Resistor in parallel and Power - Question

[kagliostro]

Today I was to the local electronic shop looking for a (round and relatively small) 300R 10W resistor

They have not so much components and had only a 330R ceramic resistor that I got also if it is a bit too long for my board

I got it also if it is 330R (waiting for a correct one) and I've think to put a 3.3k 2W resistor in parallel to have 300R as result

After that I remembered that may be I've some 10W 3.3k resistor that are of the correct lenght and if I put a 330R 2W resistor

I'll obtain the desired 300R

But I've a doubt, if I use a 10W 330R + a 3.3K 2W resistor the larger current follows the lower resistance path, so no dissipation problems

and if I use a 3.3K 10W + a 330R 2W resistor what will happen ? Is this an easy way to burn the 330R 2W resistor ??

Many Thanks

Franco

[DummyLoad]

and if I use a 3.3K 10W + a 330R 2W resistor what will happen ? Is this an easy way to burn the 330R 2W resistor ??


possible to leak much smoke.


can you get three 1K 5W? that with 3.3K 1W or 2W would work.


--pete

[kagliostro]

Thanks DummyLoad

I must give a look to my stock to see what I've or use the 10W 330R + a 2W 3300R

this way I think I've no problem

Franco

[sluckey]

What is the purpose of the resistor. 300Ω may not be such a critical value. Perhaps any value between 200 and 400 will be acceptable?
 

[kagliostro]

Ciao Steve

the resistor is for the cathode of the 1625 on the schematic you have seen (modified AA764 + 5C1)


Franco

[sluckey]

In that case I definitely would not zero in on 300 as the magic number.

[kagliostro]

Grazie Steve

In that case I definitely would not zero in on 300 as the magic number

       May be I must think to a more flexible solution .....

Franco

[HotBluePlates]

... if I use a 10W 330R + a 3.3K 2W resistor the larger current follows the lower resistance path, so no dissipation problems

and if I use a 3.3K 10W + a 330R 2W resistor what will happen ? Is this an easy way to burn the 330R 2W resistor ?? ...

Let's solve the problem and find out.

How much voltage do you expect across the resistors?  Pulling a number out of the air, I'll guess 15v.

Resistors in parallel see the same voltage.
15v / 3.3kΩ = 4.5mA
15v * 4.5mA = 68mW

15v / 330Ω = 45.5mA
15v * 45.5mA = 0.68W

Do you expect more than 50mA of idle current (plate & screen)?  If no, it looks like 2w is plenty for either resistor and 10w is overkill (but might be cooler for a nearby cap).

Ofcourse you could have calculated as 15v² / Resistance, and found power directly.  But knowing the implied current might be helpful in some cases.

[DummyLoad]

was wondering the same thing... usually we double up on wattage or even 3X if there's big swing in current demand.


--pete

[kagliostro]

Many Thanks Bill & Pete

I think there will be ~18V across the resistor(s), it is the cathode resistor of a 1625 tube in an SE amp

here schematic



(There is a typo, bass cap is .1 NOT .01)

and ... yes, 10W was planned as to be a safe situation, in an SE amp I don't think there will be a large swing in current demand

--

I've this nice (and smaller 3.3K resistors that fits better the board)

respect to the 330R ceramic resistor I got yesterday



But when Steve told

In that case I definitely would not zero in on 300 as the magic number.

I started wondering a different solution (more flexible)

a 220R 10W resistor + this russian 220R ceramic pot and this give me a range from 220R till 440R to perfectly tailor the bias





but I'm still wondering about what to do (there are also other solutions, one is to find a lot of correct sized resistor in various values  )

Many ideas but well confused 

Franco

[HotBluePlates]

But when Steve told

In that case I definitely would not zero in on 300 as the magic number.

I started wondering a different solution (more flexible)

a 220R 10W resistor + this russian 220R ceramic pot and this give me a range from 220R till 440R to perfectly tailor the bias

You probably won't know the "ideal resistance" until you actually build the amp.  You could use the wirewound pot, but I suspect once you get it close to a happy value of resistance, you probably won't change it much ever again.  In that case, it seems a waste to use the pot.  But if you already have the pot, then cost doesn't really matter unless you have some other project where it would be critical to have.

[DummyLoad]

that's nothing more than a 807 SE. 


https://frank.pocnet.net/sheets/155/8/807.pdf


probably going to have around 14 to 15V Vk. to keep Pdiss around 22.5W @ 335V, you'll need about 75mA Ik. so 14V-15V/75mA ~~180 to 200 ohms.


--pete

[kagliostro]

Thanks Pete

I've this extract of the 807 datasheet



Seems that (with 350V B+) plate + screens are around 59mA

I don't know the exact value of  Rk (I think they refer to a fixed bias with that  -18V)

but seems that also a less W rated resistor can be used (may be 7W)

I'll prefer to use a bigger one but I'll see what I can find

Franco

[DummyLoad]

if V bias fixed is shown to be at -18V with 350V, a good approximation is that you need 18ish Volts on the cathodes with auto bias. so 18V/58mA ~~310R. use a 330R. Va-k is actually 350 - 18V = 332V but still close enough to 350V to make the approximation valid. more or less. you're cutting hairs where you don't need to. fire it up with a 330R, measure Vk and make the determination if Pdiss is where you want it. in the spec sheet it's about 20W. on a footnote, during the initial power up, monitor your screen V closely; the closer to 250V it actually is, the closer the approximation is.


--pete

[ac427v]

Unrelated detail. Bass cap in the tonestack is .01?
--Craig

[kagliostro]

Thanks Pete

Those are important info for me, I'm not very skilled and all this info give me help

----

Many Thanks Craig

is a typo, correct value is .1

Grazie

Franco