Can someone explain this tone circuit?

[ikay]

I'm doing some restoration work on a 1965 Hofner 185 and am a bit baffled by the tone circuit (see pic of schematic below).

Each tone pot has a tiny 10n capacitor going to ground. This would bleed off some very high frequencies (much higher frequencies than a standard tone pot). There is also a 'regular' size capacitor (100n or 47n) fixed in the signal path to the vol control and the output jack. As far as I can see this would let through most of the high frequencies but attenuate some of the lower frequencies.

In practice, when either tone is turned down the output signal is considerably attenuated. I guess because the 250k resistance of the tone pot has then been put in line with the signal path.

Wierd. It's quite different to a standard passive tone circuit and doesn't work as I'd expect it to.

Can someone explain to me what's going on here?

[Geek99]

if its any help my stack knob harness has just the same arrangement, so its not that unusual

[ikay]

Hi Geek99, thanks for replying.

The stack knob jazz circuit (below) is actually quite a bit different to the Hofner. Only two lugs of each tone pot are connected - the signal goes to the middle wiper lug, then through the cap to ground via the variable resistor. The third lug (the other end of the variable resistor) is left floating. With this configuration, even with the tone fully rolled off, the pot resistance doesn't impede the main signal path. 

With the Hofner circuit, when the tone is fully rolled off the full pot resistance is added to the signal path. The Hofner circuit also has 2 additional caps compared to the stack knob circuit and the main tone caps (in this case 100n and 47n) aren't where you'd expect them to be.

I've checked the wiring of the bass itself (which is original) against the Hofner circuit diagram and it matches, so that's how they intended it to be. It just doesn't function as you'd expect a regular tone circuit to function. It's all a bit odd!

Edited September 12, 2018 by ikay

[MoonBassAlpha]

The 10 nF caps might be there to bleed off radio frequency interference? As the 100nF cap is always in series it must be a resonant circuit with the pickup inductance and the pot. Would need some helpful soul to model it for us, but as you say, it isn't a current way of doing things. I had a 60s 185 with a 70s scratch plate which was slightly different, and only had 1 volume pot iirc. I didn't sketch the circuit out as it seemed to work sort of ok!

[ikay]

Hi MoonBassAlpha. Interesting thought about the 10nF cap, but if it was there to reduce radio interference I'd expect it to be permanently in-circuit, not after a variable resistor. Not that I really know what I'm talking about lol! Interesting also about the resonant circuit. I believe this acts like a band-pass filter and the bass does actually sound a bit like this.

Pic showing wiring of the later 1970s model 185 below which is a much more traditional arrangement. Maybe they eventually realised that the standard wiring was better! I'm tempted to just rewire it along these lines.

[MoonBassAlpha]

The other oddity I noticed was that there is no earth wire to the bridge or string anchor plate, with no noticeable adverse effect.

[ikay]

Yes, I'm a bit dubious about the lack of a bridge earth! The bass is in bits at the moment so I don't know if this is a problem but none of the old Hofner solid bodies seem to have a bridge earth so I'm hoping not...

[Grangur]

Hi @ikay,

What you have are 2 pickups with on/off switches. Then the 2 tone controls. The signal then goes through coupling capacitors. These are invisible to an alternating current , but will stop any dc voltage. Then it goes through the 2 volume controls.

The reason for the coupling capacitor is because, although we humans think of these as 2 separate circuits feeding a jack socket, electric current doesn't see it this way.

If one volume is turned up and the other turned down, current from 1 pickup can feed back through the volume of the other pup and interfere with the signal from the other pickup. 

[ikay]

Hi Grangur, thanks for that. Can you please clarify a few things:

1. Are the 100n and 47n caps before the two vol controls the coupling caps? It just seems a bit odd that they're different values and that the values are the same as regular tone caps.

2. The two 10n caps are positioned where you'd expect to find the tone caps (ie. bleeding high freq to ground), but why are they such tiny values? I'd expect the cap values to be  the other way round, with the tone caps being 100n and 47n and the coupling caps being 10n.

3. The way the tone pot is wired, when the tone is fully rolled off, the full 250k resistance of the pot is in series with the signal path which significantly attenuates the output signal. Why would they do this?!

Thanks, Ian

 

[rmorris]

16 hours ago, Grangur said:

The reason for the coupling capacitor is because, although we humans think of these as 2 separate circuits feeding a jack socket, electric current doesn't see it this way.

If one volume is turned up and the other turned down, current from 1 pickup can feed back through the volume of the other pup and interfere with the signal from the other pickup. 

You're right that the circuits aren't separate. But the coupling capacitors don't do away with interaction apart for blocking dc - and that isn't an issue here.

Although they will affect the degree of interaction depending on frequency and pot' settings.

[Grangur]

Just now, rmorris said:

You're right that the circuits aren't separate. But the coupling capacitors don't do away with interaction apart for blocking dc - and that isn't an issue here.

Although they will affect the degree of interaction depending on frequency and pot' settings.

What we don't know is if the capacitors are electrolytic. If they are, then the current flow is directional. At the time of posting I was in a bit of a rush and didn't want to get into the finer detail of, then, what does an electrolytic capacitor look like.

Electrolytics have a "+" on one leg.  However, as at all times, I reserve the right to be proved wrong by someone more knowledgeable than I. Maybe you do, @rmorris ?

(not having a swipe, merely asking) 

The other thought is, if the pups are out of sync - one outputting +ive as the other goes -ive, then between them, the 2 voltages could maybe create a form of DC?

[Grangur]

6 hours ago, ikay said:

Hi Grangur, thanks for that. Can you please clarify a few things:

1. Are the 100n and 47n caps before the two vol controls the coupling caps? It just seems a bit odd that they're different values and that the values are the same as regular tone caps.

2. The two 10n caps are positioned where you'd expect to find the tone caps (ie. bleeding high freq to ground), but why are they such tiny values? I'd expect the cap values to be  the other way round, with the tone caps being 100n and 47n and the coupling caps being 10n.

3. The way the tone pot is wired, when the tone is fully rolled off, the full 250k resistance of the pot is in series with the signal path which significantly attenuates the output signal. Why would they do this?!

Thanks, Ian

 

I reserve the right to be totally wrong...

1 - Yes, they're the ones I was thinking of. Yes that is interesting.

2 - Agreed.

 

Edited September 13, 2018 by Grangur

[rmorris]

2 minutes ago, Grangur said:

What we don't know is if the capacitors are electrolytic. If they are, then the current flow is directional. At the time of posting I was in a bit of a rush and didn't want to get into the finer detail of, then, what does an electrolytic capacitor look like.

Electrolytics have a "+" on one leg.  However, as at all times, I reserve the right to be proved wrong by someone more knowledgeable than I. Maybe you do, @rmorris ?

(not having a swipe, merely asking) 

The other thought is, if the pups are out of sync - one outputting +ive as the other goes -ive, then between them, the 2 voltages could maybe create a form of DC?

With those values of capacitance they will almost certainly (never say never - I once disputed the existence of tantalum resistors only to be proved wrong :-) not be electrolytics (or tantalum). As it happens Aluminium Electrolytics are usually marked on the negative leg (making the other +ve by implication while Tantalum Caps are maked at the positive leg / terminal. There are also Non-Polarised Electrolytics usually marked 'NP').

Yes - I deal with capacitors a lot :-)

It wouldn't be strictly a good idea to use polarised capacitors in a passive tone / vol circuit in any case simply because you don't know what any dc bias will be - it's basically down to the following stage / amp input etc. However, in reality they are fine with a few volts 'the wrong way' as might occur in a typical audio circuit.

As for the 'form of DC' - no :-)  Although if I can work out a way to do that I might be in the money !

 

[Grangur]

my knowledge is rusty. I was in electronics about 20 years ago.

[ikay]

H Grangur, re item #3, regular passive tone wiring leaves the third lug of the pot disconnected (see the schematics in posts 3 or 5 above). The resistance of the pot is only used to control how much signal is passed through the tone cap to earth, it doesn't impede the signal path.

The Hofner circuit is different - when the tone is fully rolled off, the full resistance of the pot is directly in the signal path which significantly attenuates the output level (in addition to losing the high frequencies). The caps BTW aren't electrolytic, just regular ceramic disc tone caps.

You can clearly hear the drop in output level when the tone is rolled off in this demo (from about 3.55 to 4.25)  

 

Edited September 13, 2018 by ikay

[rmorris]

3 hours ago, Grangur said:

my knowledge is rusty. I was in electronics about 20 years ago.

Me too + the rest. Still am. Cheers.

[Grangur]

Yes, @ikay, I stand corrected - post corrected.

On the video... is it me or when he uses the tone control, does it also lose volume? This wouldn't surprise me from the circuit.

[Grangur]

2 minutes ago, rmorris said:

Me too + the rest. Still am. Cheers.

Hmm.. I now sell LED lighting, so I guess my boss would like to think I still know a fair bit about some aspects of it.

[rmorris]

5 minutes ago, Grangur said:

Hmm.. I now sell LED lighting, so I guess my boss would like to think I still know a fair bit about some aspects of it.

Well there's a lot to remember / forget 😊

Anyway, we can probably agree that it's not a great circuit idea when you lose volume with tone.

Personally I don't really use passive tone control much at all.

[ikay]

1 hour ago, Grangur said:

... is it me or when he uses the tone control, does it also lose volume? This wouldn't surprise me from the circuit.

Yes, big drop in volume, which is due to the wiring of the tone pot. Very wierd. Mine is exactly the same. I'm debating whether to leave the wiring in its original state and just not use the tone pots, or rewire it in a more conventional way.

[rmorris]

Unless you're keeping it as an historic artifact then I say rewire it.

 

[ikay]

Yes, it's a player not a collectors item so I think you're right! I'd still like to know what the thinking was behind the original wiring though. Hofner have always been a bit 'individual' on the wiring front but this is a curious one.

[MoonBassAlpha]

Oh yeah, the other Hofner oddity is that the tone controls operate in the opposite sense to everyone else's, 10 is treble rolled off, 0 is wide open. Those crazy Germans!

[Unknown_User]

On 13/09/2018 at 09:53, ikay said:

Hi Grangur, thanks for that. Can you please clarify a few things:

1. Are the 100n and 47n caps before the two vol controls the coupling caps? It just seems a bit odd that they're different values and that the values are the same as regular tone caps.

2. The two 10n caps are positioned where you'd expect to find the tone caps (ie. bleeding high freq to ground), but why are they such tiny values? I'd expect the cap values to be  the other way round, with the tone caps being 100n and 47n and the coupling caps being 10n.

3. The way the tone pot is wired, when the tone is fully rolled off, the full 250k resistance of the pot is in series with the signal path which significantly attenuates the output signal. Why would they do this?!

Thanks, Ian

As I understand it (and I'm not necessarily sure I do, so someone please correct me if the following is a load of nonsense), smaller capacitor values block more of the bass frequencies.  With that in mind:

1. The capacitor on the bridge pickup has the smaller value, so will pass less of the bass frequencies through to the output.  Whereas you'll get more low end allowed from the neck pickup through the 100nF capacitor.  Which makes sense I reckon.

2. If these 10nF capacitors are there to bleed high frequencies to ground, then wouldn't you need low value capacitors here otherwise they'd be letting more of the low frequencies out to ground too?

3. I guess that means that the potentiometer acts to cut the output from that pickup, rather than boost it?  If you were going to boost it instead then wouldn't you need some sort of active component, by which I mean a separate DC voltage going into a transistor with the AC from the pickup?  So cutting the sound from the pickup makes more sense than boosting it from a "how much do you need to throw at it to get it working" perspective.

 

I'm still very much in the early stages of learning all this, so I would appreciate any corrections to that!

[tauzero]

On 24/09/2018 at 09:52, Unknown_User said:

As I understand it (and I'm not necessarily sure I do, so someone please correct me if the following is a load of nonsense), smaller capacitor values block more of the bass frequencies.  With that in mind:

1. The capacitor on the bridge pickup has the smaller value, so will pass less of the bass frequencies through to the output.  Whereas you'll get more low end allowed from the neck pickup through the 100nF capacitor.  Which makes sense I reckon.

2. If these 10nF capacitors are there to bleed high frequencies to ground, then wouldn't you need low value capacitors here otherwise they'd be letting more of the low frequencies out to ground too?

3. I guess that means that the potentiometer acts to cut the output from that pickup, rather than boost it?  If you were going to boost it instead then wouldn't you need some sort of active component, by which I mean a separate DC voltage going into a transistor with the AC from the pickup?  So cutting the sound from the pickup makes more sense than boosting it from a "how much do you need to throw at it to get it working" perspective.

 

I'm still very much in the early stages of learning all this, so I would appreciate any corrections to that!

Capacitors of whatever value present a higher impedance to lower frequencies than higher frequencies - the impedance doubles when the frequency halves. To give you some idea of values, at 100Hz (somewhere round E, 2nd fret of D string, with apologies for any inaccuracy) 10nF would be an impedance of 160k, and 100nF would be a tenth of that. So imagine a resistor of 320k replacing the 10nF capacitor when you're playing bottom E, 160k when you're playing an octave up, and 80k another octave up, and you'll see how the circuit behaves.