Target output power at the secondary is 2W. That is 1W per each side of the primary. Not going to push it that hard, just want the headroom to maintain as much linearity as possible. Assume Vcc=12V, Vb=0.7V, using equation [1] calculate RL = 64 ohms for each of the primary windings.

64x2=128 ohms total for the primary. So impedance ratio is 128:50, or 2.56:1. Which means a 1.6:1 turns ratio. (8+8):10 works, so would (4+4):5. Go with lower number to keep leakage losses lower.

Calculate minimum number of turns to have inductive reactance at lowest operating frequency to be at least 4x the impedance connected to that winding. Assume for starters an FT50-43 ferrite core which has an A_L of 523. Equation [2] says for each primary at 4 turns yields an inductance of 8.4uH. Equation [3] says 8.4uH at our lowest operating frequency of 7,000,000Hz has a reactance (X_L) of 368 ohms. 368 divided by 64 ohms is 5.75x. Similar check for 5 turns on secondary yields 13uH which has an X_L of 575 ohms at 7MHz, which divided by 50 is 11.5x. So both sides are good.

Calculate core saturation and volumetric loss. With P=2W and 50 ohms, Vrms at the secondary is 10V using equation [4]. With n=5, A=0.136cm^2 as the cross-section area of a size 50 core, f=7MHz, equation [5] yields 4.7mT. Ferrites saturate around 300-400mT, so no danger there. According to (1), at 7MHz rule of thumb is to stay below 57 gauss, or 5.7mT. 4.7mT is close but below, so assume OK for now.

[1]

2W

50Ω

5t

13uH

1W

64Ω

4t

8.4uH

1W

64Ω

4t

8.4uH

HF Transmitter RF Preamp Calcs 08May2025

[2]

[3]

[4]

[5]

(1) https://www.vkham.com/Info/ferro/tut_3.html

Transformer T3

Transformer T3

R_L per transformer T3 calcs is 64 Ohms. Design target transducer gain (G_t) for this stage is 18dB. We are barely conducting for Class AB operation, so choose i_e=5mA.

Need to determine the emitter degeneration resistor r_d (R15 on the schematic), feedback resistor R_f (R17 on the schematic), and Z_in and R_S (they will be the same here, so we can impedance match the secondary on transformer T2 to the input of Q4).

This is an iterative process using the three equations shown below, where we input G_t=18dB, R_L=64 Ohms, and i_e=5mA. For the formulas below, use R_d=26/i_e + r_d. After several iterations of those equations in Excel, settle on these values: r_d (R15 on the schematic)=2.7 Ohms, R_f (R17 on the schematic)=620 Ohms which yields a G_t of 17.9dB (close enough to 18dB), Z_out of 66 Ohms (close enough to the R_L of 64 Ohms) and Z_in of 75 Ohms. So use R_s equals 75 Ohms for calculating the turns ratio for transformer T2.

Note that the equations below are DIFFERENT from the equations given on page 2.25 of EMRFD for the same three values. The reason is, the EMRFD equations include the effect of transistor Beta over frequency. If you plug these same inputs into the EMRFD equations and set the frequency really low (but not zero!) they give the same results.

HF Transmitter RF Preamp Calcs 08May2025

Transistor Q4 (also applies to transistor Q5)

From “fba_with_simple_model.pdf” by Wes Hayward, 16Jun2009:

R_L=Z_out=64Ω

Q4

R15 = r_d= 2.7Ω

i_e= 5mA

R17=R_f= 620Ω

R_s = Z_in=75Ω

G_t=18dB

From transistor Q4 calculations, transformer T2 needs a secondary impedance of 75 Ohms. Need the same impedance for Q5’s secondary side too. For the primary side impedance, we are designing transistor Q3 to have 200 Ohms impedance at it’s collector (load) side.

75x2=150 ohms total for the secondary. So impedance ratio is 200:150, or 1.33:1. Which means a 1.15:1 turns ratio. There is not a reasonable integer turns ratio that works for 1.15, however using a very slightly higher ratio of 1.167:1 we can use 7:(3+3). A 1.167:1 turns ratio means a 1.36:1 impedance ratio. So 1.36x150 = 204 Ohms - close enough!

Calculate minimum number of turns to have inductive reactance at lowest operating frequency to be at least 4x the impedance connected to that winding. Assume for starters an FT50-43 ferrite core which has an A_L of 523. Equation [2] says for each secondary at 3 turns yields an inductance of 4.7uH. Equation [3] says 4.7uH at our lowest operating frequency of 7,000,000Hz has a reactance (X_L) of 207 ohms. 207 divided by 75 Ohms is 2.8x. Similar check for 7 turns on primary yields 25.6uH which has an X_L of 1127 Ohms, which divided by 200 is 5.6x. Secondary is low, consider increasing windings, but that will increase losses.

Calculate core saturation and volumetric loss. Here I’m gonna cheat a bit and just use the V_RMS from my LTSpice simulation. At -12dBm of drive, V_RMS is 1.59V on the primary and 1.0V on the secondary. With n=7 on the primary, n=3 on the secondary, A=0.136cm^2 as the cross-section area of a size 50 core, f=7MHz, equation [5] yields 0.5mT for the primary and 0.8mT for the secondary. Ferrites saturate around 300-400mT, so no danger there. According to (1), at 7MHz rule of thumb is to stay below 57 gauss, or 5.7mT. We are well below that concern level.

​

HF Transmitter RF Preamp Calcs 08May2025

[2]

[3]

[5]

(1) https://www.vkham.com/Info/ferro/tut_3.html

Transformer T2

200Ω

7t

25.6uH

75Ω

3t

4.7uH

75Ω

3t

4.7uH

Transformer T2

Transistor Q3

This is a common emitter amplifier with both emitter degeneration and collector to base feedback. Starting assumptions: I_e=50mA and V_e at idle of 1.1V, R_LPF = 10Ω (R12), C_LPF = 100nF (C8). I_e is kinda chosen arbitrarily, want a value high enough to get decent gain and higher IMD, but not too high as to waste excessive power as heat. Also V_e, want a value around 10% of V_DC (12V in this case).

Set the DC biasing first. R_B3 = V_e/I_e=1.1/.05 = 22Ω (R11). V_b is then ~0.7V more (1.8V). For the DXT2222A, the h_fe as is common for BJT’s is not precisely defined. So use a rule of thumb geometric mean of reasonable min and max values. I’m using sqrt(100*300)=173. So then I_b=I_e/h_fe = 0.05/173 = 289µA.

Want IR_B2 to be at least 10x larger than I_b, so arbitrarily choose 15x to get resistor values close to standard values, which gives IR_B2 = 4.3mA. R_B2 = V_b/IR_B2 = 1.8V/.0043A = 415 Ohms. Use 390Ω (R9).

Calculate other circuit values: IR_B1=IR_B2+I_b = 4.6mA. I_c = I_e-I_b = 49.7mA. IR_LPF = I_c+IR_B1 = 54mA. I chose 10Ω for IR_LPF (which is a bit arbitrary, want a low resistance value here to make a decent low pass filter), so Vlpf=12-(0.054*10)=11.46V. R_B1=(Vlpf-V_b)/IR_B1 = (11.46-1.8)/0.0046 = 2089Ω. Use 2kΩ (R8).

Will use the equations on page 2.25 of EMRFD with the following additional input parameters: target gain = 14dB, R_S=50Ω (want to present 50Ω impedance to the preceding stage, R_L=200Ω (this is what was assumed for transformer T2 primary), can calculate the following: Rdeg=47Ω (R13), R_FB=645Ω (Use 680Ω (R10), gain=14.3dB, Zin=59Ω and Zout=184Ω. That’s close enough to design targets.

BLUE shading indicates values and component number on amplifier schematic

Transformer T1

From transistor Q3 calculations, transformer T1 needs a secondary impedance of 50 Ohms. For the primary side impedance, we are designing transistor Q1 to have 200 Ohms impedance at it’s collector (load) side.

So the impedance ratio is 200:50, or 4:1. Which means a 2:1 turns ratio. An easy way to accomplish this is via a trifilar winding, with the first two windings used for the primary and phase connected as shown in the illustration below.

Calculate minimum number of turns to have inductive reactance at lowest operating frequency to be at least 4x the impedance connected to that winding. Assume for starters an FT50-43 ferrite core which has an A_L of 523. Equations [2] and [3] say for our lowest operating frequency of 7,000,000Hz, 3 turns is sufficient to reach at least 200Ω (that’s 4x 50). But for the primary we need at least 6 turns (3+3) to reach 800Ω (that’s 4x 200), so go with 7 turns.

Calculate core saturation and volumetric loss. Here I’m gonna cheat a bit and just use the V_RMS from my LTSpice simulation of the whole circuit. Both the primary and secondary RMS voltages are well below those on transformer T2, plus T1 has more turns. So even less of a concern here.

PRIMARY

200Ω

14t (7t+7t)

42uH

SECONDARY

50Ω

7t

42uH

[2]

[3]

[5]

Transistor Q2

This is a common emitter amplifier with both emitter degeneration and collector to base feedback. Starting assumptions: I_e=30mA and V_e at idle of 1.2V, R_LPF = 10Ω (R5), C_LPF = 100nF (C3). I_e is kinda chosen arbitrarily, want a value high enough to get decent gain and higher IMD, but not too high as to waste excessive power as heat. Also V_e, want a value around 10% of V_DC (12V in this case).

Set the DC biasing first. R_B3 = V_e/I_e=1.2/.05 = 40 (choose 39Ω (R6)). V_b is then ~0.7V more (1.9V). For the DXT2222A, the h_fe as is common for BJT’s is not precisely defined. So use a rule of thumb geometric mean of reasonable min and max values. I’m using sqrt(100*300)=173. So then I_b=I_e/h_fe = 0.03/173 = 173µA.

Want IR_B2 to be at least 10x larger than I_b, so arbitrarily choose 17x to get resistor values close to standard values, which gives IR_B2 = 2.9mA. R_B2 = V_b/IR_B2 = 1.9V/.0029A = 655 Ohms. Use 620Ω (R3).

Calculate other circuit values: IR_B1=IR_B2+I_b = 3.1mA. I_c = I_e-I_b = 29.8mA. IR_LPF = I_c+IR_B1 = 33mA. I chose 10Ω for IR_LPF (which is a bit arbitrary, want a low resistance value here to make a decent low pass filter), so Vlpf=12-(0.033*10)=11.67V. R_B1=(Vlpf-V_b)/IR_B1 = (11.67-1.9)/0.0031 = 3152Ω. Use 3.3kΩ (R2).

Will use the equations on page 2.25 of EMRFD with the following additional input parameters: target gain = 11dB, R_S=50Ω (want to present 50Ω impedance to the upstream bandpass filter, R_L=200Ω (this is what was assumed for transformer T1 primary), can calculate the following: Rdeg=39Ω (R6), R_FB=491Ω (Use 510Ω (R4)), gain=11.3dB, Zin=62Ω and Zout=170Ω. That’s close enough to design targets.

BLUE shading indicates values and component number on amplifier schematic