I am looking at making a power supply with a TNY263 TinySwitch at the center.
Here is the datasheet.
How can a power supply accept a range of inputs, 85-265VAC in this case, and output a constant 5VDC with only a single transformer.
There is an example Constant Current Battery Charger on page 7 of the datasheet. From what I understand of the circuit the AC comes in on the left, the bridge rectifier fully rectifies the input, some capacitors smooth out the signal and send it through the transformer where it exits and is controlled further to output a constant 5V.
What is the role of the TinySwitch in this circuit? How can it enable a transformer with a set number of input and output turns to output a constant usable DC voltage?
Does the TinySwitch register the input voltage and then pulse that to create a lower output voltage? Similar to pulse-width modulation.
Best Answer
Most low/med power supplies use a transformer but it is used in a non-conventional way. It's called a fly-back transformer and operates as follows.
The primary winding is connected to the rectified DC supply for a short period of time. Current ramps up to some value and this value is dependent only on the primary inductance and the incoming rectified and smoothed supply voltage - it does not depend on anything connected to the secondary.
How does that happen you might ask - well, the secondary output is half wave rectified but the winding polarities (see dot notation) are such that when voltage is applied to the primary, a negative voltage appears on the secondary - this means the half wave rectifier does not pass anything thru to the load.
OK so far?
The primary inductance and current together conspire to store energy and when the transistor (that switches the primary) goes open circuit, that energy is released into the secondary load thru the diode rectifier: -
If the load needs (say) 10 watts then the energy transferred per cycle is 10 watts divided by switching frequency. If the load resistance suddenly halved, then to maintain the same output voltage requires twice the power and so the "thing" that controls the power transistor in the primary keeps the primary attached to the DC supply a little bit longer. The primary now takes in twice the energy it took previously and after a small glitch in the output voltage has passed (due to the load discontinuity and the feedback system taking a finite time to respond), the device continues to be in regulation. Here's a little picture that shows the feedback system: -
Turns ratio is less important than primary activation duty cycle - this is the "new turns ratio" and most switching devices might operate from 1% to over 50% duty cycle.
As far as energy is concerned, a 50:1 change in primary activation time means a 2500:1 change in the energy that can be transferred to the output.
Given that loads might vary between 5% and 100% and input rails might vary over a range 3:1 (9:1 for energy), the total change in energy demand might by 20*9:1 = 180:1 and is easily covered by the duty cycle considerations.
It is the main transistor switch for the primary and part of the feedback system that monitors the output voltage so that it may adjust the duty cycle. This diagram shows the top switch but it's pretty much the same as the tiny switch: -