Question about function and signal generators

5vrms whats vpp of function generator

Hi folks,

I have a couple of questions

What is the difference between a function generator and a signal generator? (no this is not a joke awaiting a punchline )

Is it possible to use one of these to power a 13.56MHz antenna upto 4 Watts?

When I look at datasheets for function generators I don't see any information on how much power it can supply to its output.

The highest dBm output on a signal generator i have seen so far is 20dBm (I assume i need around 36 dBm to output 4 Watts)


Are my requirements too demanding?


Kind Regards

Cat

"Signal generator" is a broad term, but it's commonly used to describe an RF sinewave generator. It's output level is typically given as dBm into 50 or 75 ohms. 20 dBm is a common maximum level.

A function generator typically gives you a choice of waveforms such as sine, square, triangle, and sawtooth, and is typically limited to several MHz maximum. It's output level is usually given in volts peak-to-peak, or volts RMS into no load or into 50 ohms (read data sheet carefully). Some instruments allow you to specify dBm, but that's less common. 20V peak-to-peak into no load (10Vpp into 50 ohms) is a common maximum level. You can calculate the watts from that.

4 watts is more than most RF signal generators or function generators. Maybe you can find a special "high output level" generator, or add a power amplifier.

echo47 said:

20V peak-to-peak into no load (10Vpp into 50 ohms) is a common maximum level. You can calculate the watts from that.

Click to expand...

Thanks, but how can i calculate the power from the above details?

power = Vpp^2/(2*R) = Vrms^2/R

Hi !

Considering simmetrical voltages (positive and negative swing)

For sinewave: 10Vpp = 5Vpeak = 5/1.41 = 3.5Vrms over 50ohms load
Power = 3.5^2 / 50 = 0.245 W

For squarewave: 10Vpp = 5Vpeak = 5Vrms over 50 ohms load
Power = 5^2 / 50 = 0.5W

For trianglewave = 10Vpp = 5Vpeak = 2.88Vrms over 50 ohms load
Power = 2.88^2 / 50 = 0.166 W

For sawtoothwave = 10Vpp = 5Vpeak = 2.9Vrms over 50 ohms load
Power = 2.9 ^2 / 50 = 0.168 W

Cheers

So, using the equations above. If i wanted 4 Watts output at 50 ohms i would need 20Vpp ? is that correct?


Cat

Boy, your equation should say:
power = Vp^2/(2*R) = Vrms^2/R
or
power = Vpp^2/(8*R) = Vrms^2/R

To get 4 watts into 50 ohms, you need 40 Vpp.

Quick sanity check:
40 Vpp is 20 Vp which is about 14.14 Vrms.
power = V^2/R = 14.14^2/50 = 4 watts.

Thank you echo27 for the correction. It is my mistake.

OK, let me just think about this.

The antenna I have is resonant at 13.56MHz at 50 Ohms Impedance.

But the antenna also has its own additional resistance (to bring down the Q factor between 10-30)

So......by putting R = 50 ohm in the above power equations surely cannot be right. Am i missing something here?


Regards

Cat

If the antenna is 50 ohms, then putting "50" in the power equation is the right thing to do. Why does that feel wrong to you? If you measure the antenna with an impedance meter, it will look approximately like a 50 ohm resistor at that frequency.

Additional resistance??