Clearly you have to load the power supply to get an accurate voltage 
measurement. You could use a purely resistive load for a current of about 100mA 
and that would give you an accurate voltage measurement. Using the formula R = 
V/I, select a resistor that would give you close to 100mA. Remember the power 
dissipated in the resistor must be considered so given P=VI, a 100mA load will 
dissipate 1/2 watt, so the resistor size is important. 

Now Gerald says the power supply needs to be 5V, but I think he is being overly 
restrictive here. The datasheet for the TPS65217C shows the AC input should be 
between 4.3V and 5.8V, so my way of thinking is 5.2V should be fine as long as 
there are no noisy pikes that cause the voltage to exceed 5.8V. However, Gerald 
is the original designer and he has set the input voltage parameters for a 
reason. 

Regards,
John




> On Jul 18, 2016, at 7:20 AM, [email protected] wrote:
> 
> first thanks for your reply.
> 
> i am measuring without the BB connected since i want to know if i am able to 
> provide the required voltage..Also if i am not mistaken the BB white has a 
> voltage regulator so there is no reason to count with the BB connected 
> because there will be a voltage drop if i supply more (but not too much). Am 
> i wrong?
> In any case i can give it a go and measure with the BB connected..
> But again..does anyone know what is the supposed voltage to be supplied?Is it 
> the one i mentioned in my previous post?(5V +/-.1 ? ) Cause as i have already 
> said it seems that my bb works when provided with 5.26V and not with 
> 5V..(hope i won't destroy the board..)
> 
> Thanks again
> 
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