How this circuit works (Baby Huey type) - diyAudio
 How this circuit works (Baby Huey type)
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 14th November 2008, 05:24 PM #1 diyAudio Member   Join Date: Sep 2008 How this circuit works (Baby Huey type) Hi, I found this during some browsing about the Baby Huey circuit, and it's pretty interesting. http://www.triodeel.com/compact.html I understand how the inverse of the signal is present on the cathode, but I don't understand why this would produce the same amplitude as the other tube. Does that make sense? Using some math, and totally made up numbers: Quiescent state: Grid voltage = .3 volts Cathode voltage = 1.3 volts Plate current = 10 mA Cathode resistor = 130 Ω Alright, that's easy enough. Since it's in quiescent state shouldn't the other tube be conducting equally? Grid voltage = 0 volts Cathode voltage = 1.3 volts Already we can see that since the grid to cathode voltage is not the same, different amounts of current will flow... What's wrong here? Thanks!
 14th November 2008, 05:49 PM #2 diyAudio Member     Join Date: Nov 2005 What value of grid current did you use?
 14th November 2008, 05:52 PM #3 diyAudio Member   Join Date: Sep 2008 Hm? Not sure what you mean. What would that affect? Thanks!
 14th November 2008, 05:56 PM #4 diyAudio Member     Join Date: Nov 2005 Well... some amount of DC grid current must be flowing to arrive at a 0.3 grid voltage...
 14th November 2008, 05:58 PM #5 diyAudio Member   Join Date: Sep 2008 Yeah, I get that for sure, but what does that affect in terms of tube conduction? Cheers!
 14th November 2008, 06:03 PM #6 diyAudio Member     Join Date: Nov 2005 The tubes conduct primarily based on the voltage difference between the grid and the cathode.
 14th November 2008, 06:06 PM #7 diyAudio Member   Join Date: Sep 2008 Yeah, I know. I'm just confused about what you said about grid current, and how that's relevant here. If one grid is at .3 volts at quiescent state, and one is at 0 volts, and the cathode voltage is the same, how can conduction be equal?
 14th November 2008, 06:11 PM #8 diyAudio Member     Join Date: Nov 2005 You are saying that there is 0.3 Volts at one of the grids... that implies a current flow does it not? Where does the 0.3 come from... and how did you arrive at that?
 14th November 2008, 06:25 PM #9 frugal-phile(tm) diyAudio Moderator     Join Date: Oct 2001 Location: Victoria, BC, NA, Sol III Blog Entries: 5 This amplifier has been discussed before -- here i think. What i understand is that the price for using the output pair as the phase splitter is that it is only capable of delivering half the power of a more conventionally driven push-pull pr. dave __________________ community sites t-linespeakers.org, frugal-horn.com, frugal-phile.com ........ commercial site planet10-HiFi p10-hifi forum here at diyA
 14th November 2008, 07:30 PM #10 diyAudio Member     Join Date: Dec 2001 Location: Hickory, NC Having a tail resistor, instead of a CCS, will cause some assymetry between the output tube drives. The cathode driven tube won't get quite as much drive signal as the grid driven tube due to some AC current loss to the tail resistor. The grid input signal to the 1st tube will generally have to be twice as large to drive both output tubes (compared to normal dual inverted drives). But should be capable of about the same power output as normal (with doubled drive) if the assymetry is not so large as to cut down the power significantly from the second tube. I think this scheme also increases the output impedance (higher damping factor then) due to effective resistance in the cathode circuit. Each tube sees approx. 1/gm resistance in its cathode circuit from the other tube and the tail R in parallel. Normal configuration would have the cathodes grounded. Don __________________ Dark matter is DEAD! McGaugh, Lelli, and Schombert, Physical Review, Sept 2016. 3D time is the last man standing. When all other possibilities have been eliminated, no matter how improbable it may seem ...

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