diyAudio (http://www.diyaudio.com/forums/index.php)
-   Tubes / Valves (http://www.diyaudio.com/forums/tubes-valves/)
-   -   Voltage divider to reduce 6v dc to 2vdc (http://www.diyaudio.com/forums/tubes-valves/118205-voltage-divider-reduce-6v-dc-2vdc.html)

 duderduderini 26th February 2008 09:58 AM

Voltage divider to reduce 6v dc to 2vdc

Hi Guys
I want to try a #30 triode in my preamp set up and have 6V available for the heaters. Is it feasible to use a divider to tap off at 2Vfor its heater at 65mA.?
What values of resistor should i use? In a floating heater supply which draws no current its resistances in the 100's of kOhm range but I thought I should seek advice re this.
If we are dropping 4 volts at .065amp then thats 0.26 watts so I should use 2 watt resistors.
Feel free to jump right in and point out what I have overlooked.
Other than Batteries, I just cant get a hold of a small enough tranny . At this point it is an experiment so if its succesful I will build a purpose specific regulator board for this
Thanks

 Nordic 26th February 2008 10:16 AM

Just some basics.. not going to comment on if it is usefull to your project....

When you have two resistors in series, regardless of size the same current will flow through each one...

when you have resistances in series, the voltage will be divided in the same ratio as the resistances....

for example if you have a 4ohm and 2ohm in series and apply 6 volt over them, the 4 ohm will have a voltage of 4ohms over it, and the 2 ohms will have a voltage of 2V...

To prove the first observation 4V/4R = 1A and 2V/2R=1A

So now you know the ratios of the two resistor, and just need to limit the current to as much as you need....

I do not know the impendance of the thing you are trying to drive but I will assume it is powered by 2V and 0.065A, so useing ohms law I would say it is 2/0.065=30.76R

Anyway you have two things in series and you want one to be 2V @ 65mA, which means the other has to be 4V @ 65mA...

4/0.065=61.5R

Dissipation over this resistor is 4V x 65mA = 260mW, Half watt is enough.

PS.. look at PCB mount transformers for little matchbox footprint ones...

 duderduderini 26th February 2008 10:24 AM

Thanks

Hi Nordic
I am using a number 30 triode of which I have a few. I figured the tube filament would only draw what it "requires" . Wont having such low resistances going to ground be a potential for problems?
Nick

 Nordic 26th February 2008 10:58 AM

Not sure why you'd think it is low when considering the voltage...

If you have a six volt line and put the 62R resistor and fillament in series, it will dissipate it's ratio of the total voltage as heat (that .26W)... (or 4V at 0.065A)

I have not heard of 2V heater grids though...

 ErikdeBest 26th February 2008 11:29 AM

As Nordic has already given the answer as to which resistor to use, I will only give another suggestion.

The 6 is VDC, I assume. Use a LM317 set as CCS to feed the valve. The resistor value should be something as 18R (1.2V/0.065A), but you can also use a 22R in parallel with some other resistor, AOT (adjusted on test). Some people prefer the sound of DHT's fed from CCS.

 duderduderini 26th February 2008 09:28 PM

Heres the data sheet

http://www.mif.pg.gda.pl/homepages/f...s/021/3/30.pdf
This shows you the data sheet and heater requirements for the tube.
I plan on using the ixys 10M45 for the ccs for this dht.
So in effect the 2 volts is for the heaters.
If i like the sound, I will build a dedicated 2V heater supply but for now i want to just use a v divider for it.
Thanks
Nick

 All times are GMT. The time now is 05:52 PM.

Search Engine Optimisation provided by DragonByte SEO (Pro) - vBulletin Mods & Addons Copyright © 2018 DragonByte Technologies Ltd.