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PostPosted: 04 May 2018, 11:33 
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Joined: 01 Apr 2017, 19:56
Posts: 49
You have added a HUGE load (1010 R) to your 215Vdc supply. If the power supply could withstand the load, it would drain 213 mA, or 45 W ! Since it can't cope with that, the voltage sags to 50 V. In this case, the resistors dissipate 49 mA (2,47 W).

If your 215 Vdc rail is regulated by a LR8K4, it possibly went into shutdown protection mode. But I would test it to see if all is well anyway.

One way to indicate HT on would be using capacitive power supply, altough it only works on AC. You can google it for better information. I have found the following page interesting: http://www.marcspages.co.uk/tech/6103.htm.

Another way would be to use a much larger resistor in series with the LED. If it is a normal LED (not ultra bright or other variations) a 68K ohm resistor would do.


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PostPosted: 06 May 2018, 08:04 
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Joined: 02 Aug 2016, 05:26
Posts: 21
Location: France
victorzwk,

Thank you for your reply. I'm confused by the idea of adding a 68k resistor when 1010R apparently caused a problem by adding a substantial load - won't 68k be worse?
In case I wasn't clear at the outset, here's a diagram that might help.

1) Was before I tried the led indicator - all fine!
2) was when I tried a voltage divider - 10R and 1k - led worked fine but voltage dropped due to overload
3) Just to check - is this what you were suggesting?

Many thanks


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PostPosted: 06 May 2018, 12:13 
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Joined: 24 Oct 2010, 07:05
Posts: 295
Interesting opinion on divider's usage. :-) I mean what you attempted to achieve is not related to what you've done.
The third option is definitely the way to go. I have a note though: if you use LM8 to regulate to the 215V you'd better put the R+LED string before it. LM8 has nominal current on the output in the range of 10mA, so it is not nesessary to add to the existing load 3-4mA just for the LED. The 68 kOhm value shoul be reconsidered in this case, but it is simple: considering 3mA LED current, the resistance could be counted using Ohm's law: R=U/3, where R in kOhm, U - in Volts - the voltage before LR8.


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PostPosted: 06 May 2018, 12:39 
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Joined: 02 Aug 2016, 05:26
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Location: France
Quote:
The third option is definitely the way to go.

:up: Thanks to all - I'll give this a go over the next day or so...


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PostPosted: 06 May 2018, 18:06 
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Joined: 03 Feb 2018, 00:47
Posts: 41
Location: Brisbane, Australia
Just to explain a little further, the way electricity behaves in a circuit as we know it can be worked out by using "Ohms law".

https://en.wikipedia.org/wiki/Ohm%27s_law

As Poty said, R-V/I (Resistance equals voltage divide by current). You can rearrange the equation to be I=V/R (current equals voltage divided by resistance) so you can hopefully understand that the higher the resistance the lower the current (thus 68K in series with a circuit will lower the current compared to 1K). A typical LED has 2-3 volts (fixed) across it, and you design the series resistor to set the desired current. 2 volts is insignificant in terms of a 215 volt total, thus Poty ignored it in his calculation. If the primary voltage before the regulator is say 415 volts, substitute that into the equation to give you a resistor value.

Note that the use of V or U for voltage varies between countries. :)


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PostPosted: 07 May 2018, 12:47 
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Joined: 01 Apr 2017, 19:56
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As poty already answered, nr. 3 is the way to go.

But just to clarify from your question, 68k or 68.000 ohms is a much larger resistance (therefore lighter load) than your first aproach of 1.010 ohms.

And as it was suggested, it is better to tap the voltage for the LED before the LR8 regulator and calculate the resistance accordingly. If you choose to tap it from the 480 Vdc source, a 160k resistor would do. But it would need to be of larger dimension to avoid the voltage jumping across it. So use at least a 3W one. The real power dissipation would be about 1,4 Watt.

Best regards.

Emsworth wrote:
victorzwk,

Thank you for your reply. I'm confused by the idea of adding a 68k resistor when 1010R apparently caused a problem by adding a substantial load - won't 68k be worse?
In case I wasn't clear at the outset, here's a diagram that might help.

1) Was before I tried the led indicator - all fine!
2) was when I tried a voltage divider - 10R and 1k - led worked fine but voltage dropped due to overload
3) Just to check - is this what you were suggesting?

Many thanks


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PostPosted: 07 May 2018, 13:02 
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Joined: 01 Apr 2017, 19:56
Posts: 49
There are also other completely different ways to do it.

If you use a delay circuit to turn on the HT, you could actually power the LED in parallel with the Relay coil.
For a 12Vdc coil, just add a 3k resistor in series with the LED.

If you use a manual switch to turn on the HT, just change this switch for one with an extra pair of contacts and power the LED from there.

Both this ways avoid you powering a 3V device with high voltage wasting power and potentially creating problems of higher periculosity.

Regards.


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PostPosted: 04 Jun 2018, 05:52 
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Joined: 03 Feb 2018, 00:47
Posts: 41
Location: Brisbane, Australia
Peer review!!!!
Attachment:
Oddwatt_top.jpg

Attachment:
Oddwatt_bottom.jpg


Way to go yet, but so far so good.

Any obvious issues with layout? 6SL7's will have regulated DC (7806) heaters and LR8 regulated B+. EL34's 12V AC each pair in series. The centre octal socket is wired as a 12N30, but I will fit a relay to a blank octal base with 555 or baby processor operating it. Main filter caps are 100 + 100, secondary (feed to regulator) is 47 + 47. 4700uF cap in the center is fed from bridge rectified 12 v heater to a pair of 7806 regs. 10uF Poly's bottom left and right off main B+ to transformer CT's. Copper bus for signal ground. Blank board top left is turret board for the regulators and remaining components, mirror image on the right. Signal inputs will be centre front, power switch left and signal select on the right.
Enjoy,
Glenn.


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PostPosted: 04 Jun 2018, 12:41 
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Joined: 29 Jun 2016, 15:59
Posts: 66
How much current is there left to play with?

Hi,
I am almost done building my first OddBlock and I am thinking about putting some LEDs to spruce-up the project.
For each mono block I am using a 5751, two KT88s and the XPWR264 Power Transformer (Edcor Power transformer for a 120V, 60Hz. or 240V, 50/60Hz. line to 360V (180-0-180) at 350mA center tapped and 10.9V at 2A). February 22, 2012 schematics.

Obviously the number of LEDs that I can install depends on the available leftover current.
I believe some time ago I read somewhere that this design uses about 1.775 Amps and I would have about 225 mA left.
Is this correct? If so, can I use the whole 225 mA or should I leave some amperage for safety?
Any help is appreciated.
Thanks,


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PostPosted: 04 Jun 2018, 15:57 
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Joined: 08 Aug 2009, 03:11
Posts: 2229
Location: Chilliwack, BC
If you use superbright LED's, you can get the same intensity level at 600uA or so compared to a regular one at 20mA.

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* Ratings are for transistors - tubes have guidelines*
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