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Supperimposing an AC signal on a DC signal: right way or not?

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IamSmooth

Sun Jan 14 2007, 04:26AM

Registered Member #190 Joined: Fri Feb 17 2006, 12:00AM
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Posts: 1567

I have a -100v DC signal from a simple rectified source:. AC goes to transformer and then through a bridge and capacitor filter. I want to add a 5v AC signal on the -100v signal so I will have the final signal oscillate through -95v and -105v. Right now I have the 5v AC signal from another transformer secondary go through a 440uf electrolytic capacitor and connect to the -100v rectified line. The transformer's center tap is connected to the 0v line of the DC signal. I figured that the capacitor would let the AC signal through, but won't ground the DC signal. Is this approach correct? Also, is an electrolytic capacitor the right choice? I want to keep things as simple as possible.

ragnar

Sun Jan 14 2007, 04:50AM

Registered Member #63 Joined: Thu Feb 09 2006, 06:18AM
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Try this:

IamSmooth

Sun Jan 14 2007, 04:55AM

Registered Member #190 Joined: Fri Feb 17 2006, 12:00AM
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I can see how that would work nicely, but is there anything wrong with the way I presented the circuit? The reason I ask is that I was going to use the other half of the secondary for something else; if I use the new circuit pattern I will need another transformer.

...

Sun Jan 14 2007, 05:16AM

Registered Member #56 Joined: Thu Feb 09 2006, 05:02AM
Location: Southern Califorina, USA
Posts: 2445

it is too late to fix it now, but you might want to think about leaning how to use the edit button.

In any case, it is possible to superimpose a dc offset with a cap, however I am a little unclear as to where you plan to take the signal off in the circuit you posted. I am not sure how it would react it you powered it up as it is drawn, but I doubt it would do what you want.

The normal way to do it would be to have a ground one side of the ac source, and send the other through a cap (something biploar, and rated for 150v) of some value I have no idea to calculate. The other side of the cap goes through a resistive divider between a higher voltage that you want, and a lower voltage than you want. Take the output off the side of the cap connected to the resistors. But this circuit can't put out any real power, it is only good for signal.

Really for high power stuff, I think that Matt's method is best.

ragnar

Sun Jan 14 2007, 05:22AM

Registered Member #63 Joined: Thu Feb 09 2006, 06:18AM
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Heh, is my stuff only ever good for "really high power"? :P

IamSmooth, if you replace the 10Vp/p secondary in my schematic with one half of the secondary of your center-tapped transformer, then ASSUMING you're getting 10Vp/p between teh center-tap and one end, AND ASSUMING there aren't any grounds anywhere else in the circuit, then, yes, you can still use the second half of your transformer for something else.

Just pop it in the schematic in series, as pictured... but if for example your -100V supply is referenced to ground, and your centre-tap is grounded, something's gonna get hot.

P.S. Peter, I think IamSmooth knows how to edit, but probably clicked "submit" several times to an unresponsive browser.

IamSmooth

Sun Jan 14 2007, 05:34AM

Registered Member #190 Joined: Fri Feb 17 2006, 12:00AM
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Yes, I know how to edit.
Yes, I only need the signal to reference a grid. The real power is going to come from the plate end of a vacuum tube. The grid will just add a bias and will vary the current that the plate/ground pathway will pass.

Finally, I think I like your suggestion the best. I will just have to add another component to heat the filament instead of the second half of the transformer. Thanks for the help.

Carbon_Rod

Sun Jan 14 2007, 06:26AM

Registered Member #65 Joined: Thu Feb 09 2006, 06:43AM
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This naturally happens with cheap 100v secondary transformers, bridge, and a small filter cap.

Hazmatt_(The Underdog)

Sun Jan 14 2007, 06:42AM

Registered Member #135 Joined: Sat Feb 11 2006, 12:06AM
Location: Anywhere is fine
Posts: 1735

If this is just a bias supply for a tube with an A.C. signal input, then you might want to look at how tube amplifiers achieve this. It is fairly simple and straightforward. Since a milliamp or so of current is needed on the grid, you can control a wide range of bias voltages with a 1/2W potentiometer. The negative bias is one end, gnd on the other, and the wiper is your variable bias voltage. Sometimes the bias voltage is current limited with a resistor of a high value like 100K, but that is not so important, its just a design preference.
Now that you have a variable grid bias voltage, how do you modulate that voltage? Well simply put, Tube Amplifiers use D.C. blocking capacitors. The blocking capacitors can be almost any type non-polar capacitor with a moderately high voltage rating, 400-600V is typical because of 300-450V supply rails.
A single 12AX7A has more then enough amplification to modulate this capacitor to drive a larger power pentode such as a 5881, EL34, KT88, you get the idea.
For a really big tube like an 811A you will likely need more amplification and power handling then the 1 watt of the 12AX7A. Large power tubes would require a smaller pentode to drive them properly.

If you'd like to know more about tube amplification, there are a lot of guitar forums that discuss and have schematics of circuits. Hoffman Amps is a pretty good place to noodle around. There are also a lot of books on amps for guitarists, and some that are just shcmatics.

Steve Conner

Sun Jan 14 2007, 12:41PM

Registered Member #30 Joined: Fri Feb 03 2006, 10:52AM
Location: Glasgow, Scotland
Posts: 6706

Iamsmooth's original circuit won't work because the smoothing capacitor of the 100v DC supply will short out the AC signal. Blackplasma's circuit should work. The method Hazmatt described is fine for tubes operating in Class AB1 where no grid current is drawn, but the DC blocking capacitor won't allow you to drive the tube into AB2. (This was never a problem in audio amps, because audio power tubes produced plenty enough power in AB1.)

Note that the high value resistor Matt said "wasn't important" is a vital part of the circuit. It gives the DC bias supply enough resistance to allow you to superimpose the AC on it. If this resistor is too small, it loads down the AC signal from the previous stage, and if it's too big, the power tube goes into thermal runaway and dies. 100k or 220k are the only values I ever saw used in audio circuits.

Hazmatt_(The Underdog)

Sun Jan 14 2007, 09:10PM

Registered Member #135 Joined: Sat Feb 11 2006, 12:06AM
Location: Anywhere is fine
Posts: 1735

Steve, I was trying to say the exact value wasn't critical. I have seen 47K up to 330K for this resistor. It varies a lot from one design to another.

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