Anyone have the schematic for a Rode classic?

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Anyways thanks all for the pointers. If this is right, for a given triode type one can choose cathode Rk so that voltage drop across it exactly offsets grid-cathode contact potential. Then the grid will naturally have ground potential, and the grid leak resistor can take extreme high value - after all there is no current at zero potential so what does it matter? Advantage is in this arrangement is input impedance at the grid can be very high indeed. And if the grid is ac coupled, I suspect there might be signal level dependant gain, because grid-cathode is a diode charging the coupling capacitor and affecting bias. And perhaps that's another part of the mojo? Anyways for me the arrangement is out of the box thinking.

Cheers T
 
I made a post earlier about the source of "grid leak" voltage. I'll briefly repeat ... and hope it helps. The grid of a tube will acquire a negative potential simply because some of the electrons will collide with the grid "mesh wires" themselves causing a tiny external current flow. The density of the mesh (ratio of wire diameter to spacing), it's spacing from the surface of the cathode, and the plate voltage will all influence this current. Lower mu triodes result from a more open mesh and larger spacing than in high mu designs. Since the 6072 is a relatively low-mu part and it operates at only about 60 V, I'd expect very low grid current and probably would increase operating bias by only a few tenths of a volt ... which makes the operating point consistent with the plate characteristic curves (see page 2 of attachment) when you imagine another plot (for 60 V) to the right of the 100 V curve.
 

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I think you are still confused. You seem to assume grid current due to electron collisions with the grid. This is not correct for a class A valve. The grid repels the cathode electrons from contact, there is only an electrostatic interaction between the grid and the electrons, which are strongly attracted to the Anode, and do not arrive on the grid, they are just controlled by the grid electrostatics. The input impedance of a valve is essentially infinite, which could not be the case if there was any grid current, some electrometer valves have input impedances as high as 10 exp15 Ohms! The grid simply controls the thermal space-charge flow.

David CEng etc.
 
Thanks for replies. I understand points made. No dispute, grid potential in this circuit must be c. zero versus gnd. Not least because, of course , that follows from curves. And, given Rg 200M, it implies effectively no grid current.

The reasonable question remains why?

Original data sheets for many triodes in cathode bias common cathode mode, published limit max values for Rg about 1M or so. 6072 data as MisterCMRR kindly posts typ is cited as 100k. Implies there must, in typ circumstance, be material grid current. Reputable reference such as Radiotron 4th Edition 1953, sets out max budget for Ig medium mu triode (accepting MisterCMRR's point 6072 not high mu) c 1uA. Trust problem into 200M is obvious ! Also sets out how to calculate, BTW.

In all the circumstances here, its plausible there might be internal grid-cathode offset potential. And then it follows there's a 'happy place' where Rk can be arranged so that grid potential is naturally zero versus ground. No matter what Rg, including extremely high.

I don't see another reasonable explanation.

T
 
Thanks for replies. I understand points made. No dispute, grid potential in this circuit must be c. zero versus gnd. Not least because, of course , that follows from curves. And, given Rg 200M, it implies effectively no grid current.

The reasonable question remains why?

Original data sheets for many triodes in cathode bias common cathode mode, published limit max values for Rg about 1M or so. 6072 data as MisterCMRR kindly posts typ is cited as 100k. Implies there must, in typ circumstance, be material grid current. Reputable reference such as Radiotron 4th Edition 1953, sets out max budget for Ig medium mu triode (accepting MisterCMRR's point 6072 not high mu) c 1uA. Trust problem into 200M is obvious ! Also sets out how to calculate, BTW.

In all the circumstances here, its plausible there might be internal grid-cathode offset potential. And then it follows there's a 'happy place' where Rk can be arranged so that grid potential is naturally zero versus ground. No matter what Rg, including extremely high.

I don't see another reasonable explanation.

T
There are several mechanisms at work simultaneously to create grid current (and therefore the voltage on a "floating" grid - or nearly so as with 200 MΩ. Several are explained in the attached excerpt from Radiotron Designers Handbook. In general, the grid will float at the voltage at the "crossover" point (zero grid current) of curve D in Figure 2.9, For most "receiving tube" types, this is somewhere between -1 V and zero (grid to cathode voltage). But, as explained, it depends on a number of factors. It's not simple by any means. I strongly suspect that the tube's operating point was determined empirically.
 

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  • Grid Current Section iii - RDH 4th-Ed.pdf
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Yes. Radiotron's explanation may not be the most lucid, but it's all there. "If the valve is operated with its grid completely open circuited, the operating point will be at D [fig 2.10], since this is the only point corresponding to zero grid current...." i.e. typ Vg -1V or so. And that isnt affected by Rk as in conventional cathode bias. Nor by Rg. The valve will simply follow characteristic curves for its 'self' grid potential. Rg 200M being quasi open circuit, and other mics even have have Rg 1G.

Interesting and worth noting for valve substitutions in mics. I feel some measurements coming on!

T
 
Radiotron figs 2.9 and 2.10 also show typ Ig versus Vg curve, 1/gradient of which is input impedance. Slightly negative of this 'floating grid' bias is a turning point, the only condition where grid input impedance is 'infinite'. And the pieces now seem to be falling into place why this arrangement is chosen here.
 
Hi guys, slightly off topic but still on the Rode Classic PSU. I have an original 220V PSU and would like to change it to 110V for use in Canada. Is it easier to get a stepup converter (if so how many watts) or can the old 220V PSU be rewired for 110V use. My concern is the hassle dealing with another item that could cause noise (hum or cause electrical interference) in the studio. I'm not a technician but pretty handy with a soldering gun. - Jim (new guy on the forum) Thanks
 
Hi, first many thanks for showing an interest in my issue I have little history of this unit as I purchased it from an estate. I do know the previous owner brought it from overseas as the voltage is noted on the back as 220V. He was an audio engineer and may have converted it to 110v already without changing the rear label BUT it came with the 2 prong male European plug on the IEC cable which made be suspicious. I see the marked holes on the board as you pointed out and it is in the 110v position. I'm not a guy who plugs something in to 'test' without knowing it's correct. Also on the back of the pc board it appears the 220v hole has never been used unless he did a super job with his solder sucker. Anyway, a few pics for your thoughts. My other option is to plug it in. I'm looking for an educated guess or opinion, not a scapegoat. Would 110V do any damage to it anyway? Thanks for getting back. Jim
 

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Hi, first many thanks for showing an interest in my issue I have little history of this unit as I purchased it from an estate. I do know the previous owner brought it from overseas as the voltage is noted on the back as 220V. He was an audio engineer and may have converted it to 110v already without changing the rear label BUT it came with the 2 prong male European plug on the IEC cable which made be suspicious. I see the marked holes on the board as you pointed out and it is in the 110v position. I'm not a guy who plugs something in to 'test' without knowing it's correct. Also on the back of the pc board it appears the 220v hole has never been used unless he did a super job with his solder sucker. Anyway, a few pics for your thoughts. My other option is to plug it in. I'm looking for an educated guess or opinion, not a scapegoat. Would 110V do any damage to it anyway? Thanks for getting back. Jim
Hi Jiminy, it looks like the power supply is still set to 220 Volt. Remove the link (red cross) and add two links where it says "110V LINK" and you should be good to go for 110 Volt. I have included how i think the winding of the transformer is attached to the pcb. Both winding in series for 220 Volt and in parallell for 110 Volt. 20220619_001856.jpg
 

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Hi Jiminy, it looks like the power supply is still set to 220 Volt. Remove the link (red cross) and add two links where it says "110V LINK" and you should be good to go for 110 Volt. I have included how i think the winding of the transformer is attached to the pcb. Both winding in series for 220 Volt and in parallell for 110 Volt. View attachment 95388
Bjorn, thanks for getting back, that worked 100% and the old microphone sounds great. I'm looking forward to using it.

Many thanks for walking me through it. Much appreciated.



Jim
 

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