what is wrong with the formula P=UI ?

For a 50ohm RF system with 3.3v DC power.
if the output RF power is 1W, we can find output RF voltage is
u=(P*R) e 0.5=7v

i am confused: the DC power is only 3.3v, how can it offord 7v voltage?

different impedance levels

The network between the load and the transistor changes the impedance seen by the transistor output. Voltage and current scale with the impedance, power does not.

HEY FLATULENT !!

CAN YOU TELL ME WAHT DO YOU MEAN WITH :


_________________
31 OCT = 25 DEC
numbers & holidays

do you mean to say that the transistor causes non-linearity? or a phase difference? thus the formula:

P = V*I
changes to
P = V*I*cos(theta)

or do you mean that the RF imedance is 50 Ohms, but due to the transistor output, the "50 Ohm" is actually different? i doubt this is the case because the output antenna is probably a 50 Ohm matched impedance to the output transistor. which means the output transistor should exibit 50 Ohm impedance at the resonant frequency (RF).

Mr.Cool

Don't your heard about the powers triangle?
You have 3 types of power:
1. Aparent power: S=UI (units: VA)
2. Real power: P=UIcos(a) (units: W)
3. Reactive power: Q=UIsin(a) (units: Wq)
The sum in vectors are: S^=P^+Q^
and a=angle formed by S in relation of P.

You can only use P=UI in a purely DC circuit.

Depending on the power the output impedance of a RF power transistor is few Ohm ( and also with a reactive part ), thus to transform it to 50 Ohm you need a trasforming network.
For this reason is correct that at 50 Ohm you have 7 V ( with say 0,14 A rms ), but if the impedance is few Ohm, you can have 3,3 V with 0,3 A on the transistor collector.
The following reactive network will change the impedance from the few Ohm of the transistor to the usual 50 Ohm.

Mandi

In any rf system infact any ac system
you have current and voltage nodes and anti nodes
these are anti phasial to the main rf carrier {gravitons by partical procession}

so when a tuned circuit is very resonant {and tuned well} there is a powerfull voltage and current peeks at the termianls to the antenna or other radiator

so this is why you have larger than supply voltage{by great amounts} remember the transisotor oscillator is vibrating the 3.3 v to well over 15,000
for every 1 watt you make you use 1 jewel of energy per second

there is 1 jewel of energy used per 1 inch rotation of the planet relative to a geocentric static position in layers

this is relative isotropic db

so for a gain stage {1 watt is lots of energy in electromagnetics} of say a theoretical 1000 you would get 1.707 X 1000 X the input voltage as a peek voltage node

present at the rf plugs

even from 1 watt
1000mw
or 1 m jewel 1000 of a jewel per second so over sec * 60 = 60,000 volts peek per min of rotation of earth
erp = * 1.616 from a sympothetic radiator

so say your body as a sympathetic radiator of a given size {if you were quite resonant} could give anything up to 50 db gain

or over 300,000,000 million volts !!!!!

so be warned

or your fingers are toast


however the answer to your the first poster
is
simple

static gain

or biased on stage by .... bla

remember most rf gain stages will use the junction of the rf stage to develop the nessisary bias dc
most low gain stages this is the only and mixer also trusted approch to stability
esp using mosfets and other linear devices

more gainy {balsy hear them for miles } stages are derived from transistors {like LF radio stuff} basicaly nowadays becouse of there response to the atmosphere conditions {then are more in tune with the weather}

fets give good gain at lf but poor rf performance

so avoid them
3.3v = 7.6 when the gain of anrf amp is 2volts bias bce at the carrier

a vtvm will show this as 2 volts bias bce

WHO KNOWS YOU MAY EVEN TURN TO METAL OR STONE LIKE ME....!!!!

In RF you have tuned circuits. For example in a 2 order RLC series circuit the voltages at the capacitor and at the inductor terminals is Q*Vin. With Vin the input voltage at the tuned circuit. If, as normal, Q is high (exp. 20) the voltage at the reactive elements is 20 times higher that at the input.

That is wy you have higher voltages than the DC supply in the RF circuit.

In RF power amplifiers you can easy get voltages in the KV range even at moderate power levels for Q factors of 30-50.

Bastos

another example

Back in the days of valve/tube hi if amplifiers, the devices ran on several hundred volts. There was a step down transformer to the speakers which were around 3-4 ohms impedance. A signal of 100 V and 1A at the devices was reduced to 17 V and 5.5 A at the speaker.

thanks for all the reply, but i am still confused.

for example, the GSM handphone,
the DC is only 3v, but the peak to peak output RF power may be several watts(say 5 watts), the load(antenna) impedance is 50 ohm.

so, due to ohm law, we can find that the voltage on the load must be about 15v.

is it possible?

impedance transforming network

There is an impedance transforming network that makes the 50 ohm antenna look like just a few ohms at the drain of the transistor. That way 1 Vrms at the transistor is across one ohm say. That gives one watt into the lossless network and therefore at the antenna.

my understanding now:

because of the low output impedance, the output voltage of the transistor is lower than the DC bias.

also because of the higher load impedance , the voltage on the antenna is higher than the DC bias.

very close

You are very close. Another way to visualize this is to think of the topic of power line transformers. They allow a 200 V line to make other voltages higher or lower on the other side. Here is another example of how the output voltage on the load is different than the input voltage.

Taking the last reply from flatulent you can consider the tuned circuit as a trasformer, primary 230 Vac, 1 A ( 230 VA ) and the secondary 23 V 10 A ( same 230 VA ).
You can see that if you reverse the transformer from a 23 Vac ( 10 A ) you can have a 230 Vac with 1 A ( but the same 230 VA ).
In this case the secondary impedance ( 230V side ) is 230 Ohm, instead on the primary side you have 2,3 Ohm ( 23 V / 10 A ), so you have 1/100 impedance transformation.
The same for the RF circuits, only that in the case you don't use transformers but tuned circuits.

Mandi

what a interest thing!

the impedance match network play the role, which translate power from one port to another.

weather or not, the translated power is
invariableness. Depend on the different
load impedance, the voltage is certainly different.

thanks !

message for eltonjohn

The number 31 in the octal number system is equal to the number 25 in the decimal number system.

Many people would read the text as dates of the year. This is sort of like many visual puzzles where turning the drawing upside down shows another picture.

A few point's to add are:

1) DC voltage is not the input signal (RF) voltage.


2) The DUT can be 1Watt, but at what input power? 27dBm? (.5Watts)
or 0dBm (.01Watts) in?

3) P(watts) = {(Vrms)^2}/R(50Ohms)
so 3V--> 3/(2)^.5 = 2.121Vrms

Hope this helps

I think you may have so many confusions with peak power,average power, AC voltage,peak voltage(transient),average voltage,etc.

I explain it as following

In a DC supplied RF system(DC energy transformed to RF energy),the following equations hold
Pdc=Prf+Plo
Pdc=Udc*Idc
Prf=Urf*Irf


Here Udc is 3V ,the supplied voltage,Idc and Pdc is not defined
Udc=3.3V
, Urf is 7V ,
Urf=(Prf*R)**0.5=(1*50)**0.5=7.07V

Part of supplied DC/AC energy can be transformed to RF energy by RF components.

For example ,
Vdc=3.3V
Idc=10A
Pdc=30W

Irf=7V
Irf=1A
Prf=7W

The efficiency is
Prf/Pdc=7/30=23.3%


In fact, signal voltage/current can be amplifed by active component such as transistors.

You should better get to know the amplifying effect and working principal of diodes and tansistors

element 15 answered right:

U=sqrt(P*R)

This is universal formula. You must allso consider complex or vector forms of it.
sqrt=square root