# Peak output current of a Full-Wave Bridge Peak Rectifier

I am attempting to design the first section of my power supply. In this case, I am going through my calculations for the Rectifier section. I am simulating my circuit with PSpice and while most of my calculations agree with my Spice simulation, my calculated diode current is way off.

Here is the circuit I am designing: Here is my input voltage (red, across the secondary coil), and my output voltage (green, across the resistor/capacitor) Everything so far matches my calculations perfectly. Now for the part that has me stumped:

Here is my diode current (blue), load current (green), and capacitor current (red). As you can see, my simulation tells me that the total current = 540mA Alright. I'm working with: Vr = peak-to-peak ripple voltage = 1V R = load resistance = 1k-ohm C = filter = 270.8uF Vin(peak) = Secondary voltage peak = 24(sqrt2)    My calculations of the peak output current give me 855mA, while the PSpice sim gives me 540mA. I am using the equations from the Sedra/Smith book, provided in the peak rectifier section. Is there something I'm missing here? Thanks!

• what is the Vr term of 1.0? – smashtastic Mar 27 '14 at 22:36

spice will be using a diode "model". That model is likely to include things like recovery time and V-I curves.

The model may have a number of different break points that reflect the V-I curve of the real device. Have a look at the data sheet of the device and you will find the V-I characteristic curves. The forward drop will be found based on the current passing through the device and may not be exactly 0.7 V.

Also the recovery time of the diode may be having an impact in your spice model that is not shown in your equations. Time taken for the diode to start conducting fully will allow some charge to pass through to the capacitor, hence reducing the peak current.

Try creating a perfect diode model that has a vertical V-I characteristic at 0.7 V and nothing else and compare that answer to your equations.

I would be happy using the theoretical value 855 mA in sizing the diodes for peak forward current as it is above the spice simulation.

Finally real life will be different again! You will not have a capacitor as precise as 270.8 uF, a real capacitor will have ESR, a real transformer with have DC resistance, real components will heat up and there characteristic will change with temperature.

• Thank you for the explanation. I will definitely have a look at the datasheet graphs! – Nevermore Mar 28 '14 at 18:51

Your calculations is more close to the reality than PSpice

IDpeak=ICpeak+ILmax

ILmax with an 85% efficiency of the power supply, is about 38mA. The time to refueling of the 270.8μF capacitor from Vmin (31.5V) to Vmax (32.5V) is 3.5msec, resulting an ICpeak of 825mA. So the IDpeak will be 863mA.

• Thanks for the insight. It looks like I'm going to go with the calculated results and make a note of the "real life" discrepancy in my design. :) – Nevermore Mar 28 '14 at 18:53
• I can provide detailed documentation if you wish – GR Tech Mar 28 '14 at 23:09