

Practical Applications of Simple Math  Part II
June 1944 QST Article 
June 1944 QST
Table
of Contents
Wax nostalgic about and learn from the history of early electronics. See articles
from ARRL's
QST, published December 1915  present. All copyrights 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 fourpart
tutorial that covers resistance and reactance, amplifier biasing
(tubes since the ShockleyBardeenBrattain 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).Practical Applications of Simple Math
 Part II
Part II  Plate and Screen Voltages
By Edward M. Noll, EXW3FQJ
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
audioamplifier stages is that the inductance of the transformer
primary winding will provide a highimpedance 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. 6A, 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. 6B. The screen
voltage, E_{s}, is equal to the voltage drop across R_{b}.
Therefore,
E_{s} = R_{b}I_{b}.
Example: In Fig. 6B, 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 gridbiasing 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 gridbiasing
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 biassupply 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 gridbiasing, 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.
8B, 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 (300volt 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 300volt 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. 8B 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 squarewave 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 directcoupled amplifiers.
From the equivalent circuit of Fig. 10B, 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. 10B, 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



