Help understanding a few Ohm's law calculations

[jmillerdoc]

I am trying to reverse engineer a amp schematic in order to "steal" the preamp/driver section from Lessard's 2A3 amp here:

An externally hosted image should be here but it was not working when we last tested it.

I figure the current flowing through the first section of the 5687 is (2v/90R) = 0.022, or 22mA. I also figure the current flowing through the second 5687 tube section is (8v/470R) = .017, or 17mA.

Now when I try to reverse engineer the C+ taking the voltage drop through the plate resistors on both sections I come up with different C+ values.

This is how I did it and I am probably wrong, please understand I am very new to this and am just now trying to step out and figure these things out and apply things like Ohm's law, etc...

V1-V2 = drop so, V1-V2=IR where V1 = C+ and therefore C+ - V2=IR

C+ -70V=0.022*13K makes C+= 286v

but on the second section

C+ -130V= 0.017*10000 makes C+= 300

C+ cannot not equal C+

Where did I go wrong???? Am I an idiot and I just still don't get it? I am sure glad I don't have to use much math in medicine (and so are my patients!)

When I model the power supply on this schematic using a current sink for the B+ of 54mA and 39mA for the C+ the PSUD1 models the voltages to be about 335 B+ and 270 C+.

Basically you probably see what I am trying to do here, I need to figure the C+ so I can make the proper alterations in my power supply to use this front end on the 2A3 output stage I already have parts for.

First, where am I going wrong on my calculations and secondly, can anybody tell me from what they see here or possibly empirically know, what the C+ is on this schematic? BTW the power transformer is a 720VCT.

Thanks a million, it's always refreshing to find out how much of an idiot one can be.
Jeff

 

[jmillerdoc]

I am not sure the image link to the schematic I used is working properly, here is the web address to the schematic, sorry.
Jeff

Horus 2A3 Silver Schematic and Parts List - Print Version

An externally hosted image should be here but it was not working when we last tested it.

 

[jmillerdoc]

I guess I should add here that my intentions are to try to RC couple this preamp/driver stage to the output stage of the Hagtech Clarion 2A3 amp. The Clarion uses a single 5687 driver stage tht requires an input of about 4V to drive to clip. I am trying to basically "Add" another 5687 preamp stage to what Hagerman built in order to have an integrated amp that doesn't require the use of a preamp to run it.

As I see it (and am probably wrong) the Clarion 5687 driver stage has a higher gain than the Lessard 5687 driver stage. I think the gain would be much too high if I tried to just add the preamp (first stage of the Lassard 5687) stage to the Hagtech/Clarion driver stage. Is that a correct assumption?

If I am incorrect, would it be possible to simply add the preamp (1st stage of the Lessard) stage in front of the existing driver stage of the Clarion?

Also, if you are just looking at the schematic of the Clarion he shows a ECC99 for the driver but the text of the article shows that he ended up using a 5687 tube.

Here is the clarion article, sorry I can't link it to the forum as it is embedded in a .pdf file and it won't allow me to link the URL for the schematic image.
http://www.hagtech.com/pdf/clarionarticle.pdf

Thanks for you help. I have toiled over this for days now when I realized after I sourced all the parts to the Clarion that it was not an integrated amp. I should have been more careful when choosing this design.

Jeff

 

[sawreyrw]

Jeff,

A couple of things. How do you know the voltages you have circled? You need to add component values and tube types to the schematic in order for anyone to help you.

Rick

 

[Sheldon]

Where did I go wrong????

Your calculations look fine. You went wrong by assuming the voltages cited on the schematic are exact. They look like rounded values to me (70V, 130V, 2V, 8V). I would not expect the values to fall so nicely on round numbers. But they are close enough for tube amp work.

As for PSUD, it's very accurate - also assuming you have very accurate values for the transformer. If the values are approximate, your calculated values will be approximate.


Sheldon

 

[Sheldon]

As I see it (and am probably wrong) the Clarion 5687 driver stage has a higher gain than the Lessard 5687 driver stage. I think the gain would be much too high if I tried to just add the preamp (first stage of the Lassard 5687) stage to the Hagtech/Clarion driver stage. Is that a correct assumption?

The Clarion and the Lessard driver stages are within a couple of dB gain. Not worth worrying about the difference. And the output stages are close too. The gain of the first stage in the Lessard amp is about 18dB (8X), so yes you will have more than you need, but so does the Lessard amp.

If you want sensitivity closer to 2VRMS, the easiest approach would be to use a higher gain input tube (SL7, 5751, AT7etc.), and use a mosfet follower to provide enough grunt to drive the output tube. See the Tubelab site for explanation.

Or, you could integrate a 12B4 preamp into your Clarion. BTW, adding the extra grounded cathode section inverts the amp, FWIW.

Sheldon

 

[Arnulf]

Your calculations look fine. You went wrong by assuming the voltages cited on the schematic are exact. They look like rounded values to me (70V, 130V, 2V, 8V). I would not expect the values to fall so nicely on round numbers. But they are close enough for tube amp work.

... plus component values usually fall into certain range (say 5% or 1%) around specified value so you could easily end up with completely different figures by using nominally same components.

Jeff, don't worry, your schematic was obviously calculated for cca. 300V, a couple of per cent deviation and you're at your values.

I love this tagline one of our forum members is using: "Ratings are for transistors - tubes have guidelines". Don't grossly violate the guidelines for extended period of time (= more than many minutes at a time) and you'll be fine 😉

Few per cent difference doesn't mean anything, there will be more mismatch between your tube and the tube the schematic was created upon than there is difference in your calculation simply because you don't know the exact parameters of the tube (and other components) originally used there when the voltage measurements were made.