Wilder Amplification

At the request of the original author, I will be taking over the writing of the series of threads titled "How A Tube Amp Works". I will be dividing these up into individual threads to make them easier to find rather than having to scroll through a whole thread to find the information.

Before I can begin to write How A Valve Works part 2, I need to you gain an understanding of simple DC series circuits. So here is installment number 2 that will explain simple DC circuits.

There will be some math involved with this article as I explain Ohms Law and how to apply it in a series circuit. But don't worry, it's all basic math and I explain all of my math work step by step so you shouldn't get lost. If you do, you can always PM me and I can hopefully better explain it to you.

As with the last article, don't try to digest it all in one sitting. Read some, take a break, read some more...re-read if you have to.

DC Series Circuits
By Jon Wilder

Electron current flows much the same way as water flows through a hose.

Voltage is to electron current flow as pressure is to water flow.

Resistance is to electron current flow as the size of the hose is to water flow.

Resistance to current flow creates voltage as resistance to water flow creates pressure.

When you have a resistance to current flow, some of the electrons are "held back", creating a difference in the amount of electrons on each side of that resistance, which creates a difference in charges. This is because one side of the resistor has less electrons than the other, making that side more positive. This difference in charges is what we see as the "voltage drop" across the resistor. Current flow through a given resistance creates voltage. Just like placing your thumb over a hose creates a resistance to water flow, the resistance to water flow creates the pressure difference you feel on your thumb (there's more water on the backside of your thumb than there is on the front of your thumb, hence the pressure difference. Just like resistors create a voltage drop across them due to having less electrons on one side of the resistor than the other.

A series circuit is a circuit in which electrical current has only 1 path to flow. The circuit imposes a certain amount of resistance on the power supply. This resistance value and the supply voltage determines how much current will flow through the circuit, and that current flow will be the same value at all points in the circuit.

Laws Of Series Circuits

Current flow is the same at all points in a series circuit.

Total circuit resistance and supply voltage determine the amount of current flow through the circuit.

Each component in the circuit will drop a certain amount of voltage, which is known as a "voltage drop". The value of the voltage drop will appear across the component. The value of the voltage drop across a particular component in a series circuit is determined by the resistance value of the component and the amount of total circuit current.

The sum of the voltage drops across each component in a series circuit will always equal the supply voltage.

Component resistance is additive in a series circuit. In other words, the sum of each component's resistance values will equal the total circuit resistance.

A break (i.e. disconnection) anywhere in a series circuit will stop all current flow.

Ohm's Law

Some basic math skills are required in the study of electricity and electronics. The equation most used in electricity and electronics is known as Ohms Law. The main equation of Ohm's Law is -

E = I x R where -

E = Electromotive Force in Volts

I = Current flow in Amps

R = Electrical resistance in Ohms

Two of the three above mentioned values must be known before you can find the third value. The above equation can be rearranged one of the following two ways -

If Voltage and Resistance are known and you need to find current flow (Amps) -

I = E / R

If Voltage and Current (Amps) are known and you need to find resistance (ohms) -

R = E / I

In series circuits, "I" (current flow in amps) will always be a constant value unless the total circuit resistance changes.

Ohm's Law HowTo

Here are the instructions for using Ohm's Law in series circuits.

1) Multiplying total circuit current by total circuit resistance will give you total supply voltage.

2) Multiplying total circuit current by the resistance value of 1 component will give you the voltage drop across that one component.

3) Dividing the supply voltage by total circuit resistance will give you total circuit current.

4) Dividing the supply voltage by total circuit current will give you total circuit resistance.

5) Multiplying total circuit current by the combined resistance value of two or more components in line with each other will give you the voltage drop across those two or more components.

6) Adding the voltage drops of all components in a series circuit together will equal the total supply voltage.

7) The only time total supply voltage is used in Ohm's Law is when trying to find either total circuit current or total circuit resistance.

8) Once total circuit current is found in a series circuit, this value remains constant for that circuit unless -

a) Supply voltage changes

b) the resistance value of a circuit component changes

9) Measuring the voltage drop across 1 component in a series circuit, then dividing that voltage by the resistance value of that one component will give you total circuit current (current remains the same at all points in a series circuit).
Typical Series Circuit

Below is a link to a schematic of a typical series circuit.



This circuit is the same circuit used in the bias supply of a 100 watt Marshall plexi, but with a positive polarity for simplification, and is known as a "voltage divider" circuit. It is shown without the bias pot for this exampe to simulate the bias control rotated for the least voltage. When the pot is rotated fully in the direction that gives the least voltage, it is effectively taken out of the circuit. The supply voltage is divided up between the three resistors as current flows through them.

The arrows indicate the direction of current flow. The positive charge on the positive terminal of battery B1 pulls negative electrons from the negative terminal of B1 (opposite charges attract), through R3, R2 and R1, then back to the positive terminal. R1, R2 and R3 limit the amount of electrons (current) the positive terminal can pull through the circuit.

If we take a look at the circuit, we can see that we know the total supply voltage of battery B1 (80VDC) as well as the values of R1, R2 and R3 (R1 = 27K ohm, R2 = 15K ohm and R3 = 47K ohm). Since total circuit resistance = the sum of the resistance of all components in the circuit, we can add the values of R1, R2 and R3 to get our total circuit resistance.

Rt = R1 + R2 + R3 where -

Rt is the total circuit resistance in Ohms

R1 is the value of R1 in ohms

R2 is the value of R2 in ohms

R3 is the value of R3 in ohms

Since the value of R1 is 27K ohms, the value of R2 is 15K ohms, and the value of R3 is 47K ohms, we'll plug these values into the above equation to get our total circuit resistance.

Rt = 27K ohms + 15K ohms + 47K ohms

Rt = 89K ohms

So our total circuit resistance is 89K ohms. We can now use our total circuit resistance and the supply voltage to find how much current is flowing through the circuit. Since current flow is the same at all points in a series circuit, this current value will be constant throughout the remaining equations.

It = Es / Rt where -

It = total circuit current in Amps

Es = total supply voltage

Rt = total circuit resistance in ohms

It = 80VDC / 89,000 ohms (89K)

It = 0.0009 Amps, or 0.9mA (milliamps)

Now that we know the total circuit current (0.0009 amps), we can multiply this by the value of R3 to find the value of the voltage drop across R3 (voltage between points Ground and A).

Er3 = It x Rr3 where -

Er3 = voltage drop across R3

It = total circuit current in amps

Rr3 = resistance value of R2 in ohms

Er3 = 0.0009 Amps x 47,000 ohms (47K)

Er3 = 42 Volts across R3

To find the voltage drop across R2 (voltage between points A and B), we can multiply the value of R2 by the total circuit current in amps.

Er2 = It x Rr2 where -

Er2 = voltage drop across R2

It = total circuit current in amps

Rr2 = resistance value of R2 in ohms

Er2 = 0.0009 Amps x 15,000 ohms (15K)

Er2 = 14 Volts across R2

To find the voltage drop across R1 (voltage between points B and C), we can multiply the value of R1 by the total circuit current in amps.

It x Rr1 = Vr1 (Voltage across R1)

0.0009 amps x 27,000 ohms (27K) = 24 volts across R1

To check our math, we use one of the laws of series circuits.

"The sum of all voltage drops in a series circuit always equals the total supply voltage."

24 volts + 14 Volts + 42 Volts = 80 Volts

Our math checks out.
Another Series Circuit

Now let's add a fourth resistor into the circuit.



This circuit is the same circuit, but with the addition of R4. Since it's value is 25K, this adds 25K to the total circuit resistance.

R1 + R2 + R3 + R4 = Rt

27K + 15K + 47K + 25K = 114K ohms Rt (total circuit resistance)

Since the total circuit resistance went up, the total circuit current (It) has dropped.

Total Supply Voltage / Rt = It

80 Volts / 114,000 (114K) = 0.0007 amps 0.7mA (milliamps)

The voltage between ground and test point B will be the voltage drop across R3.

It x R3 = Voltage R3

0.0007 Amps x 25,000 ohms (25K) = 17.5 volts

The voltage at test point A will be the voltage drop across both R3 and R2 together. This is because when measuring between ground and test point A, you're reading the voltage across both resistors together.

To find the voltage at test point A, we must first add the resistance values of R3 and R4 together to get the total resistance between test point A and ground.

R3 + R2 = Ra (resistance to ground at test point A)

25K + 47K = 72K

Resistance to ground at test point A (Ra) is 72K ohms.

Voltage at test point A = It x Ra

0.0007 Amps x 72,000 (72K) = 50.4

Voltage at test point A = 50.4 Volts

Another way we could've done that was to find the voltage drop across R3 by multiplying the total circuit current (It) by the value of R3. Then add that value to the voltage drop across R4.

It x R3 = Voltage drop across R3

0.0007 amps x 47,000 ohms (47K) = 32.9 Volts across R3

Then we add the voltage across R3 to the voltage across R4.

Vr3 + Vr4 = Voltage at test point A where -

Vr3 = Voltage drop across R3

Vr4 = Voltage drop across R4

32.9 volts + 17.5 volts = 50.4 Volts

Voltage at test point A = 50.4 Volts

Now that we have our total circuit current, we can go through and find the voltage drops across the last two resistors.

It x R2 = Vr2

It x R1 = Vr1

0.0007 amps x 15,000 (15K) = 10.5 volts

0.0007 amps x 27,000 (27K) = 18.9 Volts

Vr2 = 10.5 Volts

Vr1 = 18.9 Volts

Now let's add up our voltage drops -

Vr1 + Vr2 + Vr3 + Vr4 = Vsupply

18.9 + 10.5 + 32.9 + 17.5 = 79.8 Volts

Vsupply = 79.8 Volts

* Note - When performing math functions on any kind of circuit with high circuit resistance, answers will be within a couple of tenths of each other due to the math involved with such tiny amounts of current. But you get the idea.

Summary

In circuit 1, we ended up with around 42 volts from test point A to ground.

In circuit 2, we ended up with around 50 volts from test point A to ground.

Circuit 1's "A" voltage was taken across a 47K ohm resistor. Circuit 2's "A" voltage was taken across both the 47K ohm and the 25K ohm resistor.

That is essentially how the adjustable bias supply in a Marshall amplifier works. Rotating the pot 1 way removes the pot from the circuit by shorting the 47K ohm resistor to ground, while rotating it the other way adds 25K ohms of resistance between the 47K ohm to ground. This decreases and increases the voltage drop between ground and the 47K ohm resistor, giving you an adjustable voltage supply.

For simplification, the transformer was replaced with a battery, and the rectifier and the filter caps in the bias supply were removed from the above circuits, but -

The rectifier is what converts the AC from the transformer to "pulsed DC".

The capacitors in the supply act as "temporary batteries" by charging on the pulses, then discharging through the bias circuit in between pulses, thereby keeping the voltage at a steady state between pulses.

The only other difference between the above two circuits and the bias supply is that the polarity of an actual bias supply is reversed to give a negative voltage. But the math works exactly the same.

__________________
Jon Wilder
Wilder Amplification

sales@wilderamplification.com

tech@wilderamplification.com

thrawn86

That one got me spinnin' a little, but I think I follow. I'll still have to go through all of them again later! Good stuff, Jon.

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Wilder Amplification

Like I said...don't try to digest it all at once. You will get confused. That's the beauty of a message board is that the information is here anytime you want to come and read/re-read.

Again I'm available to answer any and all questions you may have. But this chapter must be understood in order for me to go any further.

Glad you're enjoying the material. I'm thinking about doing this as a once a week thing in order to not feed you guys too much all at once.

__________________
Jon Wilder
Wilder Amplification

sales@wilderamplification.com

tech@wilderamplification.com

TheGummy

really enjoying these, or at least the two of them that i've seen.

are there more than this one and how a valve works part 1?

just wondering if there's any way this series of journals could be put into a sub-forum to make them easier to find and follow in order?

Wilder Amplification

Glad you're enjoying the articles. Hopefully you enjoy reading them as much as I enjoy writing them.

I've emailed Ad-wex (super mod) with a request to add a subforum so that we can group these threads, but I haven't heard from him on it yet.

I haven't finished How A Valve Works Part 2 yet. I will probably be postings these once a week to not only give me time to write the article and draw the pictures, but also to give you guys time to read these articles and give it time for the material to sink in rather than overwhelming you with it all at once.

__________________
Jon Wilder
Wilder Amplification

sales@wilderamplification.com

tech@wilderamplification.com

TheGummy

thats fine, im in no rush. I was just making sure I hadn't missed out on any. looking forward to valve part 2.

KickStart

This is great stuff. I wish someone could make these sticky.

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