# Powering LEDs light

Why is it that if I connect rectified 9V (using a transformer and diode) my LEDs (about 50 pcs) doesn't get hot, but when I connect same 9V using a cell battery my LEDs get hot quicker and even brighter? Please I need a guide to understand and solve this problem.

The voltage after the diode is slightly high ~10.8V.

And for the battery it's exactly 9V.

• Measure the voltage across the 9 V half-wave rectified supply and across the 9 V battery and edit these into your question. Commented Mar 27, 2020 at 22:03
• How much time does a device getting half-wave rectified AC actually get any voltage? Also, did you include the resistor with the battery? Commented Mar 27, 2020 at 22:07
• What brand and model multimeter are you using to measure the voltage, so that we can determine if you made the measurement with a "True RMS" meter or an average-responding type meter. (FWIW, average-responding meters are usually less expensive compared to True RMS meters.) Are you using a half-wave or full-wave bridge rectifier? Do you have a filter capacitor after the bridge?, and if so, what is the capacitance value and voltage rating for the capacitor? We can't offer good advice when a question does not provide enough specific information. Commented Mar 28, 2020 at 1:03

... I connect rectified 9v (using a transformer and diode) ...

I'm assuming you are describing a half-wave rectified voltage source without a filter capacitor (a capacitor isn't mentioned in your question).

An AC signal whose RMS voltage is $$\X\,V_{RMS}\$$ delivers the same power to an ideal resistive load as a DC voltage $$\X\,V_{DC}\$$. (NB: This equality does not hold for reactive loads.)

So what is the RMS voltage of a half-wave rectified signal? If we ignore the forward voltage drop across the rectifier diode, the RMS value of a half-wave rectified signal can be approximated as,

$$V_{RMS} = \sqrt{ \frac{1}{T} \int^{T/2}_{0} A^2\,\mathrm{sin}^2(\omega t)\,dt } = \frac{A}{2} \qquad \qquad (1)$$

where $$\A\$$ is the peak amplitude of the rectified sine wave. If we assume the voltage at the transformer secondary's winding is 9 VRMS, the peak voltage at the secondary winding is

$$A = 9\,V_{RMS} \times \sqrt{2} \approx 12.7\,V_{pk} \qquad \qquad (2)$$

Plugging result (2) into (1) we see that the half-wave rectifier power supply applies an RMS voltage of 6.35 V to the light-emitting diodes. If one includes the forward voltage drop across the rectifier diode—which is present in the real circuit, then the RMS voltage is slightly less than 6.35 V.

Note that your 9 V power source supplies roughly 50 % more voltage to the LEDs compared to the rectified power supply, which is about 6 V:

$$6\,V + (6\,V \times 50\,\%) = 6\,V + 3\,V = 9\,V$$

so it's not surprising the LEDs are hotter and brighter when you use the 9 V battery compared to the rectified supply.

One final comment. If you want to measure the RMS voltage of the half-wave rectified signal you must use a "True RMS" AC voltmeter (or perhaps an oscilloscope that has an RMS measurement function). You cannot use an average-responding AC voltmeter to measure this signal because an average-responding AC meter is designed to measure pure sine wave signals ONLY. If you attempt to measure any non-sinusoidal signal with an average-responding AC meter—e.g., you try to measure the pulsating DC of a rectified signal, the RMS value that's indicated on the average-responding meter will be incorrect.

• Isn't the formula simplier? Like: Vrms = 2/3 Vac. (Unless you want to know Vrms down to 1/100th of a volt) Commented Mar 28, 2020 at 13:06
• @Fredled, equation (1) is a correct approximation for the RMS value of a half-wave rectified signal (the pulsed DC signal) that IS NOT smoothed by a filter capacitor. The OP makes no mention of a filter capacitor; he states the power supply consists of "a transformer and diode" (diode is singular: one diode) and no other components are mentioned. The RMS value for a full-wave rectified signal is approximately A/sqrt(2) or about 0.7A. Commented Mar 28, 2020 at 17:48