Choke loading Anodes.

We all start somewhere
Post Reply
User avatar
Nick
Site Admin
Posts: 15707
Joined: Sun May 06, 2007 10:20 am
Location: West Yorkshire

#76

Post by Nick »

Yes, thats more like I would expect, a lower gain with the smaller load. Though to be precise you are not measuring mu, just the gain of the amplifier with this load and this op point. The mu of the valve is still the same as it ever was. (same as it ever was).
Whenever an honest man discovers that he's mistaken, he will either cease to be mistaken or he will cease to be honest.
User avatar
Dave the bass
Amstrad Tower of Power
Posts: 12273
Joined: Tue May 22, 2007 4:36 pm
Location: NW Kent, Darn Sarf innit.

#77

Post by Dave the bass »

Nick wrote:Yes, thats more like I would expect, a lower gain with the smaller load. Though to be precise you are not measuring mu, just the gain of the amplifier with this load and this op point. The mu of the valve is still the same as it ever was. (same as it ever was).
ISWYDT :lol:

Ah re:mu.

So the 20K loadline/lower gain scenario shows we're not at a very efficient operating point then perhaps.

The 100K load offers a higher gain.

DTB
"The fat bourgeois and his doppelganger"
User avatar
Nick
Site Admin
Posts: 15707
Joined: Sun May 06, 2007 10:20 am
Location: West Yorkshire

#78

Post by Nick »

Well, as I tried to say before, the two points have differing features, one has more gain, but less range, one will have more distortion, but more swing. Depends on what we need, the gain is also different as you say, but thats possibly less of a reason to decide between them. Of cource, the op point could be chosen any way along the 100k load line to provide greator voltage swing. Say 185v, 1.5ma. But if you draw a tangent to the grid line at that point you will see you are getting to the point where ra starts increasing.

At the end of the day, there are some thing best found by ears I think, and this is one of them. You can pick either load, or go for one between then (say 47k). Chose one, and we will get the rest of the amp stage built, then you can try it out.
Whenever an honest man discovers that he's mistaken, he will either cease to be mistaken or he will cease to be honest.
User avatar
Dave the bass
Amstrad Tower of Power
Posts: 12273
Joined: Tue May 22, 2007 4:36 pm
Location: NW Kent, Darn Sarf innit.

#79

Post by Dave the bass »

Ok, would you mind if I pick a new inbetweeny point? I'm leaning towards experimentation time ... the other 2 points are 'known' OP's. I'm going to go where no man has gone before (probably a good reason for it too!),

160V at 4mA. Right where the -8V grid line crosses.

Shall I calculate the value of Ra? HT will still be 350V.

DTB
"The fat bourgeois and his doppelganger"
User avatar
Nick
Site Admin
Posts: 15707
Joined: Sun May 06, 2007 10:20 am
Location: West Yorkshire

#80

Post by Nick »

Yep, and get the gain as well.
Whenever an honest man discovers that he's mistaken, he will either cease to be mistaken or he will cease to be honest.
User avatar
Dave the bass
Amstrad Tower of Power
Posts: 12273
Joined: Tue May 22, 2007 4:36 pm
Location: NW Kent, Darn Sarf innit.

#81

Post by Dave the bass »

Nick wrote:Yep, and get the gain as well.
<salutes>

Yes Sah, Sgt. Sah!

<turns to left and quick marches to desk with calculator and coloured pens on>

DTB
"The fat bourgeois and his doppelganger"
User avatar
Dave the bass
Amstrad Tower of Power
Posts: 12273
Joined: Tue May 22, 2007 4:36 pm
Location: NW Kent, Darn Sarf innit.

#82

Post by Dave the bass »

Sah! <salutes>

For a 160V @4mA OP...

Ra will be 47.5K (ha, uncanny! what was a poss suggestion a few posts back!!??)

Gain about 11.25. (for 8V Pk-Pk the o/p looks to be about 90V to my eyes).

Haven't got a scanner here at work :-(

Image

Next?

DTB
"The fat bourgeois and his doppelganger"
User avatar
Nick
Site Admin
Posts: 15707
Joined: Sun May 06, 2007 10:20 am
Location: West Yorkshire

#83

Post by Nick »

Ok, so we vave 350b+, a 47k load resistor, 4ma through the valve 160v on the anode amd to do that we need -8v on the grid.

So assume we are going to use cathode bias. We need to insert a resistor under the cathode that raises it up by 8v, so when the grid is referenced to 0v, then compared to the cathode the grid will be at -8v.

So you need to work out what size resistor to put under the cathode, so that when 4ma passes through it creates a 8v voltage...
Whenever an honest man discovers that he's mistaken, he will either cease to be mistaken or he will cease to be honest.
User avatar
Dave the bass
Amstrad Tower of Power
Posts: 12273
Joined: Tue May 22, 2007 4:36 pm
Location: NW Kent, Darn Sarf innit.

#84

Post by Dave the bass »

Ohms law is our friend again yes. V/I=R

8V/0.004A= 2000R = 2K ohm? 1/2Watt resistor would be OK there I think.

DTB
"The fat bourgeois and his doppelganger"
User avatar
Nick
Site Admin
Posts: 15707
Joined: Sun May 06, 2007 10:20 am
Location: West Yorkshire

#85

Post by Nick »

Yep.

Now you need to consider the cathode bipass cap. Without it the gain will be lower, and the ra will be higher.

Why the gain will be lower can be understood by considering the valve at idle, then a signal coming on.

Lets assume the signal is +1v, this will move the op point 1v along the 47k load line. So we will expect the anode voltage to change from 160v to (say) 150v (so thats a gain of 10, 1v in 10v out), and the current to increase from 4ma to 4.2ma. But, as the current is now 4.2ma, the voltage across the cathode resistor has changed from 8v, to 8.4v. So the change of voltage the grid sees for the 1v in is not 1v, its 1v - 0.4v as the cathode moves up at the same time. So the voltage on the anode doesn't change by 10v, but by (say) 6v. Of course that means the voltage change on the cathode is no 0.4v as the change in current is less than 0.2ma. But you can see the process, because the voltage on the cathode changes at the same time as the voltage on the anode, then there is feedback from the cathode resistor that reduces the output voltage.

So, to prevent that happening we want to avoid any of the signal from changing the voltage on the cathode. We can do that by using a component that has a fixed voltage drop so the change in current has no effect. Examples of what we could use here are forward biased diodes or LED's. Or batteries. But we want to us ea resistor in this position, so we prevent the signal from changing the voltage across the resistor by placing a capacitor across the resistor so any signal current will pass through the capacitor and not the cathode resistor.

Now we have to decide what size a cap we need. The capacitor is being driven from a source that has a resistance that is the resistor in the cathode in parallel with the dynamic cathode resistance of the valve. In the same was as the anode has a internal dynamic resistance, so does the cathode. And if I remember correctly the cathode resistance is given by the anode resistance (ra) in parallel with the load resistance Rl and all that divided by the mu of the valve.

So next step is work out the anode resistance from the chart at your op point, and then we can work out the cathode resistance. And from that the resistance the cap is looking into.
Whenever an honest man discovers that he's mistaken, he will either cease to be mistaken or he will cease to be honest.
User avatar
Dave the bass
Amstrad Tower of Power
Posts: 12273
Joined: Tue May 22, 2007 4:36 pm
Location: NW Kent, Darn Sarf innit.

#86

Post by Dave the bass »

Nick wrote: So next step is work out the anode resistance from the chart at your op point, and then we can work out the cathode resistance. And from that the resistance the cap is looking into.
Zoinks! that last post :shock: thats a lot to take in, I'll try. the idea of using a cap to give the signal a low impedence path to ground is the same as that 10uF cap you told me to fit when I choke loaded the 1st stage of 6EM7 yes? I

Re finding ra at my new custom OP...Right, we're gonna go through this step by step, so to find my ra at the OP I'm back to tangents yes?

DTB
"The fat bourgeois and his doppelganger"
User avatar
Nick
Site Admin
Posts: 15707
Joined: Sun May 06, 2007 10:20 am
Location: West Yorkshire

#87

Post by Nick »

Yep, tangent of the grid curve again, find its slope.

Sorry about the burst. I couldn't think of a simpler way of how to get the cathode resistance, and to ignore it and just use the cathode resistor would have been wrong.

BTW, I got the cathode resistance a bit wrong, its ( ra + Rl ) / ( 1 + µ )

Imagine it as the resistance at the top of the valve reflected to the bottom of the valve, so reduced by the factor of 1 plus the gain.

Sorry this bit is a bit more complex, but you will get there, you are starting to find what the number are (gm, µ, ra, Rl) once you recognize them it will make the formula start to look less daunting.
Whenever an honest man discovers that he's mistaken, he will either cease to be mistaken or he will cease to be honest.
User avatar
Dave the bass
Amstrad Tower of Power
Posts: 12273
Joined: Tue May 22, 2007 4:36 pm
Location: NW Kent, Darn Sarf innit.

#88

Post by Dave the bass »

Ok, now I'm treading gingerly here.

I've been plotting the tangent on my 76 curves, its the purple line. The green x and y axis are the points I take my data from.

Tangent (purple line)cuts the Anode volt line at approx 125V, I've extended it up high and picked a convenient point where the tangent crosses the 220V vertical anode voltage line. Aty that same point I've scored across my horizontal line which cuts the anode current at approx 9.75mA

So, 220V-125V = 95V.

Therefore, my ra at DTB OP is found by R= V/I = 95/0.00975 = 9743Ohms.

Errrr.... I think!

Image
"The fat bourgeois and his doppelganger"
Andrew
Eternally single
Posts: 4206
Joined: Thu May 24, 2007 2:18 pm

#89

Post by Andrew »

Bruce Rozenblit's book has all the equations in Dave.

-- Andrew
User avatar
Nick
Site Admin
Posts: 15707
Joined: Sun May 06, 2007 10:20 am
Location: West Yorkshire

#90

Post by Nick »

Yep, that looks about right Dave. So now you have

ra = 9743
µ = 13.8 (from the sheet)
Rl = 47k

So we can calculate

rk = ( ra + Rl ) / ( 1 + µ )
Whenever an honest man discovers that he's mistaken, he will either cease to be mistaken or he will cease to be honest.
Post Reply