# Power Supply Circuit ±9V from a 6V centre tap transformer

At work, there's a power supply which is used on one of our products. The guy who originally made the board has since left and none of the senior electronics guys could give me a good answer.

Here's a schematic of the PSU

There's the transformer (Myrra 45055) a bridge rectifier (KBP305G) and some 10000uF caps.

From my understanding, from the output of the transformer I've got 12V AC. With the centre tap being GND I would have expected my outputs to be +6V with respect to centre tap and -6V with respect to centre tap. What the actual circuit gives is ±9V (±0.2V), I'm unsure how I'm getting a total of 18V difference between + and - from 12V out the transformer.

I've never really dealt with transformers or AC before, is there a misunderstanding in my thinking? Or are the caps doing something more than just ripple smoothing?

• Note that the 6V is an RMS value but you rectify to the peak value which is ~17V in range Apr 27 '16 at 9:14
• and its unlikely to be exactly 6V so with a 10% tolerance it may be as much as 6.6V giving a 9.25V peak per winding ( 18.5V) Apr 27 '16 at 9:17
• @PlasmaHH I think it's starting to come back, something about 1/sqrt2? So the RMS value is essentially the average of the sin wave which when rectified and smoothed the voltage becomes the peak of the sin wave? Apr 27 '16 at 9:18
• @Hayman: Well, RMS is RMS, if you want to know more, there is a wikipedia article about it. I am seriously wondering though why your senior engineer doesn't know about it... Apr 27 '16 at 9:24
• @dim I'm just going off what I measured, I measured 5 different power supplies that all varied from 8.82V to 9.06V. It's not a design tolerance, it's just the extents of the values I got from my measurements. Apr 27 '16 at 10:53

RMS versus peak: -

RMS is the equivalent DC voltage that would dissipate the same heat in a resistor when the AC voltage is applied. You can do the math$^1$ if you want but, for a sinewave, the RMS is 70.71% of the peak OR the peak is $\sqrt2$ higher than the RMS.

When ever we talk about AC voltages (without specifying peak or peak-to-peak) we imply RMS hence, a 6V RMS signal will have a peak of 8.49 volts (nearly 9 volts). After passing through a rectifier diode this might be more like 8 volts DC on the smoothing capacitor.

However, transformers unloaded will run at a slightly higher voltage on the output so expect a volt more (or so) on the output.

$^1$ Take a 10V peak sinewave. Square the waveform - this produces a double-frequency sinewave between 0V (lowest peak) and 100V (highest peak). Take the Mean - 50V then take the Square Root - 7.071 volts i.e. 7.071 volts is the RMS value of a sinewave having a peak of 10V.

R = root, M = mean, S = square = root of the mean of the squared signal.

When you smooth the rectified output of an AC supply, you get the peak value which is root 2 (1.41) times the RMS AC value. Less any voltage drop in the rectifier diodes. Which is about 8.5 volts for 6 volts dc minus the losses. It shouldn't be right up at 9v, but it'll be well above 6. With the transformer lightly loaded, the AC value will be higher than nominal also. Generally small transformers have poorer regulation than large ones. I have a small nominal 12V (DC) PSU in front of me (from a "wall wart") which produces over 20V DC when unloaded. This can be a significant factor when rating other components. (For instance, it renders a 16V capacitor inadequately rated).