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13th August 2007, 11:12 AM  #2271  
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Join Date: Sep 2006

Quote:
Hi Charles, This assertion is quite wrong. You need to plug in the numbers and do the math. Many amps don't use RE smaller than 0.22 ohms, and most of those few that do use a lower RE operate the transistors at lower voltages. Moreover, many amps built over those years have used 200Watt TO3 devices that can achieve a lower thermal resistance from case to heat sink than TO247 devices. Take the Leach amp you cut your teeth on. IIRC, it used 0.33 RE with 85V rails and 200W TO3 devices. They probably had a thermal resistance from junction to heat sink of about 0.8 C/W. Net gm was about 1.5 S at optimal ClassAB bias with RE = 0.33. Beta = 85V * 0.0022V/C *1.5S * 0.8C/W = 0.22 This is a very comfortable value. A 100W amplifier with 50V rails and using quite small 0.1 ohm resistors with T03 output devices biased at 150 mA and having a net gm of 3 S would have a Beta of: Beta = 50V * 0.0022V/C * 3S * 0.8C/W = 0.26 This is also a comfortable value. Cheers, Bob 

13th August 2007, 11:21 AM  #2272  
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Join Date: Sep 2006

Re: Re: Thermal Bias Instability
Quote:
Hi Pooge, Unfortunately, this does not address the particular problem of local thermal bias stability for a given device. Its not that this is a bad idea, because it certainly does improve matters by taking the average and splitting the difference. However, taken from the perspective of just one device, in this kind of arrangement its ThermalTak diode has only 1/8 influence, which means it is only able to provide a little bit of the needed local thermal compensation for the single device under consideration. Cheers, Bob 

13th August 2007, 12:07 PM  #2273  
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Join Date: May 2003
Location: Northern Va.

Re: Re: Re: Thermal Bias Instability
Quote:
In that case, it would seem to me that 1/8 influence is better than NO influence if only a few of the diodes are used, the hottest one being one of the diodes not used. It seems a waste not to use all of the diodes somehow. If not taking advantage of all of them, you may as well just use a sensor piggybacked on the top of one or a few output devices. 

13th August 2007, 12:12 PM  #2274 
diyAudio Moderator Emeritus

Hi, Mr. Hansen,
This is a nonfeedback amp (at least it is what in the brochure said ), but it has servo (TL071) around VBE multiplier to maintain 0Vdc offset. The output stage is biased about 20mV accross the 0.1R/5W resistors. The outputs are 2SC5200/2SA1943, 5pairs. What is improveable from this design? 
13th August 2007, 12:58 PM  #2275  
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Join Date: May 2003
Location: Northern Va.

Re: Re: Re: Thermal Bias Instability
Quote:
If two diode strings were made, one for each sexed output group, for a complementary Vbe multipler, then the influence of each diode could be doubled. 

13th August 2007, 01:26 PM  #2276  
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Quote:
Hi Charles, This is just a matter of basic feedback theory. In a simple feedback system, closed loop gain is given by: G = A/(1A*B), Where A is the forward gain and B (Beta) is the feedback factor. The product A*B is the loop gain. For negative feedback, A*B is negative, so the denominator grows with increased amounts of NFB. When A*B is positive, we have positive feedback wherein A*B subtracts from the unity term and decreases the denominator with increased loop gain. This in turn increases G. When A*B goes to +1, the denominator goes to zero and G goes to infinity, and we have runaway or latchup. In the case I described with my formula for local thermal stability, I used A=1 for convenience and lumped all of the loop gain into Beta. When my value of Beta goes to unity, the denominator goes to zero and we have infinite gain and runaway. When Beta = 0.5, you can see that the denominator goes to half its value when there is no feedback, meaning that G goes to twice its nofeedback value. If you prefer, for the thermal feedback analysis, you can say that the input to the system is the base voltage and the output of the system is the bias current. The forward gain, A, is then the change in bias current for a given change in base voltage. This is simply the net transonductance of the combination of the output transistor and its emitter resistor. This is gm in my formula. The value of Beta, looking at it this way, then becomes equal to Vce * Theta_cs * TCvbe. Vce gets us from change in current to change in power. Theta_cs gets us from change in power to change in temperature. TCvbe gets us from change in temperature to change in voltage, completing the ‘loop”. G = gm/(1gm * Vce * Theta_cs * TCvbe) [amperes per volt] You can see that loop gain, which is what counts in determining how much positive feedback you have, is still the same as what it was in my formula. I hope this makes the origin and validity of my formula more clear. Cheers, Bob 

13th August 2007, 02:44 PM  #2277  
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Cheers,
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13th August 2007, 03:08 PM  #2278  
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Join Date: Sep 2006

Re: Re: Thermal Bias Instability
Quote:
Hi Charles, As you said to another poster: Like Rafiki told Simba in The Lion King, "look harder!" You are not hearing what I am saying. Moreover, you are not plugging in the numbers and doing the math. If you do, you will see that most amplifiers, including those using paralleled devices, come out with a Beta value that is safe. If you tell me the numbers for your ThermalTrak amp, I'll plug the numbers in for you. What RE are you using, what is your rail voltage, and how many output pairs do you have in parallel on each side? I'm guessing you are in safe territory, as your rail voltages are fairly small, if I recell from something you mentioned in an earlier post. Cheers, Bob 

13th August 2007, 05:54 PM  #2279  
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Join Date: May 2003
Location: Colorado

Quote:
"I used A=1 for convenience." Why? What is your justification? The way I see it, your formula you gave previously is A NOT B. That formula shows the gain in the *forward* direction, not the reverse direction. The gain in the negative direction (B or Beta) is given by the thermal compensation network. Given your starting point, everything makes sense. But I think your starting point is incorrect. As I said before, just look at it from an intuitive standpoint and forget the math for now. Let's say that today is hotter than yesterday and the ambient temperature goes up one degree. If the forward gain of the system (A) is 1, then the temperature of the device goes up (surprise!) one degree. I think it would be hard to argue that the forward gain of such a system is NOT 1. But your equation said that if the ambient temperature went up one degree that the positive (forward) gain due to the tempco of the system created ANOTHER degree of temperature increase for a total of two degrees. This is clearly a forward gain of 2, and NOT a reverse gain of 1. I don't even see how you could talk about the reverse gain (B) without including the feedback mechanism of the bias sensor. Of course, if you want to talk about the system WITHOUT the bias sensor, just looking at the mechanisms intrinsic to the output devices themselves (such as channel resistance tempco, Vgs threshold tempco), then you could use your formula. In that case, the Vgs threshold tempco would be in the A side of things and the channel resistance tempco would be in the B side of things. But then unless you are using lateral devices you will see that the system will ALWAYS "latch up". Which is why thermal feedback is required in the form of the bias sensor. But you have completely ignored this in your equations, and I think without adequate justification. 

13th August 2007, 06:09 PM  #2280 
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Location: Prague

B is reverse direction transfer function, it may be complex function as well, not only simple divider ratio.

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