June 1944 QST
[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. 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-W3FQJWhenever 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