MURPHY A272C Restoration.

[sobell1980]

I understand it is not an error and that this is correct fitment. This is really my first FM set that I've been involved with other than a Braun radiogram on which FM was all ok. I've seen this cap fitment before on other diagrams and wondered why? So it's the FM discriminator cap. Time to dig my books out again and so some homework. I'm I roughly correct in thinking it's some way of detection / demodulation? If anyone has a simple way and the time to explain whilst I also look it up too that would be great.
Many thanks.
Dave.

[sobell1980]

I've been reading Paul Stennings repair and restoration pages on FM circuts on this. I kind of understand it. I will need to read it several times over to fully absorb it.

[sobell1980]

I've removed the ancient mains cord and someone has earthed this set. Would this put more strain on the mains transformer?
Dave.

[PJL]

Leave the chassis earthed. It will only put strain on the mains transformer if it is already faulty. If it's already faulty you probably don't want to find out by touching a live chassis!

[sobell1980]

I have done some cold checks of the mains transformer and it all checks out ok . Windings check out ok too.
So you think when I replace the mains cable to earth the chassis again? I have heard mixed views on the forum . Some say do, some say don't.
Many thanks.
Dave.

[sobell1980]

PJL, I have done a search on the forum and seen your advice on this before in other threads. I will be earthing the chassis..

[sobell1980]

Spent a few hours on the set yesterday. Replaced the dual electrolytic capacitor and fitted a temporary mains cable as I like to fit a braided cable on completion.
I removed the large speaker from the baffle board and rigged it all up on my bench so the set could be tested.
Perfection on LW and MW.
FM is not working correctly and receiving mw stations badly on it too. I've checked my voltages of the ECC85 while set on FM. V1 a anode I have 138 volts. V1 b I have 18 volts only . Removing the valve I have approx over 200 volts on each anode. Am I safe to assume the ECC85 is dead. It is getting warm and I can see it glowing. Many thanks.
Dave.

[sobell1980]

I don't have a donor ECC85 to substitute it.
Cheers,
Dave.

[ms660]

Have you checked V1B's HT feed components, L9, R4 and S1 (Trader refs.)

Lawrence.

[sobell1980]

Hi Lawrence. I like to go away a nd mull over whst you say so sorry for the late reply.
I made a school boy error. I removed the valve and tested the voltage. But because the valve was out of circuit the resistors and circuit were not underload so any matters of high resistance wouldn't show up unless it s consumers were fitted.
R3 has gone very high and reads 1.388 mega ohms. R4 is high by approximately 10 % and measures 5.16 kilo ohms. These were of course measured out of circuit. I haven't checked the switch yet as I need to get my head around that switch and what's making and breaking circuit. But finding these resistors is a positive. Prob still find the ECC85 may still be dodgy. Any thoughts gratefully received.
Dave.

[Phil G4SPZ]

Dave, I have several spare ECC85s if you need one.

[ms660]

I would check the switch contacts (S1)

For instance, if S1 was ok and R4 measured approx. 5k as you say, and the ECC85 was pulling the anode voltage down on FM by the amount you measured then R4 would be well on fire by now....

Lawrence.

[sobell1980]

Phil, many thanks for your very kind helpful offer and I will definately bare it in mind. I need to get the working voltages sorted and the switching etc first. But I will get back to you when I've got to the bottom of it. Really appreciate your offer of help.
Dave.

[sobell1980]

Hi Lawrence.
Strangely S1 is switching . Getting a reading of 0.3 ohms across the switch when FM is selected.
I need to replace these resistors and get back to you I think.
Dave.

[sobell1980]

Would 1/4 watt resistors be suitable for R3 and R4. It's going to be difficult to work out the correct current draw with the set not working on FM to work out the wattage.
R3 and R4 are not very big in physical size.
Dave.

[ms660]

Why not check the voltages first?

You've already established that the HT rail appears to be ok, you reported that when switched to FM the voltage on V1b's anode is very low but when that valve is removed the anode voltage shoots up.

So...with the valve in and the receiver switched to FM measure the voltage on both contacts of S1 and on both sides of R4 and on both sides of L9 and on V1b's anode tag on the valve socket and see what's what.

Lawrence.

[ms660]

Re: post#35:

As regards the wattage dissipated in those two resistors (R3 and R4) you don't need to know the current flowing through them, the voltage dropped across them will do...The voltages when the receiver is working as normal are given in the Trader sheet, the HT voltage is given as 200 volts, so for AM the voltage drop across R3 is the given anode voltage (30 volts) subtracted from the HT (200 volts) which results in 170 volts, square that and divide the result by the resistors value (R3) that will give the power that's dissipated in the resistor (R3), in the above I've ignored the effect of R4 as it's small.

The same procedure for FM but this time it's R4 that's of interest, looking at the voltages in the Trader sheet the voltage dropped across R4 is the HT voltage (200 volts) minus the given anode voltage when switched to FM (172 volts) which results in 28 volts, square that and divide that result by the resistors value (R4) and that will give the power dissipated in that resistor (R4)

In all the above, P diss = Vsquared/R

Lawrence.

[julie_m]

..... Which is the same as calculating the current yourself anyway, since I = V / R (Ohm's law) and P = V * I (power law). Just if you do the division by R first, then you get I as a handy side effect. (When using a slide rule, it is usually easy to alternate between multiplication and division, as opposed to doing all the multiplications top and bottom and one final division as you would with log tables or a calculator; but you can usefully pick the order to suit -- and the answer will come out the same either way ..... Showing My Age)

[ms660]

Current isn't given in the Trader sheet measurements, hence Vsquared/R to make it easy for Dave as the relevant voltages are given in the Trader sheet.

Lawrence.

[julie_m]

Yes, I see that; but I don't think it does any harm to explain exactly why the V ** 2 / R and I ** 2 * R shortcuts apply.

Ohm's law, V = I * R allows you, if you know any two out of Voltage, Current and Resistance, to determine the third one; and Power is simply the product of Voltage times Current, so equal to either V * V / R or I * I / R. Some people just don't seem to get this straight away; it looks like you are missing a step out, or something.

But the only real difference is the order of the calculation steps: whether you multiply V by itself first and then divide by R, or divide V by R first and then multiply by V. You need to evaluate either V ** 2 or V / R (which is equal to the current I) as an intermediate step anyway; but the second is more likely to be useful. Although, the first might be preferrable if the voltage term already has a square root in it.

I also hope I'm making the maths at least a little bit less scary (because it really shouldn't be; it's just simple multiplication and division, when all is said and done).