Practical Applications of Simple Math - Part II
June 1944 QST Article
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).
June 1944 QST
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 are hereby acknowledged.
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.
Fig. 1 the plate voltage is
Ep = Eb
Example: In Fig. 1,
Eb = 250 volts. Rp = 10,000 ohms.
Ip = 10 ma. (0.01 amp.).
What is the plate voltage,
Ep = 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,
Rk in Fig. 2, as well as the drop across the plate resistor,
Rp, must be subtracted from the supply voltage in calculating
Fig. 2 the plate voltage is
Ep = Eb
- IpRp- IpRk.
= Eb - Ip(Rp + Rk).
In Fig. 2,
Eb = 250 volts.
= 25,000 ohms.
Rk = 2000 ohms.
= 5 ma. (0.005 amp.).
What is the plate voltage, Ep?
Ep = 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.
Fig. 3 the only resistance affecting the plate voltage is that of
the transformer primary winding, Rt, so
= Eb - IpRt
Ep = 250 volts. Ip =
20 ma. (0.02 amp.)
Rt = 100 ohms.
the plate voltage, Ep?
250 - (0.02) (100) = 250 - 2
Screen voltage is determined in the same manner
as plate voltage, using the screen current to calculate the voltage
drop across the screen resistor.
Ep = Eb
Es = Eb -
In Fig. 4,
Eb = 250 volts. Ip = 5
ma. (0.005 amp.).
Rp = 20,000 ohms. Is
= 2 ma. (0.002 amp.).
Rs = 75,000 ohms.
What are the plate voltage, Ep, and screen voltage, Es?
Ep = 250 - (0.005) (20,000) = 250 - 100
= 150 volts.
Es = 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,
R1, so that plate and screen currents must be added in
calculating the voltage drop across R1.
Ep = Eb - (Ip + Is)
(R1) - IpRp
Eb -" (Ip + Is) (R1)
In Fig. 5,
Eb = 250 volts. Rp = 40,000
Rs = 200,000 ohms. Ip = 2 ma, (0.002
Is = 0.5ma. (0.0005 amp.). R1 =
What are the plate voltage, Ep,
and screen voltage, Es?
Ep = 250 -
(0.002 + 0.0005) (20,000) - (0.002) (40,000)
= 250 - 50 - 80
= 120 volts.
Es= 250 - 50- (0. 0005) (200,000)
= 250 - 50 - 100
= 100 volts.
the circuit of Fig. 6-A, the screen voltage, Es, is obtained
from a tap on a voltage divider consisting of Rs and
Rb The equivalent circuit is shown in Fig. 6-B. The screen
voltage, Es, is equal to the voltage drop across Rb.
Es = RbIb.
Example: In Fig. 6-B,
Eb = 250.
Is = 1 ma. Rs = 10,000 ohms.
= 40,000 ohms.
What is the screen voltage, Es?
Es = IbRb.
Eb is equal to the sum of the voltages across Rs
Eb = RsIsr
since both Ib and Is must flow through Rs,
Isr = Ib + Is.
this value for Isr in the above equation,
= Rs (Ib + Is) + RbIb.
RbIb = Eb - RsIs
Ib(Rs + Rb) = Eb - RsIs
Substituting known values,
Es = (0.0048) (40,000) = 192 volts.
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, Eg, is the voltage developed across R2
by virtue of the current flowing through it from the bias supply.
Eg = IgR2
being the bias-supply voltage.
Screen and plate voltages are calculated as before.
Example: In Fig. 7,
Eb = 250 volts. Ec
= 100 volts.
R1 = 49,000 ohms.
= 1000 ohms. R3 = 30,000 ohms.
= 20,000 ohms. R5 = 20,000 ohms.
= 1 ma. (0.001 amp.).
Ip = 5 ma. (0.005 amp.).
What are the grid-biasing, screen and plate voltages?
= 2 volts (negative in respect to cathode).
= 250 - (0.005) (20,000) = 250 - 100
= 1.50 volts.
Es = IbR3
= 0.0046 amp.
Es = (0.0046) (30,000) = 138 volts.
8 is used to illustrate the effects of low voltmeter resistance
upon the accuracy of voltage measurements. Rm is the
With the meter disconnected, the plate
voltage will be
Ep = Eb - RpIp.
However, with the meter connected, the current, Im,
will flow through Rp. Thus, the voltage drop across Rp
will increase and the plate voltage will be lowered, especially
when the resistance of the meter is low in comparison with Rp.
The equivalent circuit with the meter connected is shown in Fig.
8-B, in which Rpi 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 Rpi
(or Rm) which is
Ep = Eb
- (Ipm) (Rp),
is the new current when Rm is connected. In other words,
Ep is the difference between the terminal voltage and
the voltage drop across Rp.
Example: In Fig.
Eb = 250 volts. Ipi =0.1 ma. (0.0001
Rp = 1 megohm (1,000,000 ohms)
= 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?
= 250 - (1,000,000) (0.0001)
= 1.50 volts = plate voltage
without meter connected.
As stated above, when the
meter is connected,
Ep - Eb - (Ipm)
Since Ipm is not known,
its value must be found before the equation can be solved. To find
Ipm, the resultant resistance of Rm and Rpi
in parallel must be found, and this, in turn, requires that Rpi
be determined. This can be done by considering the circuit before
the meter is connected. The total circuit resistance, Rt
is then given by
= 2,500,000 ohms
= 2.5 megohms.
Rt = Rp + Rpi
Rpi = Rt - Rp = 2,500,000 -
Rpi = 1,500,000 ohms = 1.5 megohms.
The resistance of the meter, Rm, 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.
Rpim, the resultant
resistance of Rpi and Rm in parallel is given
This gives the total circuit resistance in Fig. 8-B as
Rt = Rp + Rpim = 1 + 0.25 +
The new current, Ipm, is then
Ep = Eb - (Ipm)
= 250 - (0.0002) (1,000,000)
= 250 - 200 = 50 volts = voltage indicated
by meter reading.
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,
Ep = Eb - IpRp ;= 250-
= 250 - 150 = 100
When the tube is not conducting, there is no
voltage drop across Rp and the plate voltage is 250,
the same as the supply voltage, Eb.
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
Example: In Fig. 10,
= 250 volts. Ip = 5 ma. (0.005 amp.)
= 10,000 ohms. R2 = 75,000 ohms
R3 = 25,000
What are the plate voltage of the first tube,
and the grid voltage of the second tube?
drop across all resistors is, of course, equal to the applied voltage,
Eb. The voltage, across R1 is R1I1,
while that across R2 and R3 in series is (R2
+ R3) (I2), 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. R2 and R3. Then,
Eb = R1I1+ (R2
+ R3) (I2)
Since both Ip
and I2 flow through R1,
= Ip + I2
value for I1 in the preceding equation,
= (Ip + I2) (R1) + (R2
+ R3) (I2).
Substituting known values,
250 = (0.005 + I2) (10,000) + (75,000 + 25,000)
+ 50 + 100,000I2
110,000I2 = 200
I2= 0.0018 amp. = 1.8 ma.
voltage of the first tube is equal to the sum of the voltage drops
across R2 and R3.
Ep = I2
(R2 + R3) = (0.0018) (100,000)
= 180 volts.
The grid voltage of the second tube is
equal, to the voltage drop across R3.
= I2R3= (0.0018) (25,000) = 45 volts.