What's the name of this circuit?

[buckethead]

Hi guys!

While servicing an Orange guitar amp I saw this configuration in the fx loop (see attachment). I searched the internet for infos on it, but without any luck - it's hard to search for something you don't know the name of. At first it looked like a standard AC cathode follower, but a closer look reaveled something different. Does anyone have some infos about it?

Thanks and best regards,
Mathias

 

[Bo Deadly]

Yeah, it's just a follower with the send normalled to the receiver.

But the output impedance of the send is a little high (3.6K). And the output cap is pretty small. If you plugged that into some old 600 ohm gear you'd get major loss. Even into 10K you loose some lows (72Hz). So it's designed for high Z stuff only like other tube inputs and most pedals.

 

[volker]

Also anode and cathode are switched. Don't build it like that .

 

[Studio Mollan]

A question about the output impedance: How did you figure that out? I’m only asking because I want to learn this myself. 15k in parallel with 4k7 equals about 3k6. Is that part of the answer?

Yeah, it's just a follower with the send normalled to the receiver.

But the output impedance of the send is a little high (3.6K). And the output cap is pretty small. If you plugged that into some old 600 ohm gear you'd get major loss. Even into 10K you loose some lows (72Hz). So it's designed for high Z stuff only like other tube inputs and most pedals.

 

[abbey road d enfer]

A question about the output impedance: How did you figure that out? I’m only asking because I want to learn this myself. 15k in parallel with 4k7 equals about 3k6.

That's about it. The output impedance of a voltage follower is roughly equal to the inverse of transconductance in parallels with the cathode resistor. Is that part of the answer? in that case, 1/Gm= 180 ohms. So the actual output has an impedance of 15k+180 in parallels with 4.7k, which indeed results in about 3.6k.
As been noted earlier, the tube has been drawned upside-down.
Due to the voltage division, the signal is attenuated by aabout factor 4.
Even if the output was shorted, the tube would operate in its linear operation, because of the 15k in series.
Also note that although the output impedance at the cathode is quite low, it doesn't mean it can drive heavy (low ohm value) loads. The limiting factor is the idle (quiescent) current.
Here, the operating point can be estimated at about 4 mA, which means that the max current that can be delivered to the load is about 3mA rms, but linear operation would not exceed about half of that.

 

[buckethead]

Also anode and cathode are switched. Don't build it like that .

This is acutally what bothers me, I have the amp in front of me and it is really designed like that.

Why would they switch anode and cathode???

 

[JohnRoberts]

This is acutally what bothers me, I have the amp in front of me and it is really designed like that.

Why would they switch anode and cathode???

To get a different result.... (note: I am not a tube guy...).

JR

 

[Studio Mollan]

Thank you! Great answer. I think I get it, but I will have to follow up with some googling.
/
Emil

That's about it. The output impedance of a voltage follower is roughly equal to the inverse of transconductance in parallels with the cathode resistor. Is that part of the answer? in that case, 1/Gm= 180 ohms. So the actual output has an impedance of 15k+180 in parallels with 4.7k, which indeed results in about 3.6k.
As been noted earlier, the tube has been drawned upside-down.
Due to the voltage division, the signal is attenuated by aabout factor 4.
Even if the output was shorted, the tube would operate in its linear operation, because of the 15k in series.
Also note that although the output impedance at the cathode is quite low, it doesn't mean it can drive heavy (low ohm value) loads. The limiting factor is the idle (quiescent) current.
Here, the operating point can be estimated at about 4 mA, which means that the max current that can be delivered to the load is about 3mA rms, but linear operation would not exceed about half of that.

 

[abbey road d enfer]

This is acutally what bothers me, I have the amp in front of me and it is really designed like that.

Why would they switch anode and cathode???

Could you check the voltages on anode and cathode?
In order to operate, the anode needs to be more positive than the cathode OR less negative. Could it be that the supply voltage was negative?

 

[jacomart]

could i know the amplifier model please? something is not clear to me in this diagram ... or are there PNP valves?

Cheers
Jaco

 

[buckethead]

Oh boy, I'm such an idiot. It is an error in the schematic, it's a standard cathode follower, nothing special there. Probably didn't have enough sleep when looking at that pcb a couple of days ago.

Sorry for wasting your time, guys!

 

[jacomart]

Oh boy, I'm such an idiot. It is an error in the schematic, it's a standard cathode follower, nothing special there. Probably didn't have enough sleep when looking at that pcb a couple of days ago.

Sorry for wasting you time, guys!

Only those who do nothing make no mistakes. In any case here attached is the wiring diagram of the Dark Terror

Cheers
Jaco

 

[CJ]

i think it is a phase inverter  ;D

 

[Winston OBoogie]

Most folks are allergic to solid state in guitar amps, but, IMHO, a Mosfet as a source follower would do a much better job than a  triode here.  Less good, but still better than a 12A**, would be a high gm pentode.

Have they discovered positrons btw? 

 

[Bo Deadly]

An inexpensive 10K:600 transformer would have gone a long way here.

 

[Winston OBoogie]

An inexpensive 10K:600 transformer would have gone a long way here.

That too. 

 

[abbey road d enfer]

An inexpensive 10K:600 transformer would have gone a long way here.

If it's meant to replace the voltage-follower, I believe there would be some consideration to be given regarding the impedance of the point that delivers signal. I see a 1Meg pot driven by a tone-shaper that has a rather high impedance.

 

[Bo Deadly]

If it's meant to replace the voltage-follower, I believe there would be some consideration to be given regarding the impedance of the point that delivers signal. I see a 1Meg pot driven by a tone-shaper that has a rather high impedance.

That was the intention yes. Specifically I was thinking a simple cathode follower with cap into 10K primary + load resistor on the secondary would present a consistent enough impedance to parallel with whatever cathode resistor(s). With a 16:1 impedance ratio it would certainly have better drive and less sensitivity to load variations than the existing crap circuit. Some experimentation with different loads might be required to find a good load resistor for the secondary (2K-ish would probably good for both 10K and 600). Although the normalled return could not come from the secondary unless it was grounded which would be a shame. So normalling the return would require more thought and knowledge about tube gear that I do not have.

 

[abbey road d enfer]

With a 16:1 impedance ratio it would certainly have better drive and less sensitivity to load variations than the existing crap circuit.

In the context of a tube guitar amp, this circuit does not qualify as crap. It ticks all the  boxes by providing high input Z, output Z and drive capabilitycompatible with most pedals and FX racks, adds negligible distortion, and uses a tube half that would be otherwise unemployed.
OTOH, putting a low-level xfmr in a tube amp is a privileged target for magnetic interference.
Apart from the guy that drew the tube upside-down, I think these people at Orange know what they do.

 

[Bo Deadly]

In the context of a tube guitar amp, this circuit does not qualify as crap. It ticks all the  boxes by providing high input Z, output Z and drive capabilitycompatible with most pedals and FX racks, adds negligible distortion, and uses a tube half that would be otherwise unemployed.
OTOH, putting a low-level xfmr in a tube amp is a privileged target for magnetic interference.
Apart from the guy that drew the tube upside-down, I think these people at Orange know what they do.

I didn't say the whole circuit was crap. As I pointed out in my original response, the send can't really drive even a regular 10K line without LF loss which IMO is kinda crappy. My guess is that they wanted to keep it cheap but they didn't want to add silicon because marketing probably wanted to be able to say it was an all-tube design. So to add a send / return but hobble it when they could have used a transistor is just goofey.

To be perfectly honest I never saw an Orange schematic that I couldn't find an issue with. I can't produce a good example at the moment but I seem to recall seeing some REALLY goofy crap in orange schems. They're not what I would consider high end amps. They use a lot of cathode / anode style phase inverter (there's a name for that, don't recall what it is but I mean not the standard LTP) and they use cathode bias a lot. Not that that's necessarily bad. I like Supro's and the low wattage department store amps for some low-fi honk (a lot of the sound of an amp is the speaker and the room). But for high gain / high wattage those are not the best design options IMO.