Transcendent Sound

Making Music Better

 

Amplifier Output Impedance
Why It’s So Important 


Some amplifier specifications have little impact in the way an amp sounds but one has a huge impact in the way all amplifiers sound.  This is the specification of output impedance.  This article explains what it is and what it does. 

There are two major areas where output impedance affects the way a loudspeaker sounds.  One is frequency response and the other is controlling the motion of a loudspeaker cone.  Both are equally important and are affected by output impedance in a similar fashion.  The case of frequency response is simpler so let’s look at that first.  We need to understand how to “model” amplifier output impedance so its characteristics can be investigated and explained.  Modeling is a way to analyze circuit behavior with simple numerical relationships.  

A simplified electrical model of an amplifier driving a loudspeaker is shown below.





                     





There are three objects in the diagram.  The voltage source, e, and the output impedance, Ro, are inside the amplifier.  Neither exists in a physical sense.  There is no device inside an amplifier that is the voltage source.  Nor is there an output impedance component.  The circuit diagram represents a “model” of how an amplifier behaves.  The symbols represent the net effect of the behavior of the amplifier.  The third device, RL, represents a load placed on the amplifier like a loudspeaker.

 The voltage source is represented as ideal, meaning it has no internal impedance.  All of the amplifiers output impedance is represented by Ro.  The current, i, shown flowing in the loop is governed by Ohms law.  It is simply the voltage, e, divided by (Ro + RL). 

 Let’s plug some typical numbers into the diagram to see how it behaves.  Assume that the amp is a modern solid state design with an output impedance of 0.01 ohm, the load is an 8 ohm resistor, and the voltage being produced is 10 volts.  We must first determine the current, i, to understand how the model behaves.  Current i is found by dividing 10 by  (8 + .01) which equals 1.248 amps.  How is the voltage, e, distributed in the model with these parameters? 

We know from circuit theory that the sum of all the voltage drops in a loop must equal the source of the voltage.  We can calculate this by multiplying the loop current, i, by the impedances it passes through.  In this case, the voltage developed across the load, RL, is 9.988 volts.  Notice that the output voltage is slightly less than 10.  The remaining 0.012 volts was consumed by Ro.  It was consumed by the amplifier itself.  If the current loop was broken and RL was removed, the output voltage would rise to 10 volts, equal to the source.  This is because with no current flowing, there is no voltage drop across Ro

 Let’s see what happens if we raise the output impedance to 3 ohms which is a typical value for a single-ended triode amp with no feedback .  The current flow is now 10 divided by (3 + 8) which equals .91 amp.  The voltage produced across the load is (.91 x 8) which equals 7.27 volts.  This is quite a reduction over the first example.  As the output impedance of the amplifier increases, more output voltage is consumed internally.  The ramifications of this result are highly significant.

 If all loudspeakers had a constant impedance with frequency, high amplifier output impedance would only reduce voltage gain, (as shown above) but not materially affect the frequency balance of the sound.  Unfortunately, most loudspeakers have a widely varying impedance with frequency.

 An 8 ohm speaker can typically have an impedance dip to 4 ohms in the upper bass and rise to 20 ohms in the treble.  Let’s examine how output impedance can affect the frequency response of such a speaker.

 Assume an amplifier output impedance of 3 ohms and voltage output of 10 volts.  In the treble range the voltage produced across the speaker is

 10 ÷ (3 + 20) x 20 = 8.7 volts

 This is a reduction in output from a no load condition of 1.2 dB.

 Now let’s look at what happens at the upper bass range when the loudspeaker impedance falls to 4 ohms.

 10 ÷ (3 + 4) x 4 = 5.7 volts

 This is a reduction in output from a no load condition of 4.9 dB.  The sound pressure level of the speaker in the treble will then be 3.7 dB louder than in the upper bass.

 As amplifier output impedance rises, these differences in frequency response are exaggerated.  The changing speaker impedance actually pushes the output voltage of the amplifier around thereby changing the acoustic output of the speaker.  The amplifier cannot control its own output voltage.

 Why does speaker impedance change with frequency?

 The function of a speaker is to produce a constant sound pressure level for a constant drive voltage across a defined frequency range.   Power is calculated as voltage multiplied by current.  If the loudspeaker impedance drops and the amplifier voltage is held constant, then the current in the loudspeaker increases (current = voltage ÷ impedance).  If the current increases, then doesn’t the power consumed increase?  Yes!  That is exactly what happens.  If the power in the loudspeaker increases, then why doesn’t it play louder?  Because the speaker efficiency is changing.  As the speaker impedance decreases, its efficiency decreases.  All that matters is that the acoustic output remain constant. 

 Speaker enclosures are “tuned” to “load” the woofer cone at a particular frequency- the frequency of system resonance.  Look at a typical bass reflex speaker impedance graph.  There is always a very high impedance peak at system resonance.  At this frequency, the speaker efficiency goes way up.  The cone will move with great amplitude and make very boomy bass.  The enclosure and port system are tuned to mass load the cone at this frequency.  They put the brakes on the cone motion.  The wild cone motion is controlled and the acoustic output is held constant.   

 This is the way loudspeakers behave. Loudspeakers are not constant power devices.   Loudspeakers must have a constant drive voltage to provide a constant acoustic output with changing frequency.  As demonstrated by the simple mathematical relationships shown above, the only way to achieve constant voltage drive with a changing load impedance is by having very low amplifier impedance.

 Controlling Cone Motion

 A loudspeaker is a mechanical device.  It is essentially a motor.  A motor causes motion when a coil is exposed to a changing magnetic field.  When a woofer cone moves, its coil is moving in a constant magnet field.  That is the way generators produce electricity.  So how does an amplifier handle this speaker produced electric current?

 A loudspeaker can be mathematically modeled just as an amplifier can.  What happens is that the net effect of speaker motion presents itself to the amplifier as a set of complex reactive impedances (inductance and capacitance).  The amplifier output current passes through these impedances producing a voltage drop across them.  Just as in the case of changing speaker impedance causing problems with frequency response, this new voltage must be controlled by the amplifier output impedance or it will change the behavior of the cone motion. 

 The calculations are the same as shown above for frequency response but operate in reverse.  To keep the voltage produced by cone motion low, the current the speaker produces must pass through a very low amplifier output impedance.  The lower the amplifier output impedance, the lower the speaker produced voltage is.  The lower the speaker produced voltage, the less effect it has on cone motion.  A good rule to follow it that amplifier output impedance must be at least ten times lower than speaker impedance for good control of cone motion.  This is called damping factor.  Amplifier output impedance must be less than one ohm for best performance.

 Calculating Output Impedance       

 Output impedance is easy to calculate.  Let’s look at three types of amplifiers, single-end tubes amplifiers, push-pull tube amplifiers, and OTL amplifiers.  Solid-state amplifiers are similar to OTL’s.

Single-Ended
The first step is to calculate the impedances of the output devices.  One 300B has a plate impedance of 700 ohms.  If there were more than one tube, we simply divide the plate impedance by the number of tubes.  Two 300B tubes would then have a net impedance of 350 ohms.

 Next we calculate the impedance ratio between the primary and secondary windings of the output transformer.  If the primary were 3000 ohms and the secondary were 8 ohms, the ratio is then 3000 ÷ 8 = 375.  The output impedance is then the tube plate impedance divided by this number or 700 ÷ 375 = 2.13 ohms.  That’s it!  Negative feedback can be applied to reduce the impedance further. 

Push-Pull
Let’s use a pair of 6550’s with a transformer with a 5000 ohm primary.  Two 6550’s in push-pull have a combined plate impedance of about 10,000 ohms in pentode mode.  The transformer has an impedance ratio of 5000 ÷ 8 = 625.  The output impedance is then 10,000 ÷ 625 = 16 ohms.  In this case it is absolutely essential to employ negative feedback to get the output impedance down.  Amplifiers of this nature typically require about 25 dB of negative feedback to get the output impedance down to 0.8 ohms.  (25 dB of feedback reduces output impedance 18 times)

OTL
Let’s look at the Transcendent T8 as an example.  It uses 8, EL509 tubes in push-pull.  Each 509 has a plate impedance of about 150 ohms.  150 ÷8 = 18.75 ohms.  This is unacceptable.  Negative feedback is employed to achieve proper performance.  The amplifier uses 33 dB of negative feedback which reduces the output impedance to 0.4 ohm thereby achieving outstanding woofer control and the ability to drive 4 ohm speakers.  This specification was verified by Stereophile when they reviewed the amplifier.

 Different tubes will provide similar results.  The configuration of the OTL output stage does not matter whether it is series connected push-pull or balanced.  The controlling factors here are the number of output devices and the plate impedance of the devices.  A 6AS7G is a dual triode with a plate impedance of 280 ohms.  An amplifier that uses 8 such tubes will have an output impedance of  280 ÷ 16 = 17.5 ohms.  A fourteen tube amplifier will have an output impedance of 10 ohms.

 So it is absolutely essential that negative feedback be employed in the design of OTL amplifiers and tube amplifiers in general.  There is simply no other way to reduce output impedance to acceptable levels.