Test Tube in Isolation?

[Wil_Bloodworth]

All,

I'm trying to understand how the 35Z5-GT rectifier tube works... specifically how it supplies the lamp with voltage.

If I had to ask one question, it would be:  Is there a way to take the 35Z5-GT and supply it with, let's say, 35 volts A/C instead of the intended 117 V A/C and then be able to measure the voltages on the other pins (and learn something meaningful from doing so)?

I have the specifications for the 35Z5-GT from two different sources as well as a radio schematic that uses this tube.  I'm just confused by the wiring even though I know there is something going on with mixing AC and DC somehow.

The positive AC of 117V is clearly starting at pin #2 (heater) as well as one leg of the lamp, and is directly tied to pin #7 (other heater).  The design is that this continues through all the other heaters with a combined voltage usage of 118V. That is fine and I understand that.

I'm confused by pin #3 which they are calling the "heater tap".  This pin is tied directly to the other leg of the lamp AND is also tied to the incoming positive AC voltage via a 140 Ohm shunting resistor AND is then tied to pin #5 (plate).

Can someone help me understand what is going on with the lamp and how all of this works?  The tube design also lists an option without the shunting resistor which I can more easily rap my head around but the lamp placement in general makes zero sense to me.



Thank you,

- Wil

[retiredfeline]

Part of the difficulty is the way the circuit has been drawn. If you redraw the circuit, maintaining the topology, you will see that the lamp, resistor, and the part of the filament between 2 and 3 all in parallel are in series with the mains feed to the anode. The idea is to provide a convenient low voltage source for the lamp.

[Wil_Bloodworth]

The idea is to provide a convenient low voltage source for the lamp.

That's just it... I don't understand ANY of the voltage flow; regardless of how it's drawn.  This is why I want to hook up some voltage to the tube so see what's happening... in isolation.

And what is the anode... is that the plate?

- Wil

[retiredfeline]

I suggest you ignore the word "tap" and consider the two parts of the filament as separate. (Physically it may be a tap in a heater wire but for the purpose of circuit analysis you can ignore that manufacturing detail.) The main filament is heated by the mains current through the heater chain. The secondary filament ie heated by part of the current going to the anode (plate) so it has a rather odd waveform. The tube would probably work just as well with the omission of the lamp, resistor and use of this secondary filament which is why you discovered two configurations for this tube.

I prefer anode to plate. For one thing the terms anode and cathode carry over to semiconductor rectifiers.

Chances are if you can get the main filament to heat up and glow, the tube will work.

[Wil_Bloodworth]

Chances are if you can get the main filament to heat up and glow, the tube will work.

Thank you but I'm not trying to get the tube to work, I'm trying to understand how it is supposed to work.  How exactly are 5 volts coming from inside the tube to pin #3?  Is there some kind of transformer-like behavior inside the tube that is taking part of the 117 volts coming directly to pin #2 and then somehow giving ~5 volts to pin #3?

Is there no way to hook this tube up to a variable power supply to see what is going on?  Tube testers have to be doing something exactly like this or there would be no way to test an individual tube.

- Wil

[retiredfeline]

No "transformer-like" behaviour. As I said the resistor, lamp and secondary filament have the anode current pass through it. So it doesn't pass any current until the tube heats up and starts rectifying. The current is also not DC or pure sine wave AC. It's an odd waveform which consists of 60 Hz peaks when the rectifier conducts and charges the capacitor.

If you're getting hung up over this, you can leave it until you understand simpler concepts, like how a half-wave rectifier works and what the voltage and current waveforms are like in a rectifier and smoothing capacitor circuit.

[Wil_Bloodworth]

If you're getting hung up over this, you can leave it until you understand simpler concepts, like how a half-wave rectifier works and what the voltage and current waveforms are like in a rectifier and smoothing capacitor circuit.

This tube is a half-wave rectifier and that is the entire point of my post... trying to understand how it works.  If you have resources that demonstrate how this half-wave rectifier tube works, I would greatly appreciate those.

- Wil

[retiredfeline]

There are plenty of articles that explain how a diode works as a half wave rectifier. If you mean the details of how a thermionic valve works as a rectifier then refer to that valve book you found.

[TimFox]

Most of the half-wave (single diode) rectifier tubes designed for "AC-DC" transformerless series-heater string radios have a "tap" on the heater for a dial light.
If the light bulb blows, or is removed, the tube will still operate.
If you do apply 32 or 35 V (AC or DC) across the entire heater (pins 2 and 7), you should see a reasonably-low voltage between pins 2 and 3, where pin 2 to 3 to 7 is a voltage divider from that 35 V.
If you use a DC lab supply for that test, it needs to be able to supply 150 mA at 35 V;  series-string heater circuits are normally specified at the design current (0.15 A for this tube).
The reason for this connection is that the time constant for the tube heaters is much longer than that of the small filament in the dial light (nominal rating for #47 is 6.3 V at 0.15 A).
If the bulb were connected directly in series with the heaters, it would be OK after everything is hot and the series-string current has stabilized at 0.15 A, but it would probably blow before that when the series-string current is several times higher than 0.15 A.
The fancy trick with the heater tap is to allow the radio to operate when the small bulb is open-circuit.
According to the GE/Ken-Rad data sheet, with the #47 bulb you should get 32 V total from 2 to 7 at 0.15 A, and 5.5 V across the #47 bulb connected from 2 to 3.
Without the bulb, you should get 35 V at 0.15 A from 2 to 7, and 7.5 V from 2 to 3 (with no bulb connected).
https://frank.pocnet.net/sheets/093/3/35Z5GT.pdf

Note the comment at the top of the datasheet about having the plate-cathode current flow through the tapped section of the heater.
Also, the circuit diagrams at the bottom of page 2:  the resistor in series with the other heaters may not be needed (depending on the tube heater voltages), and the resistor in parallel with the panel light is needed if the rectified DC current is > 60 mA.

[Wil_Bloodworth]

Thank you Tim!  I have a variac so I can test it with A/C up to the 35 V but my DC power supply is limited to 30V so I am not sure I would have enough DC "juice" to test it.  I might try to give that a shot and see how it goes.

- Wil

[Wallace Gasiewicz]

Here is the spec sheet:
https://tube-data.com/sheets/127/3/35Z5GT.pdf
Tim is correct. The tube should work just fine without the light bulb. That feature is just an "extra " Meant for this kind of radio, which uses the whole 117 volts for the heaters of ALL the tubes.   I think the reason for the 35 volt filament voltage is just to make the entire filament series circuit to take the 117 volt house voltage.  This is kind of goofy as related to other tubes with more common 6 or 12 volt filaments. So this is a kind of specialty tube that was designed to take up the 35 volts left over from the other tubes so that the 117 volt would come out OK and no tube would be subject to higher volts than designed for. The light bulb thing with the "TAP"  is another goofy thing meant for this kind of radio. just so there would be an "ON" light.  Sounds like a lot of design effort for this feature but a lot of radios were made this way...

Most heaters are either 6 or 12 volts, the tubes which start with "6" are 6 volt filaments the tubes that start with '"12" are 12 volt filaments. ( Like 12BY7)   More "advanced" radios used 6 and 12 volt heaters with the voltage applied in parallel, not in series.  Sometimes the heater supply would have 6 and 12 volts. Sometimes the heater supply would be just 6 volts. Sometimes the heater supply would be 12 volts and then two 6 volt filaments would be connected in series...if the designer needed a specific 6 volt tube that was not available in 12 volt similar tube.
It should heat up just fine with 30 volts and will not take any excess of current, so operating the filament with 30-35 volts will limit the current. Just like a resistor does. Heaters have different resistances when cold or hot, so it is hard to correlate the cold resistance to hot resistance.But the heater should NOT be entirely open when measured cold.

[retiredfeline]

Actually the tube wasn't inserted to "take up the extra voltage" but rather families of tubes with different heater current ratings were designed for various TV and radio configurations and number of tubes so that the sum would come out to the mains voltage or somewhat under, with a wirewould resistor dissipating the difference. For example the European PCL82 is the same as the ECL82 but rated at 300 mA heater current. I've seen the P series of valves in 220 V TVs with a large number of valves.

My memory popped up the the fact that there was even a valve whose heater worked off 117V. It was the 117N7 which is a combined pentode and diode. Used in cheap record players with piezo pickups which provided enough voltage to drive the single pentode's grid.

[TimFox]

As mentioned above, with parallel connections from a low-voltage winding on a power transformer, 6.3 V heaters were "normal".  In your 6-tube radio, there were 12.6 V @ 0.15 A heaters on the 12SK7s, 12SA7, and 12SQ7 versions of the 6SK7, 6SA7, and 6SQ7 (6.3 V @ 0.3 A).  In a series connection, all heaters operate at the same current.  12.6 V/0.15 A is sufficient heater power for the "signal-level" tubes, but not enough for the higher-power tubes:  rectifier and output tubes, so higher voltage (at the same current) is required.  Typical values are 35 V and 50 V for the higher-power tube heaters.
For 6.3 V parallel heaters, a 6V6 (6.3 V @ 0.45 A) would have enough heater power for the output tube, and a 6X5 (6.3 V @ 0.6 A) would work for the rectifier.
Series-heater-string TV sets often used higher-current series strings, so the manufacturers could have more than two heater voltages available with the same guts, and many "12" tubes had a center tap to allow, for example, 6.3 V @ 0.3 A or 12.6 V @ 0.15 A (e.g., 12AX7).

[retiredfeline]

The key thing to note is that this dial lamp circuit has nothing to do with the rectification function of the valve. The tap and associated components are there to supply suitable power for the dial lamp.

Reverse engineering the technical requirements: Why didn't they use a neon bulb instead of a dial lamp? Well, a neon bulb wouldn't be bright enough to illuminate the dial. And electroluminescent illmination was in the future. Why not put the dial lamp in series with the other filaments? The cold current inrush would blow the bulb. Why not use a dial lamp suitable for 117V? Such a low power lamp would have a very thin filament and be fragile. Step down the voltage with a small transformer? Well the whole point of transformerless is to avoid a transformer. So some bright spark thought up the idea or using the anode current to power the dial lamp, It also means that dial lamp won't reach operating brightness until the valves start operating. Part of the filament and a parallel resistor are there because the radio has to continue to work even if the lamp blows.

Note that if you supply the main filament with your 30 V transformer, you will not get the normal operating voltage across the lamp. That part of the circuit uses the curent passing through the anode too. You need to draw current from the B+ rail to simulate that.

[CaptDon]

The filament (heater) is just a long orange hot piece of wire with a voltage (35 volts in your case) from end to end when passing a certain current. If we measure from one end to any point along that wire we will see a ratiometric voltage along the wire. Tap the wire in the middle and it will appear to be half the voltage applied end to end. That makes a great way to find a reduced voltage to drive a pilot light. Many of the little 9 pin miniature tubes are 12 volt rated with a center tap. You can run the tube on 12 volts or feed power to the center tap and parallel the two ends and run the tube on 6 volts (it will require twice the current of 12 volt operation, hence 12v@.3a=3.6w or 6v@.6a=3.6w see the relationship? In a line operated series filament string when you place a lamp across half of the filament it would tend to reduce the available current/voltage in that half of the filament and so a 'balancing resistor' may be placed across the other half. In any case the available current through the filament must remain high enough to provide the proper filament voltage based on I x R = V. In some series strings a pilot lamp with the same current rating is placed in series with the filament string. When the bulb burns out the filaments see an open circuit and no available power and they won't work. The whole radio is dead due to a burned out pilot bulb. One characteristic of that style wiring is the pilot light gets very bright when the radio is first turned on.

[radiogeek381]

OP: I’m not sure your question has been answered yet.

The explanation of the heater circuit has been laid out pretty clearly.  But I think you are asking about pins 5 and 8.

Let’s assume that the DC output is connected to a 5.6 k ohm resistor.

This tube has an indirectly heated cathode. That means that the major portion of the electron flow is beting emitted by the thing connected to pin 8.

When pin 5 is positive relative to pin 8, current will flow. 

If we use “convential current” things get a little easier to talk about.  So let’s switch to that:

When the 117 volt line at the top is positive with respect to the 117 volt line at the bottom, pin5 will rise to something like 112V.  Eventually, it will rise above whatever voltage is on pin 8, so current will flow from the 117 volt input, through the plate, to the cathode at pin 8 and thence through the output resistor.

Just ballparking and using my limited memory of tube curves.  The reference for this tube suggests a 20 mA plate current will result in a plate-cathode voltage of 5 V ( cognescenti, please correct me on this.). So 20 mA across 5400 ohms gets us somewhere near 112 - 5 V V RMS leaving us with about 105 V rms.  Call it 100.

Of course this is all back of the envelope calculation.  The old timers would have done something similar, then worked it out with nomograms and paper.

But suffice it to say that the diode action is taking place between pins 8 and 5.   The filament circuit is important, and others have discussed that.

hope it helps.  Corrections welcome.



What voltage?  Well

[Wil_Bloodworth]

Thank you radiogeek381!  That helps considerably.

- Wil

[TimFox]

Going back to your original question:
If your 30 V DC supply can handle 150 mA, then you can test the heater of a disconnected 35Z5 by applying 30 V to pins 2 and 7.
The current (after about 20 seconds) should be about 10% less than 150 mA, and the voltage between pins 2 and 3 should be about 10% less than the spec sheet value of 7.5 V (without dial lamp).

[Wil_Bloodworth]

Going back to your original question:
If your 30 V DC supply can handle 150 mA, then you can test the heater of a disconnected 35Z5 by applying 30 V to pins 2 and 7.
The current (after about 20 seconds) should be about 10% less than 150 mA, and the voltage between pins 2 and 3 should be about 10% less than the spec sheet value of 7.5 V (without dial lamp).

Awesome!!  I will try this.  Thank you... and Happy Thanksgiving.

- Wil

[Wil_Bloodworth]

If your 30 V DC supply can handle 150 mA, then you can test the heater of a disconnected 35Z5 by applying 30 V to pins 2 and 7.
The current (after about 20 seconds) should be about 10% less than 150 mA, and the voltage between pins 2 and 3 should be about 10% less than the spec sheet value of 7.5 V (without dial lamp).

Well, you're my favorite person today!



- Wil

[retiredfeline]

That's good, the heater works and thus it's more likely the tube will work. However your original question was:

If I had to ask one question, it would be:  Is there a way to take the 35Z5-GT and supply it with, let's say, 35 volts A/C instead of the intended 117 V A/C and then be able to measure the voltages on the other pins (and learn something meaningful from doing so)?

followed by a side track about the heater tap which was in fact a second question.

So to answer the original question, you can probably rig up the heater to 30 VAC instead of 30 VDC, and also connect the anode and you will see the heater light up as well as a DC voltage on the cathode, since there will be rectification even at 30 VAC.

[Wil_Bloodworth]

That's good, the heater works and thus it's more likely the tube will work.

It’s a brand new (quite pricey) tube so it better work! LOL.. the old one did not.

- Wil

[retiredfeline]

Ah well, you did pick an area of electronics where NOS (New Old Stock) are expensive.