Part Number: SN6505B
I'm testing the SN6505B.
In datasheet Table 3, With the SN6505B and transformer(750316029), It is said that about 3.3V(input) -> 5V 1A(output) can be used.
However, when a circuit is constructed and a 5V regulator(LM1117s-5.0) is attached to the output stage and 250mA is pulled out from the regulator, a voltage drop occurs such that the regulator can not be used.
The circuit was designed as follows.
(SN6505B -> Transformer -> Diode(1N5819) -> 0.1uF/1uF/10uF -> Regulator) , In datasheet Figure 46, 47
How can i use 1A?
In reply to Tejas Hommaradi:
When the LDO is not used and no load is applied, about 8.1V is output. At this time, if a resistor of about 50 ohms is attached, the output is only 0.7V. Is this correct? And can you suggest a transformer with a higher turns ratio? Thank you. Regards, Shin
In reply to kyungwon shin:
The above figure shows the circuit tested.
The main power supply will use a lithium polymer battery, but it is currently being tested using a DC power supply.
Therefore, the current limit on the primary of transformer need not be considered.
I would be grateful if you could confirm the problem.
Apologies for delay in response,
Thank you for providing the schematic, consider including a bulk cap at VCC near to the transformer input. Rest looks fine.
To understand the system, can you please provide the voltage & current requirement at the isolated side?
The 1A mentioned in the datasheet is the transformer drive current. After considering efficiency loss of the transformer, the net current available for the system at the isolated side will be lower than 1A.
For instance, in the referenced solution of SN6505B+ Transformer (750316029) and WITHOUT LDO will be able to supply a maximum output current of 561mA at peak efficiency of 85% (lets assume 85% across load for simplicity) of the transformer under worst case condition. Please refer the calculation below,
Transformer drive current =1A
Primary VCC =3.3V
Required output voltage= 5V
Peak efficiency of the transformer = 85% (750316029), let’s assume this is efficiency is constant at all load conditions
Based on above values,
Input power =3.3W -> (3.3V*1A)
Output power = 2.8W ->(3.3W * Transformer Efficiency)
Hence the output current available at 5V will be , 2.8W/5V = 561mA.
Considering the LDO,
With maximum drop out voltage of 1.3V, the voltage after rectification has to be atleast 6.5V.
Considering 6.5V output, the max output current will be 2.8W/6.5V= 430mA
Note, this current would be further lower than 430mA as efficiency of the transformer decreases below 85% at higher load.
Hence, under worst case condition the available load current will be lower than 430mA for the application referred. Kindly provide us the current requirement on isolated side and we can provide suggestions specific to application.
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