Majority of tube amplifiers on the market have an output transformer. Only a very small group of OTL or Output Transformer-Less tube amplifiers get away without them. You may have noticed that transistor amplifiers also rarely have output transformers.
To understand their purpose, we must first dive into a little physics on basic transformer properties. Since the purpose of this discussion is in the results and not the math, we will avoid the derivations necessary from basic magnetic and electric laws called Maxwell’s Equations. If you have never head of Maxwell before, don’t worry!
1. Transformer Basics
A. Parts of a Transformer
Transformers are very simple devices (in theory) that have three basic components:
· Primary Coil
· Secondary Coil
· Core
The primary coil is the side of a transformer that is connected to the source and the secondary is connected to the load. The core is a ferromagnetic material, which couples the primary to the secondary. A voltage on the primary will induce a voltage on the secondary scaled by a proportional factor.
The core can be made from many different materials depending on the design goals of a particular transformer. The core acts as an intermediary stage so a process of induction or magnetic coupling can take place in a controlled fashion. The core directs what is called magnetic flux (think of an invisible field of magnetic energy), which is induced into the core by the primary, through the winding of the secondary. The core helps eliminates losses.
A simple transformer can be made without a core simply by having a primary and secondary coil in close proximity in free space; however, coupling will be poor since the flux is not directed through the secondary by a core material. In a coreless transformer, the magnetic field of the primary will radiate into free space so the secondary can only receive a portion of the magnetic field energy that happens to pass through the secondary coil.
B. How Transformers Work
Current flowing through any copper wire will create a magnetic field radiating in space around the cross-sectional diameter of the wire.
The primary winding of a transformer when energized behaves in the same manner. It is wrapped around the core a number of times where the magnetic field created by the current flowing in the wire is coupled or induced into the core material. The ferrous core will now hold this energy as flux and distribute it around the path of the core equally. The core simply acts as the pathway to transfer the energy of the primary into the secondary.
The strength of the core flux is dependent on the number of times the primary is wrapped around the core. The more turns of wire around the core, the more the flux.
Like the primary, the secondary is also wound around the core, so the fluctuations of the core’s flux will then be coupled back into the copper windings of the secondary, causing a proportional current. With an ideal transformer, no energy will be lost from primary to secondary.
Transformers only couple AC or alternating current since the law of induction states that only a change in magnetic field will cause induction into a magnetic material. Since AC is constantly changing with respect to time, making a sinusoid pattern, the entire waveform will be coupled through a transformer. Any steady state or DC current will not. We can even offset the AC with large DC value and still only the AC will pass through.
This is one reason why homes are wired with AC as found in a wall socket. In the USA 120V RMS is the standard. This allows a simple and convenient means for using a power transformer.
C. Power Transformers Step Up or Down Voltages
You may be familiar with the fact transformers step up or step down voltage and current. We can find them in the wall wart boxes that plug into the wall socket to charge our cell phones, computers, and other electronic devices. Here they are taking the 120V or 240V AC from the wall and stepping it down or up to the required voltage to run the device. These are generally called power transformers.
It is easy to calculate the step up or step down voltage ratio. It is just the ratio of the number of turns of wire around the magnetic core from secondary to primary.
Voltage ratio equation: Vs/Vp = Ns/Np
· Np = number of turns in primary
· Ns = number of turns in secondary
· Vp = primary voltage
· Vs = secondary voltage
Example:
Np = 100
Ns = 200
Vs/Vp = Ns/Np = 200/100 = 2
The factor Vs/Vp = 2 for this transformer. This means that the secondary has twice the windings of the primary and will thus scale the primary voltage by a factor of two.
If Vp was equal to 120V.
Vs = 2(Vp) = 240V
D. Conservation of Power
The fact that transformers step up or down voltages is not particularly usefully for an audio output transformer.
We still need to tie in one more piece of information:
The power delivered to the primary of transformers is equal to the power available at the secondary. What goes in must come out!
This is assuming zero power losses within the transformer itself, which is a very good assumption. Most quality transformers are in the 94% to 98% efficiency range if properly designed.
Electrical power is defined as voltage (V) multiplied by current (I).
Power Equation: Power = I x V
So…
Ip x Vp = Is x Vs
The primary power must equal the secondary power
Combining this with the voltage ratio equation above yields:
Transformer Ratios: Is/Ip = Vp/Vs = Np/Ns
E. Transformer Impedance Reflection
Now we are ready to calculate the important result of output transformers with a few algebraic steps. Using good old Ohm’s Law we can find out what the resistance is of the primary in terms of the secondary.
Ohm’s Law: V = I x R
R = V/I
So looking into the primary of a transformer:
Rp = Vp/Ip
Substituting the transformer ratios:
Rp = ((Np/Ns)Vs) / ((Ns/Np)Is)
Rp = (Np/Ns)^2 x (Vs/Is)
Since Vs/Is = Rs
We get the result:
Reflected Impedance Equation: Rp = (Np/Ns)^2 x Rs
This is the result we have been looking for! It states that the primary resistance, Rp, is equal to the ratio of the windings squared multiplied by a load attached to a secondary Rs.
Based on the winding ratio, a load on the secondary is scaled onto the primary, which was derived from the relationship of voltage and current! This is called reflected impedance.
F. Basic Transformer Equations
From some simple arguments we have derived the most useful equations pertaining to transformers:
Transformer Ratios: Is/Ip = Vp/Vs = Np/Ns
Reflected Impedance: Rp = (Np/Ns)^2 x Rs
Using these equations we can go on to discuss output transformers.
2. Output Transformers
Output transformers have a completely different purpose than power transformers. Unlike power transformers, which are used to generate the correct voltages and currents to power a device, the main task of an output transformer is to match impedances from a loudspeaker to the active output devices. In a tube amp, these are the output tubes.
In changing the impedance, output transformers do step up or down voltages and currents, as they must! However, that is a secondary effect to their main purpose of load matching.
A. Tubes are High Impedance Devices
By nature, vacuum tubes are high impedance and high voltage devices. They typically operate with high voltage power supplies, by consumer standards, which can range from a couple hundred volts to over 1KV in specialized “high output” triode power tubes like the 211 and 845.
This means that a 40-Watt tube such as a KT88 biased at 70% of its maximum dissipation for an idle power of 28 Watts and a 500V plate voltage will have only 56mA or 0.056 Amps of current idling through the tube. At maximum signal swing for full power this could be up to a couple hundred milliamps of current.
Since we know that R = V / I….
At idle the impedance of the above KT88 tube is 500V/.056 = 8929 ohms
As we can see form this example, a large voltage and small current mean high resistance. This is why tubes are characterized as high impedance devices.
In contrast, an output MOSFET transistor with a 24V power supply can have well over 1A of signal current through it!
B. Speakers Demand High Current
Say we want to put 60 Watts of power through an 8-ohm loudspeaker. Let’s first find the necessary drive voltage across the load.
P = V^2 / R
V = Sqrt(P x R)
V = Sqrt(60 x 8)
V = 22 Volts
This means we need 22V of signal across an 8-ohm loudspeaker to achieve the necessary 60-Watts of power.
How about the current through the speaker?
I = V/R
I = 22V/8ohms
I = 2.75Amps
Compare these results to the KT88 tube example above. The KT88 can deliver the 60-Watts in a push-pull amp, but not directly. It cannot supply nearly enough current into an 8-ohm load. Its output impedance is way too high!
C. Enter the Output Transformer
Since transformers can reflect impedances from secondary to primary we use this to our advantage.
We can directly tie an 8-ohm speaker to the secondary of a transformer and, through the square of the turns ratio, reflect a much higher impedance onto the primary side. This primary is coupled to the output tubes. Thus, the output signal is transformed from high impedance to low, so a speaker can be driven effectively. Both the primary and secondary of the transformer will still see the same power, but now we have the low voltage high current we need to drive a loudspeaker.
Example:
For a particular amplifier we need to drive an 8-ohm speaker from the secondary of an output transformer. If we have a transformer with 500 turns on the primary and 25 turns on the secondary what would the primary impedance be?
Reflected Impedance: Rp = (Np/Ns)^2 x Rs
Rp = (500/25)^2 x 8 = (20)^2 x 8
Rp = 400 x 8
Rp = 3200 ohms
We just designed a transformer that has a primary impedance of 3200-ohms with an 8-ohm load attached across the secondary.
Our tube will be much happier than connecting the 8-ohm speaker directly!
D. Output Transformers Set the Load on the Output Tube
From the above example you should be convinced that that tube will see 3200-ohms thanks to the turns ratio of the transformer. Thus, this sets up the load on the tube. To determine if a 3200-ohm load is acceptable you need to pull out the data sheets for the specific tube you want to use.
The “correct” load however is much more complicated than meeting a specific number for a specific tube. It is up to the designer to find the right compromise since this load impedance from the transformer will set up the operating points of the tube as it amplifies signals.
This means that the load impedance on the tube will set up bias points and determine power, distortion characteristics, and output impedance. Commonly in power amplifiers, most designers choose a transformer impedance to load the power tube for maximum power.
When paralleling output tubes, the load impedance is generally halved to achieve twice the power as before since twice the signal current will be produced for the same signal voltages as with one tube. Adding more tubes in parallel cannot increase the maximum current unless the transformer impedance is lowered.
However, transformer impedance can be picked to achieve better distortion figures and lower output impedance, which is often critical in non-feedback amplifiers. This will however come at the expense of efficiency and output power.
E. Transformers Impedance Ratio Determine the Amplifier Output Impedance
As mentioned above, the output impedance of an amplifier is directly related to impedance ratio of the transformer. Just as a speaker connected to the secondary will reflect a load onto the primary, any impedance on the primary side of the transformer will reflect a load onto the secondary. Since secondary is considered the output of the amplifier, the reflected impedance from primary to secondary is the output impedance!
Say a tube has a nominal impedance of 7500 ohms and is connected to a transformer with a 3200-ohm primary when the secondary is loaded with an 8 ohm load.
The impedance ratio of the transformer is simply: 3200/8 = 400
Any load on the primary is reflected over to the secondary by a division of 400.
So, the Output impedance = 7500/400 = 18.75-ohms
This output impedance is way too high, so we can try to increase the impedance ratio of the transformer to divide the plate resistance down to a smaller number.
However, if a 3200-ohm primary gave us max power for a specific tube, increasing the ratio for lower output impedance will begin to reduce the available power.
We can also employ negative feedback to help reduce the impedance. This is commonly done; however, the topic of feedback is complex and should be saved for another day.
F. What Makes a Good Output Transformer?
Unlike power transformers, which operate exclusively at 50 or 60Hz frequency from the wall, output transformers, line-stage, and other transformers used for impedance matching must operate under all expected frequencies. For audio, human hearing is 20Hz to 20KHz. We want a transformer that can extend significantly beyond this range for good stability and sound reproduction without altering the frequencies transmitted through them.
These alterations include phase, (the output leading or lagging in time from the input), amplitude, and distortion to the passed signal. An ideal transformer
would have none of these problems. However, the electrical equivalent model of a real transformer is extremely complex. At some frequency extreme, both low and high, these nonlinearities will begin to dominate.
Winding technique, materials, and design all play a role in the quality of a transformer. It takes years of experience and practice to design and build high quality output transformers. It is for this reason that output transformer are almost always the single most expensive item in a tube amplifier and their quality cannot be scarified if the goal is a great sounding amplifier.