How to convert an amplifier or receiver from incandescent lamps to LED lamps
I use a Dual CV 1460 amplifier for my office/workshop and not just because it sounds good; I like the looks too.
Mostly the reason I like the way it looks is because of the big, analogue VU meters it has on the front. But they were lit with incandescent lamp bulbs and I had already replaced them once when they all went again. So this time I have replaced them with LEDs which should last a lifetime. In this post I will document how I replaced them including the calculations.
The original incandescent lamps were rated at 12V/110mA as can be seen on the Service Manual circuit diagram. The current is irrelevant to us as that will now be dictated by the resistance of the new LEDs.
Figure 1. Circuit diagram from Service Manual.
I bought the LEDs to replace incandescent lamps in a Harman Kardon HK 730 receiver and had a few left over. I could see that I had written on the packet I kept them in “12V/1W”.
I do not like the lamps on my equipment to be bright as I find it intrusive so I put one of the LEDs across the output of a power supply and gradually turned the voltage up. The LED lamp came on at around 7.0V (note there are several LEDs within each lamp hence the high value of 7V) and was at a brightness I liked at 8.0V so with 3 in series I would need 24V dc to power them.
The original lamps were powered by 35V ac and they were also in series giving a voltage of 35 / 3 = 11.66V across each – so a little under-rated which is a good way to increase their lifespan.
Figure 2. My drawing of the original lamp circuit.
LEDs conduct (and therefore give off light) only when a positive voltage is applied to its anode. This means an ac voltage would cause the LED’s light to flicker at around 50Hz (mains frequency) and that can be seen by the human eye especially when one’s eyes are moving. So I didn’t want that and to get rid of it I decided to use a bridge rectifier giving full-wave rectification. With full-wave rectification the ac waveform output from the bridge is approximately 1.414 times the input less a volt or two for the voltage drop across the two conducting diodes (the exact voltage depends on the diode and how much current is being drawn). So off-load (no current being drawn) the output voltage from the bridge is 35Vac x 1.414 = 49.49Vdc (off load – remember the voltage will drop a little across the diodes when conducting, usually around 0.6V for silicon and around 0.3V for germanium but who uses them anymore). I used an electrolytic capacitor to smooth any ripples on the voltage output, it doesn’t need to be exact, anything between 47µF and 470µF is fine so long as the rated voltage is well above what it has to cope with – I would have preferred to use a 100V cap but 63V is Ok.
Figure 3. My drawing of the new lamp circuit.
In the same way that the original lamps had a third of the input voltage across them, the LEDs will also drop a third of the voltage giving them each about 16.5 volts – too much. Some of the voltage has to be dropped across a resistor (RL) to get rid of the excess.
As said above, the LED lamps had “12V/1W” written on their packet.
Watt’s power law says: W = V x I
Therefore: I = W / V = 1 / 12 = 0.08333A = 83.333mA or to put another way 83.333×10-3A
NB: Since all three LEDs and the resistor RL are all in series, the same quantity of current flows through each of them.
So we know that the new LED lamps should have a current through them of 83.33mA at 12V. But since I would be using them at 8V, the current/wattage used by them would change. Why is it important to know the current or wattage? The voltage applied to the LEDs is too high so must be reduced. This is done by using a potential divider network which is a fancy name for what you see in Figure 3 above; the output from the bridge rectifier (at 49.49V) is across the resistor RL and three LEDs all in series (the capacitor is only used to smooth the voltage peaks, it can be ignored in these calculations). The voltage of 49.49 volts is divided across each LED and the resistor. For the sake of ease, lets say the voltage is 100V. If the LEDs were 200Ω each and RL 400Ω then there would be 2V dropped across each LED and 4V dropped across RL. The voltages dropped would be the same if the resistances were 2000Ω and 4000Ω (respectively) but the current would drop to 1/10th. This is why the current/wattage/volts rating of the LEDs is important; it tells us what their resistance is and from that we can work out the resistance of RL.
But we aren’t running the LEDs at 12V and being a semiconductor, their resistance will not be the same at 8V (it is notable that the resistance change is non-linear). However, when I put one on the power supply, I had a multimeter measuring the current flow and saw that at 8V it was 3.5mA – or thereabouts but it was a good starting point to calculate the LEDs’ resistance.
RLED = V / I = 8 / 0.0035 = 2285.7Ω
RLEDTOTAL = 2285.7 x 3 = 6857.1 = approx. 6857Ω
We want 8V across each LED which totals 24V across all three. That leaves 25.49V (49.49 – 24) across RL. So RL has to drop 25.5V. We can now use Ohms Law to work out the resistance of RL:
RL = V / I = 25.5 / 0.0035 = 7282.85Ω which is approx. 7K3Ω.
But as already stated, the resistance of semiconductors changes depending on the voltage applied and it changes in a non-linear fashion and there is also the tolerance of the resistor to take into account so . . . having constructed the circuit and turning on the amplifier, the LEDs looked a little dull. When I measured the voltage across the LEDs they were 7.82V, 7.8V and 7.77V – all too far away from 8V so I wanted to get the voltage up closer to my ideal of 8V. This meant I needed to reduce the resistance of RL so it has less voltage dropped across it, leaving more to drop across the LEDs.
What if I used the next common value down at 6K8Ω? Using the formula to work out the potential divider network:
(RLEDTOTAL / RLEDTOTAL + RL ) x V = Voltage across RLEDTOTAL
( 6857 / 6857 + 6800 ) x 49.49 = 24.848V across all three LEDs.
24.848 / 3 = 8.283V across each LED (on average) which is perfect. They look great, just right to me (brightness-wise).
Close up of power conversion circuit.
Distance view of the lamp PCB.
Close up of power conversion circuit.
The LEDs stood off from the substrate by about 5mm.
I soldered wire to the LED’s two connectors.
Bird’s eye view of power converter and LED.
CV 1460 amplifier using LEDs, in situ.