Stereo Light on a Sherwood S-7200

[redpackman]

Just picked up a nice one-owner Sherwood S-7200. Cleaned out the scratchy controls. Put in a new light or two. It operates beautifully and sounds great. Everything works on it, except for two things, of course, the power switch is bad, so there's a click-wheel switch in the power cord. Works great, and, of course, I can always turn it on and off with a power bar.

The only other malfunction on it is the stereo light. The stereo works great in the FM tuner. Very clear separation. Nice tuner. But while the FM function light works nicely the stereo indicator light itself was burned out.

Here's what puzzles me: there's a constant +35v DC current to one of the leads of that light. When in FM and a stereo signal is received, the other lead is grounded, thus there's a circuit, and presumably an intact bulb would glow. When there's no stereo or we're in another mode (AM, Aux etc) that second lead that was grounded, is open, thus no circuit and thus no stereo light. So that function is working well, opening and closing a circuit.

What I don't get is the 35v reading. 35 Volts DC for a stereo indicator light??? I've been told from a good source that the bulb that's supposed to go in there is rated at 8v .05A. On 35V? What am I missing?

A schematic for the S-7200 can be found at: https://www.hifiengine.com/manual_library/sherwood/s-7200.shtml

 

[chazix]

What I don't get is the 35v reading. 35 Volts DC for a stereo indicator light??? I've been told from a good source that the bulb that's supposed to go in there is rated at 8v .05A. On 35V? What am I missing?

The power source for the stereo indicator is not a "stiff" power supply - under load (that is, when the lamp is lit), the voltage will drop substantially.

That being said, there does still seem to be a discrepancy. The schematic shows a 270 ohm series resistor in the power source for the indicator. If the lamp used as the indicator ran 0.05A of current, then the voltage drop would be 13.5V, and that would leave over 20V to appear across the lamp. So, I think the info you got about either the voltage or the current rating for the lamp must be incorrect. The math would work out pretty nicely for an 8V 0.1A lamp. (But that seems a bit hefty for a stereo indicator. Is the lamp installed in a metal socket?)

(If it was me, I would probably use an LED to replace the stereo indicator, if the cosmetics were satisfactory. In that case, the 270 ohm resistor would also be replaced with one that would limit the current to the right neighborhood for an LED. A 2700 ohm 1 watt resistor would be about right, but it depends on the LED efficiency.)

Cheers,

chazix

 

[redpackman]

The power source for the stereo indicator is not a "stiff" power supply - under load (that is, when the lamp is lit), the voltage will drop substantially.

That being said, there does still seem to be a discrepancy. The schematic shows a 270 ohm series resistor in the power source for the indicator. If the lamp used as the indicator ran 0.05A of current, then the voltage drop would be 13.5V, and that would leave over 20V to appear across the lamp. So, I think the info you got about either the voltage or the current rating for the lamp must be incorrect. The math would work out pretty nicely for an 8V 0.1A lamp. (But that seems a bit hefty for a stereo indicator. Is the lamp installed in a metal socket?)

(If it was me, I would probably use an LED to replace the stereo indicator, if the cosmetics were satisfactory. In that case, the 270 ohm resistor would also be replaced with one that would limit the current to the right neighborhood for an LED. A 2700 ohm 1 watt resistor would be about right, but it depends on the LED efficiency.)

Cheers,

chazix

The lamp is an axial lamp with leads attaching to the circuit board. I'm going to check the 270 ohm resistor. Perhaps it has shifted over the years.

Update: I checked the 270 ohm resistor. It is a 5watt "sand" resistor....fairly large rectangular bloc. It tests within tolerances. I have now wired in two diodes, that do work at the appropriate time. I also put in a resistor in series before them. So here's how the measurements go: the 270 ohm 5 watt resistor has around 35v DC going in and around 24v going out. Then I put in a 480 ohm 1/4 watt resistor in the line leading to the LED's. It, of course has the 24 v. going in and around 7v going out. That voltage lights the LED's nicely. I should be satisfied, BUT the 480 ohm resistor, after the "stereo" light has been glowing for awhile, gets pretty warm. I can touch it, but I couldn't hold it between my fingers for very long. It's not enclosed but open to the air so it can radiate the heat energy OK, but is that acceptable? I've noticed too that the 270 ohm resistor and the 330 (5 watt "sand" - same size and shape as the 270 ohm resistor ) resistor feeding it get pretty warm. I can touch them, but they're more than warm. I couldn't hold my finger on them. The 330 ohm resistor was checked and was around 340 ohms.

Do I need to get a higher wattage 480 ohm resistor to replace the 1/4w one I have in line now?

 

[Spe@kerBox]

I have a 7200 on my bench now, and the DC voltage on the stereo lamp is 7.5V with respect to chassis ground.

 

[chazix]

Then I put in a 480 ohm 1/4 watt resistor in the line leading to the LED's. It, of course has the 24 v. going in and around 7v going out.

That means the resistor is developing about 0.6W. It should be at least a 1W part.

It also means the LEDs are passing over 35mA, which is eye-catching in at least one way. Might want to check the specs for the LEDs and see if they're supposed to be able to run that much current.

Maybe it would be helpful to state the formulas involved:

Wattage (W) developed by a resistance (R) that has a certain voltage (E) across it: W = (E squared) / R. This is a form of what's called Watt's Law.

Current (I) through a resistance (R) that has a certain voltage (E) across it: I = E / R. This is a form of what's called Ohm's Law.

Cheers,

chazix

 

[redpackman]

S

That means the resistor is developing about 0.6W. It should be at least a 1W part.

It also means the LEDs are passing over 35mA, which is eye-catching in at least one way. Might want to check the specs for the LEDs and see if they're supposed to be able to run that much current.

Maybe it would be helpful to state the formulas involved:

Wattage (W) developed by a resistance (R) that has a certain voltage (E) across it: W = (E squared) / R. This is a form of what's called Watt's Law.

Current (I) through a resistance (R) that has a certain voltage (E) across it: I = E / R. This is a form of what's called Ohm's Law.

Cheers,

chazix

So you would recommend a 480 ohm 1 watt resistor?

 

[chazix]

So you would recommend a 480 ohm 1 watt resistor?

Well, not really, I think 480 ohms is too small a resistance, leading to too high a current through the LEDs. But the worst that will result from that is that the LED longevity might be less than LEDs are known for. And your eye is king here - if the LED brightness looks perfect to you with the current setup, then by all means stick with 480 ohms, but do change it to a 1 watt part.

 

[redpackman]

I have a 7200 on my bench now, and the DC voltage on the stereo lamp is 7.5V with respect to chassis ground.

Any ideas as to why my S-7200 has 24v instead of 7.5v?

 

[Spe@kerBox]

I measured from the live pin (to chassis ground) when the light was on. So it was under load. The other pin had a few tenths of a volt on it so current could flow. If your bulb is out you are not measuring under load and the voltages may be a bit different but would not account for 24VDC. When I get a chance I will measure the unloaded condition and report back. The service manual may also help you. The only other thing I can think of is a bad power supply or failed component somewhere.

 

[chazix]

Meaning no disrespect, but there might be some over-thinking going on here. I don't believe that there is any observation in the thread that could not have been fully explained by Georg Simon Ohm in the 19th century. (Well, OK, you might have had to help him out a little with 20th century schematic symbols and "What on earth is a DMM?"...)

To be a little less tongue-in-cheek: Keep in mind that there is a large (relatively) resistance series-connected in the supply line for the stereo lamp. (Also, the OP has added even more series resistance, in conjunction with replacing the incandescent lamp with an LED.) Because of the series resistance, the voltage that one would measure at the supply terminal doesn't change a little depending on whether the lamp is on or off, it changes a lot (relatively).

Cheers,

chazix

 

[redpackman]

Thank you for the help.

Where does it say "DMM?" If I wrote it, and I can imagine just such a thing, it's a typo.

So, are you saying, if I took out the diodes mentioned above and put in an 8v .05A lamp it would work fine, in spite of the 35v being fed into it and that the power administered to it would drop to a level manageable by the 8v lamp?

I know you guys who are experienced with this are rubbing your head at this stupid question and asking, "Where do these guys come from?" but you're talking to a definite newbie. Thanks.

 

[chazix]

So, are you saying, if I took out the diodes mentioned above and put in an 8v .05A lamp it would work fine

That's the general drift. Getting down to details, though, it needs to be an 8v 0.1A lamp. That's assuming that you retain the original 270 ohm sand resistor (and remove the added 480 ohm resistor).

The reasoning is like so: When no current is flowing (lamp off or absent), the maximum available voltage is 35v. To use an 8v lamp, there needs to be a voltage drop of 27v across the 270 ohm resistor. That will work out just right if the current flowing when the lamp is on is 0.1a (270 ohm times 0.1a equals 27v). So, the 8v lamp's current rating needs to be 0.1a.

Again, no disrespect was meant, and I hope my attempt at levity didn't rankle. (The "DMM" business was supposed to paint a funny picture of a 19th century dude getting a look at a Digital MultiMeter. I'm better at Ohms Law than I am at humor...)

Oh - and if you want to use something different from an 8v 0.1a incandescent lamp, it's possible to do so, by also changing the resistance in the circuit to something different from 270 ohms. I can help figure out what's needed, given the voltage and current ratings of the lamp or LED that you want to use.

Cheers,

chazix

 

[redpackman]

Thank, you, thank you for the clear and detailed response. That's precisely what I needed. I'm sure there are some (most) guys on this board who find such instruction as obvious as "Now when you drain the oil from the auto's engine, you must then put back a proper amount of oil before you start the engine up...and don't forget to put in the drain plug before you put the new oil in."

But I'm not one of them. Thanks, I'll look for a properly rated lamp and have at it.

 

[redpackman]

OK, final report: As you guys said, the circuit supplying the light may have 35 volts in it, but once the circuit senses an FM Stereo signal it closes the ground circuit which until that point has remained open. At least that's the way I understand it. The light goes on because now it's finally grounded, and when I took a reading of the actual voltage going to the light, it was 8 volts DC. I got an 8 v. 100mA bulb from Dwojo and it works perfectly. This electrical operation is above my head, but the light works and the voltage, indeed seems right. When I saw 35 volts supposedly going into an 8 volt light, I thought it would go up like a fuse, but it doesn't.

Thanks to all of you who helped and were patient with me.