Modifying the headphone out jack on 145 for line signal?

[rjd2rjd2]

Hi all-apologies if this is somewhere in the archives of the forum(I couldn't find it). I have a 145 tube wurly, and would love to use the headphone jack output as a line level output. However, I get a nasty thump in the >50hz range, as well as a signal that is SO hot it is tough to keep from distorting. I saw the VV breakout box thing, but it's discontinued. I am very comfortable soldering and knocking up a simple circuit, but I am NOT having an easy time finding a schematic,  or even values for components. Does anyone have a simple schem for a breakout box that can live between the headphone jack, and an amp/DI for this model? Thanks so much if so!!!

[beginnersluke]

I'm not an expert on the 145, or much of anything in this life, really.

That said, since no one has responded, I'll share some of my non-expert ideas (so you know, listen to me at your own risk).

You could try something like a 500 ohm pot across the signal wire to adjust the level to what you really need.

You may need a resistor around 100 ohms (and a small capacitor like .1uF) bridging the signal and ground. (This would change the impedance of the signal.)

(I can draw a schematic tomorrow if you'd like me to.)

If I were doing this, I'd be tempted to insert a diode to not accidentally send voltage back to the piano accidentally.

I hope that helps.

Luke

[beginnersluke]

By the way, if you used a potentiometer like that, you'd need something that can handle more power than like 1/2W. Looking at a 200A schematic, it looks like the resistor they have on the headphone out is a 5W. You can get a 5W CTS pot for like $4 or $5. (You could measure and calculate what you'd need, but 5W is a standard value that will cover it, unless the 145 is more of beast than I thought!)

Luke

[rjd2rjd2]

Thanks, Beginners' Luke!

for the most part, what you're describing sounds fairly simple. So, in essence, a 5W pot in line with the signal pin of the headphone, as well as a 100 ohm R>.1uf cap>diode(I have a bunch of 1418's lying around) from tip to sleeve at the jack, with the polarity reflecting the voltage coming from tip and to sleeve, so to speak? All seems simple enough. I don't mind knocking that up and trying it; whatever load the amp wants to see when the headphone jack is engaged has to be maintained internally, right? Thanks!

[pianotuner steveo]

No, those values are too low. 100 ohms won't help enough. And the volume control already acts as the trim pot (in a way) so adding another doesn't help much, even if it is the correct value.
I don't remember the exact values, but I know the resistor (not the pot)  needs to be a pretty high value, (ohms) and I believe it should be a 5w. Years ago (40?) I got detailed instructions directly from Wurlitzer on how to do this, but somehow, during a move, it got lost. I really wish
I could remember the values.

Do you want to completely eliminate the headphone jack or take a signal from it and add a second jack somewhere? I personally would add a second jack.

Also, be sure you piano no longer has the "death cap". If there is any stray voltage coming out of that jack, you could fry whatever you are connecting to.
(See previous posts regarding the "death cap")

And of course, you are adding any resistors in series between the + (hot) wire and the output jack. You are not adding resistors in parallel across the + and -, this will short out the amp. (If a resistor is too small, it could short out the amp)

[pianotuner steveo]

The thump sounds like another issue,  possibly related to the vibrato circuit.

[beginnersluke]

"You are not adding resistors in parallel across the + and -, this will short out the amp."

Can you explain a bit more about this?

If you look at the schematic of the headphone jack on the 200A, for example, when you plug in the headphones, you're activating an 8ohm resistor (5W) between the signal and ground. My understanding was that this replaces the load of the speakers that the amp is "looking for".

A line-in has a much higher impedance than a headphone, so you may need to provide a load to the amp with a similar resistor between the signal and ground to mimic the load of the headphones.

I guess you're saying that's wrong, but can you say more about why.

By the way, you may be right that 500 ohms is not enough attenuation., though I do see value in a separate control here (esp since we're unsure of a proper value), but it's definitely a thing where YMMV.

The thump may be related to the impedance of the line-in (high) and the "missing" load. I'd be curious if it goes away once he gets a good line-out circuit. If it's not present on the speaks or headphones, I'd hope that's the case. (That said, you may well be right that it's something separate.)

Thanks,

Luke

[beginnersluke]

Oh, I should add this for the original poster. /If/ you do use a pot, make sure you also have a resistor in-line with the pot on the signal path.

I looked up a 145 schematic and it looks like when you connect headphones you connect a 8ohm resistor (again to mimic the speaker load) between signal and ground.

The reason for adding a resistor in line with the pot is to protect the amp from shorting if the cable gets shorted. It's not that difficult to short the tip and sleeve on a 1/4 instrument cable. So if you have a resistor in series on the signal, even if the cable is shorted you still have more resistance than the 8 ohm resistor already in your circuit.

In the real world, your pot probably won't go down to 0ohms, but an extra 500ohm resistor won't attenuate the signal much, and it will help protect your amp so it's definitely worth it.

(Also, I think I meant to say 5000 ohm potentiometer in the original post, not 500. (You may not need to attenuate much, but that will give you a good range to get the signal you need.)

pianotuner steveo knows more than me, and like I said early on, take what I say with a grain of salt because I don't claim expertise, but the above seemed worth mentioning if you build a circuit. (I actually just learned this and why you see this resistor in these circuits through thinking about and researching your question here.)

[Alan Lenhoff]

Not the cheapest solution, but it works nicely for taking the output from the headphone jack on my 140B and running it into my Twin Reverb. 

https://www.sweetwater.com/store/detail/ProRMP--radial-prormp-1-channel-passive-re-amping-device?mrkgcl=28&mrkgadid=3303549796&rkg_id=0&product_id=ProRMP&campaigntype=shopping&campaign=aaShopping%20-%20SKU%20-%20Live%20Sound%20%26%20Lighting&adgroup=Live%20Sound%20Accessories%20-%20Radial%20-%20prormp&placement=google&adpos=1o1&creative=285688985997&device=c&matchtype=&network=g&gclid=EAIaIQobChMIivf-5NvE4gIVg4bACh1_Yww2EAQYASABEgJVkPD_BwE

It also met another requirement for me:  I didn't want to to alter the piano in any way.

Alan

[pianotuner steveo]

Luke, the 5w resistor in the headphone circuit is correct. Yes, that simulates the load of a speaker, and does connect in parallel.. I was saying that adding another resistor to attenuate the signal should be in series. You need a very high impedance resistor here, but not a 5w. I believe a 1/2 w is fine here.
If I remember correctly, a small cap is added between + and -  near the resistor you will be adding.

Does anyone here remember the value of the resistor used here? I think it's really high, like 50k, or even more.

[pianotuner steveo]

Alan, how did you connect the piano to this? It looks like input only uses a 3 prong DIN.

[Alan Lenhoff]

Alan, how did you connect the piano to this? It looks like input only uses a 3 prong DIN.

That's correct. I bought one of these to run between the piano and the re-amp box: https://www.sweetwater.com/store/detail/PXM105--hosa-pxm105-5-foot. The box has a standard 1/4" output to the amp.

Alan

[pianotuner steveo]

Did you have to make a cord to adapt to the 3 prong, or buy a premade 1/4" to 3 prong?

[Alan Lenhoff]

I bought the exact cord in the Sweetwater link I posted yesterday.

Alan

[pianotuner steveo]

Oops, missed it! Sorry!