Output jack and impedance ideas

[basschops1528]

Hey guys,

I've been fooling around with a couple drawings on what I'd like to do for the output on my 1959 SLP.

When I originally drilled the chassis I made four holes in the back and I feel I should use them all, rather than just two and dummy the other two.

Two ideas I had I'm posting for you guys to see. I know the first one should be fine and makes sense. The second one I may need a little help finding the right resistors to do it.

They're just two different options I'm playing around with. Advice and opinions welcome and thanks in return.

[PRR]

Don't switch the NFB line. Pick a tap and leave it there.

The Line Out (for 100 Watts at the 8 Ohm tap) should be a 10K pot and 10K+470 resistor divider.

[basschops1528]

Ok cool! so the "NFB tap" could be instead soldered with the 8 ohm secondary tap on the bottom left of the DPDT?

[sluckey]

Ok cool! so the "NFB tap" could be instead soldered with the 8 ohm secondary tap on the bottom left of the DPDT?

Yes. Same goes for the line out circuit.

[2deaf]

100 watts at 8 ohms is 28.3V RMS, which is way too much for a line out, so you need to make a voltage divider using resistors.

The divider is in parallel with the speaker, so the resistance of the divider has to be way larger than the impedance of the speaker in order to have a negligible effect.

The input impedance of a device connected to your line out is very likely to be low-Z with an input impedance of 10K or less.  You will want to make the resistance of the portion of your divider that is in parallel with the input of the device much smaller than the anticipated input impedance so that differences in device input impedances have a minimal effect on your divider.

Since this is a line out, it should have some sort of line level.  Since you have a pot. for one of the resistors in your divider, you may as well go with +4dBu (1.228V RMS) because you can turn that down to -10dBv if you have to.

A 500 ohm pot. is readily available and is also significantly lower resistance than the typical input impedance of a device.  If you are only going to run the amp at 100W output, then an 11K resistor from the 8 ohm tap to the 500 ohm pot. to ground will give you the +4dBu at the wiper with the pot. at max.

10watts (which a lot of people interpret to be half as loud) is about 9V RMS into 8 ohms, so you would need about a 3K resistor in series with the 500 ohm pot to get +4dBu at the wiper at max.  You can still get +4dBu at 100W by turning the pot. down, so this combination may be more useful for you.  There is only 8 ma flowing through this circuit at worst case, so a 1/2 watt resistor and a 1/4 watt pot. are fine.

The voltages on the different taps are all different at the same output.  Switching taps at the same output will change the NFB and your voltage divider.  That's why PRR and sluckey told you not to do it.   

[jeff]

just wondering, why a switch AND seperate  4/8ohm jacks?

Do you intend to hook up a 4 ohm cab to the 4 ohm jack AND a 8 ohm cab to the 8 ohm jack and use the switch to switch between cabs?

If not, and you only use one cab at a time, you can simply wire two "output" jacks and use the switch to switch between the 4 or 8 ohm taps to match the cab you are using. Use the 3rd hole for line out jack and the 4th hole for line out level.

[basschops1528]

You guys gave me a ton of info and I appreciate it. More or less I need to fill four jack holes and I thought a line out would be an interesting way to make the amp versatile. The dpdt switch idea was to keep outputs exact so you can't mix a four and an eight ohm cab unless you can't read. But if you flip the switch then the amps unloaded anyway. It was more or less for preventing more than one tap selected at a time and utilizing at least two of them. My big question is if wired with the NFB and line out to the bottom left on the second schematic, when switched to four ohm ( dual 8 ohm) output, wouldn't both the eight and four ohm taps be in use and cause some kind of shorting or load imbalance problems. I'm assuming you can't use two (or even three ) taps at the same time.   

[basschops1528]

Thought I'd throw these pics up so you guys can see what I worked on today. I drew what I thought to be the Line out section from what 2deaf had suggested, reconfigured where IT and the NFB go, and came up with a better solution for the 4 and 8 ohms jacks that I think will be more preferable.

FYI, the cabs are marshall 1960A and B rated at something like 300W mono.

[2deaf]

if wired with the NFB and line out to the bottom left on the second schematic, when switched to four ohm ( dual 8 ohm) output, wouldn't both the eight and four ohm taps be in use and cause some kind of shorting or load imbalance problems.

The load from the NFB and Line Out is insignificant when compared to the speaker load, so there is no problem.

If a picture is worth a thousand words, a quick scribble's gotta be worth at least fifty.

[basschops1528]

Thanks man! I think we are in agreement between my drawing and yours, which hopefully makes me look a little more competent! haha

[basschops1528]

  If you are only going to run the amp at 100W output, then an 11K resistor from the 8 ohm tap to the 500 ohm pot. to ground

10watts (which a lot of people interpret to be half as loud) is about 9V RMS into 8 ohms, so you would need about a 3K resistor in series with the 500 ohm pot to get +4dBu at the wiper at max.  You can still get +4dBu at 100W by turning the pot. down, so this combination may be more useful for you.  There is only 8 ma flowing through this circuit at worst case, so a 1/2 watt resistor and a 1/4 watt pot. are fine.

 

I'm a little confused here though. Are I using two resistors and a pot? or just one resistor and a pot? your drawing shows one resistor but the quote sounds like conflicted.

[sluckey]

In reference to your 'math.png' speaker drawing... Both schematics are exactly the same and both will yield  R_T = 4Ω

[basschops1528]

I just figured out the problem.. I will redraw the 16 ohm out and post it for verification

[2deaf]

I'm a little confused here though. Are I using two resistors and a pot? or just one resistor and a pot? your drawing shows one resistor but the quote sounds like conflicted.

There are two different examples.  Each one has just one resistor and one pot.  I gave one example that has a drawback in an attempt to illuminate certain criteria when selecting resistor values.

Example one:  This one is the same as my scribble, except the 3K resistor is replaced by an 11K resistor.  Now you have an 11K to 500 voltage divider.  At 100W RMS, 28.3V RMS is across the divider from the 8 ohm tap to ground and 1.23V RMS is across the 500 ohm pot.  1.23V RMS is +4dBu which is the larger of the two standard line levels.  The drawback is that when you turn the amp down, you can no longer get +4dBu out of the Line Out.

Example two:  Exactly the same as my scribble with a 3K to 500 voltage divider.  This one can also get +4dBu at 100W RMS simply by turning the pot. down.  It can also get +4dBu at 10W RMS, whereas example one cannot.

[basschops1528]

So the 3k resistor more or less let's a hotter signal through across the entire wiper, whereas the 11k causes too much attenuation at full pot resistance?

[basschops1528]

Sluckey,

Here is an updated schematic.

[PRR]

> two resistors and a pot? or just one resistor and a pot?

I text-posted a plan with two R and a pot. 2deaf describes a plan with one R and pot.

The results are very nearly identical. Negligible load on the amp, low-low impedance to the line output. (Small difference in output voltage, but either should be fine.)

A difference comes if you don't have 500r pots handy. (Or 1K pot and 6K resistor.) IME, 10K is a more commonly handy pot value, which leads to two resistors (like 10K and 470r) to get the cut-down of voltage and impedance.) But that IS twelve-cents added expense.

[2deaf]

So the 3k resistor more or less let's a hotter signal through

That part seems right, but the rest of the sentence doesn't quite make sense to me.

Attached is my interpretation of what PRR is talking about and my two examples.

[2deaf]

Attached is a wiring diagram for a Marshall 1960 cabinet which may be of interest to you.

[PRR]

> my interpretation of what PRR is talking about

No. That form does not give low output impedance (can be over 2K, which might be OK, but we can do better).

This is what I was thinking. Output is 450 Ohms for any pot setting. The 10K:470r ratio might be adjusted on test (if all the way up is not enuff for your soundperson, try changing 10K fixed to 5K or 2K).

[2deaf]

This is what I was thinking. Output is 450 Ohms for any pot setting. The 10K:470r ratio might be adjusted on test (if all the way up is not enuff for your soundperson, try changing 10K fixed to 5K or 2K).

Cool.  Glad you clarified this.

[jeff]

Just to clarify what I meant.
Most amps have either seperate jacks for 4 or 8 ohms and NO switch and you use which ever jack matches your cab
OR the amp has one set of jacks and a switch that switches those jacks to either the 4 or 8 ohm taps.

Having seperate 4 and 8 ohm jacks AND a switch is unnessecary and may lead to confusion for most players

If you're trying to fill 4 holes you have line out jack(1) line out level(2) main output(shorting jack)(3) and ext output(non shorting)(4)
 

[HotBluePlates]

just wondering, why a switch AND seperate  4/8ohm jacks?

... The dpdt switch idea was to keep outputs exact so you can't mix a four and an eight ohm cab ... 

I agree with Jeff that the switch is unnecessary; basically more $ spent for another part. I'd suggest that if you have multiple speaker jacks someone can always plug in some odd combination of speakers.

Simple wiring: You have an 8Ω tap, so connect that to your 8Ω jack(s). Multiple jacks just means they get put in parallel. 4Ω tap goes to the 4Ω jack(s). Negative feedback wire goes to whichever tap/jack was stock for that amp.

If you plug 2x 8Ω speakers into the 2x 8Ω jacks, total load seen by the amp is 4Ω; you'd need to plug 2x 16Ω speakers into the 2x 8Ω jacks to get 8Ω total load with parallel jack wiring. Same scenario, all numbers halved, for thinking about the 4Ω jacks.


"What happens if I plug an 8Ω speaker in the 8Ω jack and a 4Ω speaker in the 4Ω jack?"

Each tap receives power and is loaded by a speaker which matches its labeled impedance. Overall, the 2 loads are in parallel and half the labeled impedance is reflected to the OT primary.

In other words, this situation has the same net effect as having 2 parallel speakers on a single tap (like 2x 8Ω speakers on an 8Ω tap).


"What happens if I plug an 8Ω speaker in the 8Ω jack and an 8Ω speaker in the 4Ω jack?"

Same as above, the two load are in parallel but now the resulting reflect primary impedance is something between half-value and nominal value. So it's actually a bit closer to having the "right-load on right-tap".


Additionally, I think that when you try it you'll find there's little effect unless the amp is cranked up. It won't damage anything.

[basschops1528]

... main output(shorting jack)(3) and ext output(non shorting)(4)

What do you mean by shorting and non shorting jacks?

[basschops1528]

This is my final product based on ALL the information you guys have given me. I decided to go with two output jacks, with a choice between 4 ohm mono, or parallel 16 ohm mono cabs (or a single 8 ohm I suppose), switched between a SPDT, and a line out with level according to PRR. My only questions that remain are: what are suitable wattages for the resistors and pot, should I lower the 10k resistor for a smaller output imp. (I'm not picky, it's really just an added function at this point, but I want versatility if I ever want to record, you know?), and what did jeff mean by this...

main output(shorting jack)(3) and ext output(non shorting)(4)

[jeff]

A shorting jack shorts the tip to ground when nothing is plugged in. There is a "switch" built into it. When nothing is plugged into it a little tab makes contact with the tip shorting it to ground. This is usually found on tube amps to short the OT in case the amp is accidentaly turned on and no speaker is plugged in(it's not used in solid state amps. very bad for SS).

It's like the type used for input jacks. When you pull your guitar cord out of the input, the jack shorts to ground quieting the amp.

Non shorting jacks do not have this switch and do not short the tip to ground with nothing plugged in. Tip goes to one lug ground goes to the other, no switch.

The reason you use one shorting jack and one non shorting jack is beacuse if they were both shorting jacks you would HAVE TO have something plugged into BOTH jacks or the OT would be shorted.

 In other words the main jack shorts with nothing plugged in but when you plug into it it no longer shorts, so you MUST use the main, The EXT does not short so you can either have something plugged into it or not depending on if you need it or not.
 

[2deaf]

My only questions that remain are: what are suitable wattages for the resistors and pot, should I lower the 10k resistor for a smaller output imp. (I'm not picky, it's really just an added function at this point, but I want versatility if I ever want to record, you know?).

1/2W is fine for the resistors and pot.  PRR suggested a 5K or 2K in place of the 10K fixed resistor.  Values in that range will give you more versatility while having minimal impact on the output impedance.

[basschops1528]

Thanks for all the help guys, I'm going to go with the last drawing I posted but change to a 3k or 5k resistor.

It turns out that the linear 10k I have is rated 1/2 W, but the log one is 1/4 W, and I'm pretty sure I want to use the log. Will it stand up?

And If I'm using Cliff style jacks, how would I wire the 4 ohm jack so that it is shorting to ground when not plugged in? (but not the 8 ohm jack!) Jeff, would you have a schematic

on hand?

[sluckey]

And If I'm using Cliff style jacks, how would I wire the 4 ohm jack so that it is shorting to ground when not plugged in? (but not the 8 ohm jack!)

You don't want to do that.

[jeff]

The only time you want to short any tap is if NO speaker is plugged into the amp at all.

In other words if you're using the 8 ohm tap the 4 ohm tap cannot be shorted and vise versa.

[basschops1528]

I meant to say, how would I wire a Cliff style so the SPK jack shorts with no speaker plugged in ( but not the EXT jack!).

[sluckey]

I'd do it like this...

[basschops1528]

Sluckey, you're the man! Doug should put that diagram next to the one you drew on how to wire Hi/lo inputs on the general hookup page.

[basschops1528]

1/2W is fine for the resistors and pot...

It looks like the Alpha LOG pot I'm buying is only rated 1/4 watt (http://www.alphapotentiometers.net/html/24mm_pot_3.html).

Should I go with a linear instead? I have a feeling it won't work right..