[Table
of Contents]These articles are scanned and OCRed from old editions of the
ARRL's QST magazine. Here is a list of the
QST articles I have already posted. As time permits, I will
be glad to scan articles for you. All copyrights (if any) are hereby acknowledged. |
Recognizing that many people were reluctant
to approach the theoretical aspect of electronics as it applied to circuit design and analysis, QST (the
American Radio Relay League's monthly publication) included equations and explanations in many of their
project building articles. Occasionally, an article would be published that dealt specifically with how
to use simple mathematics. In this case, the June 1944 edition, we have the second installation of at
least a four-part tutorial that covers resistance and reactance, amplifier biasing (tubes since the
Shockley-Bardeen-Brattain trio hadn't invented the transistor yet) oscillators, feedback circuits, etc.
I do not have Part I from the May 1944 edition or Part IV from the August 1944 edition, but if you want
to send me those editions, I'll be glad to scan and post them
(see
Part III here).
See all available
vintage QST articles.
Practical Applications of Simple Math - Part II
Part II - Plate and Screen Voltages
By Edward M. Noll, EX-W3FQJ
Whenever the d.c. plate current flows through any resistance placed in the plate circuit of a vacuum tube
as a load or coupling medium, it is obvious that the voltage at the plate will be less than the supply voltage because
of the voltage drop across the resistance.
In
Fig. 1 the plate voltage is
E
_{p} = E
_{b} - R
_{p}I
_{p}.
Example: In
Fig. 1,
E
_{b} = 250 volts. R
_{p} = 10,000 ohms.
I
_{p} = 10 ma. (0.01 amp.).
What is the plate voltage, E
_{p?}E
_{p} = 250 - (10,000) (0.01) = 250 - 100 = 150 volts.
Since true plate voltage is the voltage between plate and cathode, the voltage drop across the cathode resistor,
R
_{k} in Fig. 2, as well as the drop across the plate resistor, R
_{p}, must be subtracted from the
supply voltage in calculating plate voltage.
In
Fig. 2 the plate voltage is
E
_{p} = E
_{b} - I
_{p}R
_{p}- I
_{p}R
_{k}.
= E
_{b} - Ip(R
_{p} + R
_{k}).
Example: In Fig. 2,
E
_{b} = 250 volts.
R
_{p} = 25,000 ohms.
R
_{k} = 2000 ohms.
I
_{p} = 5 ma. (0.005 amp.).
What
is the plate voltage, E
_{p}?
E
_{p} = 250 - (0. 005) (25,000 + 2000)
= 250 - (0.005) (27,000) = 250 - 135
= 115 volts.
One advantage of transformer
coupling between audio-amplifier stages is that the inductance of the transformer primary winding will provide a high-impedance
load for the tube at audio frequencies, while the d.c. resistance of the winding is sufficiently low to cause only
a small drop in d.c. plate voltage.
In
Fig. 3 the only resistance affecting the plate voltage is that of the transformer primary winding, R
_{t},
so
E
_{p} = E
_{b} - I
_{p}R
_{t} Example: In Fig. 3,
E
_{p}
= 250 volts. I
_{p} = 20 ma. (0.02 amp.)
R
_{t} = 100 ohms.
What is the plate voltage, E
_{p}?
E
_{p} = 250 - (0.02) (100) = 250 - 2
= 248 volts.
Screen voltage is determined in the same manner as plate voltage, using the screen current to calculate the voltage
drop across the screen resistor.
E
_{p} = E
_{b} - I
_{p}R
_{p} E
_{s}
= E
_{b} - I
_{s}R
_{s}
Example:
In Fig. 4,
E
_{b} = 250 volts. I
_{p} = 5 ma. (0.005 amp.).
R
_{p} = 20,000 ohms.
I
_{s} = 2 ma. (0.002 amp.).
R
_{s} = 75,000 ohms.
What are the plate voltage, E
_{p},
and screen voltage, E
_{s}?
E
_{p} = 250 - (0.005) (20,000) = 250 - 100 = 150 volts.
E
_{s} = 250 - (0.002) (75,000) = 250 - 150 = 100 volts.
In the circuit of Fig.
5 both plate and screen currents flow through the common resistor, R
_{1}, so that plate and screen currents
must be added in calculating the voltage drop across R
_{1}.
E
_{p} = E
_{b} - (I
_{p}
+ I
_{s}) (R
_{1}) - I
_{p}R
_{p} E
_{s} = E
_{b} -" (I
_{p}
+ I
_{s}) (R
_{1}) - I
_{s}R
_{s}.
Example:
In Fig. 5,
E
_{b} = 250 volts. R
_{p} = 40,000 ohms.
R
_{s} = 200,000 ohms. I
_{p}
= 2 ma, (0.002 amp.)
I
_{s} = 0.5ma. (0.0005 amp.). R
_{1} = 20,000 ohms.
What are the plate
voltage, E
_{p}, and screen voltage, E
_{s}?
E
_{p} = 250 - (0.002 + 0.0005) (20,000)
- (0.002) (40,000)
= 250 - 50 - 80
= 120 volts.
E
_{s}= 250 - 50- (0. 0005) (200,000)
= 250 - 50 - 100
=
100 volts.
In
the circuit of Fig. 6-A, the screen voltage, E
_{s}, is obtained from a tap on a voltage divider consisting
of R
_{s} and R
_{b} The equivalent circuit is shown in Fig. 6-B. The screen voltage, E
_{s},
is equal to the voltage drop across R
_{b}. Therefore,
E
_{s} = R
_{b}I
_{b}.
Example: In Fig. 6-B,
E
_{b} = 250. I
_{s} = 1 ma. R
_{s} = 10,000 ohms.
R
_{b} = 40,000 ohms.
What is the screen voltage, E
_{s}?
E
_{s} = I
_{b}R
_{b}.
Since E
_{b} is equal to the sum of the voltages across R
_{s} and R
_{b},
E
_{b} = R
_{s}I
_{sr} + R
_{b}I
_{b}.
Also, since both I
_{b} and
I
_{s} must flow through R
_{s}, I
_{sr} = I
_{b} + I
_{s}.
Substituting
this value for I
_{sr} in the above equation,
E
_{b} = R
_{s} (I
_{b} + I
_{s})
+ R
_{b}I
_{b}.
Transposing,
R
_{s}I
_{b }+ R
_{b}I
_{b}
= E
_{b} - R
_{s}I
_{s}I
_{b}(R
_{s} + R
_{b}) = E
_{b} - R
_{s}I
_{s}
Substituting known values,
Then,
E
_{s} = (0.0048) (40,000) = 192 volts.
In
the circuit of Fig. 7, both screen and grid-biasing voltages are taken from voltage dividers. In the case of the divider
in the grid circuit, the voltage division is in exact proportion to the resistance values of the divider sections,
since it is assumed that the grid is biased so that no grid current flows. Therefore, the grid-biasing voltage, E
_{g},
is the voltage developed across R
_{2} by virtue of the current flowing through it from the bias supply.
E
_{g} = I
_{g}R
_{2}
being the bias-supply voltage.
Substituting,
Screen and plate voltages are calculated as before.
Example: In Fig. 7,
E
_{b} =
250 volts. E
_{c} = 100 volts.
R
_{1} = 49,000 ohms.
R
_{2} = 1000 ohms.
R
_{3} = 30,000 ohms.
R
_{4} = 20,000 ohms. R
_{5} = 20,000 ohms.
I
_{s}
= 1 ma. (0.001 amp.).
I
_{p} = 5 ma. (0.005 amp.).
What are the grid-biasing, screen and plate
voltages?
E
_{g = }
= 2 volts (negative in respect to cathode).
E
_{p} = 250 - (0.005) (20,000) = 250 - 100
= 1.50 volts.
E
_{s} = I
_{b}R
_{3}
= 0.0046 amp.
E
_{s} = (0.0046) (30,000) = 138 volts.
Fig.
8 is used to illustrate the effects of low voltmeter resistance upon the accuracy of voltage measurements. R
_{m}
is the meter resistance.
With the meter disconnected, the plate voltage will be
E
_{p} = E
_{b}
- R
_{p}I
_{p}.
However, with the meter connected, the current, I
_{m}, will flow
through R
_{p}. Thus, the voltage drop across R
_{p} will increase and the plate voltage will be lowered,
especially when the resistance of the meter is low in comparison with R
_{p}. The equivalent circuit with the
meter connected is shown in Fig. 8-B, in which R
_{pi} is the internal resistance of the tube which is assumed
to be constant with a change in plate voltage. The new plate voltage desired is the voltage across R
_{pi}
(or R
_{m}) which is
E
_{p} = E
_{b} - (I
_{pm}) (R
_{p}),
where I
_{pm} is the new current when R
_{m} is connected. In other words, E
_{p} is the difference
between the terminal voltage and the voltage drop across R
_{p}.
Example: In Fig. 8,
E
_{b}
= 250 volts. I
_{pi} =0.1 ma. (0.0001 amp.)
R
_{p} = 1 megohm (1,000,000 ohms)
R
_{m}
= 1000 ohms per volt (300-volt scale).
What is the true plate voltage with the meter disconnected and what
voltage will be measured by the meter when it is connected?
E
_{p} = 250 - (1,000,000) (0.0001)
= 250 - 100
= 1.50 volts = plate voltage without meter connected.
As stated above, when the meter is connected,
E
_{p} - E
_{b} - (I
_{pm})
(R
_{p}).
Since I
_{pm} is not known, its value must be found before the equation can
be solved. To find I
_{pm}, the resultant resistance of R
_{m} and R
_{pi} in parallel must be
found, and this, in turn, requires that R
_{pi} be determined. This can be done by considering the circuit
before the meter is connected. The total circuit resistance, R
_{t} is then given by
= 2,500,000 ohms
= 2.5 megohms.
Also,
R
_{t} = R
_{p}
+ R
_{pi} R
_{pi} = R
_{t} - R
_{p} = 2,500,000 - 1,000,000
R
_{pi} = 1,500,000
ohms = 1.5 megohms.
The resistance of the meter, R
_{m}, is given as 1000 ohms per volt. Since
the meter has a 300-volt scale, its resistance is 300,000 ohms, or 0.3 megohm.
R
_{pim}, the resultant
resistance of R
_{pi} and R
_{m} in parallel is given by
R
_{pim = }
This gives the total circuit resistance in Fig. 8-B as
R
_{t} = R
_{p} + R
_{pim}
= 1 + 0.25 + 1.25 megohms.
The new current, I
_{pm}, is then
I
_{pm = }
Then
E
_{p} = E
_{b} - (I
_{pm}) (R
_{p})
= 250
- (0.0002) (1,000,000)
= 250 - 200 = 50 volts = voltage indicated by meter reading.
Example:
In the case of Fig. 9, it is assumed that the grid is to be fed a square-wave pulse. Compare the plate voltage when
the tube is conducting a current of 15 ma. with the effective plate voltage when the tube is idle and not drawing
plate current. The plate resistance is 10,000 ohms.
When the tube is conducting,
E
_{p}
= E
_{b} - I
_{p}R
_{p} ;= 250- (0.015) (10,000)
= 250 - 150 = 100
volts.
When the tube is not conducting, there is no voltage drop across R
_{p} and the plate
voltage is 250, the same as the supply voltage, E
_{b}.
Fig.
10 illustrates another use for the voltages divider. The coupling circuit shown is that commonly found in direct-coupled
amplifiers. From the equivalent circuit of Fig. 10-B, it will be seem that the plate of the first tube is connected
at one tap on the voltage divider, while the grid is connected at another tap less positive. It is assumed that the
grid of the second tube is biased, by the voltage drop across its cathode resistor, so that the grid does not draw
current.
Example: In Fig. 10,
E
_{b} = 250 volts. I
_{p} = 5 ma. (0.005 amp.)
R
_{1} = 10,000 ohms. R
_{2} = 75,000 ohms
R
_{3} = 25,000 ohms.
What are
the plate voltage of the first tube, and the grid voltage of the second tube?
The total drop across
all resistors is, of course, equal to the applied voltage, E
_{b}. The voltage, across R
_{1} is R
_{1}I
_{1},
while that across R
_{2} and R
_{3} in series is (R
_{2} + R
_{3}) (I
_{2}), bearing
in mind that no current is being drawn from the tap, marked G in Fig. 10-B, so that the same current flows through.
R
_{2} and R
_{3}. Then,
E
_{b} = R
_{1}I
_{1}+ (R
_{2} +
R
_{3}) (I
_{2})
Since both I
_{p} and I
_{2} flow through R
_{1},
I
_{1} = I
_{p} + I
_{2} Substituting this value for I
_{1} in the preceding
equation,
E
_{b} = (I
_{p} + I
_{2}) (R
_{1}) + (R
_{2} + R
_{3})
(I
_{2}).
Substituting known values,
250 = (0.005 + I
_{2}) (10,000) + (75,000 + 25,000)
(I
_{2})
= 10,000I
_{2} + 50 + 100,000I
_{2}
110,000I
_{2} = 200
I
_{2}= 0.0018 amp. = 1.8 ma.
The plate voltage of the first
tube is equal to the sum of the voltage drops across R
_{2} and R
_{3}.
E
_{p} = I
_{2}
(R
_{2} + R
_{3}) = (0.0018) (100,000)
= 180 volts.
The grid
voltage of the second tube is equal, to the voltage drop across R
_{3}.
E
_{g} = I
_{2}R
_{3}=
(0.0018) (25,000) = 45 volts.
Posted 12/28/2012