Just doing some AC testing on an OT, using an Agilent sig gen to input a sine wave at varying freqs, and measuring AC volts and amps. Plotted the resultant R=E/I vs frequency, but don't understand the results. I was hoping to be able to determine the winding impedances (for an unknown xfmr, for ex). Attached is my testing and graphs. Can anyone explain what I did/did wrong? Thanks!
You don't say what load was applied to the secondary. Unloaded readings on an OPT will be affected greatly by strays and are generally meaningless. The secondary colors are standard so the black is the common, yellow is probably 4 ohms and green likely 8, but I have seen other values. Note 4 ohms is not the CT of an 8 ohm winding.
Coming from a P-P 6V6 amp the primary is probably in the 6600 to 10000 ohm CT region. Book standard is 8K.
Put an 8 ohm (or close) load between black and green then repeat your tests. If your generator cant drive this, use a higher load impedance, but stay below 100 ohms. Most OPT's are measured at 1K.
For simplicity, I use 60Hz at say about 120 volts....but I am not recommending this to unexperienced testers.
Coming from a P-P 6V6 amp the primary is probably in the 6600 to 10000 ohm CT region. Book standard is 8K.
Put an 8 ohm (or close) load between black and green then repeat your tests. If your generator cant drive this, use a higher load impedance, but stay below 100 ohms. Most OPT's are measured at 1K.
For simplicity, I use 60Hz at say about 120 volts....but I am not recommending this to unexperienced testers.
Tubelab-thanks! I've got a lot to learn about impedance.... I was testing the primary and secondary windings separately in my tests, and using THEM for the load (to apply AC at different freqs, and measuring the AC current).
- You say "Most OPT's are measured at 1K" this is 1K Hertz?
- I was using a Variac (at low-low voltages), driving the secondary and reading the primary, to get a voltage ratio, which squared is the impedance ratio, per www.geofex.com/ampdbug/outtrans.htm . Got close to expected impedances, but realizing this was at only one freuqency (60Hz), I went to a sig gen.
- I just want a method to determine the impedances of unknown OPTs....
- You say "Most OPT's are measured at 1K" this is 1K Hertz?
- I was using a Variac (at low-low voltages), driving the secondary and reading the primary, to get a voltage ratio, which squared is the impedance ratio, per www.geofex.com/ampdbug/outtrans.htm . Got close to expected impedances, but realizing this was at only one freuqency (60Hz), I went to a sig gen.
- I just want a method to determine the impedances of unknown OPTs....
Yes.
An OPT does not have an inherent "impedance". It has an impedance ratio, which as you state is the square of the turns ratio.
In an ideal world the transformer will function for all applications that need the same ratio....IE, a transformer rated with a 6600 ohm primary and an 8 ohm secondary will work as a 3300 to 4 ohm transformer and a 1650 to 2 ohm transformer....In the real world this holds true over a small range due to inductance limitations on the low frequency end, and capacitance limitations at the high frequency extremes.
I have a bunch of cheap OPT's designed for guitar amps that are rated for 6600 ohms input and have the usual 4, 8 and 16 ohm taps on the secondary. The data sheet states that they are designed for 80 VA at 80 Hz. They will support about 25 watts from 30 Hz to 20 KHz when used as a 6600 ohm OPT. However wiring an 8 ohm load to the 16 ohm tap reflects a 3300 ohm load to the tubes. In this configuration I can run about 60 watts at 30 Hz through these OPT's before they saturate. The main reason for this is the lack of primary inductance. If there aren't enough windings and enough metal in the transformer, it will saturate at low frequencies. Running the transformer at a lower impedance than it was designed for can sometimes help the low end.
Think of the OPT as a lever that is greatly off center. The fulcrum doesn't move, so the ratio doesn't change, but the beam may bend if you try to lift too much!
Googling can turn up methods for measuring the primary and secondary inductances, the leakage inductance, and some of the associated capacitances, but those numbers will not adequately predict the performance of a given OPT in a given circuit. I measure the ratio of an unknown OPT and mark it. Testing in your chosen circuit is the best way to figure out what it will do.
Your explanation helps tremendously! That a speaker load "reflects" a load to the tube(s) based on the ratio (the off-center lever...) makes sense. I'm having to fight my DC instincts to see it as a resistive load when it's not. I need to clarify something. If I put 1 VAC on the secondary of an unknown OPT and get 28.72 VAC from the primary, this voltage ratio is 28.72:1 (and you can call this the "turns ratio"), and it squared is 825:1 (the impedance ratio), meaning an 8 ohm speaker would "reflect" 8*825 = 6600 ohm load on the circuit?
Yes, that is correct. Given a clean 1 volt source it will usually work that way.
A real transformer has stray inductance and capacitances associated with it, and measuring any winding without a load on it can invite errors (especially if the source has significant distortion). Placing a resistor across the primary, even a large one, say 10 to 30 K may help to reduce these errors, but in some setups it may not matter.
The OPT primary is a coil of wire wrapped around a metal core. It has a DC resistance based only on the resistive losses in the copper wire. Usually this is a few hundred ohms. You can measure it with an ohmeter.
The action of the coil and the core will give the primary a certain amount of inductance. This inductance is not constant. It will vary with the DC current and the applied AC signal. The variation is minimal over a fairly large region where the core is relatively linear. The transformer acts like a lever in this region.
Technically, the inductance has an associated "inductive reactance" which acts like a variable AC resistor, and it is effectively in parallel with the primary. This variable AC resistance is proportional to the applied frequency....It goes down as the frequency goes down. You can see why a transformer works better at 1KHz than say 20Hz. This variable resisrance steals more and more of your signal at lower frequencies. It is inversely proportional to the number of turns of wire, and the size of the core. This shows why you need big transformers to get big bass.
If a sufficiently large signal at a low enough frequency is applied to a transformer primary, saturation occurs. This means that the core can not be further magnetized. We have left the linear region and the lever has bottomed out. One end is on the ground and no more weight (current) on the bottom end will raise the other end. At this point the OPT's inductance drops toward zero, leaving only the DC resistance to control tube current. An amp, particularly one with GNFB will try to apply more current. The output tubes will red plate and other bad stuff may happen.
Back in the 60's there was an old saying. "Never play bass through a bandmaster" I watched this saying proved correct way back then. I didn't have a clue then, but now I know that the tiny OPT was saturating and 6L6's finally fried.
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