Power Supply Design w/Transformer - Not Enough Current

Question

I am testing the following power supply which steps down from 230VAC to 12VDC.

When testing without T1 and supplying 39VDC instead of the rectifier, the switch-mode power supply (TPS5401) outputs 12V and 50mA can be drawn continuously. When testing with T1 and the bridge rectifier, the bridge rectifier outputs 39VDC and unloaded the switch-mode output is 12V but upon trying to draw 50mA from the 12V, the output drops to around 2V with the bridge rectifier remaining at 39VDC. How can I ensure that the transformer is suitable for my current draws?

Transformer Specifications:

1. 230V-15V AC, 1.8VA. Measured (unloaded) : 28.2V AC
2. For 230VAC, no load peak voltage, Vsec peak = 35.5V
3. For 230VAC, full load peak voltage, Vsec peak = 19V at 70 degrees Celsius

Thanks

Answer 1

As the value of the rectified DC input to your buck regulator stays the same @ 39V, it's unlikely that the transformer is at fault.

You wish to draw 12V @ 50mA from the regulated output, i.e. 0.6W.

A transformer rating of 1.8VA is more than adequate.

I calculated the ripple voltage on the storage cap and likewise get a 3.8V drop during the diode non-conduction period,

Something else is amiss.

Answer 2

Consider the waveform below:

As you can see, the transformer only supplies current during the portion of the waveform during which the output voltage of the rectifier is greater than the voltage on the capacitor (during time Tc). You must supply an average of 50 ma during operation, so the instantaneous current in the transformer will be a large multiple of this value, with the multiple being at least T/(2*Tc). Since the current must be supplied at or near Vpeak, the VA value of the transformer must be increased accordingly. You are supplying 12 volts at 0.05 amps or 0.6VA, and your 1.8VA transformer is probably marginal or inadequate.

You are seeing only a small ripple on the output when loaded, but this is occurring when your device is not really supplying power. Capacitor current is C*dv/dt, and you have 110 uF. At 50 ma, dv/dt = 455 volts per second. Since the full-wave rectification is 120 Hz in the US, the voltage should drop up to 3.79 volts between each charging peak at this capacitance. In the 50Hz world the peaks are 1/100 of a second apart, so the drop would be as high as 4.55 volts. Increasing the capacitance reduces ripple, but it also reduces Tc and drives instantaneous current higher, meaning larger transformers if you want to stay efficient.