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Old 13th Dec 2018, 2:51 pm   #1
Diabolical Artificer
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Default Long Tailed Pair voltage question.

Something just occured to me, the amps I've just built use a minus 78v rail to feed the cathodes of a LTP, both grids are grounded, one directly, one through a 680k R , how can the cathode swing positive with reference to ground/0v? It doesn't make sense, surely any voltage drop across a resistor or CCS should still be minus whatever volts, or am I being dim? The maths using ohm's law doesn't jive either.

If for instance we have a 5k6 tail R to set both cathodes at 7mA each (-5v ish Vgk - it's never bang on as per loadline), total 14mA, R x I = 78v , therefore with all the voltage dropped across the resistor, the cathodes should be at 0v not the 5-7v I'm getting. This is really bending my mind, I must be missing something obvious.

Andy.
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Old 13th Dec 2018, 3:50 pm   #2
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Default Re: LTP voltage question.

The valves will auto-bias by passing sufficient current via the anodes for the cathodes to become positive.

I think we need to see a circuit diagram.
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Old 13th Dec 2018, 6:01 pm   #3
G8HQP Dave
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Default Re: LTP voltage question.

Why can't the cathodes swing positive? They have a (conventional) current arriving from the anodes, which will develop a voltage across the cathode resistor. It is only negative feedback which constrains the cathode voltage to be near 0V, otherwise it could be anywhere. Valve action means that it will normally be a bit positive, if the grids are grounded.

You need to think about what grid-cathode voltage plus the approximately known anode voltage will produce a current in the region of 7mA. Then remember that it is grid-cathode voltage, not grid voltage.
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Old 13th Dec 2018, 6:39 pm   #4
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Default Re: LTP voltage question.

Quote:
Originally Posted by Diabolical Artificer View Post
If for instance we have a 5k6 tail R to set both cathodes at 7mA each (-5v ish Vgk - it's never bang on as per loadline), total 14mA, R x I = 78v , therefore with all the voltage dropped across the resistor, the cathodes should be at 0v not the 5-7v I'm getting. This is really bending my mind, I must be missing something obvious.
If you have a -78V rail, the 'tail' resistor is 5.6kΩ and the cathodes are sitting at 5V positive of 0V, then there is 83V across the tail resistor and the total tail current is 14.8mA not 14mA.

As PGL says, the valves will self-adjust themselves to satisfy both their own Ia/Vg characteristic and the V/I characteristic of the cathode circuit.
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Old 14th Dec 2018, 7:27 am   #5
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Default Re: LTP voltage question.

Light dawns, I was thinking about the cathode in isolation but of coarse as you point out PJL, Dave and Paul it's connected to the anode, so therefore there is 430v ish (HT350v) across the valve as a whole; the cathode could be anywhere from minus 78v to plus whatever volts as you all point out.

Sometimes you get stuck in a cul de sac of thinking, after you've navigated your way out, you think how did you miss that?

Yours sheepishly, Andy.
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Old 14th Dec 2018, 8:29 am   #6
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Default Re: LTP voltage question.

Remember that your constant current source isn't really a constant current SOURCE. it's really negative power supply with a controlled resistance (transistor). The resistance is controlled to try to keep the current fairly constant within the limitations of your negative power supply and of the availability of current and voltage from the valves.

Consider your CCS transistor replaced by an ammeter, a variable resistor and a trained monkey. It can only turn the knob on the variable resistor and can't make energy out of nowhere.

Your CCS transistor is a similar feedback arrangement. Emitter degeneration is feedback, it's just that the loop isn't obvious on paper.

David
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Old 15th Dec 2018, 10:15 am   #7
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Default Re: LTP voltage question.

Quote:
Your CCS transistor is a similar feedback arrangement. Emitter degeneration is feedback, it's just that the loop isn't obvious on paper.
I Like that.

There is another scenario too, that doesn't seem obvious on paper.

It is when you have some system comparing a value with a "constant". Which in the obvious paper diagram case is a reference voltage, fed into one input of a comparator.

When it is not so obvious is when the comparator is something like a base-emitter voltage drop on a BJT and it is a "2 wire" comparator.

There are many circuits that use this technique, but it confuses some when they are looking to find where the reference voltage is and where the comparator is.
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Old 15th Dec 2018, 11:42 am   #8
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Default Re: LTP voltage question.

If like me you are a bit slow sometimes with acronyms:-

LTP = Long Tailed Pair. (I had to google it)
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Old 16th Dec 2018, 9:06 am   #9
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Default Re: Long Tailed Pair voltage question.

Thanks for the explanation David, can you expand a bit on emitter degeneration?

Actually, the CCS/LTP combo is a bit odd, it isn't behaving entirely as it should on paper or in theory or more probably my understanding of it is at fault.

The CCS is simple one tranny job, 6.7v zener on the base with a dropper r is series and an emitter R made up of a 220r and 1k pot.

Adjusting the pot so Re is reduced, and therefore Ik increased ( and Vgk also falls as does Va) results in the negative -80v (-78v or whatever) dropping as it was only designed to supply about 10mA. This increases the OP power because this rail is also the neg bias for the OP valves.

That's not the odd thing though, increasing Re should result in less cathode current, Vgk should go up, which it does to a point, however there comes a point where it will not go up anymore. I have increased Re to 10k, the valve bias point sticks at about 6v. I have a suspicion this is because of the zener being 6v7.

Another thing of note. If one draws a loadline for a valve and choose a bias point, midway say, then Va should be X volts; it never is. I put this down to the mutual characteristics graph being a work of fiction, to a point, for an ideal valve, and as we know all valves arn't equal or the same. But Va is always less.

Another interesting thing to observe is to scope the OP of the LTP (thought everyone knew that one Colin, sorry for the confusion) {Terry Pratchett has some interesting things to say about "what everyone knows" but that ones for the virtual pub) If you twiddle the pot on the CCS a few things happen, the amplitude of the two OP's of the LTP change, though one less so, OP of the amp as a whole is changed and I dare say a few other parameter's are effected.

So I've found that there is a sweet spot where OP of the LTP is well balanced, OP of the amp as a whole is optimum, as is THD. It's quite tricky to hit this sweet spot with a one turn preset but it's probably not worth mucking about with a ten turn pot.
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Old 16th Dec 2018, 12:32 pm   #10
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Default Re: Long Tailed Pair voltage question.

Think of a standard NPN transistor, emitter grounded, a bit of bias to the base and a load resistor to a power supply connected to the collector.

The base looks like a diode to ground. Voltage current obeys the diode equation. Once you get up to about 0.6v the base current goes up like gangbusters. The transistor tries to pull down Hfe times more current than is going into the base. As current increases, Hfe increaases to a peak then falls away, the collector current times the load resistance gives a voltage drop, the collector voltage falls and eventually the transistor collector voltage is saturated low and the load current is set by the load resistance and the power supply voltage. The transistor can't pull down any more current than that.

The voltage gain is as much as the transistor can give into that load, the distortion is poor, input impedance is low. Essentially the transistor is used flat out for gain.

Now let's stick a resistor in the emitter in its connection to ground.

We raise the base voltage, the transistor starts to turn on and pass current. The emitter resistor drops voltage and this voltage robs the base emitter junction of some of the incoming voltage. We have a feedback mechanism, where collector current is fed back as an opponent to input voltage applied to the base. The circuit applies Vin to the base, the transistor only cares about base-emitter voltage. As the transistor turns on it pulls its own emitter voltage up, and reduces the effect of Vin.

The emitter resistor is common to *two* circuits. It's in the collector>load>supply>ground>emitter circuit and it's in the >input>base>emitter>ground>other end of input source circuit

And by this commonality, it links them.

With degeneration, the voltage gain is reduced and stabilised (and less temperature dependent) The distortion is improved, but the output swing cannot be as much because of the drop in Remitter.

It's called degeneration. It's feedback and it's used a lot, but it's invisible to many people who haven't thought it through.

You can now have a good laugh at some of those transistor amps penned by people who hate feedback on theological grounds. They use it all over the place, but aren't aware! It's only the overt sort that they oppose.

And there's cathode degeneration in valve circuits. Cathode decoupling caps remove it for AC but leave it at DC.

David
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Old 17th Dec 2018, 4:03 pm   #11
G8HQP Dave
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Default Re: Long Tailed Pair voltage question.

I have had arguments with people who insist that degeneration is not feedback. This allows them to design/make/buy 'feedback-free' amplifiers which may contain and/or finish with some sort of follower. They are quite impervious to logic.

Quote:
That's not the odd thing though, increasing Re should result in less cathode current, Vgk should go up, which it does to a point, however there comes a point where it will not go up anymore. I have increased Re to 10k, the valve bias point sticks at about 6v. I have a suspicion this is because of the zener being 6v7.
The zener is in the CCS. Vgk is in the LTP, so unless you have some unusual circuit there is no direct link between them.

If the valve is approaching grid cutoff then Vgk cannot exceed the cutoff point, and cannot actually reach it. Unless your CCS is open circuit it must pass some current. If the valve is passing some current then it is not cut off, so Vgk has not quite reached the cutoff point.

If you want to push a valve beyond cutoff then you have to apply voltages to it, not expect voltage drop across resistors to do it - if the valve is conducting then it is not cutoff. Hence no matter how huge a resistance you put in the cathode circuit (or a proxy in the CCS emitter circuit) you cannot cut off the valve.
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