J
Jacques St-Pierre
- Jan 1, 1970
- 0
Hi,
Is-it possible to calibrate a RF Power Meter using DC current?
I suppose a direct relation can be compute between RF Watt and DC Watt. I
have a RF power meter that I need to check and recalibrate. I have no way to
do it using RF signal, I have no source, nethier than any reference. So I
assume that since the principle is only the measure of a voltage pickup from
a 50 ohms load, it can be done using DC. The meter did react to DC, but
return value around half of what it suppose to be, correct or not, I am not
sure.
What I did; since the load is 50 ohms, I applied 316ma to it using 15.8vdc
source. P=R(I*I) so 50 * (316ma*316ma) = 5 watts. The meter indicate
different value on different scale, so It must be out of alignment, and I
never get 5 watts on any scale.
What I don't exaclty now, is the relation between RF watt and DC watt, since
RF signal is AC signal and should probably be view as RMS watt.
Any explanation or correction will be welcome.
Bye
Jacques
Is-it possible to calibrate a RF Power Meter using DC current?
I suppose a direct relation can be compute between RF Watt and DC Watt. I
have a RF power meter that I need to check and recalibrate. I have no way to
do it using RF signal, I have no source, nethier than any reference. So I
assume that since the principle is only the measure of a voltage pickup from
a 50 ohms load, it can be done using DC. The meter did react to DC, but
return value around half of what it suppose to be, correct or not, I am not
sure.
What I did; since the load is 50 ohms, I applied 316ma to it using 15.8vdc
source. P=R(I*I) so 50 * (316ma*316ma) = 5 watts. The meter indicate
different value on different scale, so It must be out of alignment, and I
never get 5 watts on any scale.
What I don't exaclty now, is the relation between RF watt and DC watt, since
RF signal is AC signal and should probably be view as RMS watt.
Any explanation or correction will be welcome.
Bye
Jacques