USB PC oscilloscope recommendation

[c_mitra]

As I see it, it is used to load the oscillator to 50R, so that measurements are done with the oscillator loaded at 50R.
Then any voltage measurements with the 1M scope probe will actually refer to 50R.
Is that so confusing?

What I am saying is that if a commercial module is to be measured, there might be no need to put the 50R feed through resistor, because these modules are already matched to 50R, in other words their output impedances are 50R. So paraleling another 50R (the feed through) will give voltages at 25R load, not 50R.

Yes, it is *confusing*

Many power RF oscillators cannot work well without a load and if the power is *reflected back* to the circuit, there is often possibility of damage to critical components.

Most small signal (low power) RF oscillators will not work with a 50E load (because it will get overloaded) but that is a different story.

Any 50E load permanently connected to the circuit as a load is dissipative and simply *wastes energy*. The energy was meant to be delivered to somewhere else.

But anyway, commercial RF units are likely to have 50E input and output impedance at specified frequencies; they can be (*not always*) easily connected with another module with 50E input/output impedance. Impedance matching means input and output must be complementary (complex conjugate) and the reactance must balance out. But I do not see how that matter with a scope probe measurement of a voltage.

 

[neazoi]

Yes, it is *confusing*

Many power RF oscillators cannot work well without a load and if the power is *reflected back* to the circuit, there is often possibility of damage to critical components.

Most small signal (low power) RF oscillators will not work with a 50E load (because it will get overloaded) but that is a different story.

Any 50E load permanently connected to the circuit as a load is dissipative and simply *wastes energy*. The energy was meant to be delivered to somewhere else.

But anyway, commercial RF units are likely to have 50E input and output impedance at specified frequencies; they can be (*not always*) easily connected with another module with 50E input/output impedance. Impedance matching means input and output must be complementary (complex conjugate) and the reactance must balance out. But I do not see how that matter with a scope probe measurement of a voltage.

See it that way: If I connect the oscillator (or small transmitter) to the scope and load it with 50R, I get a voltage level of 1vpp (for example). Then if I remove the oscillator from the scope and connect it to a 50R antenna I should expect the oscillator to deliver this same 1vpp to the antenna, since it is loaded with the same load value. Isn't that true?

Now there are two possibilities:
The first is that the oscillator may be of high impedance (mostly the case with simple buffered oscillators) and coupled with a capacitor to the output.
The second is that the oscillator may be already matched internally (using a step down transformer or a resistive internal load fixed to the oscillator).

In both case the oscillator must be connected to a 50R antenna after the measurements on the scope have been done and the power delivered to this antenna must match the power measured on the scope.

How would you measure in each case?

 

[c_mitra]

See it that way: If I connect the oscillator (or small transmitter) to the scope and load it with 50R, I get a voltage level of 1vpp (for example). Then if I remove the oscillator from the scope and connect it to a 50R antenna I should expect the oscillator to deliver this same 1vpp to the antenna, since it is loaded with the same load value. Isn't that true?

Yes and no. The oscillator will deliver approx the same power to the antenna (yes). But if the antenna cannot deliver the same power out into the space, part of the power will be reflected back into the circuit and the *apparent impedance match* between the oscillator and the antenna serves no purpose (no).

A 50E load is a dummy dissipative load that will not reflect back (unless capacitance and inductance mismatches are present).

Somewhere along the way, the simple AC circuit has been taken over by messy electromagnetic theory. You need to measure the power the oscillator delivers with RF power meters that have suitable load within for reliable measurements. Measuring the voltage with an oscilloscope is not accurate any more (50R or no 50R) good indication of usable power.

The first is that the oscillator may be of high impedance (mostly the case with simple buffered oscillators) and coupled with a capacitor to the output.

Sources with high output impedances cannot deliver power efficiently (in general) but you can interface them with another receiver with matched impedance and it will still transfer 50% power happily (even if the matched impedances are not 50R)

The second is that the oscillator may be already matched internally (using a step down transformer or a resistive internal load fixed to the oscillator)

We need to consider the final output impedance of the RF amplifier. Radio engineers use power meters for this purpose, not oscilloscopes.

In both case the oscillator must be connected to a 50R antenna after the measurements on the scope have been done and the power delivered to this antenna must match the power measured on the scope.

How would you *measure* (what) in each case?

If you are interested in voltage readings, you may do so without disconnecting the antenna.

 

[neazoi]

Yes and no. The oscillator will deliver approx the same power to the antenna (yes). But if the antenna cannot deliver the same power out into the space, part of the power will be reflected back into the circuit and the *apparent impedance match* between the oscillator and the antenna serves no purpose (no).

A 50E load is a dummy dissipative load that will not reflect back (unless capacitance and inductance mismatches are present).

Somewhere along the way, the simple AC circuit has been taken over by messy electromagnetic theory. You need to measure the power the oscillator delivers with RF power meters that have suitable load within for reliable measurements. Measuring the voltage with an oscilloscope is not accurate any more (50R or no 50R) good indication of usable power.



Sources with high output impedances cannot deliver power efficiently (in general) but you can interface them with another receiver with matched impedance and it will still transfer 50% power happily (even if the matched impedances are not 50R)



We need to consider the final output impedance of the RF amplifier. Radio engineers use power meters for this purpose, not oscilloscopes.



If you are interested in voltage readings, you may do so without disconnecting the antenna.

Thank you very interesting points!
So you are saying I cannot measure power (by converting the Vpp values I measure on the scope) with the scope at 50R?

 

[Aussie Susan]

If you are actually wanting to measure the effective output of a transmitter into an antenna, then I suggest you read up on SWR.
Also a 50 ohm dummy load is typically a large wire-wound resistor so that it can take a high power output of the transmitter- my dummy load can take up to 50 watts (not that I really push it that high but it can be useful to check the basic operation of the final RF amplifier).
Unfortunately a real antenna does not appear to be purely resistive because it will reflect the RF signal back to the transmitter. If the combination of the antenna, the transmission line between the transmitter and the antenna and the actual RF frequency mean that the reflective wave is EXACTLY in phase with the transmitted wave then it will appear resistive. However it will either lead or lag the transmitted wave and so you will see an inductive or capacitive impedance which will alter the apparent power sent to the antenna. This is why you measure the SWR (which is displayed as a complex number to show the resistive and inductive/capacitive components separately) and then use an antenna matching circuit to counter the inductive/capacitive component so that the transmitter sees a purely resistive load. (Note that this does NOT mean the full power of the transmitter is sent out of the antenna - the transmission line will absorb some of the power.)
The power delivered by a transmitter needs to take into account the total impedance of the load, not just the resistive component. As the scope will measure the voltage but not the current, it cannot give you a direct value for the power.
Susan

 

[neazoi]

The power delivered by a transmitter needs to take into account the total impedance of the load, not just the resistive component. As the scope will measure the voltage but not the current, it cannot give you a direct value for the power.
Susan

As I see it, you measure the voltage accross a 50R resistor, which is used as the load for your transmitter. If the transmitter cannot deliver that much current, there will be problems (oscillation will stop, or transmitter loaded too much). If the transmitter can deliver the power to the 50R though, the voltage will drop until some point and you measure THIS voltage at the ends of the 50R resistor. I do not consider the complex things happening on the antenna, just a simple connection of the transmitter (buffered oscillator) to the scope.

However, it is not clear to me yet if I have to include this load resistor on the 1M scope, if the output port of the equipment I measure already is matched to 50R (It may not be necessarily resistive match).

 

[Aussie Susan]

I suspect I know understand your confusion: a circuit is generally designed to have a specific output impedance which means that it is designed to work into a specific load that presents the impedance to the circuit.
That does not mean the circuit itself has that actual impedance built in to it. Rather it is designed to work as specified when an external load is connected to it.
Therefore, if you simply measure the unloaded output voltage then it need not be representative of how the circuit will work under load.
In your case it sounds like you will need an external 50ohm load resistor connected to the output and you can then use your scope probe to measure the voltage.
However please keep a few things in mind. Firstly, the load you connect must be able to take the power output of the device. If you normally drive a 100W load then your 50ohm dummy load must also take 100W. You might be able to reduce the power output of the transmitter to something that your dummy load can withstand.
Second, if you do then remember that not all transmitters are linear in their performance. It sounds like you have designed your own circuit in which case you will know if this is the case for you.
Thirdly, the power output working into a real load is frequency dependent whereas working into a purely resistive dummy load is not. What you measure may not represent what you will get under normal working conditions.
Susan

 

[neazoi]

I suspect I know understand your confusion: a circuit is generally designed to have a specific output impedance which means that it is designed to work into a specific load that presents the impedance to the circuit.
That does not mean the circuit itself has that actual impedance built in to it. Rather it is designed to work as specified when an external load is connected to it.
Therefore, if you simply measure the unloaded output voltage then it need not be representative of how the circuit will work under load.
In your case it sounds like you will need an external 50ohm load resistor connected to the output and you can then use your scope probe to measure the voltage.
However please keep a few things in mind. Firstly, the load you connect must be able to take the power output of the device. If you normally drive a 100W load then your 50ohm dummy load must also take 100W. You might be able to reduce the power output of the transmitter to something that your dummy load can withstand.
Second, if you do then remember that not all transmitters are linear in their performance. It sounds like you have designed your own circuit in which case you will know if this is the case for you.
Thirdly, the power output working into a real load is frequency dependent whereas working into a purely resistive dummy load is not. What you measure may not represent what you will get under normal working conditions.
Susan

Thanks a lot Susan, everything is clear now!