering of the internal linear regulator input.

The schematic for DCR010505 is

I s your circuit the same as the above schematic?

Tom

0 Haroldo Amaral over 10 years ago in reply to Tom Guerin

Expert 1105 points

Yes, my circuit is the same.

I put the Cfilter of 1uF, but the circuit just work when I inserted the capacitor of 100nF connected to the 0V pin and the isolated ground. This is strange because it's like a short in the capacitor pins (0V and isolated ground were the same point...), but works.

I will post a photo of the breadboard to show.