Practical Applications of Simple Math - Part II
June 1944 QST

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).

Practical Applications of Simple Math: 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_pI_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_pR_p- I_pR_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_pR_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_pR_p

E_s = E_b - I_sR_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, R₁, so that plate and screen currents must be added in calculating the voltage drop across R₁. 

E_p = E_b - (I_p + I_s) (R₁) - I_pR_p

E_s = E_b -" (I_p + I_s) (R₁) - I_sR_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.). R₁ = 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_bI_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_bR_b.

Since E_b is equal to the sum of the voltages across R_s and R_b,

E_b = R_sI_sr + R_bI_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_bI_b.

Transposing,

R_sI_b + R_bI_b = E_b - R_sI_s

I_b(R_s + R_b) = E_b - R_sI_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 R₂ by virtue of the current flowing through it from the bias supply.

E_g = I_gR₂

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.

R₁ = 49,000 ohms.

R₂ = 1000 ohms.   R₃ = 30,000 ohms.

R₄ = 20,000 ohms.   R₅ = 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_bR₃

     = 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_pI_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_pR_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.)

R₁ = 10,000 ohms. R₂ = 75,000 ohms

R₃ = 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 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,

E_b = R₁I₁+ (R₂ + R₃) (I₂)

Since both I_p and I₂ flow through R₁,

I₁ = I_p + I₂

Substituting this value for I₁ in the preceding equation,

E_b = (I_p + I₂) (R₁) + (R₂ + R₃) (I₂).

Substituting known values,

250 = (0.005 + I₂) (10,000) + (75,000 + 25,000) (I₂)

       = 10,000I₂ + 50 + 100,000I₂

110,000I₂ = 200

I₂= 0.0018 amp. = 1.8 ma.

The plate voltage of the first tube is equal to the sum of the voltage drops across R₂ and R₃.

E_p = I₂ (R₂ + R₃) = (0.0018) (100,000)

     = 180 volts.

The grid voltage of the second tube is equal, to the voltage drop across R₃.

E_g = I₂R₃= (0.0018) (25,000) = 45 volts.