# Power supply delivering more voltage than indicated even when loaded [closed]

Yesterday, I built a very simple circuit to examine the output of an old power supply (probably used for a camera or a very old fashioned mobile phone). It read 6 V / 2100 mA, so I wired it to a 4.7 kOhm resistor and a red LED (forward voltage was 2.1 V iirc). The interesting point was that the actual voltage drop over the whole circuit (measured at both ends of the supply) was in fact around 9V even when loaded. I'm aware of the higher unloaded voltage of unregulated power supplies, but I had assumed that the simple circuit was enough to force it on its "normal" 6V. So, what's the reason for this? Is the supply just garbage (probably, but should still not be 50% higher than indicated!) or is the load just too small?

• Switchmode supply? Commented Jul 15 at 19:07
• maybe the output would be near 6 V when supplying 2100 mA Commented Jul 15 at 19:11
• Please clarify your specific problem or provide additional details to highlight exactly what you need. As it's currently written, it's hard to tell exactly what you're asking.
– Community Bot
Commented Jul 15 at 19:16
• Your load is less than 2 mA, 0.1% of the rated load current. Commented Jul 15 at 19:26
• Load the supply with 4 ohms (6 V / 1.5 A) and see what happens. Commented Jul 15 at 20:54

The most unexciting answer is that it's perfectly normal for a good old unregulated transformer supply. None of this fancy regulated linear supply or newfangled switch mode supply stuff we have these days.

They are those big bulky power supplies with mains frequency AC step down transformers, with the output simply converted to DC with a full bridge rectifier and a bulk reservoir electrolytic cap.

Because 6 VAC * sqrt(2) is theoretically about 8.5V, and AC transformers generally do output a bit higher voltage when unloaded (or with very light load such as a single LED), the output of such power supply is very likely be at around 9V.

It is extemely likely that the RMS output of that 6V supply would be 6V when loaded with about 2A it can output.

It depends on the type of power supply. For example, an unregulated full-wave bridge supply with capacitor filtering,such as this, will charge the capacitor to 1.414 times the root-mean-square (RMS) voltage coming out of the rectifier when unloaded.

This is because a sinusoidal wave has peak voltage of √2 x peak voltage. √2 ≈ 1.414... The capacitor keeps charging, and when the AC voltage drops, the rectifier prevents the capacitor voltage from dropping.

When current is drawn from that supply, voltage will drop towards the RMS AC value. This depends on the size of the capacitor, the frequency (double mains frequency, i.e., 100 or 120 Hz in Europe and Americas), and the load. Calculate the impedance of a capacitor to see how much of a drop is expected.

At 120 Hz, a 1,000 µF cap has ~1.3 Ω impedance. If the load has the same impedance, then about half the voltage above the RMS value would be dropped across the cap. A supply with 7 VAC RMS transformer outputs ~10 V peak. A silicon-diode bridge might drop ~1.2 V, leaving ~8.8 V pulsating DC peak. With the small current drawn by that LED, then, 8.8 VDC would be expected output. With a load, the voltage would drop back towards 5.8 VDC, or even lower, because of transformer impedance and core saturation.

At the rated load of ~2 A, the voltage would drop considerably, likely to the expected 6 VDC rating. Test that by placing a 3 Ω resistor across the output.