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11784

2015-10-15 12:46:31

peter_dl8ov

A Quick Math Problem

Firstly, thank you to those who responded to the problem with hidden messages, 'View Message History' was an instant cure.

Now, I have a quick math problem. An oscillator has a 50 ohm output impedance and 0 dBm output level, what will the output level be if the oscillator is connected across a 750 ohm resistive load?

Please don't just tell me the answer, tell me how to work it out.

Peter DL8OV

11787

2015-10-15 13:42:47

Will

Re: A Quick Math Problem

First query, is the 0db output level when output is open circuit or with matched load?

Cheers,
Will

11788

2015-10-15 13:55:03

n5ib_2

Re: A Quick Math Problem

Hi Peter,

Here goes...

Imagine a circuit that has a pure voltage source in series with two 50 ohm resistors. The first resistor represents the internal impedance of the source, the second one represents the load impedance where the 0 dBm is measured.

0 dBm (1 mW) into 50 ohms means the voltage across that load is 0.223 Vrms (631 mV p-p). The other 50 ohm resistance will also have 0.223 Vrms across it, so the voltage source must be those voltages added together, or 0.446 Vrms (1.26 V p-p).

Now re-imagine the circuit as that 0.446 Vrms source in series with the 50 ohm source impedance, and now a 750 ohm load. By voltage division, the voltage across the 750 ohm resistance will be 0.418 Vrms (1.18 V p-p).

Calculating the power, 0.418 Vrms across 750 ohms will be 0.000233 watts ((0.233 mW). Expressed in dBm that will be -6.3 dBm

That was fun... thanks!! Better than crosswords to keep the brain cells going
Hope I got it right... better get a GA to check my work :^))

73
Jim, N5IB

---In emrfd@yahoogroups.com, wrote :

11789

2015-10-15 13:55:06

kerrypwr

Re: A Quick Math Problem

As Will says, output can be defined as either open-circuit (EMF) or into a matched load where dBm is used.

Once you have established this, convert the 0dBm (one milliwatt) to a voltage; the problem then is just a basic voltage divider solution.

Kerry VK2TIL.

11790

2015-10-15 19:22:57

K5ESS

Re: A Quick Math Problem

I don’t think you can define a 0dB output level across an infinite resistance (open circuit). I think Jim has it right in his post.

Mike

11791

2015-10-15 21:20:38

kerrypwr

Re: A Quick Math Problem

I thought that was what I said;

"... output can be defined as either open-circuit (EMF) or into a matched load where dBm is used".

Perhaps I didn't make it clear.

EMF is the open-circuit voltage of a generator output; it is twice the voltage that appears across a load that equals the generator impedance.

dBm should really be always specified as xdBm (zohms) but 50 ohms is so common that we generally just say xdBm.

Since Peter just asked for guidance I didn't go into detail.

His problem is a "voltage" one as Jim showed (I didn't see Jim's post before I wrote mine).

It only gets to power at the end.

I haven't checked Jim's maths; he seems to know what he is doing! :)

I do think, though, that the final power would be best left as mW or even in voltage form; -6.3dBm is what the power would be in a 50-ohm system whereas the 0.418v is across 750 ohms.

Kerry VK2TIL.

11792

2015-10-15 23:09:55

Will

Re: A Quick Math Problem

I agree that Jim put it very clearly..

Jim’s probably correct in his assumption that measured across 50 ohm load was the intended meaning .

I'll stop digging myself a bigger hole at that.

Cheers,
Will

11793

2015-10-15 23:19:24

peter_dl8ov

Re: A Quick Math Problem

Thank you for the replies. In case anyone is wondering where the problem comes from I have a 400 MHz oscillator with a specified output impedance and level of "0 dBm 50 ohms" and I want to use it to drive two AD9951 DDS chips. The clock pin input impedance of one chip is 1500 ohms so two in parallel will be 750 ohms.

At 25 Euro each I don't want to kill them by overdriving the clock pin which is specified as 0dBm maximum, a strange figure to use when you consider that 1500 ohm input impedance.

Peter DL8OV

11794

2015-10-16 00:48:01

kerrypwr

Re: A Quick Math Problem

I looked-up the data sheet for the AD9951 and I understand your problem.

As you said, it specifies REFCLK input impedance of 1500 ohms and a maximum REFCLK input of -15 to +3dBm with a typical value of 0dBm.

The impedance environment in which the dBm figures are to be measured is not specified.

I'm no engineer, indeed I'm not very smart but that seems rather ambiguous to me as it obviously did to you.

Since dBm is rarely specified in a 1500-ohm system, we might assume that they mean 0dBm in a 50-ohm system but we all know that assumptions are usually dangerous.

Only AD could really resolve this; perhaps you could ask them (and we would be interested in the answer).

Incidentally, Wes wrote a good paper on the use of dB & dBm; I'm glad that I kept a copy before he closed his website.

Kerry VK2TIL.

11795

2015-10-16 00:56:44

Claudio Girardi

Re: A Quick Math Problem

Hello Peter,
as already said, the REFCLK maximum power level is very likely referred to a 50 ohm termination and it comes actually from the maximum input voltage the pin should see.

Looking at the AD9951 datasheet, they say that, when you use the REFCLK as input you should put a capacitor in series, as it is self biased ad 1.35 V (page 12). On page 2 they show the equivalent input circuit of the pins, with the classical protection diodes to VDD and GND, and the usual recommendation to avoid overdriving the inputs and forward-biasing the diodes (REFCLK is likely powered by AVDD but the circuit is likely the same).
So the maximum peak swing you can have on the pin before starting to forward biasing the diode is 1.8V-1.35V=0.45V (you could likely go a little higher than 1.8V, the absolute maximum ratings say 2.2V max on 1.8V inputs, but let's stay on the safe side)
If you have a sinewave with a peak voltage of 0.45V its RMS voltage is 0.45V/sqrt(2))=0.318V. This voltage applied to a 50 ohm termination resistor gives a power of (0.318)^2/50=2mW, which corresponds to 3 dBm, exactly the maximum REFLCK input power specified on page 4 of the datasheet.

So, as a practical solution in your case you could put a 100 ohm termination resistor at the input of each DDS (before the coupling capacitor so they do not change the pin bias, of course), and run two equal-length 100 ohm lines to the 50 ohm oscillator output. In this way the DDS inputs will see a peak voltage close to 0.3 V (the input capacitance has some effect at 400 MHz also, so the match won't be perfect).

73 de Claudio, IN3OTD

----Messaggio originale----
Da: emrfd@yahoogroups.com
Data: 16-ott-2015 6.19 AM
A:
Ogg: Re: [emrfd] A Quick Math Problem

Thank you for the replies. In case anyone is wondering where the problem comes from I have a 400 MHz oscillator with a specified output impedance and level of "0 dBm 50 ohms" and I want to use it to drive two AD9951 DDS chips. The clock pin input impedance of one chip is 1500 ohms so two in parallel will be 750 ohms.

At 25 Euro each I don't want to kill them by overdriving the clock pin which is specified as 0dBm maximum, a strange figure to use when you consider that 1500 ohm input impedance.

Peter DL8OV

11796

2015-10-16 00:59:11

John

Re: A Quick Math Problem

Since this is at 400 MHz, I would want a 50 ohm transmission line from
the oscillator to the DDS chips. 50 ohm microstrip on the circuit board.
To terminate this line with 50 ohms, I would put a 53.6 ohm resistor at
the DDS end of the line. 53.6 in parallel with the 750 ohm total of the
DDS inputs = 50 ohms.

However, the method to solve your original question:

Can't solve this sort of problem using db's directly, so
assume the generator outputs 1 volt terminated in 50 ohms.
The generator itself would look like a 2V output with a series 50 ohm
resistor between the ideal generator and the output connector. The 50
ohm load then puts the two 50 ohm resistors in series creating a 2:1
voltage divider., a 1V output.

So with a 750 ohm load instead of 50 ohms, the voltage divider would
have be 50 in series with 750.

The output V would be 2V divided by 750/(759 + 50) = 2 * (750/800) = 1.875V.

Substitute the true voltage value of 0 dBm for the 1V in the example and
calculate.

Make sure both the generator and the DDS are both talking about the same
measurement, RMS, not peak or peak to peak. Seems strange to me
to specify an IC input pin in terms of RMS, rather that a peak voltage,
as it's the peak voltage that destroys. A non sinusoidal waveform could
have a peak value 10 times or more the RMS value if it has a low duty cycle.

John
KK6IL

11797

2015-10-16 04:33:48

k1rf_digital_stev...

Re: A Quick Math Problem

See: dB or not dB? - Application Note - 1MA98


			dB or not dB? - Application Note - 1MA98 dB or not dB?
			View on www.rohde-schwarz.us	Preview by Yahoo

Regards,
"Digital Steve", K1RF

11798

2015-10-16 14:49:38

kb1gmx

Re: A Quick Math Problem

That is not a DB question at al, its more.

All logic (chips) have a input voltage max and min. If you exceed this

the device dies or maybe latches up. If the source has the right driving

voltage then there is the issue of ringing as overshoot or undershoot

could easily be Vmax +too_much or Vmin- even_more (like -5V poof!).

Also most logic devices expect 0V to some positive (or rarely some -value)

swing to the signals. Usually in the range of ground to supply rail.

That comes from working with digital logic of all forms.

Add to that working in the RF realm of 400MHZ where a improperly

terminated (or sourced) transmission lines have standing waves (SWR!)

and if that wave also know as ringing its the device improper operation

is nominally the result.

So getting a 400mhz clock from a source to the DDS or some chip is

a non trivial thing. One does well to see what the chip manufacturer

suggests.

Allison

11799

2015-10-16 15:18:30

Bill Carver

Re: A Quick Math Problem

Allison's comment is well taken.

My approach would be to put a good 50 ohm termination right at the DDS
chip, of course with a blocking capacitor so you don't screw up the DC
bias on that pin.
For two DDS, use a 50 ohm power splitter on the clock oscillator so you
can use plain 'ol 50 ohm coax (though that might be .086" diameter
hardline if everything is in one box). Everything is properly
terminated, the two DDS clock inputs are semi-isolated from each other,
and you can test a DDS on the bench with 50 ohm instrumentation before
folding into an integrated assembly.

Bill W7AAZ

11800

2015-10-17 01:50:04

peter_dl8ov

Re: A Quick Math Problem

11801

2015-10-17 08:06:14

Bill Carver

Re: A Quick Math Problem

So far Bill I like your approach the best, although I will replace the
hardline with some PTFE 50 ohm coax and a 49.9 ohm resistor will
probably be close enough. I emailed Analog Devices last night but I do
not expect an instant answer because of the weekend.

Peter DL8OV

What you're thinking sounds OK. I think that's what PA3AKE did on his
AD9910/AD9912 boards, and that was with 1 GHz clock!
Bill

11802

2015-10-17 10:33:43

Lasse Moell

Re: A Quick Math Problem

/Lasse SM5GLC

17 oktober 2015 17:06:03 +02:00, skrev Bill Carver bcarver@safelink.net [emrfd] :

So far Bill I like your approach the best, although I will replace the
hardline with some PTFE 50 ohm coax and a 49.9 ohm resistor will
probably be close enough. I emailed Analog Devices last night but I do
not expect an instant answer because of the weekend.

Peter DL8OV

What you're thinking sounds OK. I think that's what PA3AKE did on his
AD9910/AD9912 boards, and that was with 1 GHz clock!
Bill

11803

2015-10-17 11:20:28

Bill Carver

Re: A Quick Math Problem

If you intend to drive several DDS from a single ref source, I would try
to have some isolation between the inputs, as I have seen spurs on the
output from PLL chips driven from the same reference source.
/Lasse SM5GLC

I agree. A splitter will provide isolation between DDS chips, but only
if the source (oscillator) impedance is very close to 50 ohms. At 400
MHz the only way to ensure that is to put a UHF-rated attenuator right
at the splitter input. Probably 10 dB would be the minimum. At that
point you might have 20 dB of isolation between the two DDS clock pins.

Individual buffers to the clocks would be another way to improve
isolation, but the low high MMIC amplifiers don't have much reverse
isolation.

Maybe use both? IE, 10+ dB 50 ohm pad into the splitter, then two MMIC,
one on each splitter output, to the DDS chips. The MMIC gain would
compensate for the 10 dB pad and splitter losses so the clock source
doesn't have to have a very large output power.

Bill

17 oktober 2015 17:06:03 +02:00, skrev Bill Carver bcarver@safelink.net
[emrfd] <emrfd@yahoogroups.com>:

So far Bill I like your approach the best, although I will replace the
hardline with some PTFE 50 ohm coax and a 49.9 ohm resistor will
probably be close enough. I emailed Analog Devices last night but I do
not expect an instant answer because of the weekend.

Peter DL8OV

What you're thinking sounds OK. I think that's what PA3AKE did on his
AD9910/AD9912 boards, and that was with 1 GHz clock!
Bill

11804

2015-10-17 12:32:02

Andy

Re: A Quick Math Problem

"I don’t think you can define a 0dB output level across an infinite resistance (open circuit)."

True, but that doesn't stop people from trying.

Years ago they used to talk about "dBm" signal levels with broadcast and pro audio signals that were not terminated. In ancient history, all those signals had 600 ohm loads, and dBm was correct. Then they stopped using the loads, so sources were "600 ohms" but loads were bridging. But they kept calling it dBm's for a while. Eventually (I think it took years) enough critical mass had changed to using "dBu". But you still run into someone who calls it dBm even though the power is much lower.

Andy

11805

2015-10-17 16:35:09

Russell Shaw

Re: A Quick Math Problem

11814

2015-10-19 14:32:21

n5ib_2

Re: A Quick Math Problem

Good point, WIll.

I assumed (always a bad thing) that the load was matched.

Jim, N5IB

11828

2015-10-22 11:16:51

peter_dl8ov

Re: A Quick Math Problem

OK Bill and others, I may have a possible solution to the dual DDS driver problem after Analog Devices finally replied.

"For premium performance the drive to the DDS clock pin should be between 900mV P-P and 1V P-P"

So..................

Start with a 400 MHz oscillator with 0dBm output. Feed this into a -6 Db coupler (AKA Return Loss Bridge) to give two -6 dBm outputs. Amplify this signal using two GALI 84 chips to give a nominal 16dBm and feed it down the coax lines to the DDS chips. 16 dBm is too high so at the DDS end fit a 13 dB attenuator to each line to reduce the signal to 3dBm or 0.91V P-P.

Comments?

Peter DL8OV

11829

2015-10-22 14:58:48

Bill Carver

Re: A Quick Math Problem

The 20 dB of reverse isolation of the GALI84, plus the attenuator, will provide good isolation (of course the GALI84 supply lines need to be very well decoupled, etc).
If the oscillator is a 50 ohm source the coupler isolation will add more isolation. If some (say 3 dB) of the 13 dB pad is moved to the splitter input the splitter isolation may be improved (by 6 dB, 2X the attenuator) if the oscillator is NOT such a good 50 ohm impedance. A single attenuator in that location would also be a good place to fine-tune the amplitude of both clocks.

But as we say, that is "nit picking". I think your solution should do the job. Glad you got the info from AD.

Bill W7AAZ