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From: dagmargoodboat on 6 Jan 2006 19:34

Winfield Hill wrote:
> dagmargoodboat(a)yahoo.com wrote...
> >
> > Winfield Hill wrote:
> >> I question the need for R5, which at any rate must not be too
> >> big, or drop more than a few volts. I question it because Q2's
> >> collector is a high-Z current-source output, most likely with a
> >> low capacitance, much smaller than Q1's base.
> >
> > Here's the rationale for R5:
> > In my case Q2 was a 2n3906, which shows about 2pF Ccb _typical_
> > with a few volts Vce in the Motorola Small Signal Transistor book.
> > That 2pF was about double the _max_ input capacitance of Q1 alone...
>
> Whoa, interesting, what part was Q1?

MRF571. Sweet transistor. You're right though, I mis-read the data
sheet. Actual Cin is spec'd at 1.4pF _typical_, which would seem to
make the effect of paralleling Q2(c) slightly smaller than I indicated.

That 1.4pF figure, however, was reported at 5mA and f=1MHz. Under
bias and near 1GHz the 1.4pF figure no longer holds: the input shifts
to being just a tad inductive, best as my foggy recollection interprets
these S-parameters.

As I recall it, the non-adjustable input network used caps on the
order of 2.2pF, 8.2pF, 6.8pF, etc., so a possible 2pF extra was
significant, unwelcome, and adding R5 seemed a cheap way to ignore it
and avoid manufacturing variations.

Cheers,
James Arthur

From: Andrew Holme on 7 Jan 2006 04:26
dgc wrote:
> Ian I am the OP of this post and, after reading all the replies, am
> certain I am corresponding with people whose knowledge beta is well
> beyond mine. Appoligies for leaving out critical data in my post. The
> circuit is indeed RF (7 MHz). I am trying to squeeze 2 watts rms out
> of a two stage arrangement the first of which is a crystall
> oscillator and the final an NTE235 NPN in common emitter arrangement.
> I am trying to maintain linearity in the output waveform and my old
> BKPrecision oscilloscope indicates I am failing miserably at this.
> The drive from the oscillator is running about 1 volt peak (i know
> this is high) which may be part of the problem. The drive is brought
> in from a 1 turn winding off the oscillator tuned circuit torroid
> inductor. I had 500 mA of standing DC on the 235 at one point and did
> indeed achieve the best output waveform at this level of bias. The
> heat sinked 235 was still running pretty hot at this level so I
> backed off. I now have a larger heat sink (3 X 6) 1/32 aluminum which
> will be better I'm sure. Frankly I may have harmonics in the output
> waveform as well. I'm no expert in deciphering oscope waves unless
> they are pretty clean. Output load is 50 ohms, which if the equation
> V0^2 / 2Po is correct would indicate 50 ohms is too high for 2 watts
> (another problem). Would this be compounding the linearity issue? I
> got a kick out of your post. You bet, I'm the kinda guy that wants
> all the peak to peak voltage available from the supply.

It is normal to use a low-pass filter network at the output of a transmitter
to remove the inevitable harmonics; although you can certainly minimise
harmonics by making the circuit as linear as possible; however, simple
2-stage oscillator-PA transmitters of the type you describe are typically
operated with the PA stage in class C i.e. no standing bias. You don't need
a linear PA unless you're amplitude modulating the drive.

As for the RL = V^2/Po issue, where RL is the load resistance "seen" by the
collector, you need an impedance transforming (matching) network at the
output which makes the 50 load "look" like the desired RL value to suck-out
the required amount of power. The transistor then needs sufficient drive to
make the requisite amount of AC collector current available.

It is normal to make the peak voltage swing at the collector close to the
power supply Vcc to maximise efficiency, but it sounds like you might be
willing to live with less than ideal efficiency. You could get 2W out with
less than a 12V swing.

One final point: it helps if you have a way to smoothly adjust the drive
level from the oscillator - perhaps by running it off a seperate variable
power supply. If the output level changes by sudden jumps, and does not
vary smoothly as you adjust the drive level, you have spurious oscillation
problems.


From: Ian Bell on 7 Jan 2006 05:47
dgc wrote:
>
>
> Ian I am the OP of this post and, after reading all the replies, am
> certain I am corresponding with people whose knowledge beta is well beyond
> mine. Appoligies for leaving out critical data in my post. The circuit is
> indeed RF (7 MHz). I am trying to squeeze 2 watts rms out of a two stage
> arrangement the first of which is a crystall oscillator and the final an
> NTE235 NPN in common emitter arrangement. I am trying to maintain
> linearity in the output waveform and my old BKPrecision oscilloscope
> indicates I am failing miserably at this. The drive from the oscillator is
> running about 1 volt peak (i know this is high) which may be part of the
> problem. The drive is brought in from a 1 turn winding off the oscillator
> tuned circuit torroid inductor. I had 500 mA of standing DC on the 235 at
> one point and did indeed achieve the best output waveform at this level of
> bias. The heat sinked 235 was still running pretty hot at this level so I
> backed off. I now have a larger heat sink (3 X 6) 1/32 aluminum which will
> be better I'm sure. Frankly I may have harmonics in the output waveform as
> well. I'm no expert in deciphering oscope waves unless they are pretty
> clean. Output load is 50 ohms, which if the equation V0^2 / 2Po is correct
> would indicate 50 ohms is too high for 2 watts (another problem). Would
> this be compounding the
> linearity issue? I got a kick out of your post. You bet, I'm the kinda
> guy that wants all the peak to peak voltage available from the supply.

I am not really an RF chap, although I did do some in my youth with tubes,
so don't take what comes next as coming from an RF expert. If I wanted 2
watts out of the final stage then I would certainly be considering
operating it in class C, in which case DC biassing does not come into it.
The classic class C circuit has an resonant circuit in the collector to
remove a lot of the harmonics although some will still remain. These are
usually removed by a subsequent LC filter circuit. As for 2 watts into 50
ohms, this implies an RMS output voltage of 10V so you would need a 30V
supply - do you have this? If not the classic way to overcome this is with
an RF transformer arrangement.

HTH

Ian

From: Andrew Holme on 7 Jan 2006 08:36
Andrew Holme wrote:
[snip]
> As for the RL = V^2/Po issue, where RL is the load resistance "seen"
> by the collector, you need an impedance transforming (matching)
> network at the output which makes the 50 load "look" like the desired
> RL value to suck-out the required amount of power. The transistor
> then needs sufficient drive to make the requisite amount of AC
> collector current available.

I missed a factor of 2 in the above equation. It should be: RL = Vcc^2 /
(2*Po)

Although Po = Vcc^2 / (2*RL) is the maximum power available for a given
supply voltage and collector load, I also find it helpful to think about Po
= 1/2 * Icpk^2 * RL since the transistor is a current-output device.


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