EMI RS124

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Hi,

Copy of a comercial unit is not done, to much trouble in the past about that kind of thing including lawsuits etc.

But recreation of a old unit, why not.

So let's see what you need schematic ? do you have that ?

Also there is more info on this site about the RS124, also on the Drip site is/was helpfull info on the forum pages.

Check the origanl compressor, wich is adapted to te RS124.

And from the build post, you can derive, power transformer, input and output, tube sockets etc.

Kind regards

Kind regards Dirk Jan
 
I've finally finished the bulk of soldering on my RS124 and powered it up for the first time today. Nothing went bang, and it's passing audio, but there are a few things that aren't working.

First though, some pics:

68zuyd.jpg


2ikpqvr.jpg


I'm using the original Peerless input tx and a Sowter output tx, and a Hammond PT. Once I have it working as stock, I plan to add an attack control. The attack pot is on a push/pull switch so that pushed in it'll have the fixed attack, and pulled out you can control the attack with the pot.

But I need to get it to work properly first...

As I say, it passes audio but:
-  I don't hear any compression happening (will check the control voltage part of the circuit).
- The meter is completely pinned to the right (there's a chance I've worked out the shunt wrong)
-  first thing I would like to get sorted is that my voltages are a lot higher than they should be according to the schematic.

I'm using the hand-drawn schematic that Winston O Boogie posted here ages ago.

The amount of DC I'm getting after the bridge rectifier is 363VDC, whereas I should be getting 255VDC. I suppose if I lower this to be closer to what the schematic asks for, that should lower all my other voltages across the board.

I'm using a Hammond 269BX which has a centre tapped secondary of 300VAC, and I've left the CT floating. I wonder if I've used too large a transformer?

It has taps on the primary side for 220V, 230V and 240V. I currently have it wired up to 230V. Maybe if I wire it to 240V on the primary it would lower the output on the secondary?

If anyone has any thoughts, it would be much appreciated!

 
letterbeacon said:
I'm using a Hammond 269BX which has a centre tapped secondary of 300VAC, and I've left the CT floating. I wonder if I've used too large a transformer?

you put the 269BX in this circuit unchanged???
 

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Changing to the 240v tap won't lower the voltage anything like enough.  You could use a dropper resistor to lower the voltage after the bridge rectifier.

For the meter shunt resistor personally I would use a pot or trimmer.  Just makes it easier.

I like you front panel it looks authentic.
 
Thanks Rob.

Yes I think changing the dropping resistors will be the easiest way of doing it. The capacitors I have before the dropping resistors are rated at 560V, so should be able to handle the extra voltage.

Here's the relevant section of the schematic:
2a4tx12.png


The two dropping resistors are R13 and R14. R13 looks to be dropping 110V, so I need it to drop 210V. That seems like a lot!

How could I calculate what sized resistor to put in there? My Ohms Law is a little rusty. I think I need to calculate the current draw of the compressor, but how do I do that?

 
Wait a sec - it was a typo! I meant to say I have the Hammond 369AX - it's 250VAC CT.

So how come I'm getting such a high voltage reading at the output of the bridge rectifier? I guess it's because the current draw isn't that high?
 
letterbeacon said:
Wait a sec - it was a typo! I meant to say I have the Hammond 369AX - it's 250VAC CT.

So how come I'm getting such a high voltage reading at the output of the bridge rectifier? I guess it's because the current draw isn't that high?

Because the rectified d.c voltage is 1.414 times the rms a.c voltage
 
Thanks Rob, ah yes of course it is. This is my first foray into DIY for about a year, so I'm a little rusty on things like that.

So this is how I'm working out how to drop the voltage by an extra 100v. Would someone be able to check to see if I'm going about this the right way?

On the schematic, there are 2 dropping resistors a R13(a 10k) and a R14 (a 6.8k), both dropping 110v each. I need them both to drop 210v each. Here's my working out:

R13 (the 10k)
V/R = I
110V / 10k = 0.011A

Therefore I know that the resistor has to drop 210V with 0.011A of current.

V/ I = R
210V / 0.011A = 19.09K

Therefore I think a 20k dropping resistor will work there.

R14 (the 6.8K)
V/R = I
110V/ 6.8k = 0.016A

Therefore I know the resistor has to drop 210V with 0.016A of current.

V/ I = R

210V/ 0.016A = 12.98k

There fore a 12K dropping resistor will work here.

Have I got that right? I'm a bit rusty on the theory, so any help will be much appreciated!
 
I'm a little confused now with all the different schematics and values for the Emi and original altec versions.

there is the original altec, the modifications by nyd, then some 436b schematic (that seems fake) and the WB original emi schematic, all having different values for the psu section. Could someone give me a tip which would be best to start with?
I want to have controls for the input and output as well as attack and release.

Actually i wanted to go for the layout attached, but it turns out to have (again) different (and maybe also wrong?) values than the other schematics..

 

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letterbeacon said:
What do you mean? Looking again the voltage reading could be 195v rather than 145v, so perhaps it's not dropping 110V...

On the 436c diagram that point is 210v with no compression & 290v with 20dB compression.
On another emi mod 436b diagram it is 182v under no compression.

So 195v seems more likely.
 
re: losing 100v.

Yes, a resistor will drop the extra volts and convert it into heat.  Remember the heat, as it's a real factor and you need to get rid of it.
power is current squared times R.

The fly in the ointment here is that the current isn't quite constant, so the voltage drop will change as the current changes, and if you look at the
schematic, you can see that voltages change depending on the amount of gain reduction...

the original supply is a full-wave doubler,  which charges the last filter capacitor after the rectifier and before the dropping/decoupling resistors
to about twice the peak of the input voltage.  Further, they used a selenium rectifier which has a several volt internal drop.

Looking at the EMI schematic,  I see that the voltage that changes is the plates of the input tube, which makes sense since its grid
is driven negative by the gain reduction control voltage varying the plate current. More gain reduction means grids more negative means less plate current. 
The input tube, V1 is the only current drain on  R13. With no gain reduction, that's  60v dropped thru 10k,  6mA. the threshold control is responsible for about 700uA of
that current.  With  however much gain reduction altec used for the measurement, the other voltage is 20v, dropped thru 10k, or 2mA.
Clearly, we are seeing that the doubler's current regulation isn't too stellar.  I think you can surmise that V1 is not quite cut-off under heavy GR conditions.

In theory, V2 is running  class A, so it's plate current shouldn't change on a drive dependent basis, and its grid voltage is not tied back into the GR control voltage,
so presumably its plate current is constant. The cathode voltage of V2 is a clue to how much plate current flows, 3.2v thru 220R is 14.5mA.

Adding the two currents says the drain on the supply is somewhere between 20.5mA and 16.5mA, a 4mA swing. Also note that the rectifier output voltage varies 10v depending on GR conditions. Let's ignore that for now.

So, your dropping resistor has to tolerate an approximate 4mA  current swing.

With your 250v transformer (it's 250v CT, right?; I'm too lazy to look back in the thread), the bridge rectifier delivers 352v (approx) to the filter capacitors. and we need 255v, which is 97v drop at 20.5mA, 4865R.
That's about 2W, so your dropping resistor should be a 5W part, and it will get warm. It should be wirewound. Maybe a mil-style part that bolts to the chassis is best. Let the chassis help get rid of the
heat. Of course, you'll need to adjust the value to the nearest standard value, which will depend on the resistor tolerance. It looks like you need to use 5k (that's what's available at Mouser),
and the 10W part is considerably less money. 

https://www.mouser.com/ProductDetail/ARCOL-Ohmite/HS10-5K-F?qs=sGAEpiMZZMtbXrIkmrvidNw3CpujCJIJrQVroexDPy%252BzUvjXwMA15w%3D%3D

under GR conditions, the current changes to about 16.5mA, and this means about 80v drop thru the resistor, 352 - 80 == 271v. That's only 6v from what the EMI schematic says.  I'd try it that way.

The effect of the value difference (4865R to 5000R) is about 5v more drop. The other standard value is 4k7, and that's a 96v drop at 20.5mA. That's better, I'd go that direction. These parts are also available with axial leads
which might be easier to deal with, but do ensure that there's enough space around the resistor so it's heat output doesn't cook something else.

You could use a zener diode to further improve the regulation, but that's another topic for another day.

OH! I see you used a Hammond 269BX transformer, which you say is 300v ct, so the filter capacitor voltage will be 1.414 times 300, 424v. You can grind the math, and see the effect. the dropping resistor will get hotter. You may need that zener.

 
Firstly, thank you very much for your very informative post - it's made things much clearer to me and I really appreciate that!

Secondly, I do in fact have a 369AX not the BX. That was just a typo from me earlier in the thread.

So what you're saying is that rather than replace R13 and R14, I should insert a dropping resistor right after the rectifier to bring the voltage down to approximately 255V, which then supplies R13 and R14?

This makes total send - thanks a lot!
 
ok, I took the time to see what the 369AX is about. so it's 250v ct as I originally thought, and it's 115mA current output. Note that transformers are measured at rated current, so the 369AX will deliver more secondary voltage when operated under its rated current.  the extra current can be used to your advantage via a bleeder resistor.

The point of trying to get close to what Altec did is to try to establish the same operating conditions, and since the stages operate from the same common point, it makes sense to try to get the standing voltage at that point the same as it is in the Altec.

You might also investigate using a bleeder resistor across the output of the dropping resistor. The effect here is the bleeder current swamps the variation in current caused by the input tube, which then minimizes the effect on the voltage drop in the series dropping resistor.  In the old days, when a voltage regulator represented a lot of extra circuitry, a bleeder resistor was a common solution to the problem of minimizing the effects of dynamic current swings on an unregulated supply.
 
I've just installed a 5.6k 25W resistor directly after the bridge rectifier and I have some slightly strange (to me) results.

The voltage is now much lower than I anticipated.

The voltage after the bridge but before the new resistor is 284V whereas before I installed the resistor I measured 363V after the bridge.

After the new 5.6k resistor the voltage reads 198V. This then means that all the other voltages after the 5.6k are all around 20% lower than they should.

Based on the maths that rickc very kindly worked out, I thought that a 5.6k resistor will drop approximately 100v, bringing the rest of the voltages in line through the rest of the compressor, but it appears now that the voltage before the new 5.6k has lowered to lower than I need it to be. I don't understand this...
 
how large are/is the first filter capacitor? did you stick with the bridge rectifier?
since you're in 50hz land, the filter capacitors should be somewhat larger, 15-25% because of the lower ripple frequency.

I'm guessing that maybe the filters are a bit undersized, and we're seeing the change in voltage due to load.

rather than agonizing over it, you can just adjust the value of the dropping resistor to get the right voltage.
 
The filter caps are 50uf, so perhaps that might be the issue. I'll change C7 to 80uf like the schematic, but C8a will have to stay at 50uf as it's shares a can with C8c.

In the meantime I think I have a few 3W axial resistors I can put in parallel with the chassis mounted 5,6k to bring the resistance, and hopefully the voltage drop down.

Thanks for your help with this.
 
I've put a 6.8k in parallel with the 5.6k dropping resistor and now by voltage afterwards is 240VDC, which is close enough to the 255VDC it says on the schematic. As it's an old design, I reckon within 10% should be fine.

The meter is is still pegged to the right and I'm don't think it's compressing.

My meter is 100uA, the schematic calls for a 200uA. I worked out that I need a 1K shunt across it, but I've used a preset pot and dialled in 1k, with the thinking that I can fine tune the needle to zero if needed.

I don't think it's the shunt though.

The compressor doesn't seem to be compressing as when I turn up the input pot instead of getting more compression, the signal starts to distort.

On the schematic, when I have 20db compression, I should get a rise of 10V after the diode bridge, and the voltage at the plates of V1 should rise by 100V.

I can't measure the amount of compression because my meter isn't working, but when I increase the input pot so I get a rise of 10V at the diode bridge (although actually after my new dropping resistor) the plate voltages on V1 only rise by 10V, not 100V.

I've double checked my wiring of the sidechain but to no avail.

Does anyone have any thoughts at all?
 

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