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Meta-Electronics 101-Part 2: A C Theory
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The Triode as an AC Amplifier
This section will look at the ability of a triode to amplify an AC voltage(signal)
and how an amplifier will distort the signal if improperly biased.
We will first observe the effects of AC amplification by a triode amplifier, connected as a common cathode, and we will determine the gain.
Next, we will observe how distortion can be caused by improper bias voltage.
In the AC voltage amplifier, a signal voltage AC is fed to the grid circuit (grid-cathode).
The signal appears as a larger voltage on the plate circuit (plate-cathode). The signal is coupled to the grid-cathode circuit by a series RC circuit, which will also contain the the battery for biasing the tube. The name of it is C- and can be seen in the circuit shown at the bottom of the page.
A load resistor (RL) is connected between the power source (B+ or Vpk) and the plate. This will cause the plate voltage to vary inversely with the plate current. A signal voltage (Vgk) will cause the plate current (Ip) to vary which we call "ip", around the quiescent current, that which flows with and is set by the bias. Now "ip" causes Vpk to vary which we call "vpk".
The wave form or shape of of "vpk" should be identical to "vgk", and if it is not the same we call it distortion. Of course, if the signal is amplified, its amplitude is greater, so we do not say it is distorted if that is the only change. For DISTORTIONLESS operation the tube should be biased with a negative voltage near the middle of the linear range.
Another way to distort the signal, even though the tube is biased in the middle, is to make the peak-to-peak signal so large that the tube is driven into saturation on the positive swing and then into or below cutoff on the negative half. In other words, overdriving with too large a signal will cutoff part of the output signal.
A graph showing grid voltage vs plate current for a 12AU7a is shown below the schematic. You can graphically compute the gain of the amplifier by picking your bias point, in this case minus 3 volts, and then plotting an input signal along the bottom axis. Then draw the corresponding lines at a 90 degree angle to the grid voltage to get the plate current. Then compute the plate voltage swing by using ohms law.
Example: A 4 volt p-p signal is plotted on the bottom axis using our minus 3 volts bias voltage as the center. According to the graph we will get almost 4 ma of plate current swing. To convert this into the plate voltage swing, just multiply 4 ma by our 22 K load resistor and you get 0.004 * 22,000 = 88 volts p-p.
So our gain can be estimated to be 88 volts plate/4 volts grid = 22.
You can find the same type of graphs in the tube handbooks or online. The graph used below has been stripped down for clarity, you will find many more plots using various plate voltages which form a family of curves.
This section will look at the ability of a triode to amplify an AC voltage(signal)
and how an amplifier will distort the signal if improperly biased.
We will first observe the effects of AC amplification by a triode amplifier, connected as a common cathode, and we will determine the gain.
Next, we will observe how distortion can be caused by improper bias voltage.
In the AC voltage amplifier, a signal voltage AC is fed to the grid circuit (grid-cathode).
The signal appears as a larger voltage on the plate circuit (plate-cathode). The signal is coupled to the grid-cathode circuit by a series RC circuit, which will also contain the the battery for biasing the tube. The name of it is C- and can be seen in the circuit shown at the bottom of the page.
A load resistor (RL) is connected between the power source (B+ or Vpk) and the plate. This will cause the plate voltage to vary inversely with the plate current. A signal voltage (Vgk) will cause the plate current (Ip) to vary which we call "ip", around the quiescent current, that which flows with and is set by the bias. Now "ip" causes Vpk to vary which we call "vpk".
The wave form or shape of of "vpk" should be identical to "vgk", and if it is not the same we call it distortion. Of course, if the signal is amplified, its amplitude is greater, so we do not say it is distorted if that is the only change. For DISTORTIONLESS operation the tube should be biased with a negative voltage near the middle of the linear range.
Another way to distort the signal, even though the tube is biased in the middle, is to make the peak-to-peak signal so large that the tube is driven into saturation on the positive swing and then into or below cutoff on the negative half. In other words, overdriving with too large a signal will cutoff part of the output signal.
A graph showing grid voltage vs plate current for a 12AU7a is shown below the schematic. You can graphically compute the gain of the amplifier by picking your bias point, in this case minus 3 volts, and then plotting an input signal along the bottom axis. Then draw the corresponding lines at a 90 degree angle to the grid voltage to get the plate current. Then compute the plate voltage swing by using ohms law.
Example: A 4 volt p-p signal is plotted on the bottom axis using our minus 3 volts bias voltage as the center. According to the graph we will get almost 4 ma of plate current swing. To convert this into the plate voltage swing, just multiply 4 ma by our 22 K load resistor and you get 0.004 * 22,000 = 88 volts p-p.
So our gain can be estimated to be 88 volts plate/4 volts grid = 22.
You can find the same type of graphs in the tube handbooks or online. The graph used below has been stripped down for clarity, you will find many more plots using various plate voltages which form a family of curves.
Attachments
Electronics 101
Experiment - Common Cathode Amplifier Page 1
OBJECTIVES
1. To observe how a small change in DC voltage on the control grid is amplified by a triode vacuum tube and appears as a large change in voltage on the plate.
2. To locate the linear portion of the amplifiers characteristics for distortionless amplification.
MATERIALS AND EQUIPMENT
Power supply: 0-400VDC (B+) (Vpp), 0-125 VDC- (C-)(Vgg), and 6.3 VAC (A).
Voltmeter: VTVM
Resistors: 22,000-ohm 1/2 watt.
Tube: 12AU7 (similar tubes: 6SN7 and 6J5)
INTRODUCTION
A vacuum tube is a device which converts small changes in grid voltage into changes in plate current, not unlike the transistor. In order to change plate current changes into variations in plate voltage we must insert a load resistor between the plate and supply voltage. Thus, there is a variable resistor--the tube--in series with the load resistor. The plate voltage at any time is according to KIRCHOFF's law: Vpk = Vpp - VRL and VRL = Ip x RL .
If Vgg, the value of the grid bias, becomes more negative the current through the tube will decrease causing the voltage drop across the load resistor (RL) to decrease allowing the plate voltage to rise. Similarly, if Vgg, the grid, becomes less negative (or we could say, more positive) plate current (Ip) will increase so Vpk will decrease. The actual changes in current and voltage will depend upon the characteristics of a particular tube, the load resistor and the power supply voltage.
It is desirable for an amplifier to have a constant gain over its operating range.
That is, a change of 1-volt on the grid should produce the same plate voltage change, no matter whether the change is from -3 to -4 or -7 to -8 or -6 to -5. If the gain varies for different amplitudes of input voltage then the amplifier signal will be distorted or change in waveform from the original signal. This is generally undesirable, and to avoid it we must limit the input signal within the linear part.
PROCEDURE
1. Connect the circuit shown. Wire up the filament first by twisting the hook-up wire 3 turns per inch and placing it down against the chassis out of the way.
2. Have the teacher check your wiring.
3. Set Vpp at 200 volts (check it often).
4. Set Vgg at zero volts.
5. Measure and record Vpk: Ip: V-RL:
6. Repeat steps 3,4 and 5 for for every voltage shown for grid voltage (Vgg).
CALCULATIONS
1.Let us again prove Kirchoff's law that states that the sum of the voltage drops equals the applied voltage. In this experiment, where it is purely resistive, just add V-PK to V-RL and record in TABLE A on line "TOTAL V." This should be very close to 200 volts which you set as the B+ to begin with.
2. Now lets again prove that a series circuit has the same current throughout.
This method is also the one you must sometimes use in trouble shooting circuits.
Take the voltage drop across the resistor (RL) and divide it by its actual resistance, which was 24,000 ohms. Record this in the next to last line on front page. These answers should be very close to those of the meter.
3. Now to determine the gain (amplification) we must use change in voltage because we are dealing with so called static values. Our formula is:
Av = a change in DC on plate
a change in DC on grid
To use this, merely take one plate voltage and subtract it from the larger and be sure to use the same grid voltage for the particular plate voltage and subtract them. The easiest here is to use only 1 volt grid change.
From your data draw graphs of Ip versus VgK on graph A.
TABLE A
GRID VOLTAGE
VgK 0 -1 -2 -3 -4 -5 -6 -7 -8 -9 -10 -11 -12 -13 -14 -15
Ip ma 6.0 5.4 4.8 4.0 3.6 2.8 2.4 2.0 1.5 1.2 .8 .5 .2 .1 .05 .01V-PK 62 80 95 108 120 138 147 155 160 168 185 190 195 198 199 200V-RL 139 120 105 92 80 62 53 45 40 33 16 10 6 2 1 0TOTAL 201 200 200 200 200 200 200 200 200 200 201 201 200 201 200 200Ip (calc)6.2 5.6 4.9 4.1 3.7 2.9 2.4 2.0 1.4 1.3 .85 .51 .22 .15 .07 .00GAIN 0 18 15 13 12 12 10 8 7 8 7 5 5 3 1 0
Experiment - Common Cathode Amplifier Page 1
OBJECTIVES
1. To observe how a small change in DC voltage on the control grid is amplified by a triode vacuum tube and appears as a large change in voltage on the plate.
2. To locate the linear portion of the amplifiers characteristics for distortionless amplification.
MATERIALS AND EQUIPMENT
Power supply: 0-400VDC (B+) (Vpp), 0-125 VDC- (C-)(Vgg), and 6.3 VAC (A).
Voltmeter: VTVM
Resistors: 22,000-ohm 1/2 watt.
Tube: 12AU7 (similar tubes: 6SN7 and 6J5)
INTRODUCTION
A vacuum tube is a device which converts small changes in grid voltage into changes in plate current, not unlike the transistor. In order to change plate current changes into variations in plate voltage we must insert a load resistor between the plate and supply voltage. Thus, there is a variable resistor--the tube--in series with the load resistor. The plate voltage at any time is according to KIRCHOFF's law: Vpk = Vpp - VRL and VRL = Ip x RL .
If Vgg, the value of the grid bias, becomes more negative the current through the tube will decrease causing the voltage drop across the load resistor (RL) to decrease allowing the plate voltage to rise. Similarly, if Vgg, the grid, becomes less negative (or we could say, more positive) plate current (Ip) will increase so Vpk will decrease. The actual changes in current and voltage will depend upon the characteristics of a particular tube, the load resistor and the power supply voltage.
It is desirable for an amplifier to have a constant gain over its operating range.
That is, a change of 1-volt on the grid should produce the same plate voltage change, no matter whether the change is from -3 to -4 or -7 to -8 or -6 to -5. If the gain varies for different amplitudes of input voltage then the amplifier signal will be distorted or change in waveform from the original signal. This is generally undesirable, and to avoid it we must limit the input signal within the linear part.
PROCEDURE
1. Connect the circuit shown. Wire up the filament first by twisting the hook-up wire 3 turns per inch and placing it down against the chassis out of the way.
2. Have the teacher check your wiring.
3. Set Vpp at 200 volts (check it often).
4. Set Vgg at zero volts.
5. Measure and record Vpk: Ip: V-RL:
6. Repeat steps 3,4 and 5 for for every voltage shown for grid voltage (Vgg).
CALCULATIONS
1.Let us again prove Kirchoff's law that states that the sum of the voltage drops equals the applied voltage. In this experiment, where it is purely resistive, just add V-PK to V-RL and record in TABLE A on line "TOTAL V." This should be very close to 200 volts which you set as the B+ to begin with.
2. Now lets again prove that a series circuit has the same current throughout.
This method is also the one you must sometimes use in trouble shooting circuits.
Take the voltage drop across the resistor (RL) and divide it by its actual resistance, which was 24,000 ohms. Record this in the next to last line on front page. These answers should be very close to those of the meter.
3. Now to determine the gain (amplification) we must use change in voltage because we are dealing with so called static values. Our formula is:
Av = a change in DC on plate
a change in DC on grid
To use this, merely take one plate voltage and subtract it from the larger and be sure to use the same grid voltage for the particular plate voltage and subtract them. The easiest here is to use only 1 volt grid change.
From your data draw graphs of Ip versus VgK on graph A.
TABLE A
GRID VOLTAGE
VgK 0 -1 -2 -3 -4 -5 -6 -7 -8 -9 -10 -11 -12 -13 -14 -15
Ip ma 6.0 5.4 4.8 4.0 3.6 2.8 2.4 2.0 1.5 1.2 .8 .5 .2 .1 .05 .01V-PK 62 80 95 108 120 138 147 155 160 168 185 190 195 198 199 200V-RL 139 120 105 92 80 62 53 45 40 33 16 10 6 2 1 0TOTAL 201 200 200 200 200 200 200 200 200 200 201 201 200 201 200 200Ip (calc)6.2 5.6 4.9 4.1 3.7 2.9 2.4 2.0 1.4 1.3 .85 .51 .22 .15 .07 .00GAIN 0 18 15 13 12 12 10 8 7 8 7 5 5 3 1 0
Attachments
Electronics 101
Experiment - Common Cathode Amplifier Page 2
5. Plot a graph on Graph B with the plate voltage against the plate current, (Ip).
6. Locate the linear portion of the curve and mark the lowest value of Ip with the letter "A". Now find the highest Ip with the curve still straight and mark it with the letter "B". This is the linear section. To amplify without distortion, we must keep our peak-to-peak AC signals within and between "A" and "B".
To do this the best way to operate an amplifier would be in the middle. Now
locate the grid voltage that lies half between and mark letter "M" on the curve directly above this. This is the most desired grid voltage for a class A amplifier.
RECORD the following information about point "A" Plate voltage V-PK145 vdc
Grid V-gK -6 Plate current Ip 2.4 ma
about point "B" Plate voltage V-PK 52 vdc
Grid V-gK 0 Plate current Ip 6.3 ma
Point "M" is the center of the linear portion, or should we say, the voltage which we will bias the tube for amplifying alternating current. It is -3 vdc
Questions
1. Define grid bias.
2. Define cutoff point
3. Define saturation.
4. Explain in your own terms, "amplification factor." ("u" as in mu) Look up
in a tube manual the "mu" for the 12AU7. What is it?
5. Explain how the control grid can affect plate current.
6. Explain how the plate voltage is caused to vary as a result of plate current
variations.
7. Why won't the plate and grid emit electrons like the cathode?
8. Draw the schematic diagram for the following tubes. Include the filaments.
Diode Triode Tetrode Pentode
Label all the the electrodes of
the pentode with their correct names.
Experiment - Common Cathode Amplifier Page 2
5. Plot a graph on Graph B with the plate voltage against the plate current, (Ip).
6. Locate the linear portion of the curve and mark the lowest value of Ip with the letter "A". Now find the highest Ip with the curve still straight and mark it with the letter "B". This is the linear section. To amplify without distortion, we must keep our peak-to-peak AC signals within and between "A" and "B".
To do this the best way to operate an amplifier would be in the middle. Now
locate the grid voltage that lies half between and mark letter "M" on the curve directly above this. This is the most desired grid voltage for a class A amplifier.
RECORD the following information about point "A" Plate voltage V-PK145 vdc
Grid V-gK -6 Plate current Ip 2.4 ma
about point "B" Plate voltage V-PK 52 vdc
Grid V-gK 0 Plate current Ip 6.3 ma
Point "M" is the center of the linear portion, or should we say, the voltage which we will bias the tube for amplifying alternating current. It is -3 vdc
Questions
1. Define grid bias.
2. Define cutoff point
3. Define saturation.
4. Explain in your own terms, "amplification factor." ("u" as in mu) Look up
in a tube manual the "mu" for the 12AU7. What is it?
5. Explain how the control grid can affect plate current.
6. Explain how the plate voltage is caused to vary as a result of plate current
variations.
7. Why won't the plate and grid emit electrons like the cathode?
8. Draw the schematic diagram for the following tubes. Include the filaments.
Diode Triode Tetrode Pentode
Label all the the electrodes of
the pentode with their correct names.
Attachments
yeah the dude stole my spray skirt on the n. fork payette last summer, what up?
ok tube experiment is complete,
is there a moderator who can clean this thread?
thanks.
cj
is there a moderator who can clean this thread?
thanks.
cj
Here is a Lab Experiment on vacuum tubes dealing with CATHODE BIAS.
It is from a high school electronics class back in 1971,
you can use a 12AU7a for this experiment,
It is from a high school electronics class back in 1971,
you can use a 12AU7a for this experiment,
