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Gito
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« Reply #175 on: July 17, 2010, 10:19:45 PM »

Hi

Back to bias, if we used Stu schematic (grounded CT for RF),And if You looked it carefully The "Zener" diode is in Parallel with the base Emitter of MRF 454.
Three diode in Parallel

So omitting this "Zener" ,the transmitter will still work,but why using this "Zener"
since if the transistor gets warm ,the bias voltage is changing ,and the colector current start to rise, So there must be a second active devise to stop it ,its the "Zener" that have a negative cofiecient. and it must have a  thermal contact with these transistor.
 
So the Bias current is totally depended of R4,and the internal impedance of The Transistor.
And since the bias current is taken from B+ ,variying the base current is also variying th base voltage to some extend.
 So in actullay the zener voltage and the MRF transistor,must have the same treshold voltage (because manufactoring. proses) which the diodes start to cunduct.,a slight diference, will make different current flowing through each diodes.

remember to used equalizing resistor in parallel diode conection.

Just a though.

Gito.n




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« Reply #176 on: July 18, 2010, 12:41:50 AM »

Guys, if Bob can't get a clean output with 900 ma of bias, then bias is not the problem. The shunt diode type is not all that critical because this is a sloppy circuit anyway. Based on amps I have played with, diode temperature compensation will have bias change a good amount as the heat sink heats up. A good thermal contact is pretty critical. The diode will do its job if the stage doesn't run away. Also I saw similar waveforms in the ERB amps before I added more primary turns to the output transformer to get it down to 160 meters from 40 meters. Most MRI amps work down to 40 and I have modified a number of different ones. I have seen a poor collector waveform effect the base wave shape.
Running the DC through the transformer makes the primary reactance even lower.
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Gito
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« Reply #177 on: July 18, 2010, 04:14:45 AM »

Hi

Not saying the schematic diagram is wrong,I don't have the right to say so,but these diagram is bothering me ,when the CT is grounded for RF ,direcy connect1ng the output of T1 to the bases and the Emitter is directly grounded to ground,and if the internal resistance of MRF 454  is 1 Ohm,
Then a 0.1 volt apearing  at T1 output will give .... 0.1 volt : 1= 100 ma.base current
          0.2 volt apearing  at T1 output will give ..... 0.2 volt : 1 = 200ma base current
In these case the MRF454 is in conducting state   becuase the biasing of it.

So the voltage of secondary winding must supply is  around 0 to 0.2 volt. to make the base current flow around 100ma and 200 ma So the turn ratio of T1 is very impotant,when using 16 t0 1 ratio ,it  ,make  a 3.2 reflecting volt in the primary .if the primary has a 50 ohm input than the current flowing trough it is 3.2 : 50 = 0.06 A,so the watt needed is 0.06 X 3.2 = 0.19 watt?
This does not make sense to me.

If turn ratio 50 to 1,than You get a 10 volt in the primary of T1,than with a 50 ohm  input load  the current is around 10 :50 = 0.2 A so the watt needed 0.2 X 10 volt is 2 watt ,more likely,
It 's more make sense to drive the amplifier with a 2 watt power output.

I must be wrong. But where ..,can You tell me?

Gito.N


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Gito
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« Reply #178 on: July 18, 2010, 09:44:54 AM »

 Hi

Now I realised having an Idea that the base Emitter resistance is 1 ohm is wrong,(where did I get it ?) ,all what I wrote is based on this asumption.

first I begin to think if the resistance is 1 ohm ,having a  0.1 voltage on it gives a 100ma current.
Second  it used a 12 ohm in parralel ,what the use of it,since the base emitter is "one ohm".It still give a resistance close to 1 ohm.

So the base emitter resistance must have a enogh resistance,at least at rf .
With it than the transmitter circuits make sense.
The purpose of the 12 ohm resistor in my opinion is a constant load for the RF driver.

the parralel input impedance is 1 ohm when the Transsistor delivered 80 watt output,
that means at that time the base emitter is taking current,as we know Impedance ,say for instant  of a transmitter  is B+ divide  current = impedance(the formula is depending on the operation class,)
so is the input impedance of the transitor when taking the max current (base current} to delivered a 80 watt input power of the transistor.
So when there's no current flowing trough the base,it's impedance must be higher.

 

Gito.n
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W1RKW
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« Reply #179 on: July 18, 2010, 12:14:53 PM »

Guys,
Ok put a 10uf right on the C-tap.  The noise dropped about 25 to 30%.  Probed the grounds and the noise is the same all over the PCB however, when I probe the heatsink which the PCB is screwed to by 9 screws the heatsink exhibits very little noise. It's nearly flat line.  Figuring the anodizing had something to do with it I scraped the anodizing off of a spot and probed again no change.  I wonder why the heat sink would be nearly clean but the traces that the screws go threw are not.

BTW, absolutely no change in the output with the extra 10uf.
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Bob
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« Reply #180 on: July 18, 2010, 01:14:25 PM »

Bob

I think it is time to test the modulation linearity of the amp.

[The output is not a sine wave... but the fraction of the total output power in the fundamental component of the output is not too bad... and one will always need a low pass filter of some kind on the output of this kind of amplifier. I.e. one would not expect the output to be a clean sine wave, unless you pass it through a low pass filter]

If you decrease the amplitude of the sine wave at the input (before the input attenuator) by a factor of 2, how much does the amplitude of the output (across the dummy load) change.

Note: it would be best to make this measurement with some kind of low pass filter between the output of the amp and the dummy load. You wil need one eventually... but if you don't have one now, you can put a low pass filter between the dummy load and the scope by making a simple 5MHz RC filter. For example: If the scope probe capacitance is specified at 15pF to ground, put 2200 Ohms in series between the dummy load and the probe. If the scope probe capacitance is specified at 22pF to ground, then put 1500 Ohms in series between the dummy load and the probe.

Stu
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« Reply #181 on: July 18, 2010, 01:22:18 PM »

Let me ask you this.  I've constructed a 50ohm 4dB attenuator at the output of the Retro75.  Can I assume that a 4dB decrease at the input of the amp will result in a 4dB decrease at the output?

I do have LPFs. No problem there.
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Bob
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« Reply #182 on: July 18, 2010, 01:49:56 PM »

"Let me ask you this.  I've constructed a 50ohm 4dB attenuator at the output of the Retro75.  Can I assume that a 4dB decrease at the input of the amp will result in a 4dB decrease at the output?"

Yes...

As you make this test... it is important that the impedance that the Retro is looking into doesn't change... and it is important that the source impedance that the amplifier sees... i.e. looking back toward the Retro... doesn't change.

If you are using the Retro in this test, it would be even better to just modulate the Retro at
1kHz... and to compare the RF output of the Retro (trace 1) to the RF output of the amp (trace 2), while sweeping the scope at around 0.2 ms/div.

If all is well, the envelope at the output of the amp will look like the modulated envelope at the output of the Retro.

If you see flattening on positive peaks at the output of the amp... then reduce the power level at the input of the amp with the attenuator.

If you can find an input power level at which the envelope of the RF at the output of the amp tracks reasonably well with the envelope of the RF at the output of the Retro... then you're "good to go"... except, perhaps, for the power output of the amp being a little lower than you were hoping for.

Stu


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« Reply #183 on: July 18, 2010, 01:55:35 PM »

OK fine business.  I will make a go of it and report back shortly.  

No problem on the lower power output.  It will still be better than 2 watts.

Now go an enjoy the rest of the weekend.  I've taken a boat load of your time with this.
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Bob
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« Reply #184 on: July 18, 2010, 03:26:18 PM »

Stu,
Despite the linearity of the sinewave (carrier) the input and output appear linear to each other.  That's a good sign.
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Bob
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"Season's Greetings" looks okay to me...


« Reply #185 on: July 18, 2010, 04:04:27 PM »

Bob

Okay...

Then whatever the Retro would sound like on the air... the amplified signal should sound the same.

Maybe it's time for an on-the-air comparison.

Stu

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« Reply #186 on: July 18, 2010, 05:08:51 PM »

I'm wondering if I should knock the gain down some more so output is about 45w.  Output is about 70watts carrier right now with the attenuator.  Having the attenuator in place effects the receive a tad on the Retro which isn't spectacular to begin with. Or should the attenuator be used  when in transmit only rather than mess with the gain.
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Bob
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"Season's Greetings" looks okay to me...


« Reply #187 on: July 18, 2010, 06:33:44 PM »

Bob

I would suggest that you put the attenuator between the relay and T1... so that it only is in the path on transmit.

Use a 50 Ohm pi attenuator made from 3 non-inductive resistors:

100 Ohms to ground on each side (2 watt or larger resistor on the input side of the pi, 1/2 watt or larger on the output side of the pi)

~67 Ohms (1 watt or larger) on the horizontal part of the pi.

The attenuation will be around 10dB from the output of the Retro to the input of the amp.

On transmit, the load on the Retro will be close to 50 Ohms, even if the impedance looking into the amp (T1) is higher or lower than 50 Ohms. The attenuator will do a good job of isolating the Retro from the amp.

Stu
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« Reply #188 on: July 19, 2010, 04:00:56 PM »

Hi Stu,
This begs a question and I'm sure this has been discussed here.  In this day and age is it possible to get non-inductive resistors?  2 watts or higher, carbon types, seem like a thing of the past.  Are metal film resistors considered non-inductive? 

Anything wrong with putting the attenuator at the Retro instead.  Space is a premium on the amp.  It would be easy to put a relay in place on the Retro and switch the attenuator in and out during rcv and xmit modes.
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Bob
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"Season's Greetings" looks okay to me...


« Reply #189 on: July 19, 2010, 04:09:03 PM »

Bob

Ok to put the attenuator any place convenient... and to use a TR switch

Metal film resistors are also non-inductive. I've been using 5W and 2W Xicon metal oxide resistors for things like terminations and attenuators.

I used 3 of them (150 Ohms, in parallel, on the input and the output; and 37.5 Ohms in series) to make a 50 Ohm 6dB attenuator. With a 50 Ohm termination on the output, the attenuator looks like 50 Ohms at the input (measured with my MFJ antenna analyser) at frequencies up to 30MHz. It was still pretty good at 148MHz (56 + j23 Ohms)

Stu
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« Reply #190 on: July 19, 2010, 04:30:10 PM »

OK on the resistors.  I don't have what I need on hand but Mouser seems to have a very good selection of metal oxide types at various power levels.  An order will be on the way. 
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Bob
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His fear was when I turned it on for the first time life on earth would come to a stand still.
Gito
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« Reply #191 on: July 20, 2010, 06:29:54 PM »

Hi

A beter circuit using MRF 454 ,with UNGROUNDED CT/WITHOUT CT at all in T!
using LIMITING RESISTOR R1.R2

using negative feed-back from T2(L3 at T2 to Q1 and Q2.

Using biasing circuit ,that proofs the importance of the biasing point
See the complete article inAN-762 sharp 300.

It's also wrote
The biasing source is ajustable from 0.5 t0 0.9 volt,which is sufficient from Class B to Class A operation
In class B the bias voltage is equal to the Transistor Vbe and there's no  COLLECTOR IDLING CURRENT PRESENT (except small collector -emitter. leakage Ices)

Gito.n


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