Pages: [1]   Go Down

Author Topic: AL811 Operate LED repair  (Read 346 times)

AD7C

  • Member
  • Posts: 85
AL811 Operate LED repair
« on: March 25, 2023, 05:55:18 PM »

I have an AL811 (not H) that does not have a working "Operate" LED.  The amp does work great, keys, produces full power, switches function. Just the LED doesn't work. So this weekend I decided to repair it.

Problem 1: I found the schematic for the AL811 and identified the LED in the schematic. However, the limiting resistor before the LED is R15. R15 is not listed in the parts. Great.

Problem 2: When I pulled the cover and looked at the current LED, wiring, and resistor what is there is not what is in the schematic. In the schematic there is a series resistor. What I found is a parallel resistor of 30 ohms (orange/orange/black) across the Anode/cathode. Huh? That seems to be the way the 811H schematic does it. But I have an 811 not 811h.

FYI: I de-soldered the LED and it tested good.

So what's going on here? I believe the operate line is as 12V so a 20mA led should have a 500 ohm or so series resistor to feed it correctly.  But I have a parallel 30 ohm resistor.

73
Rich
AD7C
Logged

SWMAN

  • Posts: 2116
    • HomeURL
Re: AL811 Operate LED repair
« Reply #1 on: March 25, 2023, 07:26:40 PM »

 Orange, orange, black is 33 ohms.
Logged

W1QJ

  • Member
  • Posts: 3344
Re: AL811 Operate LED repair
« Reply #2 on: March 25, 2023, 08:35:23 PM »

Well thanks for finding yet another mistake in the schematic!!  You found a resistor with an LED across it because I do believe that is the way it is!!  The idea of making the LED work is to produce about a 1.5v voltage drop across the resistor when the relay is keyed up.  The resistor value is based on the nominal keying voltage and the current drawn by the relay(s).  As I recall the nominal relay voltage is about 15v so let’s say when keyed the relays draw 100ma.  So you want a resistor that drops 1.5v  at 100ma.  The value based on these numbers would be 15 ohms.  If the relay draws 50ma the resistor would be 30 ohms, so 33 ohms would make perfect sense.  The key is that the LED must be put across the resistor in proper polarity.  It the polarity is backwards it won’t light up.  So maybe the LED is on backwards.
Logged

AD7C

  • Member
  • Posts: 85
Re: AL811 Operate LED repair
« Reply #3 on: March 26, 2023, 11:44:56 AM »

Fixed. LED was backwards.

I was unaware you could drive an LED with a parallel resistor. I always thought an LED had to have the resistor in series.

Thanks
-Rich
Logged

K6JH

  • Member
  • Posts: 701
Re: AL811 Operate LED repair
« Reply #4 on: March 26, 2023, 02:09:07 PM »

Fixed. LED was backwards.

I was unaware you could drive an LED with a parallel resistor. I always thought an LED had to have the resistor in series.

Thanks
-Rich


Still should be some limiting element upstream or all a parallel resistor will do is add to the current.
Logged
73
Jim K6JH

W1QJ

  • Member
  • Posts: 3344
Re: AL811 Operate LED repair
« Reply #5 on: March 26, 2023, 04:07:15 PM »

Fixed. LED was backwards.

I was unaware you could drive an LED with a parallel resistor. I always thought an LED had to have the resistor in series.

Thanks
-Rich

No, not really,the normal voltage for an LED is in the neighborhood of 1.2v to1.5v if you have a 1.5v battery you can hook an LED across it with nothing else.  If the LED can only see a source voltage that it wants and that’s all that is available putting a resistor in will only create a voltage drop that you technically wouldn’t want.  An LED draws about 18ma.
Logged

W9IQ

  • Member
  • Posts: 8862
Re: AL811 Operate LED repair
« Reply #6 on: March 27, 2023, 06:00:33 AM »

An LED has an exponential current to voltage curve. A small change in voltage across it can result in a large change in current through the LED. Here is a typical red LED I-V curve:



You can see that the upper voltage drop across this LED will be about 2 volts. Contrast this to a silicon rectifier diode with an upper voltage drop of about 0.6 volts. But the LED cannot handle nearly the current of a rectifier diode. Its total power dissipation is often limited to 40 milliwatts or so for a small, panel type device. LED specifications vary widely based on number of dies, die size, LED color, bonding frame, bonding wires, encapsulation and mounting technique so check the datasheet for the particular LED in question.

The amount of light emitted from a particular LED is proportional to the current through the LED. For this reason, designers are generally more interested in controlling LED current. The voltage drop across the LED is then essentially a byproduct from this current.

When the LED is to be supplied by a voltage source with a known maximum voltage, it is common to use a series resistor to limit the LED power dissipation and to control the brightness. This makes the circuit operate in a more nearly linear fashion since the series resistor is limiting the current through the LED if the voltage source is relatively stable.

When the peak current through a series circuit is known, it is possible to properly bias the LED such that it is operating within nominal specifications using a paralleled resistor. The parallel resistor value must ensure that the voltage across the LED does not allow too high of current to flow through the LED. Because the resulting current through the LED can change rapidly with a slight change in voltage across the parallel resistor, the designer must be certain to limit the voltage to well less that the 2 volt forward drop limit and well below the maximum power dissipation limit. A maximum voltage of 1.8 volts or so would be a reasonable bias point based on the above curve.

I am still shaking my head at the thought that MFJ can ship an amplifier with the operate LED wired in backwards. How pathetic...

- Glenn W9IQ
« Last Edit: March 27, 2023, 06:07:15 AM by W9IQ »
Logged
- Glenn W9IQ

God runs electromagnetics on Monday, Wednesday and Friday by the wave theory and the devil runs it on Tuesday, Thursday and Saturday by the Quantum theory.

VE7RF

  • Member
  • Posts: 1608
Re: AL811 Operate LED repair
« Reply #7 on: March 27, 2023, 07:59:05 PM »

Ameritron obviously does not test anything that leaves their factory.   That's why they have a massive  amount of returns. I can't see how they can make any money doing business that way.

A few years ago, one fellow was hired by his local HRO store....and his exclusive job was to unbox, and test each Ameritron amp sold...before it left the store. The problem was, he had too many of them that blew up...and then required extensive, various repairs.  Plan B, unbox each amp, take it all apart 1st, check and find all the miswired components, components wired in backwards, missing components,  poor soldered joints, stuff not even soldered and everything else you can think of, including band switches wired wrong, and tank coils leaning over to one side.  That alone reduced the blow ups to almost zero.

Next issue is alignment on a lot of their gear.  Pots are either set mid range, or fully CW...or fully CCW. 
Their entire business model is flawed.   IF it's designed correctly, built correctly, aligned correctly, and tested correctly, packaged and boxed up correctly, it should leave the factory to a new owner...and the item sold never comes back.
Logged

K8AXW

  • Posts: 7391
    • HomeURL
Re: AL811 Operate LED repair
« Reply #8 on: March 29, 2023, 09:01:56 AM »

If the owner of MFJ allows gear to leave his factory(s) FUBAR, one an only imagine what kind of working conditions his workers must tolerate!
Logged
A Pessimist is Never Disappointed!
Pages: [1]   Go Up