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AB2EZ
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"Season's Greetings" looks okay to me...


« Reply #150 on: July 15, 2010, 03:41:49 PM »

Spice simulation

I simulated the behavior of the amplifier for the case where the center tap is delivering bias from a current source (high impedance to ground at RF). I.e., the center tap does not have a bypass capacitor for RF.

Please refer to the schematic (attachment 1).

In this approximation, each transistor (one on the left and one on the right) is modelled as a diode in parallel with a 10 Ohm resistor. The fixed bias is set at 80mA... i.e. half the maximum current that the existing bias supply can deliver. I.e. (13.8 V - 0.7V) / 82 Ohms = 160mA total for both transistors. Without expalining why, I believe that the bias supply will deliver its maximum current when the amplifier is operated without a center tap RF bypass capacitor.

The diodes in the simulation are realistic, not ideal. So they represent an actual forward-biased emitter-to-base junction... at least to some approximation.

I assumed that the input transformer (4:1 turns ratio) transforms the 50 Ohm source (transmittor) output impedance to 50/16 Ohms ~ 3 Ohms.

Therefore, I am representing the secondary of the transformer as a voltage source with a 3 Ohm source resistance.

Since Bob's amplifier is currently delivering 50 volts peak to a 50 Ohm dummy load, the output current into the dummy load is 1A peak. Taking into account the 1:4 output transformer... the peak RF current (not including resting current) being produced by each transistor is 4A.

The RF portion of the base burrent is, therefore, 4A/100 = 40mA peak.

Still referring to attachment 1 (the schematic)... I adjusted the RF voltage source to produce a peak base current of 50mA... i.e. allowing for 10mA of DC base bias current. The required RF voltage was 0.55 Volts peak.

Attachment 2 shows the simulation of the voltage on the base of one of the transistors (i.e. positive side of the diode to ground): blue waveform. Note that this waveform is at least somewhat similar to what Bob is seeing on his scope.

Attachment 2 also shows the base current in one of the transistors: red waveform. Note that the peak base current is (as I targeted it to be) 50mA. The minimum base current (with the RF applied) is 0mA.

All of this is as I expected, for the case where the center tap does not have an RF bypass to ground. It's easier to simulate the waveforms rather than talk one's way through what they should be... but the result is the same.

Next, I will modify the schematic to simulate the case where there is an RF bypass to ground

[After I modified the schematic to add the RF bypass to ground (see my post #154)... I added the green curve in attachment 2 to show the voltage on the center tap in the absence of any bypass capacitance. Note that it is about 155mV peak to peak... just as Bob reported when he measured the voltage from center tap to ground. This strongly suggests that there is no RF bypass capacitance present in Bob's amp... even though they are physically attached to the board. Sounds like a broken trace.]

Stu

 


* Slide1.JPG (33.74 KB, 960x720 - viewed 975 times.)

* Slide2.JPG (89.14 KB, 960x720 - viewed 995 times.)
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« Reply #151 on: July 15, 2010, 03:45:10 PM »

Bob

An electrolytic would suffice. I'm pretty sure that the problem is simply that the 470uF capacitor is not really there (for whatever reason)

An RF choke would serve little purpose... because R4 (82Ohms) is already a high impedance relative to the stuff on the other side of it... but it can't hurt.

Stu
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« Reply #152 on: July 15, 2010, 03:46:16 PM »

Bob just measure the power supply current don't lift anything. I was just interested in the efficiency
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« Reply #153 on: July 15, 2010, 04:01:01 PM »

Frank,
About 8.7amps according to the PS current meter.
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« Reply #154 on: July 15, 2010, 04:08:45 PM »

Here's the simulation with a bypass capacitor from CT to ground. Note that I'm doing the simulation at 1000Hz (for convenience), instead of 3.885MHz. It doesn't make any difference... provided I scale the capacitor properly. I used a 1 Farad capacitor in the simulation... which scales to 257uF at 3.885 MHz.

The voltage on the primary side of the transformer is the same as it is in my earlier simulation (no center tap)... because the total voltage across the secondary (end-to-end) is the same.

Note the base voltage waveform. It is not symmetircal (as I thought it would be)... because of the effect of the series resistance of each of the pair the RF sources that represent the output of the input transformer. [Hmm... that sounds strange]

Note the increase in the peak base current (and thus the peak collector current) v. the previous case where there was no bypass capacitor from CT to ground. The peak base current is more than twice as large ... which means that the RF output power of the amplifier would be 4 times as large (e.g. 200 Watts instead of 50 Watts)

The brown trace is the voltage across the capacitor... essentially flat, as expected.

Stu


* Slide3.JPG (40.81 KB, 960x720 - viewed 1002 times.)

* Slide4.JPG (82.06 KB, 960x720 - viewed 967 times.)
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« Reply #155 on: July 15, 2010, 04:25:43 PM »

Yanked the 470uf out and replaced it with an 820uf @25v. Also installed a 2.2uf next to it.  Center tap V has decreased to about 50mVpp and to the point  where the scope  won't trigger properly.  Waveform is  dim and not pretty.  

Output no change.  Still got the near squarewave output.

Current draw from the PS  has increase slightly  ~9.1amps.

BTW, feed thru's between top and bottom are OK.
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« Reply #156 on: July 15, 2010, 05:14:26 PM »

After I modified the simulation schematic to add the RF bypass to ground... I added the green curve in attachment 1, below, to show the voltage on the center tap in the absence of any bypass capacitance. I.e. I set the bypass capacitance to zero by removing it from the circuit.

Note that it is about 150mV peak to peak... just as Bob reported when he measured the voltage from center tap to ground. This strongly suggests that there is no RF bypass capacitance present in Bob's amp... even though they are physically attached to the board. Sounds like a broken trace.


* Slide2.JPG (89.14 KB, 960x720 - viewed 962 times.)
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« Reply #157 on: July 15, 2010, 05:24:48 PM »

Bob

If you are drawing 9A at 13.8 volts... then the electrical input power to your amp is 124 Watts.

Is the output across the dummy load still a 50 volts square wave?

This corresponds to 50 watts of RF output power.

That sounds about right if there is still headroom on the collector voltage.

I'm also wondering if the bypass capacitor at the output of the on-board power supply filter is doing its job.

With RF on... what is the waveform, including the DC  voltage, at the center tap of the output transformer or at the point where the on-board DC filter connects to the center tap of the output transformer. Measure the DC level with your scope, rather than a voltmeter.

Stu
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« Reply #158 on: July 15, 2010, 05:48:21 PM »

Bob
Gito
et al.

Attached are the simulated currents on the base of each of the two transistors for

Case 1 (attachment 1) No bypass capacitor on the input transformer center tap

Case 2 (attachment 2) Bypass capacitor on the input transformer center tap

In either case, if you flip over one of the currents, and add them together... you get a nice sine wave. The current is larger with the input RF bypass capacitor in place, but the presence of the capacitor doesn't impact on the shape of the current (at least to the degree that the schematic in the simulation model is accurate...i.e.  in the absence of various parasitic inductances and capacitances... which probably aren't very important at 3.885 MHz if the amplifier is capable of operating at 30MHz).

Therefore... why does the output voltage across the dummy load look like a square wave?

That leads us from the input circuitry to the output circuitry. Again... is the center tap of the output transformer properly bypassed to ground... with respect to RF?

Stu


* Slide5.JPG (74.87 KB, 960x720 - viewed 981 times.)

* Slide6.JPG (86.25 KB, 960x720 - viewed 976 times.)
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« Reply #159 on: July 15, 2010, 06:05:36 PM »

Hi

here I attached a class C Transmitter as You see at the equivalent schematic,
The current flows to the diode is uncontrolled  ,a  1 volt ac will makes the current flow of 1A in the base emitter of MRF  ,since the base emitter resistant is 1 ohm (like Stu wrote) 1 volt ac makes the base emitter to conduct( over 0.8 V),so 1; 1 = 1 A,How hard You tried ,the current that will flow ,even a 0.8 volt voltage (the turning on of the transistor) will give a 0.8 : 1 = 0.8 A and what happens when the gain is 100,the collector current will be 80 A?

What the difference of class C and class B ,the bias voltage,in class B the Ct is not directly grounded for DC voltage,but if You bypass it for RF ,than it's like grounding the CT  for RF and what happens, it work just like the Class C transmitter(for RF)

If you don't bypass for RF ,but bypass for DC only(500uf in the Schematic) ,than the circuit in the center ,look at the equivalent circuit ,at one half cycle ,in this case the upper Transistor is conducting, the current  trough the base emitter is limited by R6 (10 0hms) ..

the botom picture is a modified circuit,if the CT is grounded use a R1 and R2 limiting resistor.

If You used the the original circuit  DO NOt BYPASS THE CT FOR RF ,instate use an RFC or Ferrite RING  between bias supply to CT.to block the RF to ground ,it's my advised.

Or use a limiting resistor between T1 and base  of the MRF 454 if the CT is "grounded" for RF look at the bottom picture.


Gito.n


* IMG_1836.jpg (542.01 KB, 2048x1536 - viewed 986 times.)
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« Reply #160 on: July 15, 2010, 06:36:55 PM »

8 amps sounds like there may be the wrong core material in the output transformer.
MRI amplifiers don't work below 40 meters because they use type 61 material in the transformers. The firat thing you observe is high power supply current and crappy waveform. Again it may be worth checking it at a higher frequency. at 8 amps you should be making a clean 50 watts output
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« Reply #161 on: July 15, 2010, 07:00:33 PM »

Frank
Bob

Core saturation of the output transformer sounds like a plausible explanation for the shape of the output waveform... but why doesn't the saturation go away (or become less) when Bob reduces the peak-to-peak input drive signal?

If the transformer is saturating at the current that leads to 50 volts peak output... why doesn't Bob see something that looks more like a sine wave when he reduces the input dirve voltage by (for example) 6dB?

Stu
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« Reply #162 on: July 15, 2010, 07:51:34 PM »

HI

The input wave may looks good a sine wave,since a representation of the output of the driver, but what if the base to emitter current 0f the MRF 454 saturated ?

look at the pictures ,and analysis of the circuit I attached above
I wonder If I'm wrong tell me ,that I'm Wrong.
And if I'm Right ,tell me that I'm right.

Gito.n
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« Reply #163 on: July 15, 2010, 08:19:18 PM »

Frank
Bob

Here are some even more interesting simulations for the entire amplifier:

Attachment 1 is the schematic used in the simulation. The transistors are similar to the MRF's, except that they have a current gain of 20 (instead of 100)

Attachment 2 is the collector current in each transistor (you need to flip one, and add them to get the total current in the primary of the output transformer) for the case where the center tap of the output transformer is at RF ground (i.e. the center tap bypass capacitor is sufficiently large)

Attachment 3 is the collector current in each transistor when the center tap RF bypass capacitor on the output transformer centre tap is small  (scaled to .003885 uF).

Look familiar?

Check the bypassing on the output transformer CT. I'm beginning to wonder if there is a systemic problem on the circuit board that is causing an open circuit on these bypass capacitors.

Stu


* Slide5.JPG (42.29 KB, 960x720 - viewed 979 times.)

* Slide8.JPG (66.01 KB, 960x720 - viewed 983 times.)

* Slide9.JPG (71.13 KB, 960x720 - viewed 929 times.)
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« Reply #164 on: July 15, 2010, 08:54:10 PM »

Hi

Stu , I looked at your stimulation picture You used a 3 ohm (R1) in series with the power generator source , so when the peak voltage is 0.8 volt the Diode is conducting trough 3 ohm trough internal resistance of diode depending on the internal resistance of the diode (in this case 1 ohm) total 4 ohm what happen? a 0.8v divided by 4 Ohm = 200ma,that's why the wave is not good.Can the signal supply it?
If You used a signal generator with a signal that' below 0.8 volt ,the diode won't conduct,so what you see is a sine wave across the 10 ohm resistor,naturally  a sine wave.

 The  base emitter current is to high ,as the transmitter have a gain of 100( ?) than the collector current is 20 A.(saturated?)
As the signal generator needs at least 0.8 volt for the diode to conduct.,How hard You try if the internal resistance of the diode is 1 ohm ,You will end with a 200 ma diode current .
if R1 is 10 ohm You got a 80ma diode current ,changed it to 100 Ohm than You got a 8 ma  diode current.

So this diode current is actually the current of base to Emitter of MRF 454 that wiil be amplified by 100 (collector current).


Gito.n

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« Reply #165 on: July 16, 2010, 09:40:52 AM »

Gito

My suggestion is that: as you think about these things, you should separate (in your thinking) the average values of the voltages and currents... such as: the average base current, the average base voltage, etc.... from the RF waveforms (with their average values subtracted out)... such as: the RF voltage on the base of a transistor (with its average value subtracted out)

For example: there is a certain average base current, flowing in each transistor that must be provided by the bias supply and/or the 10 Ohm resistors.

In the simulation results that I have posted, I have set the total average current flowing from the bias supply to be 160mA. That's what I set it to be, for the purpose of those simulations. In the simulation results I posted, I used a pair of 80mA current sources (one from base to emitter of each transistor) to represent the total current from the bias supply. I have other simulations that I have done, where I have used a center tapped transformer to supply the bias... but, again, for the simulations I have posted so far: I placed an 80mA current source from base to ground (emitter) on each transistor. What the simulator calculated is based on those two 80mA current sources. The average base current in each transistor must be 80mA minus the average current flowing in its corresponding 10 ohm resistor. That is Kirchoff's current law.

According to the simulation, when RF input is applied, the base current on each transistor is varying from 0mA to 88mA (in my latest simulations)... with an average value of around 30.5 mA.  If I turn off the RF, the average base current in each transistor is around 28.5 mA. That's what the simulator predicts for the particular transistors I am using in the simulation.

Likewise, there is a certain average base voltage on each transistor.

When no RF is applied, this average base voltage, for the transistors I used in the simulation, happens to be around 515 mV. That is, of course, consistent with the average base current (with no RF applied) being around 28.5mA. I.e. 80mA of average bias supply current (per transistor) - 28.5mA of average base current (per transistor)= 51.5 mA of current flowing through each 10 Ohm resistor. 51.5 mA x 10 Ohms = 515 milliVolts.

When RF is applied, the base voltage varies from 382mV to 578mV. It is not a perfect sine wave, and the average value is not exactly in the middle. The average value is 495 mV. i.e.when RF is applied, the average value of the base voltage drops from 515mV to 495mV. Thus the average voltage across the corresponding 10 Ohm resistor drops by 20mV... corresponding to a 2mA reduction in the average current through that 10 Ohm resistor... and a 2mA increase in the average base current (from 28.5 mA to 30.5 mA).

That's the way those particular transistors behave... as per the simulation results... with a fixed 80mA current source from base to ground (emitter) of each transistor.

Note that, with RF applied, the peak-to-peak voltage (i.e. not including the average value) on the base of each transistor is 578mV -382mV = 196 mV peak-to-peak (not a perfect sine wave). That is the peak-to-peak RF drive voltage that the secondary of the input transformer must deliver.

Since the transformer has a 4:1 turns ratio... the peak-to-peak voltage on the primary of the input transformer must be (in this simulation, with these components): 784mV.

Stu


 
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« Reply #166 on: July 16, 2010, 11:13:24 AM »

Bob
Frank
et al.

I'm learning a heck of a lot from the simulations.

Playing around with a few parameter values... and observing the results of the simulation:

1. It is very important that the center tap of the output transformer have a good RF ground. I'm using a value that corresponds (when scaled to 3.885MHz) of 10uF for the output transformer center tap bypass to ground. If it's missing (bad circuit board trace, bad capacitor, or whatever) one sees the output waveform that Bob has been observing.

2. If you want to get high power out... you need to have enough average current from the base bias supply. When I used 82 Ohms... the output power was limited because the transistors end up with too low of an average base voltage*... and they operate deep into class C. When I switched to 33 Ohms in the simulation... everything was much better.

*The reason that the average base voltage drops is because: as you increase the RF input drive to get up to the target output power level, more and more average base current has to be provided by the associated 10 Ohm resistor. In order for the 10 Ohm resistor to provide that average current, the average voltage across the 10 Ohm resistor has to decrease. E.g. to increase the average base current by 30mA, the average voltage across the 10 Ohm resistor (which is also the average base voltage) has to drop by 300mV.
 
Stu.
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« Reply #167 on: July 16, 2010, 12:01:05 PM »

Stu,
I think the transformer is just saturating so the turnoff voltage is distorted into a square wave. Also Bob has blown a set of devices so seems to me the load is collector load is wrong causing excessive current when the drive is increased.
Increasing the DC current through the promary will drive the reactance even lower  
Easy to verify or prove me wrong by increasing the frequency.
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« Reply #168 on: July 16, 2010, 02:00:21 PM »

Frank

In my latest simulation (schematic and results attached below), there are no transformers to saturate. The inductors in the simulation are generic Spice models, and they behave as ideal components.

However, when I make output center tap bypass capacitor (C1 in the schematic) too small... look at what happens to the collector current waveforms: compare attachment 2 (C1=0.001uF) to attachment 3 (C1=10uF bypass)

[All capacitor values are quoted by scaling the frequency in the simulaton to 3.885MHz]

I agree that Bob will have to be cautious about increasing the average base current. Too small a value of R4 will enable the desired level of peak RF output power during transmit...but will mean that the transistors are drawing too much current when the RF input is removed. I.e. they will idle at too high a current. Too large a value of R4 keeps the idling current down... but starves the transistors for base current during transmit. The simulations were done with R4=82 Ohms.

This is very much like trying to build a transformer-coupled push-pull linear amplifier with tubes that draw no plate current unless their grid is positive with respect to the cathode.

I think we agree that properly biasing this type of amplifier is a PIA.
 
Stu


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« Reply #169 on: July 16, 2010, 07:19:25 PM »

Hi

Stu,I admit, You are good in Explaining,and make a big input to me.But I think there's a  big difference
Between by passing the CT and un bypasing the CT,

When You Bypass the CT (grounded for RF) . There's two individual circuit,the upper circuit and the bottom circuit ,that can be  separated from each other ,
Since there's no Connection to Ground to make the current flow,it must take it current from the bias supply. when the bases of the transistors takes current(to make a complete circuit)

But if You un bypass the CT .The RF voltage and current  is flowing around a complete  circuit so there's no need to take the current from the bias supply
 the current is flowing through the Base of the MRF 454 from the bias supply .(transformer .to upper transistor trough R6 ...back to transformer)

So since it does not take it's bias current from the Bias supply ,The bias voltage more stable.

I have proven that without CT "grounded" the current can be make to flow in the bases of the transistors,if the CT is not bypass for RF.("grounded")
It needs no other source to make the current flow in the bases of the transistors.

So after all we are back to the biasing topic.
enough bias to make the transistor work as class B,it is stable?
In case 1 where the CT is "grounded" like Stu wrote it takes current from the bias supply so with a "big" current flowing in the bases of MRF 454 like 100 ma, than the bias voltage  with 33 ohm will drop to 33 X (0.1 + 0.3)= 13.2 volt since the power supply is 12 volt,than What is the class of this transmitter working?
The 0.3 is the current Flowing from B+ through R4(33 ohm) trough R5 and R 6 in parallel to ground.

And in the second case  if we un ground the CT for RF ,the Bias voltage is more stable ,since it does not  take the current from the bias supply.

Than the only thing that we must do is Variable bias supply that's stiff enough and well regulated to feed the CT with it, omitting the  Zener diode and R4.

and if we used it ,since it's stable and can supply enough Current without changing it 's output voltage.
Just set the power supply to it's intended voltage to make The transistor work as class B,or AB2.

With these bias supply both circuit ,grounded and ungrounded CT.I believed will work


Gito.N
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« Reply #170 on: July 17, 2010, 10:10:28 AM »

Output is what appears in the attachment. Sorry for the blurred image.  Some improvement from doing the following:

Dropped R4 to 50ohms.  At 13.6V, bias is sitting at .701V.  Current draw from PS is 940mA (standby mode).  Transmit current has increased from about 8.5A to 11.3A.  Bias does drop to .647V.  Previously, bias was around .660V with a 300mA draw from the PS at 13.6V (standby).    I don't believe I can go any lower with R4 unless the bias method is changed.

Checked bypassing.  Tried different values on the output side. Made no difference.  Resoldered all feed thru's and compared grounding to schematic.  Don't see any abnormalities.


* P1010004.JPG (184.7 KB, 1280x960 - viewed 941 times.)
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« Reply #171 on: July 17, 2010, 10:34:40 AM »

Bob

Can you post the following two pictures?

1. Dual trace: output RF at dummy load (same as last post) and the voltage waveform on the output transformer's center tap.

2. Dual trace: output RF at dummy load (same as last post) and the voltage on either transistor's base

Stu
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« Reply #172 on: July 17, 2010, 10:59:11 AM »

Revised
Here you go Stu.  At first I had them mixed up but they are correct now.

Image 1. Dual trace: output RF at dummy load (same as last post) and the voltage waveform on the output transformer's center tap.

X10 Probe
Channel 1: Center tap of Output Transformer  (lower trace)
0.2V/div
Channel 2: Dummy Load  (upper trace)
2.0V/div


Image 2. Dual trace: output RF at dummy load (same as last post) and the voltage on either transistor's base.

X10 Probe
Channel 1: Base of output transistor   (lower trace)
50mV/div
Channel 2: Dummy Load   (upper trace)
2.0V/div

One additional tidbit.  The input power is approximately 1watt.

And found a source for the original bias diode (1n4997).


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« Reply #173 on: July 17, 2010, 12:33:50 PM »

Bob

That's a big signal on the center tap of the output transistor (about 1.7 volts peak). Remember, the B+ supply is 13.8 Volts.

What causes this noise?...

It could be a number of things... but, when each transistor tries to draw collector current, it is helpful if that transistor's current can all flow up to the collector from the center tap ... rather than flowing out of the other transistor's collector (or not having anywhere to flow from).

In my latest simulaton (just making little tweaks to the model)... if you really have 10uF from CT to ground (as per the schematic)... the voltage waveform on the CT should be essentially invisible.

Here's a posibility. Could it be that the physical distance from the center tap to the 10uF bypass capacitor is too long... and therefore the inductance of that path is coming into play? This is a relatively low impedance point (compared to what we're used to with tube designs).

At 3.885 MHz, and around 10 Amps of amplitude... it only takes 4 nanoHenries (0.1 Ohm of inductive reactance) to produce a 1 volt signal on the center tap (the waveform will be at twice 3.885 MHz)

Can you add a 10uF capacitor that is physically closer to the center tap ... and going to ground close to where the emitters go to ground? [10uF presents a reactance of 0.004 Ohms at 3.885 MHz]

Stu
 
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Stewart ("Stu") Personick. Pictured: (from The New Yorker) "Season's Greetings" looks OK to me. Let's run it by the legal department
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« Reply #174 on: July 17, 2010, 12:43:01 PM »

it could also be the way the scope probe is grounded.
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