Pioneer SX 434 dial pointer lamp replacement - UK Vintage Radio Repair and Restoration Discussion Forum

spaceford · 28th Mar 2017, 9:21 pm

I need to replace the grain of wheat bulb in the dial pointer in a Pioneer SX 434.

It should be 8v 50ma but in the UK, I can only find 12v 80ma.

Would this work or would there be issues?

Grateful for any advice.

David G4EBT · 28th Mar 2017, 10:34 pm

I think they'd be a bit dim. You'd probably be better using a 6V 'GOW' bulb with a series resistor to drop the excess 2V - assuming you have space to fit a low wattage resistor in line to the bulb. The bulbs at the link below (from a UK supplier) are 6V @ 65mA, so to drop 2V that would be 2V/.065mA = 30.7R. Nearest preferred value resistor is 33R. (0.5W would be fine: 2V @ 65mA is 0.13W).

http://www.ebay.co.uk/itm/5x-Clear-W...-/121861801568

Hope that might help.

Radio Wrangler · 28th Mar 2017, 11:24 pm

How about a tiddy LED of the colour of your choice with a resistor. That way you can keep the current down and twiddle the brightness to taste without going over the original current.

David

spaceford · 29th Mar 2017, 9:05 am

Thank you both.
I'm a bit of a noob with this though. If I used a 33r resistor or (I'm guessing) a variable resistor for an LED, where does it actually go? Do you solder the ends of the bulb to it and kind of bridge the wire?

Also, if a 0.5w 33r resistor is ok, would 0.25w be better?

TonyDuell · 29th Mar 2017, 6:51 pm

No, the resistor goes in series with the bulb (or LED). You connect one of the bulb wires to wherever one of the original bulb wires was connected. Then connect the other bulb wire to one end of the resistor. The other end of the resistor goes to wherever the other original bulb wire went.

An LED is polarised. You have to wire it the right way round or it won't work at all. And you must have a resistor in series with it or it will be ruined. I would start with, say, a 560Ohm resistor and see what sort of light intensity you get. A light bulb (of course) can be connected up either way round. As can the resistor (in either case).

[To the more experienced people, do we know that the lamp supply in this unit is DC, and not AC straight from a mains transformer? The latter will complicate using an LED. I don't have the service manual]

The power rating (watts) for a resistor is the maximum power it can disipate without damage. In this case of the 6V bulb, David has shown the power disipated in the resistor is 0.13W. So a 0.25W resistor would be OK (0.25>0.13). So would any higher wattage resistor, provided there is space for it. In general it is better to use (sensibly) over-rated components, I would use a 0.5W resistor here simply because they are common and not too large.

ms660 · 29th Mar 2017, 6:58 pm

According to the schematic the bulb supply is AC from a separate winding if that's any use.

Lawrence.

TonyDuell · 29th Mar 2017, 7:01 pm

If the supply is AC, then it is more complicated to use an LED, and I think as Spaceford says he has little knowledge it is going to be easer to use the 6V bulb and 33 Ohm resistor.

spaceford · 29th Mar 2017, 7:22 pm

Thanks for all the advice from everyone. I will go with the 6v bulb and 0.5w resistor. LED sounds tricky for now!

Radio Wrangler · 30th Mar 2017, 12:18 am

The LED is still a diode, and a 6v winding should not over-voltage it in the reverse direction. One will probably work, and with 560 Ohms in series, it's not going to damage the transformer. An opportunity for an experiment.

If you want to do the full job, then you can use two LEDs (one wired across the other, but in the opposite direction) Just use one 560 Ohm resistor and both LEDs will light.

David

David

TonyDuell · 30th Mar 2017, 5:10 am

Spaceford, this will not mean much to you, so ignore it and use the bulb. But I feel I have to comment...

It's not a 6V transformer winding, it's 8V or thereabouts. Amd that is the RMS voltage, so the peak voltage is a little over 11V. Since the LED passes no current in reverse, the series resistor drops no voltage so there will be an 11V peak reverse voltage across the LED.

I have just looked up a white LED on the Farnell site. The data sheet says that the Absolute Maximum Rating for reverse voltage is 5V.

So connecting such an LED + resistor to the 8V transformer winding will exceed the maximum reverse voltage for the LED. It might work. It might destroy the LED. It might work for a bit and fail. I agree it will not damage the transformer.

If you want to use an LED then either connect a pair in inverse parallel (as you suggest) or put a diode (a 1N4148 is fine) in inverse parallel to the LED.

I feel that given Spaceford's lack of experience here, a bulb is going to be simpler. And there is no reason not to use one.

28th Mar 2017, 9:21 pm	#1
spaceford Retired Dormant Member Join Date: Jun 2012 Location: Bristol, UK. Posts: 9	Pioneer SX 434 dial pointer lamp replacement I need to replace the grain of wheat bulb in the dial pointer in a Pioneer SX 434. It should be 8v 50ma but in the UK, I can only find 12v 80ma. Would this work or would there be issues? Grateful for any advice.

28th Mar 2017, 10:34 pm	#2
David G4EBT Dekatron Join Date: Feb 2003 Location: Cottingham, East Yorkshire, UK. Posts: 5,766	Re: Pioneer SX 434 dial pointer lamp replacement I think they'd be a bit dim. You'd probably be better using a 6V 'GOW' bulb with a series resistor to drop the excess 2V - assuming you have space to fit a low wattage resistor in line to the bulb. The bulbs at the link below (from a UK supplier) are 6V @ 65mA, so to drop 2V that would be 2V/.065mA = 30.7R. Nearest preferred value resistor is 33R. (0.5W would be fine: 2V @ 65mA is 0.13W). http://www.ebay.co.uk/itm/5x-Clear-W...-/121861801568 Hope that might help. __________________ David. BVWS Member. G-QRP Club member 1339.

28th Mar 2017, 11:24 pm	#3
Radio Wrangler Moderator Join Date: Mar 2012 Location: Fife, Scotland, UK. Posts: 22,902	Re: Pioneer SX 434 dial pointer lamp replacement How about a tiddy LED of the colour of your choice with a resistor. That way you can keep the current down and twiddle the brightness to taste without going over the original current. David __________________ Can't afford the volcanic island yet, but the plans for my monorail and the goons' uniforms are done

29th Mar 2017, 9:05 am	#4
spaceford Retired Dormant Member Join Date: Jun 2012 Location: Bristol, UK. Posts: 9	Re: Pioneer SX 434 dial pointer lamp replacement Thank you both. I'm a bit of a noob with this though. If I used a 33r resistor or (I'm guessing) a variable resistor for an LED, where does it actually go? Do you solder the ends of the bulb to it and kind of bridge the wire? Also, if a 0.5w 33r resistor is ok, would 0.25w be better?

29th Mar 2017, 6:51 pm	#5
TonyDuell Dekatron Join Date: Jun 2015 Location: Biggin Hill, London, UK. Posts: 5,222	Re: Pioneer SX 434 dial pointer lamp replacement No, the resistor goes in series with the bulb (or LED). You connect one of the bulb wires to wherever one of the original bulb wires was connected. Then connect the other bulb wire to one end of the resistor. The other end of the resistor goes to wherever the other original bulb wire went. An LED is polarised. You have to wire it the right way round or it won't work at all. And you must have a resistor in series with it or it will be ruined. I would start with, say, a 560Ohm resistor and see what sort of light intensity you get. A light bulb (of course) can be connected up either way round. As can the resistor (in either case). [To the more experienced people, do we know that the lamp supply in this unit is DC, and not AC straight from a mains transformer? The latter will complicate using an LED. I don't have the service manual] The power rating (watts) for a resistor is the maximum power it can disipate without damage. In this case of the 6V bulb, David has shown the power disipated in the resistor is 0.13W. So a 0.25W resistor would be OK (0.25>0.13). So would any higher wattage resistor, provided there is space for it. In general it is better to use (sensibly) over-rated components, I would use a 0.5W resistor here simply because they are common and not too large.

29th Mar 2017, 6:58 pm	#6
ms660 Dekatron Join Date: Apr 2011 Location: Cornwall, UK. Posts: 13,454	Re: Pioneer SX 434 dial pointer lamp replacement According to the schematic the bulb supply is AC from a separate winding if that's any use. Lawrence.

29th Mar 2017, 7:01 pm	#7
TonyDuell Dekatron Join Date: Jun 2015 Location: Biggin Hill, London, UK. Posts: 5,222	Re: Pioneer SX 434 dial pointer lamp replacement If the supply is AC, then it is more complicated to use an LED, and I think as Spaceford says he has little knowledge it is going to be easer to use the 6V bulb and 33 Ohm resistor.

29th Mar 2017, 7:22 pm	#8
spaceford Retired Dormant Member Join Date: Jun 2012 Location: Bristol, UK. Posts: 9	Re: Pioneer SX 434 dial pointer lamp replacement Thanks for all the advice from everyone. I will go with the 6v bulb and 0.5w resistor. LED sounds tricky for now!

30th Mar 2017, 12:18 am	#9
Radio Wrangler Moderator Join Date: Mar 2012 Location: Fife, Scotland, UK. Posts: 22,902	Re: Pioneer SX 434 dial pointer lamp replacement The LED is still a diode, and a 6v winding should not over-voltage it in the reverse direction. One will probably work, and with 560 Ohms in series, it's not going to damage the transformer. An opportunity for an experiment. If you want to do the full job, then you can use two LEDs (one wired across the other, but in the opposite direction) Just use one 560 Ohm resistor and both LEDs will light. David David __________________ Can't afford the volcanic island yet, but the plans for my monorail and the goons' uniforms are done

30th Mar 2017, 5:10 am	#10
TonyDuell Dekatron Join Date: Jun 2015 Location: Biggin Hill, London, UK. Posts: 5,222	Re: Pioneer SX 434 dial pointer lamp replacement Spaceford, this will not mean much to you, so ignore it and use the bulb. But I feel I have to comment... It's not a 6V transformer winding, it's 8V or thereabouts. Amd that is the RMS voltage, so the peak voltage is a little over 11V. Since the LED passes no current in reverse, the series resistor drops no voltage so there will be an 11V peak reverse voltage across the LED. I have just looked up a white LED on the Farnell site. The data sheet says that the Absolute Maximum Rating for reverse voltage is 5V. So connecting such an LED + resistor to the 8V transformer winding will exceed the maximum reverse voltage for the LED. It might work. It might destroy the LED. It might work for a bit and fail. I agree it will not damage the transformer. If you want to use an LED then either connect a pair in inverse parallel (as you suggest) or put a diode (a 1N4148 is fine) in inverse parallel to the LED. I feel that given Spaceford's lack of experience here, a bulb is going to be simpler. And there is no reason not to use one.