If I'd only known earlier ... things that stumped me for far too long

[Deke609]

I thought I'd start a thread for people to post about aspects of electronics principles, circuit design and amp building that stumped them b/c they were missing something basic.  Why? Because I think it will be very useful for others. If something stumped you -- no matter how basic or advanced -- I bet there are a bunch of other people who are or will be be stumped by the same thing. And because not everything is spelled out in the usual texts and online sources for learning about this stuff.

Let me get the ball rolling with a case in point. Until an hour ago, I did not understand that where a resistor is connected to a "source" voltage potential (battery or even 0V chassis ground), but no current flows, the non-connected lead of the resistor on the far side of the "source" has the same voltage potential as the source -- no voltage potential is "dropped" by the resistor; and instead the voltage potential of the source is somehow "translated" or "passed" to the far end of the resistor (I have no idea what the appropriate terminology is for how the resistor acquires the same voltage potential as the source). Here's an experiment that I just did to verify this: take a household battery -- e.g. an AA -- and measure the voltage potential across its -ve and +ve terminals (I measured 1.55V for my AA battery). Now connect a large resistor to the +ve end (I used a 68K), but leave the -ve terminal of the battery and the other lead of the resistor disconnected (do not connect them to anything). Now measure the voltage potential across the battery's -ve terminal and the end of the lead of the resistor not connected to (farthest from) the battery: it will be nearly equivalent to the voltage potential you previously measured (I measured 1.54V). I believe the tiny drop in measured voltage is a function of the voltage meter forming a closed loop with the two measured ends (but only by way of the extremely high r-value resistor contained in the meter). [Edit: As i think more about it, this experiment may not prove the principle since it involves closing a circuit where the intention is to measure the conditions of an open circuit - but I am pretty sure the principle is correct).

Not knowing this until recently, I have never been able to make sense of the voltage potentials of circuit elements through which no current flows. Example: cathode biasing stumped for more than a year. I understood how a cathode resistor raised the voltage potential of the cathode above ground as soon as current began to flow through the tube, but what boggled me was how the grid was set to 0V (chassis ground) potential. Looking at the grid leak resistor, which is always much larger than the cathode resistor, I figured the grid leak resistor should be making the voltage potential of the grid much more positive than the cathode. So I figured there must be some direct connection between grid and 0V somewhere earlier in the circuit (e.g., through a volume pot). But I could never find one. Nope. It's the grid leak resistor itself (PB tried to explain this to me in the last few months, but I didn't really get it).

I still don't really understand the physics or theory of how this works, but having tested it for myself, I trust that it is the case (Plus, as I finally concluded earlier today, this just had to be the case for cathode biasing to work).

Although some people have posted online to the effect that Ohm's Law tells you this, I don't agree. Or at the very least, it is not an intuitive implication of Ohm's Law. 

If I'd only known earlier ...

Pleas share your "If I'd only known" milestones in learning - no matter how basic or advanced. I'm sure someone will be grateful.

cheers, Derek

 

[mcandmar]

The key i found was to start thinking about current and not voltage.  Its a bit like how people always talk about engine power in horse power, when in reality it is the Torque that does the work, power is how quickly the work gets done.

 

[Deke609]

The key i found was to start thinking about current and not voltage.

@mcandmar - can you say more about this, or give an example where thinking in terms of voltage rather than current helped make sense of something?

For me, thinking of voltage and current as always linked helps. For example, I've been thinking more about the issue I discussed in my first post: establishing 0V reference with a grid leak resistor and to my embarrassment it occurred to me that, despite my earlier protestations to the contrary, you can actually understand this pretty well with Ohm's Law if you grasp the basic relationship between voltage and current: a instantaneous difference in voltage potential between two conductively connected points produces an instantaneous current that, absent a perpetual generator of different voltage potentials (everlasting battery or never-failing mains voltage), will tend to cancel the voltage difference. Accordingly, if the opposite sides of the grid leak resistor have different voltage potentials, a current will be induced to flow, and the voltage difference will be dropped across the resistor. So if no current is flowing between electrically connected components, that means there is no difference in voltage potential between them. In retrospect I don't know how I managed to so over-complicate the issue.  That simple idea took me years to figure out! Then again, I've always been a slow starter and only start to pick up steam later on. When the concept of "pi" was first introduced to me in grade 6 math, it stopped me mentally in my tracks for years (the memory of that day in class still conjures a sense of the intense cognitive dissonance I felt at the time). I don;t think I got over it until we learned calculus (whatever that integral summation was where you set lim n = infinity). As put to me by my grade 6 math teacher, I understood the the "pi" proposition as comprising two seemingly contradictory ideas: (a) there is a fixed (i.e., finite and unchanging) ratio between radius and circumference; but (b) the ratio is impossible to pin down because it's value shoots off to the right of the decimal place without limit and without repeating.  I could not reconcile  "finite" with "keeps going to right of the decimal place". I didn't appreciate that a finite number or distance or whatever can be infinitely divided and that the limit (end value) can be approached infinitely without contradicting the limit's property of being finite and fixed.

Anyway ... not sure that there was any point to the pi story above.  ;D  Maybe it helps  explain why a belabor basic concepts to death. It's just what I do. Hah!

cheers, Derek

 

[Deke609]

Reflected Loads!

For years the idea of reflected loads through a transformer has baffled me. All that stuff about "what the tube sees" and "what the load sees". Until today with the help of Jac at EML.  What seemed totally counterintuitive to me, is now so obvious that I have to laugh at my own thick-headedness. I had no clue that the impedance of the primary varies with the magnitude of the resistive load on the secondary - in fact, when that possbility first occurred to me (this morning), I rejected it as plainly crazy. But Ohm's Law should have told me it was true. In case anyone else is baffled by the concept of reflected load, just work through the following simple problem that Jac posed to me:

Suppose you have a 10:1 step-down transformer fed by 120VAC on the primary.  On the secondary, you have a lamp that draws 2A at 12VAC. What is the resistance (impedance) of the load on the secondary (i.e., the lamp)? What is the reflected load on the primary (its impedance)?

Solutions: Ohm's Law gives us the impedance/resistance of the lamp. R = V/I = 12V/2A = 6 ohms.  The law of conservation of matter/energy (or whatever the equivalent is called in electrical theory) tells us that power in = power out (assuming, for the sake of simplicity, that the transformer is "perfect" and no energy gets gets lost to heat, motion, etc.). Power dissipated by secondary = Vs x Is = 12V x 2A = 24 Watts = power in at the primary.  Since P = V x I, and we know that 120V drops across the primary, that means that 24W = Vp x Ip = 120V x Ip. So primary current Ip = 24/120 = 0.2A (which we could have figured out more simply by seeing that since the voltage step-down ratio is 10:1, the current step-up ratio must be 1:10). So what's the resistance/impedance of the primary when the secondary is loaded by 6 ohms? Back to Ohm's Law: R = V/I = 120V/0.2A = 600R.  Which just happens to confirm the formula that the "impedance transfer ratio" aka "reflected load" is (turns ratio)squared, or in this case, 10:1 x 10:1 = 100:1. So a 6 ohm load on the secondary of a 10:1 transformer is "reflected" on the primary 100 times bigger, as 600 ohms.

And here's what really blew my mind today: if a different lamp instead draws only 0.5A (b/c it is 4 times more resistive than the original lamp), the primary impedance is 4 times bigger, 2K4 ohms! Why do I care? Because I've been using all my BH amps configured for 16 ohms output impedance with 200 ohms headphones - usually without a 16 ohm resistor in parallel. That means my output tubes have been loaded way more heavily than PJ's design intended. I will use a 16 parallel resistor from now on (or as close as I can get to 17.5 ohms to average out to a 16 ohm effective load).  ;D

And here's how I finally stumbled upon this (those of you who already understand the principle of reflected load may find this funny): I couldn't make sense of the wiring instructions for my new Lundahl output transformers.  They are stated to be configurable for three different primary impedances: 2K6, 4K5 and 9K7. And for each primary impedance there are three possible secondary output impedances, 4, 8 or 16 ohms.  So I figured there must be 3 different wiring schemes for the primary, and 3 different schemes for the secondary. Nope. The primary is wired only one way, regardless of what primary impedance you want, and the secondary has 4 wiring options. Huh. That baffled me, and got me thinking that maybe the Lundahl OT was some kind of crazy autoformer, with options to wire different winding sections in combinations of series, parallel and in/out of phase. I was prepared to live with that mystery until I noticed that one of the wiring schemes, scheme "C", was the common wiring scheme for three different setups: 9K7 and 16 ohms, 4K5 and 8 ohms, and 2K6 and 4 ohms. What the hell?  So I stayed up really late last night trying to make sense of this - most which time was spent going over the datasheet again and again looking for the 3 missing primary wiring schemes that I was sure just had to be there. No luck, still stuck. So I went to bed really confused. This morning, I took a crack at it again, and the only way I saw for the same wiring scheme to result in three different primary/secondary impedance setups required something truly bats&@t crazy: the impedances must depend on the load. Madness! So, unsure of how to wire the new OTs, I contacted Jac at EML to verify what the datasheet was telling me.  And I ended the email by noting (mentally cringing as I typed) that it appeared that the Lundahls used the load on the secondary to determine the primary impedance. "Man", I thought, "is Jac ever going to think me a moron for suggesting such a thing."

Happily, Jac, like the BH crew, suffers fools with patience and grace. And that's how I figured out that what I thought was entirely implausible, or at best some weird idiosyncracy specific to my Lundahl OTs, is in fact the basic working principle of all transformers. Huh! If I'd only known earlier ...

cheers, Derek

 

[2wo]

Nicely put, they say when you can explain it simply to someone else you know that you understand it yourself.

I think the  electrical term you're looking for is Kirchoff's law, basically the sum of the currents entereing a node equals the current exiting the node or something like that.

If it's any consolation, When I got a set of Lundell outputs years ago I fell into the same trap and went through the exact same thought process to get the penny to drop

 

[Deke609]

Thanks John.

If it's any consolation, ...

No consoling necessary. I've made peace with the fact that learning about this stuff is going to be an ongoing lesson in humility.  ;D
To put myself on a better footing, and to force myself to learn the basics properly, I recently enrolled in a distance learning "electronics technician" certificate program from a respected local college (in Canada the term "college" has a different meaning than in the U.S. - here, it usually refers to a technical or vocational training school).  I haven't written an exam in decades - should be fun.