D J 8 W X

jo43sv Longwave site

Radiation and power output

1. EM (HM) Field:

Reactive Near Field (1 : 2π)λ = 350m.

Radiating Near Field (1 : 2π)λ to 4λ = 350m to 8800m.

Far Field = > 8800m.

2.Radiation

EIRP (effective isotropically radiated power) product of power supplied to an aerial and its gain (hypothetical).

ERP (effective radiated power) product of power supplied to an aerial and its gain relative to a half wave dipole.

EMRP (effective monopole radiated power) product of power supplied to an aerial and its gain relative to a short vertical aerial mounted on a ground plane.

3. Norm

The Radio Regulations define:

 

In a perfectly conducting plane area

1kW EMRP means 300mV/m FS in 1km distance

1kW ERP means  222mV/m

1kW EIRP means 173mV/m

 

 

Measuring the fieldstrength (FS) the radiated power is determined by

 

Peirp = 0,0111( E • d)2 -P in W, E in mV/m, d in km-

 

0, 0111(300 • 1) = 1000W EMRP

0, 0111(222 • 1) = 547W ERP

0, 0111(173 • 1) = 332W EIRP

                                                          

If EMRP, ERP and EIRP have to be compared the following factors are in use:

 

0,547 = 547/1000 (ERP/EMRP)

0,61 = 332/547 (EIRP/ERP)

1,64 = 547/332 (ERP/EIRP)

1,83 = 1000/547 (EMRP/ERP)

3,01 = 1000/332 (EMRP/EIRP)

 

 

The overall view:

ERP = 1,64EIRP 

ERP = EIRP/0,61

ERP = EMRP/1,83 

ERP = 0,547EMRP

EMRP = 3,01EIRP 

EMRP = 1,83ERP

EIRP = ERP/1,64

EIRP = EMRP/3,01

 

4.Efficiency of aerial

Efficiency according to the progr "grndwav" = 0,06%.

( Rloss = Rrad/efficiency = 0,05/ 0,0006 = 83Ω. Rrad see loadcoil )

5.From TX pwr to radiated pwr

a)Ropex 156W

b)home brew 600W

EMRP = TXpwr • efficiency

a,)EMRP = 156W • 0,0006 = 94mW (ERP = EMRP/1,83 = 51mW)

b,)EMRP = 600W • 0,0006 = 360mW (ERP = 197mW)

6.From aerial current to radiated power

a,) I = 1,5A aerial current. 1,52  • 0,05 = 112mW EMRP

b,) I = 2,8A aerial current. 2,82  • 0,05 = 392mW EMRP

Remark: The HF thermo current meter is not calibrated.

And now the truth:

7. From FS to radiated power

FS measured with calibrated receiver ESH3 and loop HF12-Z2 (R&S equipment for standard FS measurements).

P eirp = 0.0111(E • d)2 - Watt, mV/m, km - .

FS of DJ8WX signal measured  in a distance of 1,806km:

a,) 1,5mV/m

b,) 3,2mV/m

a,) EIRP = 0,0111(1,5 • 1,8)2 = 0.0111 • 7,25 = 81mW

ERP = 1,64 • 81 = 132mW.

b,) EIRP = 0,0111(3,2 • 1,8)2 = 0,0111 • 33,2 = 368mW

ERP = 1,64 • 368 = 603mW.

Remarks abt some short transmitting tests:

I modified the aerial, see "New arrangement of aerials" < arrangementofaerials>

The FS in 1,8km distance changed from 3,2mV/m (70dBµV/m) to 4,5mV/m (73dBmycroV/m) using the home brew TX:

b,) EIRP = 0,0111(4,5 • 1,8)2 = 0,0111 • 65,6 = 728mW

ERP = 1,64 • 728 = 1,194W.

Newly change: The power output has been increased. The FS changed from 4,5mV/m to 5,4mV/m (74,6dBuV/m):

b,) EIRP = 0,0111(5,4 • 1,8)2 = 0,0111 • 94,5 = 1,05W

ERP = 1,64 • 1,05 = 1,7W.

TX power output = 2020 • 0,0006 = 1,2kW.

ERP has to be 1,0W .

So TX power output decreased.

Measurement of power output

Measurementequipment:

Dummy load DC - 2GHz "SPINNER" 1kW(30dB) 50Ω, reflectionfactor 0,04.

Meters R&S URV5 V, W, dBm, dBV display and oscilloscope OS5020P.

Insertion unit R&S URV-Z2, 9kHz - 2GHz, 10V.

 

 

 

 

The power output of the two transmitters

a)Ropex

b)home brew

have to be examined.

1. With an oscilloscope the voltage is measured at the transmitter output and on the assumption that the impedance is 50Ω the power output is calculated.

1.1 TX output is connected via 80m RG213U to the aerial system. The system is tuned to maximum of aerial current:

a,) 147 Vpkpk • 0,707 = 104Vrms

I = U/R = 104 / 50 = 2,1A

N= U • I = 104 • 2,1 = 218W

b,) 283Vpkpk • 0,707 = 200Vrms

I = 200 / 50 = 4A

N = 4 • 200 = 800W

1.2 TX output connected to dummy load "SPINNER" Z50Ω:

a,) 125Vpkpk • 0.707 = 88,4V rms

I = 88,4 / 50 = 1,768A

N = 1,768 • 88,4 = 156,3W

b,) 248Vpkpk • 0,707 = 175Vrms

I = 175 / 50 = 3,5A

N = 3,5 • 175 = 612W

2. Measurement with URV equipment:

a,) 155W

Check: I = √ (N/R) = √ (155/50) = √ 3,1 = 1,76A

U = I • R = 1,76 • 50 = 88Vrms.

b,) 605W

Check: I = √ (605 / 50) = 3,46A

U = 3,5 • 50 = 173Vrms

 

Summary.

155W and 605W are the correct results. Calculations from FS to radiated power and via efficiency to TX output power are (almost) confirmed by those results. So what about the results 218W and 800W? Unfortunately I can not offer an absolut value for the TX output current because the HF current meter is not calibrated. The Vpkpk measurements are correct. The other factor, the impedance, must be wrong. I changed the matching point at the TX side of the loading coil again and again. The situation did not alter.

 back to

index.htm

 

 

Web Counter
Web Counter