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Old 8th Jul 2012, 4:58 pm   #1
sobell1980
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Default Low tension help with Roberts RMB please.

Hi all,

I finally got round to replacing the big can electrolytics in my RMB, dual can C27 and C28 and C25 and C26 for the low tension. Hoping this was going to help raise my low tension to the valve heaters as before I was measuring 1.10 volts at the heaters. Now replaced this has risen slightly to 1.2 volts but still not good enough. I've taken some measurements of voltages for those that would like to help.

From the LT secondary winding I have 2.60 volts AC to chassis and this is from both sides and measured at the point where the secondary winding joins MR2.

At connection C26 positive side after rectification by MR2 I measure 1.8 volts and at L17 before it passes through the choke. The choke measures 2.4 ohms as per the data sheet and LT secondary where it connects to MR2 is 0.3 ohms as per the data sheet.

After Choke L17 I'm getting 1.2 volts at positive side of C25 and then the same reading at the valves themselves.

Is it possible that the voltage being received by L17 is too low before it even passes through L17 and then obviously being dropped lower as it passes through. It can't be L17 at fault as it resistance measures ok. Secondary winding for LT measures ok at 0.3 ohms. There are no measurements available to check out or test MR2 if this is causing too much of a drop.

One thing I did notice was that with the voltage adjustment screw set at the highest voltage rating (its correct position) 240 to 250 volts the primary of the mains transformer only measured 197 ohms compared to the data sheet which says PRI TOTAL 219 ohms. But surely if this was lower it would pass to higher current and voltage to the low tension circuit.

Is there any way of working out what the voltage should be at the secondary winding and then work out each section through? Is it topping up the resistances and divide by the mains voltage etc etc. Hope one of you experts can shed some light.

Many thanks,

Dave
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Old 8th Jul 2012, 6:42 pm   #2
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Default Re: Low tension help with Roberts RMB please.

The primary resistance of the mains transformer will depend on the number of turns, the diameter of the spool and the diameter of the wire. It looks as if the drawing die for the wire had worn and you got slightly fatter wire than standard.

The current is determined mostly by the inductance and the load, the wire resistance will have little effect unless the transformer is very inefficient.

The low heater voltage is probably due to the selenium rectifier feeling its age, use any 1A silicon diodes or diddy bridge to replace. Check that the resulting voltage is not too high.
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Old 8th Jul 2012, 6:47 pm   #3
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Default Re: Low tension help with Roberts RMB please.

I would concur.

peter
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Old 8th Jul 2012, 7:01 pm   #4
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Default Re: Low tension help with Roberts RMB please.

So what you are saying is perhaps the data sheet is slightly incorrect and was dependant on tolerances of mass produced transformers?

So if I measure the primary mains transformer at 197 ohms and the secondary LT at 0.3 ohms that makes a total resistance of 197.3 ohms. 240 volts divided by 197.3 gives a current of 1.2 amps AC before it is rectified at MR2.

So how do I work out the correct voltage drop? I measure 2.6 VAC before rectification at MR2. It would just be nice to know from the mains transformer through to the rectifier to calculate each voltage drop correctly to see at which point it is dropping too much.

The problem I have here is that part of the circuit is AC and the other DC. Does the same maths apply as it drops 2.6 VAC before being rectified to 1.8 volts DC out of MR2? Is there a calculation or resistance test across MR2 I can do? I'd like to eliminate it by diagnosis rather than parts substitution but I'm sure you're right with your prognosis.

Dave
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Old 8th Jul 2012, 7:06 pm   #5
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Default Re: Low tension help with Roberts RMB please.

On both of the RMB's that I have worked on, the filament supply is not as documented in the Trader sheet 1037. On these sets the filament winding was actually centre tapped to earth, and the bridge rectifier fitted did not have the -ve end connected to ground - in other words the filament supply works as a conventional two diode bi-phase rectifier. In addition there was a small resistance wire ballast resistor in circuit (it looks like a wire link), arranged so that it could be shorted out if the filament voltage was too low.

The 0.6V voltage drop across the choke seeems ok as the valve current totals 0.25A and the choke resistance measures 2.4 ohms

If you don't have the resistance wire link to short out, I suggest you try replacing the bridge rectifier with a couple of 1N400x diodes, assuming the filament rectifier is actually bi-phase as I described above.

Ron
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Old 8th Jul 2012, 8:06 pm   #6
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Default Re: Low tension help with Roberts RMB please.

Yes i only have 3 wires from mr2 there is nothing to ground so off the mains it only runs through two diodes on mr2. Not sure what you mean by ni phase?


Yes, I only have three wires from MR2. There is nothing to ground so off the mains it only runs through two diodes on MR2. Not sure what you mean by "bi-phase"?
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Old 8th Jul 2012, 9:05 pm   #7
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Default Re: Low tension help with Roberts RMB please.

You cannot just add primary and secondary resistance, the transformer has a ratio of about 100:1 so the impedance ratio will be 10,000:1 so 200R on the primary will be equivalent to 0.02R on the secondary but remember that the primary also has to supply the HT.

If you have 2.6V AC then the peak voltage will be 2.6 * sqrt(2) = 3.6 V.
Knock off 0.6V for the diode drop, you get 3.0V DC if the reservoir capacitor is very large. With 0.6V dropped across the choke, that leaves 2.4V which is way above what the valves require so you should add a resistor between the rectifier and the reservoir capacitor to get the right voltage under load. The resistance will be much smaller than a simple calculation would indicate since the current comes in pulses.
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Old 8th Jul 2012, 9:12 pm   #8
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Default Re: Low tension help with Roberts RMB please.

Sobell1980 - Do you have the coiled resistance wire somewhere in series with the filament circuit positive output as I described? If it is fitted, check the voltage drop across it.

A bi-phase rectifier is a rectifier using two diodes and a centre tapped transformer to achieve full wave rectification. I was comparing this with the bridge rectifier, which does not need a centre tapped transformer. The latter is shown on the Trader sheet 1037, but not implemented on your set.

Trevor re post #6 - I believe the OP has 2.6V ac across a centre tapped winding - i.e. 1.3V ac for each half of the winding.

Ron
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Old 8th Jul 2012, 11:28 pm   #9
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Default Re: Low tension help with Roberts RMB please.

If the TX is 1.3-0-1.3 V a pair of silicon diodes like 1N4007 will only give around 600mV DC out, after allowing for the diode drops and the choke resistance on a 5.6R load (250mA @ 1.4V).

At 2.6-0-2.6 V you get about 1.6V which is too high.

A bridge on 2.6V ends up with only 1.1V.

It's a mess!

2.6-0-2.6 V biphase is about the best but will need some series resistance (unless there's enough in the transformer windings).

We need a few more hard facts.....

What is the transformer winding type and voltages for sure?

What are its winding resistance(s) on the LT secondary?
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Old 8th Jul 2012, 11:51 pm   #10
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Default Re: Low tension help with Roberts RMB please.

I have read post #1 again and the OP does say 2.6V to chassis from both sides of the winding. So it is 2.6-0-2.6 V. Apologies to Trevor and Chris.

Ron
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Old 9th Jul 2012, 7:25 am   #11
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Default Re: Low tension help with Roberts RMB please.

I am having a little difficulty here following what you are meaning with all the figures, or at least some of them. I understand I'm getting 2.6 Volts AC at the LT secondary. I still don't understand how you managed to work out the mains transformers resistance what do you mean by 200R? How do you know there is a diode drop of 0.6 volts and how did you convert the 2.6V AC TO DC? Surely if there are two diodes being used that would then account for a 1.2 volt drop? I'm still very new to all this and please be patient with me as I'd rather diagnose correctly than just replace parts on a whim and not fully understand what's going on.

Some one asked what's the resistance of the LT secondary, it's 0.3 ohms as in my first post and measures 2.6 volts. The primary of the mains transformer measures 197 ohms where the data sheet states 219 ohms. I have 102 Volts AC before MR1 which rises to 128 V DC after MR1 which then drops to 111 V DC after the R14 resistor which I know is high and have put down to the valves not working correctly due to the LT being low.

A few more questions what do you mean by TX?

How did you work out 5.6R load and not sure what mean by "R" and how you got the figure 5.6R.

How did you work out the 250 milli amp valve current draw? I'm sorry if I'm stating obvious things here but just very keen to learn and understand. If someone could teach me the maths you have all just used behind this it would be a huge help. Especially the AC to DC conversion etc

Dave
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Old 9th Jul 2012, 10:57 am   #12
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Default Re: Low tension help with Roberts RMB please.

Hi Dave

Lots of questions!
1 Transformer.
Some people abbreviate this to Tx, I don't like this as Tx in my mind is transmitter.
I used 200R for the resistance of the primary since it was near enough to the data sheet and what you measured and was easier on the maths. I used R for resistance, if you know how to find omega on the computer, let me know and I will use that.

2. Voltage drop.
You will get very little voltage drop in the transformer so I havn't worked it out - too lazy. There will be >0.6V drop across a silicon diode, that is just how they are. If you have a bi-phase supply (centre tapped winding) with one diode passing current, then the other, you will get a drop of 0.6V or so. If you have a bridge rectifier with the current going through two diodes at any one time you will get a drop of 1.2V as you say. From what others have said, you may have either circuit in your PSU.

The rectifier passes current on the peak of the waveform which is 1.414 times the AC RMS voltage. It charges the reservoir capacitor to hold charge till the next peak. The capacity is not infinite so there is some reduction in voltage before the next pulse. It is therefore difficult to know the exact voltage which will be obtained.

The inductor smooths the current flowing and so looks at the average voltage input, not the peak. It will have a voltage drop accurately assessed by its resistance.

3. Current taken.
I took the correct curent drain from what others have said. You could look up the valve data and tot up the current taken by all the valves.

4. Fault finding.
How I agree with you about finding a fault and putting it right. I cringe when I see someone has changed all the capacitors and resistors and, surprise, surprise, now it doesn't work. Why don't they just go and buy a new radio?
In your case, the fault is in the heater supply, get this right first and the high tension may then be OK, it is certainly near enough to get the set going initially.
Remember that there are tolerances on all components and they will deteriorate over time. When I started playing with circuits, the resistors had a tolerance of +/- 20%. You could get closer tolerance resistors but they cost the earth. Now +/- 1% seems the norm.

Hope this helps,
Trevor
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Old 9th Jul 2012, 11:02 am   #13
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Default Re: Low tension help with Roberts RMB please.

As Trevor's explained, "R" is just a way of saying ohms, and is easier to type on a keyobard than Ω.

It's very common in modern circuit diagrams, and is sometimes used instead of the decimal point, e.g. 0R6 means 0.6Ω.

You're getting very close now, Dave, keep at it.

Nick.

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Old 9th Jul 2012, 11:36 am   #14
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Default Re: Low tension help with Roberts RMB please.

Thanks guys so 0.6V of a volt drop or less is standard across any type resistor?
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Old 9th Jul 2012, 11:56 am   #15
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Default Re: Low tension help with Roberts RMB please.

Nick,

So how do you find omega?

Trevor
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Old 9th Jul 2012, 12:03 pm   #16
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Default Re: Low tension help with Roberts RMB please.

If you've got WORD, Ω is available under INSERT, SYMBOL...

Insert it into a document, then copy/paste into the forum post.

There may be a better way, but that's what I do when I can be bothered. Otherwise "R" is a lot quicker.
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Old 9th Jul 2012, 12:06 pm   #17
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Default Re: Low tension help with Roberts RMB please.

Or you can cheat and copy (control + C) it from another thread, then paste it where you want it (control + V)!
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Old 9th Jul 2012, 12:07 pm   #18
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Default Re: Low tension help with Roberts RMB please.

Quote:
Originally Posted by sobell1980 View Post
Thanks guys so 0.6V of a volt drop or less is standard across any type resistor?
No, across any type of silicon diode
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Old 9th Jul 2012, 12:08 pm   #19
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Default Re: Low tension help with Roberts RMB please.

Quote:
Thanks guys so 0.6V of a volt drop or less is standard across any type resistor?
NO. The volt drop across a SILCON diode is 0.6V.

The volt drop across a resistor depends on its value and the current flowing through it. You can calculate the volt drop using Ohms Law. V=IR. Where V is the volt drop, I the current and R the resistance.

So if 2 amps flow through a 50R resistor the volt drop is 100 Volts.
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Old 9th Jul 2012, 12:20 pm   #20
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Default Re: Low tension help with Roberts RMB please.

Not resistor, silicon diode (aka silicon rectifier). The 0.6V or so is a feature of the physics of the device material. For a Schottky diode it's about 0.4V and Germanium about 0.2V.

A rough model for a silicon diode is a perfect diode (zero reverse current and zero forward voltage) in series with a -0.6V source and a resistor.

I got 5.6R by dividing the nominal filament (heater) Voltage of 1.4V by the total filament current 250mA to get a value to use when I simulated the heater supply in a "Spice" based circuit simulation package (a nice lazy way to try out different circuits and look impressive on here!).

So, you have 2.6-0-2.6 V which is good! You need to replace the Metal Rectifier with a couple of silcon diodes (anodes to winding ends, cathodes joined together and fed to the smoothing capacitor/inductor point.

This will probably give a slightly high voltage (my simulation reckoned about 1.6V) so you need to add about 1R of series resistance somewhere. Putting it between the diode cathodes and the smoothing capacitor will probably do.

We'll all sort this out eventually between us, it's one of those things for us old radio nutters! (Terrier with rat sort of thing )
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