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Measuring unwanted grid current

Hi. I something of a beginner.

Often it can be the passing of unwanted grid current that makes a tube undesireable for service. That would often be associated with a tube described as "soft" or "gassy".

If a tube is seriously gassy, I believe plate current will tend to "run away". It's also possible, I think, to consider a tube gassy if grid current amounts to too many microamps.

I think that with a tube like a KT88, some small unwanted grid current could be acceptable, but at any rate I am assuming that anyone wanting to purchase such a tube would like to know, if the data is to hand, how many microamps of unwanted grid current is flowing.

My KT88 seems to be passing (at the moment - I may "cook it") some unwanted grid current. And I want to know how to measure it correctly.

I want someone to please explain how to measure grid current and why it is measured the way described.

I have a test setup. Originally created to "cook" the tubes if required.

I have 310V feeding a triode connected KT88. I have put in a 270R cathode bias resistor. Grid is connected to the negative end of the bias resistor, via a 1K0 resistor, or directly. I have a switch that can short out the 1K0.

With the setup:

Plate voltage: 310V
Plate current: 90mA
Cathode bias: 24.4V

Since the curent stabilises at 90mA, I presume I don't have a tremendously gassy tube.

When I switch out the 1K0 resistor in the grid circuit, no change in plate current occurs.

Now, I understand that grid current is often calculated by measuring voltage across a resistor in the grid circuit. Whether this is the only acceptable way I don't know. I have a 200uA meter and currently it is in series with the grid circuit. In other words the grid connects to one end of the cathode resistor via the meter. I muse this is perfectly acceptable way of measuring the grid current. If not I want some to say so.

So, a meter in series with the grid circuit is how I'm measuring unwanted grid current. And actually it comes out the same whether the 1K0 is in circuit or not. And the KT88 seems to passing about 4uA grid current.

As to this 4uA value: If it stays at this value, I don't whether it makes the tube unfit for purpose - or what.

Anyway, I am interested in the measurement of grid current.

Also, why does grid current flow if g1 is -24.4V negative WRT k?

Good you're checking this and investigate. You'll learn a lot when doing so and understand a little more about tubes.
Everything you're doing and your measurements are correct.
But you must determine if the grid current is negative or positive. The grid current caused by gas ions makes the grid positive.

The normal grid current caused by the space charge around the catode, makes the grid negative.
That's the reson they specify a max "grid leak" resistor in the datasheets, so this current not interfere with the normal bias.

100KOm I think is minimum for a fixed bias KT88 and some 270KOhm for a cathode bias, were the gid current induced extra bias is of less importance.

If you want to include a gas test in your rig, you can keep the 1KOhm fixed in place (to ensure no HF oscillations) and add a switchable 470K resistor in series with the grid.
If the current rises when switched in, you have gas in the tube. If nothing happends or the anode current is slightly lower, the tube is fine.

Don't throw away a little gassy tube. The getter will still work when you're burning in the tues in your rig. So a gassy tube can "repair it self".
But to much gas and if the getter mirror turns white in the edges, nothing can save that tube.

I've seen many cheap chinese power tubes, that are eating up the getter mirror in a few hundred hours. This i probably not caused by a leaky tube, but gas is released from not clean and gassy components inside. Like gas trapped in the micas or in poor quality anode metal sheet. This gas evaporates when the tube get varm and shortens the life substantially.

Let me see if I can understand unwanted grid current:

Non gassy tube
Electrons from the cathode are speeding out towards the plate. Since the grid is negative (we have cathode bias), these electrons pass by the grid, so do *not* produce an electron flow from cathode to grid, through external grid circuit, back to cathode. Thus no unwanted grid current.

Gassy tube
Electrons speeding out towards plate, hit some gas molecules, knocking off electrons. This creates positive ions. These positive ions are attracted towards the grid and make the grid positive.

Problem: I cannot see how the ions make the grid positive, when it's significantly negative WRT cathode because of cathode bias.

There are two examining situations: a) where there is no bias whatsoever, I mean no cathode or fixed bias. Then who have a situation where we have fixed or cathode bias.

I'm trying to measure unwanted grid current with cathode bias. If this is possible, then I don't understand how these positive ions create a current.
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Positive ions make the grid positive when they land there because the current flow goes through the grid bias resistor. This is typically 100's of K. Ohm's law!

This might not be a good time to introduce a complication, but electrons can still land up on the grid even when it is a bit negative. Electronic grid current disappears below about -1V to -1.5V on the grid. How can a negative grid attract electrons? Two reasons: contact potential (the grid and cathode are made of different metals), heat (the electrons have random thermal energy from the hot cathode so some can overcome a small repulsion).
Lets say there is no fixed or automatic bias. A gassy tube means positive ions land on g1. g1 is connected to k, via a resistance.

Will electrons flow through the external resistance? I guess. From k to g1. Probably also from k to g1 inside the tube as well.

Since g1 becomes positive wrt to k, electrons will want to flow towards the grid through the external grid circuit. g1 and k will be at different potentials, because g1 and k are not connected by a low resistance wire.

If g1 and k are connected with a wire, then current, in theory, wants to pass from k to g1 as before. But g1 and k would be at the same potential. So, I guess current flows, but you never really get any potential between g1 and k.

But, anyway, lets assume we have cathode bias.

Positive ions land on g1. But g1 never becomes posative, because of the negative bias on g1. So all that gassyness results in is a reduction of negative bias on g1. You cannot measure any grid current due to positive ions landing on g1.

Well, at least, that is what is bothering me. How can you measure unwanted grid current due to gassyness, when you have cathode bias?
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There has to be an explanation for this.

g1------------k___/\/\/\__ Cathode bias resistor
Grid resistor

I'm seeing the end of the grid resistor connected to g1 as being at a negative potential, due to voltage developed across bias resistor. Ions landing on g1 are not making g1 positive, just less negative.

That's how I'm seeing it. But, I'm seeing it wrongly, I must be?
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Lets say there is no fixed or automatic bias. A gassy tube means positive ions land on g1. g1 is connected to k, via a resistance.

Positive ions land on g1. But g1 never becomes posative, because of the negative bias on g1. So all that gassyness results in is a reduction of negative bias on g1. You cannot measure any grid current due to positive ions landing on g1.

Well, at least, that is what is bothering me. How can you measure unwanted grid current due to gassyness, when you have cathode bias?

You don't need to maesure the positive or negative grid currents. As they have different polarity, they can be there but can equals each other so there will be zero current resuling.

As I wrote, the best way to detect a gassy tube is to see if there is any positive anode current change when you're open/shorting the 470KOhm resistor from G1 to ground (cathode)

In a tube there is a cloud of electrons surrounding the cathode due to the heat and the cathode emission coating. This is called "space charge". This charge will transfer to G1 and make a negative grid current through the grid leak resistor. The closer the grid is to cathode, the larger the initial current will be, but it should be close to zero at 1-1,5V. (We were discussing this in an other thread)

I'm also a little confused that you could measure 4uA (seems much) with -24 v on the G1, but we still don't yet know if it is a positive or negative current.

4uV in a 470 KOhm resistor will give nearly 2V difference in the grid voltage and will absolutely be visible in the anode current, even with a catode resistor bias!

Maybe you've seen in diagrams, that many microphone and tape-head triode input stages are "missing" the cathode resistor to make a bias. Only a 10MOhm resistor from grid to ground and a capacitor to the signal source. This capacitor is charged by the "space charge" to a negativ bias of arond -1 to 1,5V
Over that knee, the "space charge" current is to low in f.ex. an ECC83/EF86 and leaks trough the 10MOhm resistor to keep it on around 1V.

But if you send a high input voltage or transients from a scratchy record, it will be rectified by the grid-cathode "diode" and give a high negative voltage to the grid/capacitor, blocking the tube for a time, by a time constant of 10MOhm and the capacitors value.
Besides saving on components, this also leaves the cathode on ground potential and it will be less disturbed by the heater voltage noise and hum.
Easy to do, but not desired in HiEnd equipment.

There has to be an explanation for this.

g1------------k___/\/\/\__ Cathode bias resistor
Grid resistor

I'm seeing the end of the grid resistor connected to g1 as being at a negative potential, due to voltage developed across bias resistor. Ions landing on g1 are not making g1 positive, just less negative.

Yes you're right. Less negative and that will show up in an increase in the anode current. Then you know that the tube has a little gas.
But after a day in your burn-in rig, it could be gone (absorbed by the getter) and perform without problems.
The getter is working all the life time to catch gas, but as I wrote before, if there is to much gas, it will be poisoned and the tube burns out.

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Do we have a situation where positive ions landing on g1 is simply said to give rise to positive grid current, a current opposite to any current caused by a negative bias, but no-one actually does measure any positive grid current?

Being a newbie and reading a about grid currents, I thought people were actually measuring unwanted grid current.
I'm confused i.e. I don't know what is confusing you. Both positive and negative grid current can actually be measured, depends on the bias and state of the valve. The net grid current is the difference between the electronic current and the ionic current. The grid current will cause a voltage drop across the grid resistor, so the grid voltage will be slightly different from the grid bias supply. The sign of this voltage drop depends on the direction of the net grid current. As the grid current will usually be small, the easiest way of measuring it is to put it through a big grid resistor and see what effect it has on the anode current.

This is all still true if the cathode is positive (due to normal cathode bias) rather than the grid negative - potential is always relative, never absolute.
It is not the absolute value of a grid current that is of interest. Only how it affects the anode current and performance of the equipment.
There is some grid current figures in some datasheets that together with other parameters, shows the end of life of a tube.
Good you're studying about tubes. Many more need to do that.....
If you not already have, here is a link to datasheets for almost all known tubes:

Frank's electron Tube Data sheets

As you can see there, there is not many data about grid currents.
About transmitting tubes and class B & AB2 amps, the tubes are often accounted with substantial grid currents, when deliberatly driven into the positive area on the G1 grid.

Try to get your hands on a sample(s) of an old "The Radio Amateur's Handbook" from the 60's. There is a lot of information about everything involving tubes and more. Good bedside reading....

The more you learn, the less you understand you know, fits good about this subject.

My confusion centred around current flowing in the reverse direction, this would be the ionic current. I'm not familiar with ionic current.

Electronic current

I understand that despite a negative bias, there will be some tiny electronic current flowing through the grid resistor caused by electrons being absorbed by g1. This has the tendency to make g1 a tiny bit more negative. Therefore this electronic current flows (conventionally) from ground (if one side grid resistor is grounded) to g1.

I can visualise electrons completing a circuit from k, to g1, via grid resistor, back to k via the cathode bias resitor (or reverse if we take conventional current flow).

Ionic current

A gassy tube gives rise to positive gas ions. These are attracted to g1. These make g1 less negative.

It is harder to see how we can have current (moving electrons constitute a current) in this case, because it's hard to visualise a closed circuit. Current in the ionic case is reversed to electronic case, because grid is made more positive. Electrons would go from ground, through the grid and out from the grid. But where do they go from there, they cannot go to k, electrons do not want to move towards k. So where do they go? I guess they simply fill in the holes of the ionic gas atoms.

On thing is for sure, you do get a grid current due to a gassy tube.

Of course, conventional current flow with ionic current is from g1 to ground.

So, I do see now that we can measure grid current due to a gassy tube.

It can be calculated by measuring voltage across the grid resistor. If that is 1M0, then each volt equals 1uA.

We can put in a micro ammeter in series with grid resistor and read it directly.

Or, we can short out the grid resistor and note the change in anode current. We can use the tube characteristic curve to infer the change in bias voltage. Each volt will equal 1uA id the grid resistor is 1M0.

When the grid resistor is shorted you observe no change in measured grid current. This being true g1 voltage has to reduce when g1 is wired directly to ground, or rise when the grid resistor is in circuit.

My 2 cents worth of understanding.
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You are almost there.

When ionic current dominates, the electrons arriving at the grid get used up neutralising the positive ions. Assume each ion carries one positive charge. Then an electron on its way to the anode must have hit it and knocked out a second electron. So now we have two electrons going to the anode, and a positive ion going to the grid.

Tot up the current/electron flows. Two electrons arrive at the anode and go through the power supply back to ground. One of these came from the ground via the cathode. There is also an electron used at the grid to neutralise the ion; this electron came from the ground too. So we have 2 leaving the ground and 2 arriving - the currents match so charge is conserved. The fact that one of the anode electrons is not the same individual as the grid electron does not matter for two reasons: all electrons are identical, and all electrons are indistinguishable.
Just one more thing:

Maximum rating for Rg-k is given as 220K. That I believe is saying the resistance between grid and cathode should be no higher than 220K.

Is that a safety feature to ensure, in case of ionic current, that anode current is limited?

Because of course value of the grid resistor, whilst not affecting the value of ionic current, does determine grid voltage rise due to that unwanted current.
Yes, the resistor value is limited in order to limit the possible grid bias shift caused by ionic grid current. A higher value resistor could allow a slightly gassy valve to shift its bias enough to draw extra anode current, heat up and release even more gas from the metal - so more grid current and eventually thermal runaway. I guess this means that provided you run the valve cool (unlikely in an output stage!) you can get away with a small increase in the resistor.

The small change in voltage may cause a small change in grid current too.