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615_ri
Junior Member
Username: 615_ri

Post Number: 17
Registered: 5-2006
Posted on Sunday, June 04, 2006 - 1:25 pm:   Edit Post Delete Post    Move Post (Moderator/Admin Only)

anyone ever try it?

i need a dummy load so i can tune my npc mod right. there is not a shoop around here that sells them.
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Rbeswick62
Junior Member
Username: Rbeswick62

Post Number: 20
Registered: 5-2006
Posted on Sunday, June 04, 2006 - 4:38 pm:   Edit Post Delete Post    Move Post (Moderator/Admin Only)

If you would like to build your own here is the link I found.
http://k4eaa.com/dummy.html
It's probably less expensive to buy one, but I found it was more beneficial for me to build mine following this tutorial.
1- I like to make things
2- I built mine with a 1% tolerance resistors
3- I was able to ohm out each of my resistors, so as to control the load to precisely achieve a 50 ohm load.
I love perfection and strive to create the best I possibly can.
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Allan
Intermediate Member
Username: Allan

Post Number: 159
Registered: 2-2006
Posted on Sunday, June 04, 2006 - 6:34 pm:   Edit Post Delete Post    Move Post (Moderator/Admin Only)

If you wanted to make your own dummy load, you would need some high wattage rated resistors, (carbon composition so they are non-inductive, and wire wound resistors will not work.) that equal 50ohms when placed in parallel.
(1) 50ohm resistor = 50ohms.
(2) 100ohm resistors = 50ohms.
(4) 200ohm resistors = 50ohms.
(10) 500ohm resistors = 50ohms.
If you used 1, 5 or 10 watt resistors the power handling capability would be:
1,5, or 10 watts
2,10, or 20 watts
4,20, or 40 watts
10,50,or 100 watts.
See the pattern?
I think that's the way you do it.....
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615_ri
Junior Member
Username: 615_ri

Post Number: 18
Registered: 5-2006
Posted on Sunday, June 04, 2006 - 7:26 pm:   Edit Post Delete Post    Move Post (Moderator/Admin Only)

any schematics?
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Allan
Intermediate Member
Username: Allan

Post Number: 160
Registered: 2-2006
Posted on Sunday, June 04, 2006 - 8:13 pm:   Edit Post Delete Post    Move Post (Moderator/Admin Only)

Do a web search. I typed in "50 ohm dummy load" and got pages and pages of drawings, schematics, how to's, pictures, doodles, and every which way to do it, from the most complex to the cheesest.
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Chad
Advanced Member
Username: Chad

Post Number: 896
Registered: 11-2004


Posted on Sunday, June 04, 2006 - 9:24 pm:   Edit Post Delete Post    Move Post (Moderator/Admin Only)

Do not use wirewound resistors, they become inductors at RF freqs. bedt bet is film resistors.

Chad
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Georgeodjungle
Intermediate Member
Username: Georgeodjungle

Post Number: 112
Registered: 1-2006
Posted on Monday, June 05, 2006 - 3:58 am:   Edit Post Delete Post    Move Post (Moderator/Admin Only)

you could use wire wound but then the math would be
(r1*r2)/(r1+r2)
for 2 resistors
or
(r1*r2*r3)/(r1+r2+r3)
for 3
o.k.o.k.
just use noninductive
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Georgeodjungle
Intermediate Member
Username: Georgeodjungle

Post Number: 113
Registered: 1-2006
Posted on Monday, June 05, 2006 - 4:07 am:   Edit Post Delete Post    Move Post (Moderator/Admin Only)

you could use wire wound but then the math would be
(r1*r2)/(r1+r2)
for 2 resistors
or
(r1*r2*r3)/(r1+r2+r3)
for 3
o.k.o.k.
just use noninductive.
what ever happend to the light bulb dummy load?
if i remember right a 12v 50w was good for 100.
& a 1255 was good for 20?
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Bruce
Senior Member
Username: Bruce

Post Number: 3904
Registered: 9-2003
Posted on Monday, June 05, 2006 - 6:36 am:   Edit Post Delete Post    Move Post (Moderator/Admin Only)

WIREWOUND ????????

NO DONT DO THAT !

And my first dummyload was a 25 watt bulb!

A GOOD ONE ARE THE OLD CANTENNAS I beleve MFJ still makes one.
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Chad
Advanced Member
Username: Chad

Post Number: 899
Registered: 11-2004


Posted on Monday, June 05, 2006 - 2:19 pm:   Edit Post Delete Post    Move Post (Moderator/Admin Only)

I have an MFJ "Cantenna" and really like it. it sits under the bench and does it's job quite well.

Chad
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Kid_vicious
Senior Member
Username: Kid_vicious

Post Number: 1757
Registered: 9-2004
Posted on Monday, June 05, 2006 - 6:20 pm:   Edit Post Delete Post    Move Post (Moderator/Admin Only)

well, i know its not supposed to be done, but i used wirewound resistors for my dummyload, and it works great.
sure, above 30mhz, it could be a problem, but the majority of people on this site who make one will only use it for CB freqs.

so, if its only to be used around 27mhz, then i say its fine. not "correct", but fine.

i used three 150 ohm 25 watt resistors.
here is a link to the thread:
http://www.copperelectronics.com/cgi-bin/discus4/show.cgi?tpc=34&post=83421#POST83421

that being said, everything Bruce and Chad are saying is correct, and they know their stuff and should be listened to.
i just know that for my purposes, building one had to be cheaper than buying one, and it was, by alot, and it gives me a 1.1 to 1 SWR at any freq. from 26-28 mhz which is all i need.
again, i am cheap and dirty, only listen to me if you are too.
just my 2 watts,
matt
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Chad
Advanced Member
Username: Chad

Post Number: 902
Registered: 11-2004


Posted on Monday, June 05, 2006 - 10:06 pm:   Edit Post Delete Post    Move Post (Moderator/Admin Only)

Honestly my MFJ is a boat-load of 2 watt FP resistors, not wirewound but standard carbon issue devices in a series parallel config. I'd snap a pic but they are soaking in ATF fluid now and it would make quite a mess!

Chad
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Bruce
Senior Member
Username: Bruce

Post Number: 3908
Registered: 9-2003
Posted on Tuesday, June 06, 2006 - 6:48 am:   Edit Post Delete Post    Move Post (Moderator/Admin Only)

WIRE WOUND ............ scary ........
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Kid_vicious
Senior Member
Username: Kid_vicious

Post Number: 1771
Registered: 9-2004
Posted on Tuesday, June 06, 2006 - 10:47 pm:   Edit Post Delete Post    Move Post (Moderator/Admin Only)

scaredy cat!
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Tech237
Moderator
Username: Tech237

Post Number: 364
Registered: 4-2004
Posted on Wednesday, June 07, 2006 - 8:54 am:   Edit Post Delete Post    Move Post (Moderator/Admin Only)

As I have said in a previous post DO NOT USE Wirewound reistsors for a dummy load (picture tech237 shivering in horror).

I cannpot find my catalog but try MCE.com. If I have the right place they have carbon film resistors = 100 ohm at 25w. Using series/paralle combinations (resistors in series add together, those in parallel [if of the same value) just divide by the number in parallel).

For example two 100 ohm resistors in parallel = 50 ohm, while two in series equal 200 ohm.
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Kid_vicious
Senior Member
Username: Kid_vicious

Post Number: 1782
Registered: 9-2004
Posted on Friday, June 09, 2006 - 11:32 pm:   Edit Post Delete Post    Move Post (Moderator/Admin Only)

on more than a few occasions; i have thought that what i was doing was, "fine", if not correct.
later, i have been shown that what i thought was fine, wasn't.

so, with that in mind, could one of you (Bruce or Tech237) explain for everyone's benefit, why the use of wirewound resistors "scares" you.

the only concern i have heard of is that they will radiate some signal, whereas the carbon film ones wont.

i would be VERY interested to learn of any dangers i am not considering.
cant wait to learn more about this.
thanks guys,
matt
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Tech237
Moderator
Username: Tech237

Post Number: 369
Registered: 4-2004
Posted on Saturday, June 10, 2006 - 10:07 am:   Edit Post Delete Post    Move Post (Moderator/Admin Only)

Kid,
My pleasure.

Wirewound resistors have a larger amount of what is called inductance, which is, basically, the ability of a current flowing through a wire to create a magnetic field around the wire, which in turns creates a second current flowing in the wire in the reverse direction to the applied current.

Now all wires will have a value of inductance, and in a straight piece of wire this is generally so small that it can be ignored. By coiling the wire up ( as in a wirewound resistor or a choke or other form of coil) you increase the amount of inductance that piece of wire has. Now this induced current flow tries to prevent the applied current flow from actually flowing, which can cause your dummy load to show a SWR reading far from what it really is. This reading may be higher or lower than the true value and which it would be is hard to predict as several other factors become involved - the actuall value of inductance (in Henries), the amount of applied current, the length of the coax attaching the load to the transmitter ( see my multi-part article on coax for a better explanation of this latter factor).

Also this effect the inductance has changes with frequency and is more of a problem at higher frequencies than it is at lower ones. This is due to inductance increasing as frequency increases.

TRhe other reason using an inductive dummy load makes me cringe is the other effect an inductor has. When power is applied to an inductor is stores energy in the form of the magnetic field. Now this energy is slow (relatively) to be stored but when power is removed the field collapses rapidly and in doing so creates a current and voltage in the inducotr many times greater than that which was originally applied. This higher voltage and/or current can cause damage to transistors in the radio.

For an example of this effect just take a look at the coil in your cars. 12VDC is applied to the coil via the points. When the points open a voltage of approx 30,000VDC is then created by the collapsing field and applied to the spark gap of the spark plug. Don't believe me? Well I for one wont recommend holding a plug lead while the engine is running - it hurts ( personal experience).

I hope this helps a little without getting too technical.
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Kid_vicious
Senior Member
Username: Kid_vicious

Post Number: 1787
Registered: 9-2004
Posted on Sunday, June 11, 2006 - 3:08 am:   Edit Post Delete Post    Move Post (Moderator/Admin Only)

thank you tech237,
dont ever worry about being too technical.
if i read something i dont understand, i research until i do. its part of my job, as i am always dealing with a new piece of equipment.
(i run computer automated winches for a las vegas theater show, and this industry likes to use whatever just came out.)

i am also a car guy since before i could drive, and the distributor coil analogy definitely makes sense to me.

i am glad to be learning this. i know what an inductor does, but didnt think that whatever inherent inductance a wirewound resistor has was enough to worry about.

so thats where im at right now; is the ammount of inductance in a wirewound resistor measurable?
is there a way to figure out mathematically whether or not this value is enough to cause problems when operated at 26-28 mhz, and lets say under 500 watts?

i realize that the theory is sound, and what you say is happening, IS actually happening, but exactly how much of an effect is it having on the SWR reading? like 3 or .3 or .03 "points"?

the part about the current flowing back into the radio does concern me.

the only thing i can say, and i know it doesnt necessarily mean all that much, but i have been using mine for about a year now, and the SWR has always been 1.1 to 1, and has never damaged a radio its been hooked to.
i have hit it with 250 watts before, but only for a few seconds, and it still seemed fine.

i know it sounds like i just dont want to have to build another one, but im actually excited to build another one, i just want to understand more of the why's and how's as i do it.

how come nobody spoke up about this when i first started posting about it a year ago?
sure, i got a few dont use them replys, but without being told WHY not to use them, it went in one eye and out the other.

again, thanks for the lesson, i hope alot of us homebrewers see this thread and take heed.
matt
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Tech237
Moderator
Username: Tech237

Post Number: 370
Registered: 4-2004
Posted on Sunday, June 11, 2006 - 12:16 pm:   Edit Post Delete Post    Move Post (Moderator/Admin Only)

Technical Time.

The inductance of a single layer air cored coil can be worked out, it can also be measured with the right sort of test equipment, by using the formula

L (uH) = (d x d) x (n x n)/(18d + 40l)

where d = coil diameter in inches
n = number of turns
l = length of coil in inches.

If the coil is wound on a core or has multiple turns the formula becomes a little more complex.

To take this a step further the energy stored in the coil is now calculated using -

W = (IxIL)/2

where W = energy in joules
I = current in amperes
L = inductance in Henrys

I'll find where I have the formula that actual works out the voltage created by a collapsing magnetic field and post it when I do.

Off the top of my head I cannot think of anyway that you could see how much effect it is actually having on the SWR and at 27MHz it may not have much at all. Maybe 833 can suggest a way.
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Kid_vicious
Senior Member
Username: Kid_vicious

Post Number: 1790
Registered: 9-2004
Posted on Sunday, June 11, 2006 - 6:21 pm:   Edit Post Delete Post    Move Post (Moderator/Admin Only)

well, i imagine its going to be pretty tough to count the turns in a wirewound resistor, and i dont feel like grinding away a layer to get to the wire.

and we're talking RF current here right?
i guess we would need to know how much RF current is equivivilent to 4 watts out.

its interesting, but im going to end up building another one anyway, so, best not to spend too much time figureing this one out.
lest we get obsessed.LOL
thanks for taking me through this,
matt

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