6 volt LT at 2 amps

[unrealdave]

I’ve acquired a few Atwater Kent sets and the battery models need 6 volts on the HT.  They use mainly 01A valves and draw a lot of current.  I tried a 6 volt battery but that dropped to 4 volts when I turned the sets on.

I think I better build a meaty HT supply, and I found a beefy looking toroidal transformer that has no colour codes written on it for the two sets of secondary’s and does not tell me what sort of power it gives out.  It’s an Antrim make and the 2 secondary’s are blue, red, yellow and grey – I joined up the blue with the red and then the yellow with the grey so they should be in parallel

I weighed it and it is just under half a kilo, then connected it up and the two sets of secondary windings gave 6.7 volts each.  When I joined the 2 secondary’s up it still gave 6.7 volts so should be man enough for the job?

I’ve got kbpc104 bridge rectifier (good for 3 amps) and an LM338T regulator (good for 5 amps) and a 470uf.

I had a thought though – if just one secondary could put out 6 volts 2 amps then I could wire the other up and drop the voltage for my 2 volt sets.
 
Apart from looking up other transformers VA rating and their weight then taking a rough guess is there anyone who knows the rating for the Antrim transformers type B2010 YK08, or another way of finding out the VA rating?

[Mark Hennessy]

I'd just go with the weight/dimensions. Half a kilo suggests 30VA or 50VA. If one winding gives 15VA, then at 6V, that's 2.5A. Both in parallel should be plenty...

If the weight suggested a smaller transformer, then we might need to get more precise. Just keep an eye on the temperature when it's running...

Antrim aren't going any more, sadly. They did a lot of custom stuff, so that part number could be specific to a particular customer/product. YK08 looks a bit like a Maplin part number, and I think Antrim did supply Maplin at one stage.

Will the sets in question be happy with AC? Saves a whole load of hassle and heat sinking if so. Guess you've already considered that...

Have you tested it at 2A though? I'd expect the output voltage to drop a bit. The regulation of a typical 30VA transformer is around 18-20%, and perhaps 15% for 50VA. You're only taking 12VA, so it won't drop as much as that, but even so...

I'd say that 6.7 volts off-load is not enough for a regulated supply. Let's assume the output drops by 10% to 6.1V. You'll lose nearly a volt in each diode in the bridge, so 2V in total (rounding up). So I think you'd struggle to get 6V after the rectifier. And the regulator will need 3V or thereabouts of "headroom". You need 9V at the bottom of the "troughs", so the average - as measured by a voltmeter - needs to be more like 10 or 11V.

You can half your rectifier losses by putting the windings in series to give 6-0-6, and use 2 diodes (or half your existing rectifier). And you can get that down to around 0.3V by using Schottky diodes (rob them from a dead PC PSU or similar). But you'd still need to find a LDO (low dropout) regulator IMHO.

470uF is far too small for 2A. Add another "0"

If only you had a 9V transformer...

[Keithstv]

From an old Maplin catalogue (Sept 98 - Feb 99) the YK08J transformer is 30VA, 15% regulation with two 6V 2.5A windings and was priced at £15.20.

Mark is right when he says that there may not be enough headroom for a 2V 2A regulator and you'll need at least a 4700uF capacitor, preferably more, for it to regulate correctly.

Keith

[Mark Hennessy]

And that's why I wish I'd kept the paper catalogues

[unrealdave]

Thanks for your input - I am going to have a look for a 9 volt one - must have one someplace

[Keithstv]

I only kept that one, it's the last one I ever got. It has a "New Low Price 10MB Compact Flash Memory card for the incredibly low price of £184.99. Couldn't even get one picture from my camera on that!

Keith

[Crackle]

Hi Dave
You may as well try it with your existing transformer first. All you are really after is a regulator to stop the rectified and smoothed supply from going over the 6 volts.
The rectified 6.7v when smoothed will be almost 9.5 volts, not quite enough for proper regulation  but if it drops to 5.5 volts under load it shouldn't matter too much.
You can add a switch to select one of two resistors in the circuit, to change between 6 and 2 volts.
It will regulate OK at 2 volts.
Alternatively you can put the 2 secondaries in series, you will have enough spare volts for the regulator to do its job then.
I have just built something similar but lower power. see, http://www.vintage-radio.net/forum/showt...p?t=134336 and it should also provide the HT and SG supplies

Mike

[Mark Hennessy]

I think you're forgetting the diode drops there, Mike. I assume 1V per diode, which is generous but not far off the mark. So I think the rectified DC will be nearer 6 or 7 volts in practice. But I'm of the "build it and see" school of thought, as trying to predict this stuff with any more precision is a lot of complication for relatively little reward. The biggest problem is the current taken by the rectifier+capacitor will be short "gulps", and these peak at some 4 to 5 times the average current (so 8 to 10 amps here). That's why the diode drops are higher than you might expect, and why you get surprising voltage drops in any resistance in the primary wiring (I-squared-R loss).

The downside of putting the 2 in series is that now we have about 14V DC after the rectifier. This means 16 watts of heat to lose in the regulator, which requires a surprisingly big heat sink for continuous operation.

A 9V transformer is a much better bet. After the bridge, you'd have around 10V DC, and around 8 watts of heat dissipation in the IC. When the mains is 10% low, you'd still have ~8.5V, which should be just about enough for the regulator.

9-0-9 with 2 diodes rather than a bridge will half the losses in the rectification, and is what I usually prefer at these low voltages.

Interestingly, for some applications, the rectifier losses are still too much, even when using Schottky diodes. The solution is to replace the diodes with MOS-FETs. Of course, you need to turn them on and off appropriately, so here's an IC just for that: http://www.linear.com/product/LT4320

This means that it's just the RDS(ON) of the MOS-FETs that cause losses, and typical values of that might be 0.05 ohms or less. I'm not necessarily advocating such a thing for this, but it's worth mentioning because it's quite interesting, I feel. In the drive for higher and higher efficiency, even the humble diode is on the radar!

[unrealdave]

Hi Guys

The 01A tubes in the American sets actually use 5 volts, its recommended to use 6 volts to allow for the voltage drop, I've also got some sets that use DE5's that are 5 volt.

I have had a look and I have got a 9 volt one - again no VA rating on it but it weighs 4.2 kilo so maybe 30va.  I will strap the two secondary's together again and that should do fine.

I don't get the 4700uf though as I have looked at loads of circuits on the web and they all use 470uf

I've already got a home built battery eliminator that works fine for LT, HT and GB - this is just a simple 6 volt LT only for my American battery sets - I've got about 8.  They knock spots off the UK ones of a similar period.

I have included a circuit that I was going to follow but use my bridge and regulator as they take more amps.  Is the increased amps why I need 4700uf?

[pwdrive]

The higher the load current the larger the reservoir needs to be in order to reduce the ripple voltage.

Lawrence.