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Zener diodes doubt
Zener diodes doubt
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Old 3rd January 2018, 12:21 PM   #1
ChurritoTierno is offline ChurritoTierno  Paraguay
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Default Zener diodes doubt

Hi!

I was repairing - modifing a guitar amp (Fender Roc-Pro 1000). It has a 12AX7 on the drive channel in a diode fashion to clip the signal.

So, the filament of the 12AX7 is supplied by a DC power supply. It consist of three 470 Ohm / 5W resistor paralleled followed by two 1N5341 zener diodes (see atachment file)

The problem with this design is it run too hot. The PCB under this three resistors are almost burned and the traces loosed hes conductivity. The same for the two diodes.

So, my plan was to change those three resistors by six (5W each one): four 1K and two 820 Ohms.

The two 1N5341 also I changed by four 1N5333 zener diodes in series (3,3V each one).

I did the modification and it runs considerably less hotty.

I wanted to check the filament voltage. Surprisingly it reads 9,8 V. (I was expecting at least 13V) Something's wrong I thougt..

So, I checked voltage across the diodes: only 2,45 V across each one.

Can someone explain me why happen this?
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Old 3rd January 2018, 01:27 PM   #2
Mooly is offline Mooly  United Kingdom
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Zener diodes doubt
Your resistor network is correct at approx 156 ohms (but check and confirm by measurement that you haven't fitted an incorrect value somewhere) and so that leaves just the zeners.

I suspect that if the zeners are OK then what you are seeing is just the result of passing a relatively low current through a high wattage part. In other words, the zener needs a lot more current to level out at 3.3 volts. It is conducting a little and pulling the voltage down.

I don't know what the heater current of that valve is but knowing that would allow you to calculate the the approx. current in the zeners and so pick more suitable devices.
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Old 3rd January 2018, 01:39 PM   #3
AndrewT is offline AndrewT  Scotland
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If you have only 2.45V across each 3.3V Zener, then they are probably not passing any significant current (<1mA?)
That means the heaters are drawing all the current that the power resistors can pass.

You have 40V-9.8V (=30.2V) across each power resistor.
each 1k passes 30.2 mA and each 820r passes 36.8mA
Total current to the Zeners+heaters = 60.5mA
this gives the heaters a hot resistance of ~162r
That might indicate that the cold resistance of the heaters is somewhere around 40r to 80r.
You can check this cold.
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Old 3rd January 2018, 09:38 PM   #4
PRR is offline PRR  United States
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12AX7 at 12V connection should be 0.150A.

40V to 12.6V is 27.4V to lose.

40V to 12.6V or 27.4V at 0.150A is 4.11 Watts to lose. Plus what you allocate to the Zeners.

Quote:
Originally Posted by ChurritoTierno View Post
...four 1K and two 820 Ohms....
That is 155.3 Ohms. 27.4V across 155.3r is 0.176 Amps. So you have more than the tube needs. Proper Zenering should put 0.176A-0.150A= 0.026,4A in the Zeners. Very sub-Watt, especially if four are used. However if the tube is pulled, the Zener has to eat the whole 0.176A or 2.3 Watts. Even so you do not need four 5W Zeners.

The heat is in the resistors, little in the Zeners.

As Mooly and Andrew say, the big Zener with low current will show low Voltage.

If 5W Zeners are available, I would use one 1N5350. 13V at 100mA, perhaps 12.75V at 10mA.

Or heck. Put 1N400x diodes (~0.6V drop) in series with the Zeners until it comes around 12V.

OTOH, the 12AX7 may work quite fine at 10V heat. Especially if it is used only as a clip diode?
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Old 3rd January 2018, 10:09 PM   #5
PRR is offline PRR  United States
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Quote:
Originally Posted by PRR View Post
............it is used only as a clip diode?
Indeed it is. Here is a clip from the plan.

This is basically a big solid-state amp. It has big +/-40V for the power stage, no other supplies except the little +/-16V supplies for opampery. So Fender dropped significant 12V power from the 40V rail.

There "is" a 12AX7-like tube with 19(?)V 0.1A heater. May be hard to get production lots, or to buy on a rainy night in a strange town before a gig. 12AL5 takes the same power as 12AX7. Today we can get a 40V-12V step-down switcher, much less heat, $9 on eBay free shipping; not 25 years ago.
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Old 3rd January 2018, 11:44 PM   #6
Enzo is offline Enzo  United States
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How about this analysis, the tube is not doing much, the resistors drop the 40v enough to heat the tube, the zeners are there to make sure it doesn;t go over the 12v the two zeners add up to. 9v is enough for the tube to conduct. it isn't amplifying, it is just acting as a pair of diodes.
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Old 4th January 2018, 09:40 AM   #7
ChurritoTierno is offline ChurritoTierno  Paraguay
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Thanks to iluminated me!.

What I conclude is that zeners really doesn't behave like "zener" (in this case).. Maybe if I change the tube with another one with a more filament "resistance" the voltage across the zeners diodes will raise to 3.3 x 4 = 13.2 V..

Am I right??

Last edited by ChurritoTierno; 4th January 2018 at 09:44 AM. Reason: explanation more ditailed
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Old 4th January 2018, 09:44 AM   #8
ChurritoTierno is offline ChurritoTierno  Paraguay
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Edit:

What I mean,... the resistors plus the filament form a voltage divider... and the output voltage of this divider doesn't reach up to or more than 3.3x4=13.2V.. beacouse of that, the zeners arrengement doesn't really perform like zeners.
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Old 4th January 2018, 10:12 AM   #9
AndrewT is offline AndrewT  Scotland
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the Zeners are probably being starved of current.
This is because the dropping resistors cannot pass sufficient current.
I showed how to determine the current that the four resistors can pass for the voltage conditions you measured.
Now go and add on another 820r resistor.
You will find a new set of voltages. Use them to calculate the total current passing through the five resistors.
See if that gets you closer.
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