# AM Amplifier Question

Discussion in 'Wireless & RF Design' started by Rydog, Apr 15, 2015.

1. ### Rydog Thread Starter New Member

Sep 4, 2011
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0
When an amplifier's specifications says that it's meant to work with a 50 Ohm load, does this mean real DC resistance or imaginary impedance? Of course, the load should be tuned such that the imaginary impedance is at a minimum. It seems to me that at high frequencies, the real resistance would be higher than 50 Ohms due to the skin effect. What do you think?

2. ### alfacliff Well-Known Member

Dec 13, 2013
2,449
428
you should match to the real impedance of 50 ohms. tune out inductive and capacitive componants and match to 50 ohms. resistance isnt really part of the system, find an impedance meter and get to matching. the better ones will tell you whether your impedance is reactive, inductive or capacitave.

3. ### Papabravo Expert

Feb 24, 2006
10,340
1,850
50 Ohms is the Magnitude of the complex impedance. In the complex plane the locus of all points with a Magnitude of 50 Ohms is a circle with the following intercepts:
1. 50 + j0
2. 0 + j50
3. -50 + j0
4. 0 - j50
And there are an infinite number of such points on that circle in the complex plane.

4. ### Tesla23 Active Member

May 10, 2009
323
67
It's meant to drive a load that looks like a 50ohm resistor. This is often not an actual resistor, but the impedance of another amplifier, microwave circuitry, an antenna etc..

The skin effect, as well as inductive effects, is important in components at high frequencies. These days with small surface mounted devices, you can get components that are well behaved well into the GHz. For an example of chip resistor RF performance look at www.vishay.com/docs/60107/freqresp.pdf

I have no idea what Papabravo is talking about - for example you would get a very nasty shock if you connected a high power amplifier designed to feed 50 ohms and terminated it in -j50 (a capacitor). All power would be reflected back to the amplifier.

5. ### Papabravo Expert

Feb 24, 2006
10,340
1,850
Impedance expressed a single, real number should ALWAYS be understood to be the magnitude of a complex quantity. It does not specify the nature of the impedance as you have observed. To terminate a transmission line in its characteristic impedance means specifically a device with the impedance of 50 + j0. Using the magnitude of a complex quantity as if it was the whole picture, has convinced an army of noobs to conclude that they can measure it with a DC multimeter. They couldn't then, they can't now, and never will be able to do that. That's what I'm talking about.

Also, the match between and antenna and the output of a transmitter or amplifier does not have to be dead nuts accurate. Most systems will operate quite happily with some amount of mismatch.

6. ### Tesla23 Active Member

May 10, 2009
323
67
Sorry Papabravo, must be another example of being separated by a common language. I've always taken 50ohms in datasheets and specs to mean 50+j0, so when the input / output impedance of an amplifier is spec'd as 50 ohms I've always assumed it was 50+j0. I still think this is the safest advice to give.

It got me thinking though, when I was driving this morning, if it is the magnitude that matters, about the locus of |Z| = 50Ω on the smith chart. It's an interesting 'curve'.

7. ### DickCappels Moderator

Aug 21, 2008
2,773
669
Let me volunteer a fact that should bring Papabravo's statements to light. RF amplifiers are designed to drive a particular impedance. Impedance is defined as including resistance and reactance.

Note that RF amplifiers are not specified to drive a given resistance.

8. ### Tesla23 Active Member

May 10, 2009
323
67
This has gone off on a strange tangent for something really simple. To answer the original question

this is neither the DC resistance nor the imaginary impedance.

If the amplifier is specified to work into a 50Ω load it means you should connect it to an impedance that is (50 + j0)Ω - simply referred to as 50Ω - at the operating frequency. This is NOT the impedance you measure at DC. So if your amplifier is operating at 1GHz, if you want the specified performance you need to connect it to a load that measures (50 + j0)Ω at 1GHz.

As Papabravo says it doesn't have to be perfectly (50 + j0)Ω, to understand how sensitive the performance is to variations in load impedance you need to look at the output impedance of the device (for linear devices). Typically a device will specify some other parameter that details how close to 50Ω the load should be, often this is the VSWR.

I only replied as there was a possibility that Papabravo's reply could be interpreted to mean that you could terminate the amplifier in j50Ω, -j50Ω, -50Ω or any random impedance with |Z|= 50Ω and expect normal operation. I'm sure he didn't mean that.

t_n_k likes this.
9. ### t_n_k AAC Fanatic!

Mar 6, 2009
5,448
783
As a concrete example it's worth looking at the input & output VSWR figures for a commercially available LNA.
The amplifier in question has nominal input and output impedances of 50 ohms.
It is useful to note from the uploaded data that perfect matching of the amplifier (either at input or output) over its nominal operating range is not perfect but more than adequate.

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10. ### Papabravo Expert

Feb 24, 2006
10,340
1,850
I did not mean to imply that all possible impedances with a magnitude of 50 Ω were equivalent for impedance matching purposes. In fact the technique requires matching with the complex conjugate. It so happens that the conjugate of 50 + j0 is 50 + j0. It was to combat the notion that a complex impedance could be measured with a DC multimeter. If the input impedance of an amplifier is 50 - j14 then the matched source would have an impedance of 50 + j14.