Project clock – with ILC1-1/8 (tube)

 

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Ok, the project is finished. It’s time to share all the necessary information.

The idea was to build a clock with the very large tubes ILC1-1/8. Each tube is 10 cm long. Quickly I created the first circuit: two tubes to show the hours, two tubes to show the minutes. To realise the seconds, I use a small IV25 tube (this tube has only dots, not more. I use 2 dots for the tact of seconds.)

Difficulties during the projekt work:
1) I got a wrong datasheet from the seller of the tubes. There were wrong values to current and voltage.

2) The different voltages:
After finished the project, I can say: you need 5 different voltages! The input voltage is 5V. For the Filament (ILC1-1/8) you need 3.3V. The Filament (IV25) works with 2V, not really more. The dots (IV25) and the Grid (ILC1-1/8) needs 20V. And, last but not least, the segments (ILC1-1/8) works fine with 30V … 40V (multiplex). Dont forget: 5V for the ATMEGA168/324! Smile.

5V to 3.3V – I use the TSR1-2433 chip. This chip is similar an 7803, but the efficiency is above 90%. This means, we don’t produce heat! First, I tested an 7803 and this chip were hot, really hot. We use 600 mA! The new chip is compatible to the 7803 (but not so cheap).

5V to 2V – and using 30 mA: I use only an pre resistor (2.2 KOhm).

5V to 20V – using the DC-DC converter XTW-SY-8

5V to 30V – using the DC-DC converter XTW-SY-8

 

2) The source driver:
In the past, I used the UDN2981 succesfully. Here too, but after a couple of hours I got ridiculous effects. I searched a long time for a solution (and the reason). A friend told me, I should use another chip, the LB1240. But this chip will not more produce here in europe. I compared the datasheets of both chips – they are equal. Only the values for the pre- resistor and pull down resistor are different.
And so I decided to use own pre-resistors and own pull down resistors.
With this changes, the circuit is ready.

 

Here is the circuit:

circuit

board_richtig

2 thoughts on “Project clock – with ILC1-1/8 (tube)

  1. Hi,

    There is something I really don’t understand here. You need to drop 5V to 2V using a resistor, and have 30mA current.

    The resistor needed should drop 3V at 30mA. That’s 3 / 0.03 = 100 Ohms. How do you get this 2.2kOhms value? This really puzzle me!

    • Hi Toni, let me explain …

      5V — 3V (Resistor) — 2V (Filament) — Ground
      So, I must calculate R=U/I — I = U/R — I = 5V/3V = 166 Ohm
      I wrote ” …The Filament (IV25) works with 2V, not really more…”. Thats why I used the next highe value 222 Ohm.

      Best regards
      Ralph

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