Generic voltage divider question...

[Gtr0]

Hello Fine Folk of MEF.

I have a question about voltage dividers... or perhaps not ??

To properly relay my question I have created a very fancy aid... please consider the following image...



when signal is passing through R26 and the switch to add R27 in parallel is not engaged (as in the image) will this connection act as a voltage divider? Will any signal be bleed off by R27?

Obviously it is a question for adding options to amplifiers... Of course I can try and figure out sonically if anything is happening, but I would rather not play head games with myself in such cases.

Thanks in advance!

[Mick Bailey]

In the diagram there is no voltage divider action. The only change is to the effective resistance value as seen when measured across R26. Now, consider if the bottom end of R26 was grounded and there was an additional resistor Rx in series with R26, and the signal was being fed into the circuit via Rx. Then there would be a voltage divider action at the junction of Rx and R26 which would be modified when R27 is switched in. That would give you the ability to switch the output level between two values.

The circuit pictured may act as a voltage divider (or not), but only in the context of other components in the circuit.

[Old Tele man]

It's an "open" path when the switch is OPEN and thus no current passes through resistor R27, however, it WILL have a voltage when probed with a voltmeter, and it will be the same as that voltage at the bottom of resistor R26.

[Helmholtz]

when signal is passing through R26 and the switch to add R27 in parallel is not engaged (as in the image) will this connection act as a voltage divider? Will any signal be bleed off by R27?

It is not a voltage divider and if the switch is open R27 has no effect at all (as no current flows through it), so you only have R26 in the circuit. When you close the switch R27 acts in parallel to R26, which results in an effective total resistance lower than each R26 and R27 (look up "paralleling resistors").

A voltage divider requires 2 resistors wired in series so that they share the same input current. The output voltage is taken from the junction between the resistors. I recommend to consult Wikipedia for # voltage divider.

[Chuck H]

I think Gtr0 was asking if the uswitched circuit would act as a voltage divider in any way. Pretty sure he didn't think the circuit in general was a voltage divider.

Gtr0, what do you want to achieve. We can help.

[Gtr0]

I think Gtr0 was asking if the uswitched circuit would act as a voltage divider in any way. Pretty sure he didn't think the circuit in general was a voltage divider.

Gtr0, what do you want to achieve. We can help.

That's exactly it - my end goal was to be reassured that in this specific situation, this would not divide any signal and bleed it off ... as, I don't know - heat perhaps. I didn't think so, however sometimes I come across some nugget of info that forces me to question what I think I know.

I have a grip on how voltage dividers in general work, I just wanted to make sure that in a case such as this (for example imagine R26 is NFB) that leaving the contact at R27 open, as it is, has no effect on the signal passing through R26... in some mysterious way that I could not understand :-) - of course in this example with NFB, assuming two resistors are of equal value that the resistor value will half. Otherwise I am aware of the various formulae in figuring series/parallel caps and resistors etc.

So, all understood.

Thank you!!

[Helmholtz]

I think Gtr0 was asking if the uswitched circuit would act as a voltage divider in any way.

Switched or unswitched the circuit (being just a single resistance) won't act as a voltage divider. As said a voltage divider always consists of a series circuit of resistances /impedances and has at least 3 terminals. A single resistor/resistance can only drop voltage (depending on current) but not divide voltage by a definite factor.
Of course any resistor could be part of some voltage divider.

[R.G.]

I didn't think so, however sometimes I come across some nugget of info that forces me to question what I think I know.

I have a grip on how voltage dividers in general work, I just wanted to make sure that in a case such as this (for example imagine R26 is NFB) that leaving the contact at R27 open, as it is, has no effect on the signal passing through R26... in some mysterious way that I could not understand :-) -

OK, in the spirit of nuggets of information...

A really common mistake made by many folks is that they don't account for the "invisible resistors". "Invisible resistors" are the internal source resistance of whatever feeds the circuit, and the internal impedance of whatever it in turn feeds. In your instance, you show only two resistors and a switch. What you don't show is what drives it and what it drives.

The unsaid likelihood is that it's a tube plate, maybe a 12AX7 plate, that drives it with signal, and a tube grid that is driven by it. A 12AX7 plate acts like it has a roughly 60K resistor in series with it (the internal plate resistance), and a tube grid is neatly nearly an infinite impedance. Sadly, that nearly infinite grid impedance is assumed to be about 1M, but may be much smaller or larger. Changing the resistor in the middle with your circuit will cause voltage divider action with these invisible resistors.

If the two resistors are large compared to 60K, you can ignore the effect of the plate resistance (and any external plate resistor) and be fine. Resistors in the range of 10X the plate resistance make ignoring the plate resistance OK. If the two resistors are small (~1/10) compared to the grid leak resistor, generally thought to be 1M, you can ignore the grid leak resistor. That means the resistor really ought to be less than 100K to be ignored. ACCKKKK!! They're probably right in the middle.

Most tube amp hackers do ignore the source and driven impedances, and get along OK. But you were looking for nuggets of understanding. You can't always ignore the source and load impedance, and you'll be better at design if you remember them, at least until you can decide they can be ignored.

[d95err]

The simple answer:

With the switch open, R27 is effectively out of the circuit and the resistance between the upper and lower part will be that if R26.

With the switch closed, the resistance will be R26 || R27 (both resistors in parallell).

As the others said, the circuit shown has no voltage divider, regardless of the switch position, only series resistance.