Heathkit sb-200 how to convert to 811a tubes

Discussion in 'Amateur Radio Amplifiers' started by K5BOO, Sep 4, 2014.

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  1. AF6LJ

    AF6LJ Ham Member QRZ Page

    You haven't posted any data, showing how you arrived atthe conclusions you did.
    Don't blow smoke up my back side Jim.

    Seriously dude you are out of your league.
     
  2. N2EY

    N2EY Premium Subscriber QRZ Page

    River in Egypt fallacy noted.

    I have posted both data and calculations in this thread. You have not been able to find a single error in anything I posted.

    I will ask again: What do you think the "series resistance" of an SB-200 transformer really is?

    If that value is plugged into the design graphs in the 1988 ARRL Handbook that you use as a reference, what does it predict for the full-wave-bridge supply with 100 uF filter and 1000 ohm load?

    The only one trying to blow smoke is you. Not me.

    It is clear to me, and to anyone who understands the technology, that the supply you and your ex built 30-odd years ago should have delivered 1000 volts or so under load, if everything was as you say it was. But it didn't. So something was not quite right.

    But you won't admit that, and instead attack the messenger for daring to point out the obvious.

    Why?

    Perhaps you don't want a code-skilled, homebrewing, FCC-tested, old-timer ham and engineer with multiple degrees to be right about something - anything - that you weren't on top of FIRST.

    In what way?

    If ANYTHING I have written in this thread is incorrect, it should be simple for someone who is "above my league" to point it out. But that hasn't happened. Anyone can see that you're avoiding the fact that I'm right.

    You make all this noise about "data" and "calculations", but you don't show anything which proves I'm wrong.

    If you could, you would. That says you can't.

    Heck, I could get an SB-200 transformer, build a full-wave-bridge supply with 100 uF filter and 1000 ohm load, get 1000+ volts out of it, and you'd probably STILL say I was wrong!
     
    Last edited: Sep 9, 2014
  3. AF6LJ

    AF6LJ Ham Member QRZ Page

    It is funny how you hold me to a higher standard of proof than you hold yourself.

    I want you to put down step by step how you arrived at your conclusions, just like you insisted I do in the SB-220 Find thread.
    You have not done that yet.
    So Where are the numbers?
    You said you did calculations, where are they.
    I want to see them all.
    Like I said you insisted that I do that.


    DO IT

    I love the "I have posted data and calculations in this thread".
    You have not. Show me, it should be a simple matter to copy and paste them here.
     
  4. N2EY

    N2EY Premium Subscriber QRZ Page

    Here's the stuff from Post #42 in this thread. It's all there. I have put the quoted stuff in bold. I used the 1988 ARRL Handbook charts in the power supply chapter, same as you did.



    Yes, I have. You just don't recognize it, for some reason.

    I'm going to bypass all the personal insults and logical fallacies you've tossed out and show that, with a good SB-200 transformer and the components you described, the output should be considerably greater than 850 volts into a 1000 ohm load. In fact it should be above 1000 volts.

    The first step is to analyze the stock SB-200 power supply.

    The SB-200 supply uses a voltage doubler operating from an 800 volt transformer secondary. 8 silicon diodes are used in each rectifier stack, and the filter capacitor string is six 125 uF 450 volt electrolytics - ordinary ones at that. Each half of the filter capacitor string is equivalent to a 41.6 uF 1350 volt capacitor - we'll use 42 uF for calculations.

    The no-load voltage of such a supply is about 800 x 2.818 = 2254 volts. But what we're interested in is what the voltage is at rated load.

    The SB-200 manual says the plate supply voltage is 2000 volts nominal (plus or minus 100 volts) at 500 mA load. 2000 volts is 2.5 times 800 volts.

    In post #23 of this thread, you posted a picture of a graph showing the performance of a full-wave doubler under various conditions. This graph is identical to that on page 6-12 of my 1988 ARRL Handbook - even down to the yellow cover.

    The one thing we don't know about the SB-200 power supply is the value of Rs - the equivalent series resistance of the transformer, wiring, supply, etc. We can use the graph to determine Rs, however.

    We have the following values for the stock SB-200 power supply:

    Eac = 800 volts
    R = 4000 ohms (2000 volts divided by 500 mA)
    C = 42 uF
    Output voltage under load = 2000 volts
    Ratio of output voltage to Eac = 2.5

    At the bottom of the graph is the variable 377 RC / 1,000,000 where R is in ohms and C is in uF.

    Plugging in the appropriate values we get a value of 63.336. We'll use 63, and look at the point where 63 crosses the 2.5 line on the vertical axis.

    What we find is that the point is above the curve for Rs/R = .001. Since R = 4000, Rs is then equal to no more than 4 ohms.

    Now let's look at the full-wave-bridge supply. As described in post #15 of this thread, it used the same SB-200 transformer, with a bridge rectifier consisting of 12 silicon diodes (3 in each leg), plus a filter capacitor string of three 100 uF 450 volt computer-grade caps. It should deliver about 1000 volts at 1 amp, but only supplied 850 into a 1000 ohm resistor.

    We have the following values for that power supply:

    Eac = 800 volts
    R = 1000 ohms
    C = 100 uF
    Rs = 4 ohms

    Flipping back to page 6-8 in the 1988 ARRL Handbook, we find a graph showing the design for a full-wave bridge rectifier.

    At the bottom of the graph is the variable 377 RC / 1,000,000 where R is in ohms and C is in uF.

    Plugging in the appropriate values we get a value of 37.7. We'll use 37.

    With Rs of 4 ohms and R of 1000 ohms, Rs/4 equals .004. We'll use the .005 curve, because it's lower, and look at the point where the curve crosses the line for 37. That point corresponds to a ratio of DC output voltage to Eac of 1.35.

    1.35 times 800 equals 1080 volts.

    Some might argue that the Rs value of 4 ohms should be doubled, because the diagram shows Rs twice. Let's make Rs 10 ohms, just to be safe, and see what happens.

    With Rs equal to 10 ohms, the value Rs/R becomes .01. We follow the curve for Rs = .01 to the point where the curve crosses the line for 37. That point corresponds to a ratio of DC output voltage to Eac of 1.27.

    1.27 times 800 volts is 1016 volts.

    But the actual measured voltage was only 850 volts, as reported in post #15 of this thread.

    So there must have been something amiss somewhere. The design calculations prove it.

    If someone can find a flaw in the above analysis, please point it out. I've purposely rounded off the various values in the direction that would result in a lower output voltage, and still get results higher than 1000 volts. So something wasn't right.

    Here are some possibilities:

    1) Line voltage drop. It was stated that the line voltage was 118 VAC, (RMS), so it is doubtful that this is the cause. However, that value is RMS, based on a pure sinusoidal 60 Hz waveform.

    2) Line voltage waveform distortion. Usually commercial power is a pretty clean sine wave, but sometimes odd loads can cause distortion, particularly at the end of a long service drop or similar situation. Loads with high crest factor can have odd effects on the mains - newer supply transformers are designed to deal with this better than older designs.

    3) High diode voltage drop under load. We normally consider silicon diodes as having less than 1 volt drop, but a defective diode may have much higher - particularly at peak currents many times that of the load current. However such a diode will usually destroy itself from internal heating, particularly if it's in a case with high thermal resistance. And since there were fewer diodes used in the bridge supply (12 vs. 16 in the original), this is even less likely. But it is still possible,

    4) Filter capacitors having high ESR and/or low value. Possible, but unlikely, because the caps were tested.

    5) Transformer saturation. This can happen with transformers not designed for loads with high crest factors, but since the SB-200 transformer was originally meant for use with a silicon-diode/capacitor filter load, this seems unlikely.

    It is possible that the transformer was manufactured incorrectly - too many primary turns, or not enough secondary turns. Without actually measuring the transformer performance, this cannot be determined.

    However, note this: The SB-200 transformer has dual primary windings, to permit operation on either 120 or 240 volt supply by series- or parallel-connection of the primaries. For 120 volt operation, the primaries are of course paralleled,

    If one of the primary windings is open, or not connected, and the supply voltage is 120, the power supply will still work normally up to about half power (500 watts or so) on a single primary. However, beyond that point, the lack of adequate primary winding VA capacity will show up as excessive voltage sag on the outputs. The result is a power supply that has poor voltage regulation even though all the components test OK.

    There's proof of my assertions, and possible explanations as to why the output voltage was more than 15% low.

    Anyone who reads this: Show me where anything I wrote in this analysis is wrong.
     
  5. AF6LJ

    AF6LJ Ham Member QRZ Page


    Why?
    Quote please and page number.
    So you have actually measured this?
    I doubt it, you just scrapped this from the manual which may or may not be correct.
    Key down in CW the voltage will be considerably less than 2KV most liley close to 1.5KV.
    Wrong again 300MF at 450V
    Since you made several assumptions which are ill informed all you did was spread the BS higher and deeper.
    You don't have the slightest idea.
     
  6. K9STH

    K9STH Platinum Subscriber Volunteer Moderator Platinum Subscriber QRZ Page

    Not to take sides!

    I just verified the following with my SB-200 driven with my SB-401 on 40-meters:

    With 246 VAC applied.

    Key up plate voltage 2400 VDC.

    Key down plate voltage 2100 VDC.

    Plate current 520 mA.

    Plate input 1092-watts.

    Output power 600-watts (measured with my Collins 312B-4 wattmeter).

    Efficiency: 54.84-percent

    The original filter capacitors, 120 uF, have been replaced with 220 uf capacitors. However, before replacing the filter capacitors the key down plate voltage was still over 2000-volts. The metering resistors have been replaced with 1-percent resistors at 2-watts.

    As such, the key down plate voltage is definitely more than 1500-volts!

    Glen, K9STH
     
  7. AF6LJ

    AF6LJ Ham Member QRZ Page

    You actually have one, that makes a difference on that point......
     
  8. N2EY

    N2EY Premium Subscriber QRZ Page

    N2EY: The first step is to analyze the stock SB-200 power supply.



    Because the analysis will reveal the characteristics of the SB-200 power transformer. It’s called “reverse engineering”.

    N2EY:The SB-200 supply uses a voltage doubler operating from an 800 volt transformer secondary. 8 silicon diodes are used in each rectifier stack, and the filter capacitor string is six 125 uF 450 volt electrolytics - ordinary ones at that. Each half of the filter capacitor string is equivalent to a 41.6 uF 1350 volt capacitor - we'll use 42 uF for calculations.

    The no-load voltage of such a supply is about 800 x 2.818 = 2254 volts. But what we're interested in is what the voltage is at rated load.

    The SB-200 manual says the plate supply voltage is 2000 volts nominal (plus or minus 100 volts) at 500 mA load. 2000 volts is 2.5 times 800 volts.




    On page 4, the SB-200 manual says that the DC input power on CW is 1000 watts.

    On page 48, in the tuneup instructions, it says the plate current should be 500 mA when in CW mode, key down.

    To get 1000 watts DC input on CW with 500 mA plate current, the plate voltage must be 2000 volts. (2000 volts times 500 mA = 1000 watts).

    In addition, in post #66 of this thread, K9STH writes:

    Also, in post #26, KD2ACO writes:

    So, just in this post, there are three references which show that the SB-200 power supply will deliver at least 2000 volts at 500 mA in stock voltage-doubler configuration.

    What evidence do you have that they're all wrong?

    N2EY: In post #23 of this thread, you posted a picture of a graph showing the performance of a full-wave doubler under various conditions. This graph is identical to that on page 6-12 of my 1988 ARRL Handbook - even down to the yellow cover.

    The one thing we don't know about the SB-200 power supply is the value of Rs - the equivalent series resistance of the transformer, wiring, supply, etc. We can use the graph to determine Rs, however.

    We have the following values for the stock SB-200 power supply:

    Eac = 800 volts
    R = 4000 ohms (2000 volts divided by 500 mA)
    C = 42 uF
    Output voltage under load = 2000 volts



    What difference does that make? You would reject my measurement as subjective, anyway.

    At least two SB-200 owners have given their readings – unsolicited. Both say the SB-200 power supply produces over 2000 volts at 500 mA plate current.

    How many independent reports do you need?



    Why do you suspect that the manual is wrong?

    Heck, all one has to do is google “SB-200 plate voltage” and you will find postings and websites from SB-200 owners which say that the normal plate voltage is 2000 volts, nominal, when loaded to 500 mA plate current. With today’s higher line voltages, 2100 volts or more is common.




    No, it won’t - not if the power supply is in good condition and the plate current is 500 mA.

    How does the SB-200 run at 1000 watts DC input on CW if the plate voltage is down around 1500 volts and the plate current is 500 mA? That’s only 750 watts DC input.

    Do YOU have an SB-200 in good operating condition? Have YOU measured the plate voltage at 500 mA plate current? I think not. I think you're just blowing smoke.

    N2EY: Ratio of output voltage to Eac = 2.5

    At the bottom of the graph is the variable 377 RC / 1,000,000 where R is in ohms and C is in uF.

    Plugging in the appropriate values we get a value of 63.336. We'll use 63, and look at the point where 63 crosses the 2.5 line on the vertical axis.


    What we find is that the point is above the curve for Rs/R = .001. Since R = 4000, Rs is then equal to no more than 4 ohms.

    Now let's look at the full-wave-bridge supply. As described in post #15 of this thread, it used the same SB-200 transformer, with a bridge rectifier consisting of 12 silicon diodes (3 in each leg), plus a filter capacitor string of three 100 uF 450 volt computer-grade caps. It should deliver about 1000 volts at 1 amp, but only supplied 850 into a 1000 ohm resistor.



    Wrong, yes, but not “again”. After many many requests and posts, you FINALLY found a typo in one of my posts!

    It is a typo of no consequence, however. The resulting capacitance of three 100 uF 450 volt capacitors in series was 100 uF, and 100 uF is used in the calculations.

    N2EY: We have the following values for that power supply:

    Eac = 800 volts
    R = 1000 ohms

    C = 100 uF
    Rs = 4 ohms

    Flipping back to page 6-8 in the 1988 ARRL Handbook, we find a graph showing the design for a full-wave bridge rectifier.


    At the bottom of the graph is the variable 377 RC / 1,000,000 where R is in ohms and C is in uF.

    Plugging in the appropriate values we get a value of 37.7. We'll use 37.

    With Rs of 4 ohms and R of 1000 ohms, Rs/4 equals .004. We'll use the .005 curve, because it's lower, and look at the point where the curve crosses the line for 37. That point corresponds to a ratio of DC output voltage to Eac of 1.35.


    1.35 times 800 equals 1080 volts.

    Some might argue that the Rs value of 4 ohms should be doubled, because the diagram shows Rs twice. Let's make Rs 10 ohms, just to be safe, and see what happens.

    With Rs equal to 10 ohms, the value Rs/R becomes .01. We follow the curve for Rs = .01 to the point where the curve crosses the line for 37. That point corresponds to a ratio of DC output voltage to Eac of 1.27.


    1.27 times 800 volts is 1016 volts.

    But the actual measured voltage was only 850 volts, as reported in post #15 of this thread.

    So there
    must have been something amiss somewhere. The design calculations prove it.

    If someone can find a flaw in the above analysis, please point it out. I've purposely rounded off the various values in the direction that would result in a lower output voltage, and still get results higher than 1000 volts. So something wasn't right.

    Here are some possibilities:

    1) Line voltage drop. It was stated that the line voltage was 118 VAC, (RMS), so it is doubtful that this is the cause. However, that value is RMS, based on a pure sinusoidal 60 Hz waveform.

    2) Line voltage waveform distortion. Usually commercial power is a pretty clean sine wave, but sometimes odd loads can cause distortion, particularly at the end of a long service drop or similar situation. Loads with high crest factor can have odd effects on the mains - newer supply transformers are designed to deal with this better than older designs.

    3) High diode voltage drop under load. We normally consider silicon diodes as having less than 1 volt drop, but a defective diode may have much higher - particularly at peak currents many times that of the load current. However such a diode will usually destroy itself from internal heating, particularly if it's in a case with high thermal resistance. And since there were fewer diodes used in the bridge supply (12 vs. 16 in the original), this is even less likely. But it is still possible,

    4) Filter capacitors having high ESR and/or low value. Possible, but unlikely, because the caps were tested.

    5) Transformer saturation. This can happen with transformers not designed for loads with high crest factors, but since the SB-200 transformer was originally meant for use with a silicon-diode/capacitor filter load, this seems unlikely.

    It is possible that the transformer was manufactured incorrectly - too many primary turns, or not enough secondary turns. Without actually measuring the transformer performance, this cannot be determined.

    However, note this: The SB-200 transformer has dual primary windings, to permit operation on either 120 or 240 volt supply by series- or parallel-connection of the primaries. For 120 volt operation, the primaries are of course paralleled,

    If one of the primary windings is open, or not connected, and the supply voltage is 120, the power supply will still work normally up to about half power (500 watts or so) on a single primary. However, beyond that point, the lack of adequate primary winding VA capacity will show up as excessive voltage sag on the outputs. The result is a power supply that has poor voltage regulation even though all the components test OK.

    There's proof of my assertions, and possible explanations as to why the output voltage was more than 15% low.


    Anyone who reads this: Show me where anything I wrote in this analysis is wrong.

    Which assumptions are “ill informed”? Be specific – I think you are just blowing smoke.

    Show where there is an error in the analysis. You can’t.

    Tell us what the value of Rs should be in the above analysis. You can't.

    The stock power supply for the SB-200 delivers 2000 volts at 500 mA, nominal. If that transformer is used in a full-wave-bridge arrangement of proper design, it will deliver 1000 volts nominal at 1000 mA.

    Half the voltage, twice the current.

    It’s that simple.

    That doesn’t mean your experience didn’t happen, or was “wrong”. It simply means something wasn’t quite right in the supply you tested 30+ years ago. I listed several POSSIBILITIES, but you keep saying I’m wrong, without any proof.

    Why? If I’m wrong, where’s the proof?
     
  9. AF6LJ

    AF6LJ Ham Member QRZ Page

    I am finished with you.
    1. You don't own an SB-200 never have.
    2. You insist on being right, you won't stop until the person you are "debating" (attacking) gives up.
    3. you have issues with females who know the subject matter as well (or better) than you.
    4. Other people have made similar comments to you regarding your need to be right.
    5. I won't even go into the comments I have received in PM's both this time and numerous times before.

    Talk to the Hand...........
     
  10. W1BR

    W1BR Ham Member QRZ Page

    Jim, Sue, a lot of hams look to us old timers for advice and guidance. I always respected both of your opinions.

    It very disheartening to see both of you at odds over a technical matter. The discussion has gone from a technical argument into bad feelings.

    Pete
     
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