why use 2 resistors in SERIES?

[SpinCycl]

I am looking at some schematics (not nessesarily wreck, though), and they show several places that have 2 resistors in series in the circuit. Most places have 2 different values and they are not high-wattage either.

Why? I can understand if that area has a really high voltage that's above the spec on just one of the chosen resistors, but most amps I have seen are not much over 400v max anyway.

[Moose]

Any chance of a specific example?

I can think of one -- the 1W series resistors in a plexi marshall power supply, but this is an obvious case of splitting the load as that's a relatively high wattage and high voltage location in the circuit. Otherwise, we'd have to see an example to hazard a guess.

-Moose

[SpinCycl]

I am looking at the "Blue Angel" by Mesa.

The ones in the power supply are most likely just like you say: to split the power so one can use resistors smaller than a volkswagon.... I can see that. I even see this done with caps to allow smaller rated caps in a high voltage place, I got that.

In one area, they use a 680 ohm with a 3.3k, but its in a signal path from the 2nd stage to the PI, which probably isnt high current or high voltage.

Then the weird one is THREE resistors in series in the FX loop section. It has a 1k, a 10k, and then a 82k. Is that circuit so precise that they needed EXACTLY 93k?

Could there me an obscure noise, RF or oscillation control reason for doing this?

[wrecked]

The reason why you would most often see two series resistors in the power supply is to divide the voltage between two caps. When you add caps in series, you get the reciprocal addition of the values and the normal addition of the voltage:

120uF, 450V + 120uF, 450V = 60uF, 900V

Well, ideal capacitors are supposed to block DC, right? If this were completely true, then this voltage addition wouldn't work because the second cap would only see the AC component of the B+ voltage (since the first one blocks the DC part of it). In reality though, there's a tiny bit of DC current leakage through the cap which would split the voltage, but the problem with this is it only splits it evenly if the leakage amount is the same for both caps. They're not often all that close, so another solution is used. Since this current can be thought of like a high value resistor in parallel to the cap, the voltage the second one sees would be decided by the relative value of each in a voltage divider. By using resistors like the 220ks in parallel to both caps, the much smaller 220k value effectively balances the "leakage resistance" between the two so the voltage rating can be shared evenly.

It sounds like the second pair you might be describing are the bias and tail resistors of the PI (they would technically be in the cathode circuit, and there's a fair amount of current flowing there). Those values wouldn't really work for a 12ax7 in that position though, so are you sure it isn't a 33k?

Finally, I've never seen the BA effects loop but I'd bet based on the values that it's attenuation. Think of it like a pot that's on a three position rotary switch. You can figure out the attenuation by the voltage divider rule.

Any more questions, and I'm glad to help!
Matt

[DutchMarco]

Any chance of a specific example?

I can think of one -- the 1W series resistors in a plexi marshall power supply, but this is an obvious case of splitting the load as that's a relatively high wattage and high voltage location in the circuit. Otherwise, we'd have to see an example to hazard a guess.

-Moose

I've been wondering about that. I know that in parallel, adding components will increase the power handling of the group of components to (the amount of components * the power handling of the item with the lowest rating). Ie a 1x15 at 15W, or a 4x10 at 40 watts. Just using some figures but this should resemble the old and newer Bassmen.

I just presumed that this worked the same way in series. Or when combining series and parallel (ie 2x12 series, in parallel with 2x12 series for a 4x12 total).

Of course this applies to the resistors inside the amp as well, not just to the noisy ones underneath!

But was my guess right? Or am I in for an expensive revalation?

[wrecked]

Hi Marco!

You're right, the power shares evenly in series or in parallel as long as the values of both resistors are the same. You can figure out all the relations with different values or networks with a couple of equations: V = IR and P = IV. The voltage and current divider rules are helpful too, but they're really just an extension of the two formulas.

Matt

[DutchMarco]

Hi Marco!

You're right, the power shares evenly in series or in parallel as long as the values of both resistors are the same. You can figure out all the relations with different values or networks with a couple of equations: V = IR and P = IV. The voltage and current divider rules are helpful too, but they're really just an extension of the two formulas.

Matt

Hey, thanks, I forgot about those rules! But it's rather a shame that the power doesn't spread equally over all components when they are of unequal value... That means I have to do a lot of calculating (I'm dreadful at math) rather than just plonking them in until you've reached the desired power handling.