Think of a triode valve. You have a cathode, an
anode and a control grid. The potential on the control grid can control
the flow of electrons through the valve, but only when it is negative with
respect to the cathode. If the grid is at zero volts or is positive, it is
unable to repel electrons and hence cannot control the current flow.
It is therefore necessary to put a standing negative potential on the grid
such that it remains at least slightly negative at the most positive
expected excursion of the grid signal. This negative potential is know as
the grid bias, or just bias. The bias also sets the quiescent current
through the valve and hence it's operating class.
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How is bias usually applied?
Think of a triode valve. You have a cathode, an
anode and a control grid. The potential on the control grid can control
the flow of electrons through the valve, but only when it is negative with
respect to the cathode. If the grid is at zero volts or is positive, it is
unable to repel electrons and hence cannot control the current flow.
It is therefore necessary to put a standing negative potential on the grid
such that it remains at least slightly negative at the most positive
expected excursion of the grid signal. This negative potential is know as
the grid bias, or just bias. The bias also sets the quiescent current
through the valve and hence it's operating class.
How is BIAS applied?
Bias is applied by two different means. Firstly, it is possible to
configure a dedicated negative supply from the amp's power unit and use
this to set the grid negative. This is usually how bias is set for output
valves and since valves differ in their characteristics, a measure of
adjustment is usually provided for each valve so that the bias voltage can
be adjusted to give identical quiescent currents in each valve. This is
very important in push-pull circuits as any imbalance can result in output
transformer core saturation and hence distortion.
Bias may also be applied by putting a suitable resistor between the
valve's cathode and earth. The current flow through the valve causes a
voltage drop across this resistor, placing the cathode at a positive
potential. This means that the grid will be negative with respect to the
cathode, providing bias. The presence of a resistor in the cathode circuit
reduces the gain of the stage, so it is normally bypassed with a capacitor
so that alternating current (the signal) sees a low impedance to earth,
maintaining the AC gain of the stage. This type of bias is
self-regulating, since an increase of valve current will cause a larger
voltage drop across the resistor which will reduce the current. This
method is invariably used in audio applications for biasing small signal
valves.
The two bias schemes can be used together in a compound bias arrangement
as in WADs KEL34 and KiT/Kat88 output stages.
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What is meant by ‘Class A’, ‘Class B’ and all that sort of
stuff?
This refers to the portion of the signal
waveform that is amplified by a valve. In Class A, the entire signal
waveform is amplified. Quiescent current is set at a level such that anode
current continues to flow even on the most negative excursion of the
control grid voltage, so in Class A, the valve never goes into cut-off. In
Class B operation, a valve has no quiescent current and only conducts
during the positive half-cycle of the signal.
This is only of practical audio use when used in push-pull configuration.
Each valve in a push-pull pair amplifies half of the signal waveform, one
valve amplifying the positive-going half-cycle and the other valve the
negative-going half-cycle. Each valve hands over to the other at the zero
crossing point. This handover gives rises to crossover distortion emerging
from the fact those tiny differences between components make it impossible
to synchronise the changeover perfectly. The payback for Class B operation
is that the circuit is much more efficient, producing more power with less
waste heat. Class AB, unsurprisingly, refers to an intermediate situation
where each valve goes into cut-off for less than half the signal cycle.
Quiescent current is set between Class A and Class B limits.
This mode allows the valves of a push-pull pair to operate in Class A up
to a certain output level, moving toward Class B as output increases. This
is a compromise often used in audio amplifiers, since crossover distortion
is more objectionable at low levels of output. You may see references to
Class AB1 and Class AB2. The suffixed numbers simply tell us whether the
control grid of the valve is allowed to swing positive and hence conduct
grid current (AB2) or not (AB1). Since both the valve itself and the
driver circuitry has to be carefully designed to permit grid current to
flow, you will not see Class AB2 operation often. Class C operation occurs
where anode current flows through the valve for less than a half-cycle.
This is an efficient operating mode but gives rise to high distortion.
It is used mainly in RF circuits where the original waveform can be
regenerated by suitable circuitry. Classes of operation beyond C denote
digital amplification and are beyond the scope of this answer.
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What is Impedance? Is it the same as resistance?
In the following formulae, we will use * to denote
multiplication, / to denote division and + and - having their usual
meanings.
Resistance and impedance are measured in the same units (ohms) but they are
not quite the same. Resistance is a measure of the ability of a material to
oppose the flow of current. It is determined from Ohm\'s famous law which
states that the resistance of a conductor is equal to the voltage across it
divided by the current flowing through it. Resistance is usually applied to
DC current circuits whereas Impedance, which is dependent on frequency, is
used in AC circuits. Ohms Law is usually stated as....
V=I*R
So a device (resistor) with a resistance of 10 ohms and having a current of
1 amp flowing through it would have a voltage of 10 volts across its ends,
thus
V=10 * 1
Resistance works identically for AC and DC.
There is another type of opposition to current flow which is dependent on
whether the applied voltage is AC or DC and, indeed, on the frequency of the
AC wave. This frequency-dependent opposition to current flow is called
REACTANCE and it arises from physical mechanisms beyond the scope of this
answer. Inductors and capacitors exhibit this reactance but differ in the
way it is determined.
For an inductor
Xi = 2*pi*f*L
where Xi denotes inductive reactance, pi is 3.142, f is the frequency of the
AC supply in Hertz and L is the inductance measured in Henries. So inductive
reactance increases with frequency.
For a capacitor
Xc = 1/(2*pi*f*C)
where Xc denotes capacitive reactance, pi is 3.142, f is the frequency of
the AC supply in Hertz and C is the inductance measured in Farads. So
capacitive reactance decreases with frequency.
IMPEDANCE is simply the sum of the reactance of a component and any
resistance it may have.
Example:
A 10 Henry inductor has a DC resistance of 200 ohms. What is its impedance
at a supply frequency of 100 Hz?
Xi = 2*pi*f*L
Substituting:
Xi = 2*3.142*100*10 = 6284 ohms
Impedance = 6284 + 200 = 6484 ohms
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What is a Push-Pull Amplifier?
Named for the configuration of the output stage.
We saw in another FAQ answer that Class B operation is efficient, but
suffers from the disadvantage of outputting an amplified, half-wave
rectified version of the original signal.
This won't do for Hi-Fi!
Fortunately, we can get around the problem by doubling up on valves. One
valve is fed with the signal to be amplified and the other is fed with an
inverted signal. During the first half-cycle, the first valve conducts and
during the second half-cycle the second valve does so. If we now invert
one of the outputs and add them together, we have neatly reconstituted the
original waveform. This is the basis for the push-pull amplifier. The
first inversion is carried out by the phase-splitter and the second by the
output transformer.
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What is a Single-ended amplifier?
Named for the configuration of the output stage.
A single-ended output stage is so arranged that the output device
amplifies the entire signal. There is no phase splitter and the output
transformer primary is used as a straightforward anode load. SE output
stages operate in Class A by definition. The circuitry of a single-ended
amp tends to be simpler than a push-pull design and the
electrical/magnetic design characteristics of the output transformer is
also more straightforward. The downside is that it becomes more difficult
to obtain higher levels of output power, as Class A is the least efficient
operating mode. The output transformer primary also has to carry the
quiescent direct current for the output valve and consequently needs to be
both physically large and constructed with an air gap in the core to avoid
saturation. These design compromises again limit the power achievable.
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What is S.R.P.P?
The best description we have found can be read
at:
http://www.tubecad.com/may2000/
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Why do some people use mF for microFarad and others use uF for
Capacitance Units?
The Greek letter mu (like a u but with a tail starting the letter) is
the correct symbol for micro. Mu isn’t available on many keyboards so m
is often used instead., as millifarads (m is used for milli in the SI
system) are not generally used in electronics (although it is a scientific
unit). Others use the keyboard letter u for the Greek letter mu as it
looks similar and can`t be confused with any other notation.
In older circuits you’ll often see mmf which is not m for milli
either but means million millionths of a farad, or pF picofarads as we
know them now.
The units are,
1 Farad (F) =
1000 milli Farad (mF - not generally used) =
1000 000 micro Farad (uF) =
1000 000 000 nano Farad (nF) =
1000 000 000 000 pico Farad (pF - old mmF)
Some capacitor values may be marked in alternative ways such as a
number followed by K being used to show thousands of pF (= number of nF).
Most formulae use the full Farad, so convert the value for
calculations. The units relate by thousands/thousandths so remembering
their names, including milliFarads, is useful for this.
Example,
10,000pF = 10nF = .01uF = .00001mF = .00000001F
All we do is move the decimal point 3 places at a time. If doing a
calculation (i.e. to find the impedance of the capacitor at a known
frequency) the last value, expressed in Farads, would be the one to use.
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What are Matched Valves and do I need them?
Each type of valve works within a set of electrical
parameters. These include anode voltage, screen voltage (if applicable) and grid
bias. When a signal is applied to the grid it is amplified by a designed amount.
These values are termed the valve's "characteristics" and are
different for each type of valve. There is a tolerance in manufacture, which
means that although they meet their type specification, they will vary somewhat
between samples.
Valves are normally checked on a valve tester, such as the AVO or Hickock range,
against a set of "characteristics". The fixed settings for the heater,
anode voltage, screen voltage, and grid bias are set on the machine.
A signal measured in mA current is applied and the resultant amplification read
as mA/V. Thus 2 variables are recorded for each valve, current, in mA, and
"slope" in mA/V. Although the valve is being measured at only 1 point
on it\'s response curve this usually gives a good indication of it`s performance
for the purpose of establishing good/bad and matching with other valves for use
in pairs or quads.
If the operator is willing, or is using more sophisticated matching equipment,
the valves can be compared at more than one point and a response curve built up
which can be compared to the valves published "characteristic curve".
In an amp the first stage amplifier valves can be matched between channels to
provide a similar level of amplification between channels and so help the stereo
performance. The main and most essential use of matching though is when using
pairs of output valves in "push pull" output stages. Here the 2 valves
must share the work equally. If one draws more current than the other under the
amps operating conditions it will cause an imbalance in the output transformer,
which will partially saturate it, and upset it's response. If the valves are
very badly matched one may draw increasing current until it fails
catastrophically.
In practice, we should aim to match our 2 variable values of mA current and mA/V
slope to within 10% to make push pull pairs. This is more critical in some
designs than others but is a usual tolerance for cathode ("automatic")
bias amps. Match current as priority, to maintain transformer balance, then
slope to match amplification.
It is here, incidentally, that we see the special need for matched quads of OP
valves in parallel push pull amps like KEL80. To share the work equally each
valve of a pair of valves on each "side" of the transformer need to be
matched, then each pair needs to match the pair on the opposite side.