Transistor oscillator.

Probably not. If the "old" transistor is significantly slower than the "new" ones then the old one will be at 90 degrees phase shift and you'll be measuring the point where the other two average 45 degrees.

You could try designing a circuit that has a consistent 135 degree phase shift and the appropriate gain and impedances on input and output, then essentially measure your geranium transistor's 45-degree current gain phase shift. This suffers from two problems:

1: A broadband 135 degree phase shift? Really? (You can solve this with a PLL, IMHO). 2: You'd be measuring, roughly, Ft/beta. This is actually probably a fairly useful number, but it's not well recognized.
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Tim Wescott
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The LC pi-network between collector and base. The author set it at 10MHz resonance. Frequency is L in parallel with the C in series or 1/L=(2pix10e6)^2 x 500p=> L=0.5uH, use 0.47uH stock part.

Reply to
bloggs.fredbloggs.fred

Isn't one of the common pathologies of bipolar emitter followers that they often end up as Colpitts oscillators, due to the parasitic diffusion capacitance between the base and emitter?

So set up the tranny in a pathological emitter follower config and see what happens?

Relevant reference from Pease:

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Reply to
bitrex

One fairly simple method is to put small resistors of equal value from the base and collector to ground, stick an appropriately biased sinusoidal current into the emitter, and use a 2_channel scope to look for the base and collector waveforms to be equal in amplitude. You might start with 10 ohms.

Cheers

Phil Hobbs

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ElectroOptical Innovations LLC 
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hobbs at electrooptical dot net 
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Reply to
Phil Hobbs

Yabbut...

What diode would you use? Schottky?

OP said germanium transistors, so, have fun with that... :-)

(Did they even make Ge "hot carrier" diodes? Want to say, even in the

40s, they were using silicon pretty well for those already.)

Tim

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Reply to
Tim Williams

Just set up a pulse test. Measure t_r or t_f or the delay (t_d or t_stg). It'll be close enough to representative of fT, at least within a gross factor.

Tim

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Is there a rock bottom oscillator that will free run at the frequency limit of whatever transistor is put in it?

I was thinking of the common base job with a capacitor strapped from collector to emitter, but am unsure what to use as a collector load.

Basically for testing how good are a collection of vintage germanium transistors.

Thanks for any help.

Reply to
Tim Williams

As long as its the transistor Ft that's the limiting factor - I doubt that saturation is going to be a problem.

Reply to
Ian Field

Having anything other than the transistor itself set/limit the frequency, defeats the object of the excercise.

Reply to
Ian Field

The common base I was thinking of is just the textbook automatic bias common emitter with the decoupling capacitor moved from the emitter to the base.

The capacitor from collector to emitter is basically the feedback capacitor - I was thinking a gimmic capacitor would probably be enough.

Usually these oscillators have an LC tank in the collector circuit, but sometimes just an RFC. What would happen if I just gave it a collector load resistor?

Reply to
Ian Field

On Friday, January 15, 2016 at 12:04:27 PM UTC-8, Ian Field wrote: ....

z

This circuit works by intentionally running at a constant oscillating frequ ency somewhat below Ft.

The series base capacitor is adjusted until oscillation just starts as indi cated by the meter. The lower the capacitor value, the higher the gain of the transistor. Assuming the transistor has a single pole with the gain dr opping off at 6db/octave the chosen capacitor value then indicates the Ft o f the transistor.

kevin

Reply to
kevin93

On Friday, January 15, 2016 at 12:10:38 PM UTC-8, Ian Field wrote: ....

There is no current gain from emitter to collector so it won't oscillate.

The normal Colpitts oscillator relies upon the resonance of the inductor with a capacitor tap give the current multiplication. In many circuits the base emitter capacitance forms the bottom half of the tapped capacitor.

kevin

Reply to
kevin93

If you put a resistor in series with the emitter, the intrinsic base resistance becomes negative. If the extrinsic resistance plus the is too small, it'll sing like a bird.

Early POR circuits were frequently just an emitter follower hung on an RC delay network. Smart people stopped doing that for a variety of reasons, one of which was that it often oscillates like crazy.

Cascading PNP and NPN followers to eliminate the V_BE offset is another good way to make oscillators.

Cheers

Phil Hobbs

--
Dr Philip C D Hobbs 
Principal Consultant 
ElectroOptical Innovations LLC 
Optics, Electro-optics, Photonics, Analog Electronics 

160 North State Road #203 
Briarcliff Manor NY 10510 

hobbs at electrooptical dot net 
http://electrooptical.net
Reply to
Phil Hobbs

Plus the what?

So, I admit that my powers of circuit design aren't great enough to see this negative intrinsic resistance thing inside my pointy little head. I had heard that this was the case with an inductive load on the emitter -- but a resistor?

Can you point to a web page, or suggest a starting point for modeling this myself? Just the good ol' pi model with an emitter resistor?

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Tim Wescott 
Wescott Design Services 
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Reply to
Tim Wescott

IIRC it's because there are two lags: high frequency beta has a pi/2 phase shift, and so does C_BE. You can find it in Hollister, for instance.

Cheers

Phil Hobbs

--
Dr Philip C D Hobbs 
Principal Consultant 
ElectroOptical Innovations LLC 
Optics, Electro-optics, Photonics, Analog Electronics 

160 North State Road #203 
Briarcliff Manor NY 10510 

hobbs at electrooptical dot net 
http://electrooptical.net
Reply to
Phil Hobbs

You're sounding a bit too imperious for my liking, so I'll leave you with your fantasies.

Reply to
bloggs.fredbloggs.fred

MHz

y,

quency somewhat below Ft.

dicated by the meter. The lower the capacitor value, the higher the gain o f the transistor. Assuming the transistor has a single pole with the gain dropping off at 6db/octave the chosen capacitor value then indicates the Ft of the transistor.

He's oblivious to the concept of current gain bandwidth product, with that being his common base Ft he's after. The circuit is deliberately run common emitter to reduce the operating frequency to the more manageable Ft/beta x

10 or so to ensure 90o phase shift on beta. The Cb is a large enough impeda nce to unload the tuned circuit as well as function as a current source dri ve of the base input with a 90o phase advance, canceling the beta 90o phase retard. At resonance, the unloaded C-L-C pi network perfectly inverts its input drive with zero attenuation. Therefore the collector voltage beta x i b x Rc is developed at the Cb node with a total 360o phase retard. And almo st all of this voltage is developed across Cb (the definition of "verge of oscillation"), so that beta x ib x Rc= ib/2pifCb, making beta x f=1/(2p iRcCb). Noting Ft= beta x f gets you to where you want to be.
Reply to
bloggs.fredbloggs.fred

I think the situation should become apparent if one analyzes the emitter folllwer topology using the hybrid-pi model, and the high frequency equation for transistor beta - beta is a complex (in the sense of complex numbers) function of frequency.

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Reply to
bitrex

And include the parasitic capacitance in the model, of course.

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Reply to
bitrex

Another circuit I have in mind is the LO used in some AM transistor radios, they have 3 windings, one each in the collector and emitter circuits - the third is tuned by the oscillator gang.

Maybe remove the tuning cup and any capacitors from an oscillator can and see what it spins at.

There's a CFL board sitting on the desk with a 3 winding toroid that might work in that type of oscillator - won't be fast enough for what I want to try with the draw full of old germanium transistors, but might be amusing to play with for a while.

Reply to
Ian Field

It might even be simpler to do it properly - a varicap tuned oscillator fed by a ramp generator.

The problem is; my best scope is down to 0.707 at 100MHz and I'm hoping a few of the transistors can do better than that.

Also I'd have to be able to switch to manual voltage control to set up markers.

Reply to
Ian Field

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