# Input current on linear power supply

I am working on a new project, and I need to build a big PCB with a lot of circuits integrated. Therefore, I need to make a power supply on the PCB.

Since I am not comfortable with SMPS design, I decided to make a linear power supply with a series regulator. I am still designing the circuit in simulation (Multisim). The other day I was measuring currents on the secondary side, and noticed the RMS current of the transformer and bridge rectifier is greater than the load current.

So far as my knowledge goes, the transformer and bridge rectifier need to handle the almost 2 A of RMS current I am getting, but the load is only around 700 mA.

How is this possible? Am I doing something wrong?  • Plotting the currents will help you understand why. Feb 5 at 13:28
• i understand the process of charge and discharge of the capacitors in the circuit, but shouldn't the current rms values be somewhat similiar? The capacitors just charge around the maximum and minimum value of the secondary ac wave, and then proceed to discharge supplying the load. Feb 5 at 13:45
• Why are you even designing a linear regulator from discrete parts instead of using a linear regulator IC? Feb 5 at 14:48
• I might need more current than 1 amp, it's a scalable project Feb 5 at 14:50
• (18 V AC leave quite a lot of headroom for the regulator.) At almost 10 W, the TIP122 will need a decent heatsink. RMS is sort of the "heating equivalent" current in resistors: resistive loss is closely related in the transformer, but not in the rectifier. Feb 5 at 18:50

Energy/Power is conserved, not current. Spikey waveforms have a high RMS value (relative to average).

Let's simplify the problem and assume that the charging current is a current spike with a duty cycle of 10%. Assuming that the voltage is the same on the input and output, if the output is 1 Amp continous, then the input must be a 10 A spike at 10% duty cycle (same energy).

Calculate the RMS of the input: 10 Amps squared is 100, averaged over one cycle is 10, square root is 3.16 Arms.

So, for this case, the input RMS current is more than 3 times the output RMS current. • Energy/Power is conserved as is charge. Feb 5 at 18:39
• So does the transformer and the diode bridge need to handle the ~2 Amps rms value? I'm almost certain they do. Another question, is how do i lower the rms current coming from the input side without affecting (at least much) the ripple on output voltage? Is there any way? Because like this the power factor is atrocious. Thank you. Feb 5 at 19:09
• @JoãoMarques - for the diode, you need to be within the RMS and the peak specifications. For the transformer, I don't know for sure, maybe you need to ask another question. Feb 5 at 23:34