Double triode solid-state equivalent

[YT2095UK]

I prefer to think of these "Holes" as States rather than an entity, but if you subscribe to the simplified Bohr model of the atom then I can see how they may be perceived as such.

I guess my point is as I touched upon with the cathode injection and the use of hexodes and heptodes for Mixers (except in the case of some VHF tuners where an ecc85 or the likes is used) is why isn`t this method proposed above used in all comercial sets, surely using a double triode would be a load cheaper than multigrid valves?

from a pragmatic perspective it makes sense.

[MichaelR]

Originally Posted by G8HQP Dave

Just to make clear what the circuit is: two triodes, with their cathodes joined and their anodes joined, each with common resistors (cathode to ground, anode to HT). The two inputs each go to one of the two grids.

If the valves are linear then nothing happens. The signal in each valve is approximately equal and opposite (i.e. antiphase) to that in the other valve (due to the cathode connection). The signals cancel in the anode circuit, so there is no output.

.

Dave I do not quite see this cancelling , I would have thought that the same signal on each grid input would have caused addition.

Mike

[G8HQP Dave]

Dave I do not quite see this cancelling , I would have thought that the same signal on each grid input would have caused addition.

The mixing comes from a second-order term. Call the grid signal G, and the cathode signal K. Then the second order term is (G-K)^2 (- because grid and cathode work in opposite ways). Expand this out to get G^2 -GK +K^2 - two lots of second harmonics and the product term -GK.

Now take the two input signals to be a and b. Each valve gets one of these on its grid, and the other on the cathode. So the first valve has -ab, and the second one has -ba, but these are equal so we get -2ab.

Alternatively, say the first valve has (a-b)^2, and the second has (b-a)^2 and multiply out. The result is the same. Basically what is happening is that we have a product of two factors and swapping the sign of either of them changes the sign of the product; we have one of each so they both have a change of sign. I think mathematically so I can't explain it without using algebra!

[G8HQP Dave]

Originally Posted by YT2095UK

I prefer to think of these "Holes" as States rather than an entity, but if you subscribe to the simplified Bohr model of the atom then I can see how they may be perceived as such.

I guess my point is as I touched upon with the cathode injection and the use of hexodes and heptodes for Mixers (except in the case of some VHF tuners where an ecc85 or the likes is used) is why isn`t this method proposed above used in all comercial sets, surely using a double triode would be a load cheaper than multigrid valves?

from a pragmatic perspective it makes sense.

I stopped believing in the Bohr atom over 30 years ago, when I learnt quantum mechanics! Interestingly, the pilot wave interpretation of quantum mechanics might reintroduce Bohr-like ideas if it manages to see off the dominant Copenhagan model (which always seemed philosophically flawed to me).

I think the double triode mixer is not used much because it has low gain (due to feedback from the cathode resistor), and poor isolation between the inputs and output. This could cause oscillator pulling and oscillator radiation in a broadcast receiver, as these typically have no RF stage and no buffer after the oscillator. Balancing out the inputs, which it does do, is not needed in a broadcast receiver. So for a simple cheap receiver, multigrid mixers are the best choice.

[MichaelR]

Hi Dave,

Thanks for the reply well written by the way. What I was referring to was your statement that if the valves are linear then the two signals cancel each other out due to the cathode connection.Linear valves do not give you 2nd order effects do they ?

If you have two signals (a) on one grid and (b) ont'other ( and they are in phase) surely you just get addition. I have not sat down and reasoned dynamically what it would settle to but I would have thought that it would have given you 2 in phase contributions to the anode current

Mike

[G8HQP Dave]

If the valves are linear then each has a signal at its grid (a or b) and the 'other' signal at its cathode (b or a). So one valve sees a total signal of (a-b) and the other valve sees a total signal of (b-a) - because the cathode has the opposite effect to the grid. Adding them together we get a-b + b-a = zero!

Connecting the anodes together like this gives us an output with just even-order components. Leaving them separate, and taking the difference as the output (as in the differential pair) gives an output with just odd-order components.

[YT2095UK]

I think I know what your saying, to get the Most out of the anode, the Cathode would have to be at a Trough of a cycle (most negative) and the Grid would have to be at a Peak of a cycle (most positive) at the same time.

[cendoubleu]

This is what I love about this and other technical forums. I asked a question which was more than adequately answered to my satisfaction, and then you guys take it and run with it. I have really enjoyed reading the subsequent discussions after my question was answered (although I hadn't looked for several days and was suprised to see the thread count up to 2 pages!). Thanks everyone for making this so interesting.
regards Charlie

[G8HQP Dave]

Yes, we do get a bit carried away at times!

[dr peppers]

Interesting discussion, getting carried away is sometimes where good ideas come from.
I spose my best description would be to say that the ecc could be loosely compared to 2 fets with the drains connected as the common cathode, and the gates as the 2 control grids and the sources as the anodes.
Dunno how you'd get a fet to self bias with diode action though.