Biasing of Tubes, part 1

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flatmap

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Re: Biasing of Tubes, part 1
« Reply #20 on: 17 Jul 2007, 04:27 am »
Well yes, I'd love to hear more about what Thévenin's theorem is about!

Roger A. Modjeski

Re: Biasing of Tubes, part 1
« Reply #21 on: 18 Jul 2007, 07:35 am »
In brief, Thevenin realized that any source could be modeled as an ideal source with a resistor in series (for a voltage source) or in parallel (for a current source). For AC one can extend the theorem to include inductors and capacitors, but let's stick to single resistors first.

Consider a 12 volt car battery vs the little 12 volt battery that is in most car remote controls that we carry on our key chain. They are both 12 volts but you can't start the car with the little battery in the remote. Why is that, they are both 12 volts? Thevenin not only answers the question but gives a very simple circuit that will predict how each battery will behave under load. Both are modeled as a 12 volt ideal voltage source. That means a theoretical 12 volt source with no internal resistance that can supply an infinite amount of current. Now he adds the internal resistance in series with that voltage source. The starter battery might have an internal resistance of .03 ohm while the little remote control battery might be 24 ohms.

So now we have a big difference in the two batteries. If you short the car battery the current will be 400 amps (12/.03=400). If you short the transmitter battery the current will be 12/24 or 1/2 amp. Big difference. Since the starter of you car can draw 60 amps in sunny California (and way more in a Minnesota winter) we can actually predict the resulting votage right at the battery terminals. Also note that corroded terminals simply add resistance thus dropping the voltage further. When my 7 year old Mustang battery started to go I simply hooked up a meter to the battery while cranking the engine and saw the voltage fall from 12 to 7 though the headlights would only pull it down to 11 or so. The battery was still a 12 volt battery, but its internal resistance had gone up markedly making it unreliable. Note also that when the starter draws the battery voltage down you have less spark which further reduces your chances of getting going. 

Once these resistances are known we can predict the battery terminal voltage under any load. Don't go looking for this resistor inside the battery, it's not a resistor but a resistance that is dependent on the surface area of the plates, condition of the electrolyte and any connection resistances within the battery. Thevenin says we can lump all those into one imaginary resistor that is in series with the ideal voltage source. Say we put a .1 amp load on each battery. The starter battery will fall .1 x .03 ohms or .003 volts.  Hardly perceptable. But the little battery will fall .1 x 24 ohms or 2.4 volts. Big difference and easily calculated. 

A little practical advise here. A very easy way to test your car battery and terminal connections is to have someone crank the starter while you have your meter probes poked well into the tops of soft lead terminals. Note the voltage while cranking. Then move the probes to outside of the clamps and measure that voltage. Whatever the difference is what you are loosing at the terminal to post interface. You can also measure separately from post to ring of the plus and then the minus. That will measure the voltage loss on that interface directly. You may find one is very good and one very bad. Leave the good one alone and fix the bad one. I never ever take a battery terminal off just because it might look bad on the outside. If it measures good then it's good and you can't make it better and will likely make it worse. Good clean terminals stay clean because they have a tight fit. Scraping, fileing and sanding destroys the fit getting you into a maintenance cycle that could be avoided.

Three other things about car batteries while we are on the subject. There is an old wives tale that you will cause a car battery to discharge if you put it on the concrete floor. I would love to hear any supportive argument for that notion. The other thing is that if you ever drop a wrench across the terminals and short them out the wrench will likely weld to the terminals and you should get out of there because the battery is likely to explode from the high current discharge. Never never use anything but distilled water. Check the levels twice a year and top them off. Many times I have seen a batter fail because one cell was only half full. No-maintenance batteries are no good and leaving a battery discharged for even a few days causes sufation and that's death to a battery.

I'll end with a question: If we want to consider how our wall outlet voltage is affected by our audio equipment loads how can we get a handle on that? One could add up all the resistances back to Hoover Dam while considering all the transformers, wire, generator winding resistances... on and on. A quite impossible task. Using Thevenin one can get the answer very simply with just a voltmeter and a light-bulb. Any suggestions?

Next time I will get into how Thevenin can be used to understand how loads affect our preamps, poweramps, signal sources, cables......and whatever else you might be curious about.


flatmap

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Re: Biasing of Tubes, part 1
« Reply #22 on: 23 Jul 2007, 07:48 pm »
Roger, thanks for this introduction to Thevenin's theorem.  I'm thinking about your question on how to deduce the thevenin resistance at the mains outlet. 

Here's my attempt to answer:   First measure V at the wall.  Then connect a lamp to the outlet and measure the voltage drop across the bulb, call it Vbulb.  I think the difference (V - Vbulb) would be the drop across the Thevenin resistance, if I'm tracking this correctly.  So we'd need to also measure the current in the circult including the light bulb = Ibulb.  Then the Thevenin resistance RT = (V - Vbulb) * Ibulb. 

Is this the right idea?

Roger A. Modjeski

Re: Biasing of Tubes, part 1
« Reply #23 on: 23 Jul 2007, 10:52 pm »
Steve,

Yes, good work, that's right. In practice I would use the heaviest load I had such as a 1200 Watt electric heater. A 100 watt light bulb doesn't make much of a dent in the voltage and we have to consider that the line is fluctuating a little. Since it's a linear system the more current the better but don't blow the fuse.

flatmap

  • Jr. Member
  • Posts: 69
Re: Biasing of Tubes, part 1
« Reply #24 on: 24 Jul 2007, 03:08 am »
So, just going with this a bit... Suppose I placed a resistive load across the speaker terminals of one channel of my amplifier and made the same set of measurements.  Then this figure for Thevenin's R,  say for a fixed sinusoidal output, would this give me what is referred to as the output impedance for the amp? 

Steve

Roger A. Modjeski

Re: Biasing of Tubes, part 1
« Reply #25 on: 24 Jul 2007, 04:46 am »
Yes Steve, that would give you what you are looking for. Just as we want to see good voltage regulation at out wall socket, we want to see good regulation at our amplifier output. I should mention here that wall socket regulation and amplifier output regulation are not dependent on one another.

Here's how I do it. Run the amp at one volt output, sine wave. Put the load of interest (8 or 4 ohms) across the amp and see how many dB the output falls. You can do the calculation to get the Thevenin resistance, but why bother, you already have the answer in dB. You can also make the test at various frequencies and see the variation with frequency. This doesn't work on high feedback solid state amps because the dB difference is virtually unmeasurable. There are other ways to get at it though.

Here are some ballpark numbers. If you want to go back to damping note that a 1 dB drop corresponds to a damping factor of about 10 and a 1/2 dB corresponds to damping of 20. If the output falls 6 dB when the load is connected you have a damping factor of 1. There the output impedance equals the load resistance.

We are told by many an author that "matching" the load to the plate resistance of a triode gives the highest power transfer. Any comments on that anyone?