## Using a 12A (112A)

### Using a 12A (112A)

I now have a couple of 12A to play with,having done a swap with Will.

There is a little idea in my bonce which I want to try but not sure of the method.

I have an anode HT of 150V (after a 6K8 resistor) and I need to work out a cathode resistor.

The grid volts need to be about -10 and the current about 7ma so how does one get the cathode volts ?

Is it then cathode volts @ 7ma to work out cathode resistor ? (using Mr Ohms law)

Also, would it need a grid stopper, and what value grid leak ?

Any ideas or suggestions welcome.

There is a little idea in my bonce which I want to try but not sure of the method.

I have an anode HT of 150V (after a 6K8 resistor) and I need to work out a cathode resistor.

The grid volts need to be about -10 and the current about 7ma so how does one get the cathode volts ?

Is it then cathode volts @ 7ma to work out cathode resistor ? (using Mr Ohms law)

Also, would it need a grid stopper, and what value grid leak ?

Any ideas or suggestions welcome.

Life is what happens to you while you're busy making other plans. John Lennon

G-Popz THE easy listening connoisseur. (Philip)

G-Popz THE easy listening connoisseur. (Philip)

- cressy
- Shed dweller
**Posts:**2909**Joined:**Fri Aug 03, 2007 7:07 pm**Location:**the great white space-
**Contact:**

i think this is right, please correct me if i get it wrong, get the curves, draw a straight diagonal line between 150v on the bottom line, and 7ma on the vertical line, pick a point on that line where the curves are about the same width apart so its at its reasonably linear. read to the top of the curve for the grid volts and there will be a voltage figure (eg -6v) the calculate v divided by ma.

for example, when i did the calcs for the 6h6n i got -6v at the top of the curve and had picked 15ma on the vertical line so it gave 6 divided by 15 so 0.4 this is 0.4k, so gave a 400r cathode resistor

i think this is right, so as i say please correct if im wrong.

using your figures phil i get 10 divided by 7 so 1.42 so 1k4 if you can get those figs off the curves

for example, when i did the calcs for the 6h6n i got -6v at the top of the curve and had picked 15ma on the vertical line so it gave 6 divided by 15 so 0.4 this is 0.4k, so gave a 400r cathode resistor

i think this is right, so as i say please correct if im wrong.

using your figures phil i get 10 divided by 7 so 1.42 so 1k4 if you can get those figs off the curves

Thanks Ant !

I can make up 1.57K from the "box" so that will give me 6.3ma which will get me going.

Now to do the cutting,drilling,filing and bleeding to get the buggers fitted and working.

Soldering iron is now plugged in

I can make up 1.57K from the "box" so that will give me 6.3ma which will get me going.

Now to do the cutting,drilling,filing and bleeding to get the buggers fitted and working.

Soldering iron is now plugged in

Life is what happens to you while you're busy making other plans. John Lennon

G-Popz THE easy listening connoisseur. (Philip)

G-Popz THE easy listening connoisseur. (Philip)

Yes-I read through that again this morning,and my head went "funny" after a couple of pages. Some of it went in though.Nick wrote:It was all in the long thread Dave worked through.

Life is what happens to you while you're busy making other plans. John Lennon

G-Popz THE easy listening connoisseur. (Philip)

G-Popz THE easy listening connoisseur. (Philip)

- Mike H
- Amstrad Tower of Power
**Posts:**18273**Joined:**Sat Oct 04, 2008 5:38 pm**Location:**The Fens-
**Contact:**

Yes I concur. Usually the starting point.cressy wrote:i think this is right, please correct me if i get it wrong, get the curves, draw a straight diagonal line between 150v on the bottom line, <snip>

If the chosen V is the B+, then the anode resistor value will be that Voltage figure divided by the left hand current figure. In other words you have to imagine your line is the resistor stretched between the V anf the mA points.

If you then manage to get the valve biased exactly halfway then it's Va and Ia will be those figures divided by 2, respectively. In other words read off the grid bias from exactly halfway along the line.

As an aside, if you can keep the Ia up, then as a general rule mu is increased and anode resistance decreased, thereby increasing gain. The limit being what's the minimum cathode bias Volts that can be tolerated. At which point I start looking for more B+ Volts, if it's an issue.

.

*I never make the same mistake twice. Instead I prefer to fill my day with a variety of different mistakes*

I might resurrect the idea of a 112A preamp.

I found an interesting variation on filament supplies from Rod Coleman on diyaudio.

http://www.diyaudio.com/forums/attachme ... _250ma.png

http://www.diyaudio.com/forums/tubes-va ... ost2683766

I found an interesting variation on filament supplies from Rod Coleman on diyaudio.

http://www.diyaudio.com/forums/attachme ... _250ma.png

http://www.diyaudio.com/forums/tubes-va ... ost2683766

G-Popz THE easy listening connoisseur. (Philip)

- Paul Barker
- No idea why I do this anymore
**Posts:**6964**Joined:**Mon May 21, 2007 9:42 pm

Having looked at flow diagram your above discussion Abbott cathode r is no longer appropriate because Rod is passing current round his reg to reduce size of cathode resistor. Follow his flow chart and buy his kit.

I used 4 112a's two stages with two channels and one common cathode resistor for all four valves. AC heating and hum dingers.

I used 4 112a's two stages with two channels and one common cathode resistor for all four valves. AC heating and hum dingers.

I can do it right or I can do it now, but I can't do it right now.