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Marantz Led light replacement project.

HomerJSimpson

HomerJSimpson

Addicted Member
I have several Marantz pieces, and I want to attempt to do the LED conversion. But instead of buying the drop in fuse style leds off Epay I'd like to try to do it my self.

My question is this: Which LED's to get. There are so many. I know that they need to 8v and 6.3v (pin) But there are so many choices. Can some offer some insight? Thanx
 
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LED's aren't like incandescent bulbs...they are semiconductors. LED's have a forward voltage rating (Vf), which is the voltage required to overcome the depletion layer junction in the diode (high brightness LED's are usually around 3V or so). Once that voltage is reached, the LED will begin to conduct (and glow) and will act like a short - so it will need a resistor in line to limit the current - otherwise you'll have a mess of real bright pretty LED's for about 3 seconds, and then a whole bunch of dead LED's.

LED's also cannot take any real reverse voltage (Vrrm), so if you have an AC voltage supply you'll have to block reverse voltage with a diode, and if you don't want the LED's flickering at 60Hz, you'll have to filter that supply.

Them's the basics.
 
OK.

Let's look at replacing one lamp with an LED to get your feet wet. If we have an 8VRMS AC supply (stock setup as in 'A' in the attached pic), and we have a high-brightness LED with a Vf of 3.2V, and we want to run 15mA through the LED, here's how it's done.

We know that we want to block reverse voltage, so we're going to have a diode in the circuit to do that. When forward conducting, the diode is going to have its own Vf to take into account, and will be about 0.7V. So the Vf of the diode plus the Vf of the LED equals about 3.9V. This is lost voltage, part of the price of dealing with a semiconductor, and for calculations to determine the value of the current limiting resistor it must be taken into account. So, the voltage available to light the LED will be the source minus the Vf of the LED and the blocking diode.

8V - 3.9V = 4.1V

We want about 15mA of current, so:

R = 4.1V/0.015A
= 273.33 ohms

270 ohms will do fine. Now calculate the power into the resistor (I²*R).

0.015A²*270 ohms = 0.0607W or about 61mW, so in this example a standard 1/4W resistor would do fine. We end up with the circuit as in example 'B' in the pic.

In reality, the calculations are skewed because once we're blocking the reverse voltage with a diode we no longer have 8VAC RMS from the transformer, but it's close enough to get the job done.

In the example above, the LED will be turning on and off 60x per second. Some people find this annoying (me included). Gives me a headache.

To eliminate this, we can filter the supply by adding a capacitor after the diode as in examples 'C' and 'D'. If the AC winding is floating, you'll need another diode to protect the polarized electrolytic capacitor from reverse voltage, as shown in example 'C'. If one side of the AC winding is grounded (commonly done), you can skip the second diode as in example 'D'.

The voltage on the cap will attempt to charge to the peak voltage, which will be the two times the square root of 2, times the RMS value, or 1.414 * 8V = 11.3V. The is the new voltage value to use in your calculations for the current limiting resistor (and of course in example 'C', you have another diode to figure into your calculations of the resistor).

The size of the capacitor must be chosen so that it can handle the ripple current of the circuit. In these instances, if we calculate for 15mA, we need a capacitor that can deal with this current, and can handle the voltage. A 16V cap will deal with the voltage, and a look at the Panasonic data sheets shows in the FC caps, even the smallest 39µf 16V cap can handle 120mA of ripple. But, I'd fit the biggest one in there that I could, and I'd go with at least a 100µf 16V cap in this example.

Clear as mud?
 

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I didn't want to go the microscope and tweezers route. Sure, it'd be nice if Don had a group purchase or sale price every now and then, but it was worth it to me to be able to just drop 'em in and go. Plus, he does very nice work. And the lifetime warranty (just how old are you, Don?) made pulling the trigger a little bit easier...
 
I'm 59! It never occured to me that my lamps will outlive me - but that is sure a possibility.

Group purchase? That never occured to me either.

I'd be willing to discuss that.

DH

BossRadio4VR said:
I didn't want to go the microscope and tweezers route. Sure, it'd be nice if Don had a group purchase or sale price every now and then, but it was worth it to me to be able to just drop 'em in and go. Plus, he does very nice work. And the lifetime warranty (just how old are you, Don?) made pulling the trigger a little bit easier...
Click to expand...
 
:thmbsp: I always try to keep in mind that "lifetime" warranties are only as good as the lifetime of the businesses or individuals offering them. But they're sure a compelling marketing hook that inspires great confidence in the product, aren't they?! :D
 
Based on experience with the LED conversion, it yields a cold, intense blue that I find emotionally harsh. In addition, the LED's produce more of a point source luminance than the original fuse-type lamps. This yields a picket-fence effect, where zones of brightness alternate with zones that are slightly dimmer.

Fred Longworth
 
That may change in a few years...several manufacturers have patented processes for LED's with a tiny array of LED clusters which are nearly omnidirectional. I do not know if any are being marketed as of yet.

As far as the color, I suspect that LED's are always going to have a bit of a cold, lifeless glow to them. We'll see.
 
Fred, you may not like it and that's your prerogative, but you have to stop spreading falsehoods about this picket fencing you talk about. It's simply not true. Frankly, it's a ridiculous assertion.

You're entitled to despise the LED display on the left and embrace the stock OEM lamps on the right.

DSCF6580.jpg


But I disagree. Wholeheartedly. :beerchug: After comparing the two, I couldn't stand looking at the stock display, with it's meters not matching the dial. And the dial being turqoise. So I did something about it.

DSCF6633Medium.jpg
 
EW - That is a great walkthrough on how to do the conversion. I will give it a try and post pic's as I do it! Thank you very much.

From the pic's posted I lke the look of the LED's better. But well see after I get one complete.
 
That greenish tone comes from aging/yellowed vellum --the whitish paper-like stuff in front of the bulbs-- adding a yellowish tone (remember blue + yellow = green). Replace it with NEW white vellum (oven baking paper makes a great substitute/replacement) and the color will be blue again, but a softer, original blue, not the slightly harsher LED blue. THEN you can make a fair comparison of which you prefer.
 
...regarding the ongoing "controversy". Adding LED's is no different than say, a classic 1955 Chevrolet with a flame paint job vs. a classic 1955 Chevrolet with the original paint. Either way, it's a classic!!!! In our case, we can all relax and have a martini..or two and crank it!!! It's a Marantz Baby, Yeahah!!!

:tresbon::music::banana::smoke::beatnik::guitar:
 
I agree.

Oh, by the way, there is no vellum paper behind the meters -- at least none that is easily accessible or worth tearing the meter apart to get to. the LED mod doesn't require you to remove the original vellum over the dial scale, either. At least, I didn't on either of the receivers shown above. For clarity, it should be noted that the 2325 (left) is all LED (dial, meter and function lamps) and the 2245 (right) is only LED on the meter and dial lamps, for now. Both receivers still have their original incandescent needle bulbs intact.

:beerchug:
 
EchoWars said:
OK.

Let's look at replacing one lamp with an LED to get your feet wet. If we have an 8VRMS AC supply (stock setup as in 'A' in the attached pic), and we have a high-brightness LED with a Vf of 3.2V, and we want to run 15mA through the LED, here's how it's done.

We know that we want to block reverse voltage, so we're going to have a diode in the circuit to do that. When forward conducting, the diode is going to have its own Vf to take into account, and will be about 0.7V. So the Vf of the diode plus the Vf of the LED equals about 3.9V. This is lost voltage, part of the price of dealing with a semiconductor, and for calculations to determine the value of the current limiting resistor it must be taken into account. So, the voltage available to light the LED will be the source minus the Vf of the LED and the blocking diode.

8V - 3.9V = 4.1V

We want about 15mA of current, so:

R = 4.1V/0.015A
= 273.33 ohms

270 ohms will do fine. Now calculate the power into the resistor (I²*R).

0.015A²*270 ohms = 0.0607W or about 61mW, so in this example a standard 1/4W resistor would do fine. We end up with the circuit as in example 'B' in the pic.

In reality, the calculations are skewed because once we're blocking the reverse voltage with a diode we no longer have 8VAC RMS from the transformer, but it's close enough to get the job done.

In the example above, the LED will be turning on and off 60x per second. Some people find this annoying (me included). Gives me a headache.

To eliminate this, we can filter the supply by adding a capacitor after the diode as in examples 'C' and 'D'. If the AC winding is floating, you'll need another diode to protect the polarized electrolytic capacitor from reverse voltage, as shown in example 'C'. If one side of the AC winding is grounded (commonly done), you can skip the second diode as in example 'D'.

The voltage on the cap will attempt to charge to the peak voltage, which will be the two times the square root of 2, times the RMS value, or 1.414 * 8V = 11.3V. The is the new voltage value to use in your calculations for the current limiting resistor (and of course in example 'C', you have another diode to figure into your calculations of the resistor).

The size of the capacitor must be chosen so that it can handle the ripple current of the circuit. In these instances, if we calculate for 15mA, we need a capacitor that can deal with this current, and can handle the voltage. A 16V cap will deal with the voltage, and a look at the Panasonic data sheets shows in the FC caps, even the smallest 39µf 16V cap can handle 120mA of ripple. But, I'd fit the biggest one in there that I could, and I'd go with at least a 100µf 16V cap in this example.

Clear as mud?
Click to expand...



OK, I did a quick test and got it to work! Now to make things complicated! I just did one LED, but I I added another, the amount of light dimmed to about half. I am thinking it would be easier if I were able to put the diode, cap and resistor between the power supply and the PCB, then solder the led's to the old fuse lamp mounts. There is 6 fuse lamps on this circuit. Can I do it this way, and only have to use one cap, diode and resistor, instead of six?
 
Oops, I may have answered my own question. If I rund them in series, I get the same amount of light, so I could use only one of each componet!


**EDIT** Well it will work, but may need to lower the resistance to get the right brighteness with 6 LEDs
 
6 LED's will require a voltage more than 6 * Vf. If your Vf is 3V, then you'd need more than 18V to forward bias the diodes.

If you run them in parallel, you need a separate current limiting resistor for each LED, since if you used only one resistor, if an LED quits then the current goes up for all the rest. Also because the LED's will not share current equally (nature of the beast). One will have a lower Vf than the rest, and try to steal all the current (and probably burn up as a result).

You could run pairs of LED's in series, and then put each set of LED pairs in parallel.
 
Bossradio4vr
Digging the post back up to ask a question. How are the LEDs changing the color of the displays? Are you using Blue LEDs or white LEDs?

Just curious.

Steve
 
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