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Author Topic: blues deville got the blues  (Read 15558 times)


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Re: blues deville got the blues
« Reply #45 on: April 08, 2014, 12:26:49 PM »
  The 13.6VAC shown on the schematic is what you should get when you use the specified input signal shown at the input jack, with the control settings described in the notes section.  All AC voltages shown are under these same conditions.
  The 58W is the rated output at full power, with what ever signal level required.


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Re: blues deville got the blues
« Reply #46 on: April 09, 2014, 05:33:51 AM »
Quote from: ilyaa
how can their power dissipation be additive if all of the current flowing through the load has to pass through the first ten resistors before it passes through the second ten?

Intuitively; because they are all equal the voltage will divide equally between the top half and the bottom half.

You can simplify this to two resistors in series, the ten in parallel above are 39r/10 = 3.9r, and similarly the bottom ten, so you effectively have two 3.9r resistors in series, 3.9+3.9=7.8 or roughly 8 ohms.  Since they are all equal the current will split equally through all of them, so the main current will split into ten times I/10 resistors, one-tenth through each resistor.

Because the current splits equally they will all equally share the power, and since we have 20 * 10W resistors that gives the whole load a capacity of 200 watts.

I've noticed that you seem to be having a bit of trouble generally with the idea of a voltage divider, so...

A single resistor R across a voltage source E (such as a battery) gives a current I according to Ohms Law;

I = E/R

Now if we divide that resistor into two equal parts the voltage on the mid point will be half the total applied voltage.

Similarly if we make the resistors unequal the voltage will distribute inversely as their resistance.

Say we connect a 1k and 2k resistor in series.  If the 1k is the upper or more positive end the voltage at their join will be 2/3rds of the supply.  If we invert them it will be 1/3rd.

This is similar to connecting a (linear) pot across our battery.  If the pot is set to the mid point the voltage on the wiper will be half of the battery voltage.  If set to 10% then the wiper voltage will be one-tenth of the supply; to 90% it will be 9/10ths.

Now if we have a fixed resistor and a pot (or a valve, or a transistor) in series the voltage on the join will depend on the relative value of the pot to the fixed resistor in series, and if we consider a valve or a transistor to be basically an electronically variable resistor in series with a fixed resistor across a supply, we should then see that the voltage on the join of the resistor and the device will depend on the incoming signal to the control element, grid, base, or gate for a FET.

What is going on in a simple voltage divider, or active amplifier stage, is essentially Ohms Law applied to two resistances in series, the Voltage Divider.

The voltage across the bottom bit is equal to the bottom bit divided by the whole, what proportion it is.

The output is the R2th bit of the total, R1 + R2;

R2 / (R1 + R2)

...and since this R2th bit of the whole resistance is also the R2th bit of the whole applied voltage the output voltage is the R2th bit of the input voltage;

Vout = Vin * R2/(R1+R2)

So now we tackle a slightly more complicated case;

The first step is to reduce this four resistor network to a two resistor network;

R2 and R3 are in series, so they simply add together to make a single resistor;

6k + 500r = 6.5k

Similarly we combine the parallel pair R1 and Rload to make a single effective resistor.  The long form general case for this is;

1/Rtot = 1/R1 + 1/R2 + 1/R3 + ... 1/Rn

The short form for only two resistors in parallel is;

Rtot = (R1 * R2)/(R1 + R2)

so we can do it either way

1/3500 = 0.00028571

1/10 = 0.1

0.1 + 0.00028571 = 0.10028571

1/0.10028571 = 9.9715104 ohms

{a reality check is that we expect 3.5k in parallel with 10 ohms to be a bit less than 10 ohms, which this is}


(3500*10)/(3500+10) = 9.97150997

{notice that my wordprocessor calculator has given slightly different results in the 5th decimal place, but this is of no practical importance in general electronics.}

So now we have a simple two resistor divider consisting of 6.5k above and 9.9715 ohms below, or 9.9715ths of the whole 6500+9.9715;

9.9715/(6500+9.9715) = 0.00153173

The applied voltage is 10 volts, so the voltage at the output is;

10 * 0.00153173 = 0.0153173, or a bit over 15mV.

There are two laws named after Russian dudes that state (what should be) the bleedin' obvious, and we've just used one of them above, Thévenin's theorem.

Essentially Thévenin's theorem says that a circuit of arbitrary complexity consisting of voltages, current sources, and resistances, can be reduced to a single voltage and resistance (which can be very helpful trying to decompose complex circuits).

Norton's theorem is similar except we end up with a current source and a resistor.

Kirchhoff's current law (KCL) - Kirchhoff's voltage law (KVL)Kirchhoff's circuit laws should be similarly intuitive; that the sum of all currents at a node is zero (the currents flowing out equal the currents flowing in - or you would have an accumulation or deficit of electricity).

The voltage law is equally simple, the sum of all voltages around a loop (circuit) is zero (all the drops equal the supply voltage - or again you would either have a surplus or shortage of electricity).

In both of these cases of course the currents and voltages have signs, so e.g. when the current inflow (+ve) is added to the current outflows (-ve), or the supply voltages (+ve) are added to the voltage drops (-ve) the sums are zero.

Now hopefully you will see that KCL applies to the central node in your dummy load, and that using Thévenin your multiple resistors can be reduced to just two in series, or just one effective resistor of ~8 ohms at 200 watts if that's what you need.

If you say theory and practice don't agree you haven't applied enough theory.