Mixing
Why are the resistors in an AM modulator necessary?
Clash Royale CLAN TAG#URR8PPP
up vote
2
down vote
favorite
In preparation for the amateur radio exam, I have been following the preparation course material by DJ4UF. In the section about modulation it presents this diagram of a simplified AM modulator.
The low frequency signal (NF) and the high frequency carrier wave (HF) are added, the diode cuts off one half-wave and the resonant circuit "recreates" the previously cut off half-wave. The result is a amplitude-modulated signal. So far so good.
What is the purpose of the 47k resistors? The explanatory text mentions that they are necessary to "add the currents in the diode". What would change if we remove them and just directly connect them to the diode?
resistors modulation am
asked 1 hour ago
ahemmetter
239110
add a comment |Â
up vote
2
down vote
favorite
In preparation for the amateur radio exam, I have been following the preparation course material by DJ4UF. In the section about modulation it presents this diagram of a simplified AM modulator.
The low frequency signal (NF) and the high frequency carrier wave (HF) are added, the diode cuts off one half-wave and the resonant circuit "recreates" the previously cut off half-wave. The result is a amplitude-modulated signal. So far so good.
What is the purpose of the 47k resistors? The explanatory text mentions that they are necessary to "add the currents in the diode". What would change if we remove them and just directly connect them to the diode?
resistors modulation am
asked 1 hour ago
ahemmetter
239110
add a comment |Â
up vote
2
down vote
favorite
up vote
2
down vote
favorite
In preparation for the amateur radio exam, I have been following the preparation course material by DJ4UF. In the section about modulation it presents this diagram of a simplified AM modulator.
The low frequency signal (NF) and the high frequency carrier wave (HF) are added, the diode cuts off one half-wave and the resonant circuit "recreates" the previously cut off half-wave. The result is a amplitude-modulated signal. So far so good.
What is the purpose of the 47k resistors? The explanatory text mentions that they are necessary to "add the currents in the diode". What would change if we remove them and just directly connect them to the diode?
resistors modulation am
asked 1 hour ago
ahemmetter
239110
In preparation for the amateur radio exam, I have been following the preparation course material by DJ4UF. In the section about modulation it presents this diagram of a simplified AM modulator.
The low frequency signal (NF) and the high frequency carrier wave (HF) are added, the diode cuts off one half-wave and the resonant circuit "recreates" the previously cut off half-wave. The result is a amplitude-modulated signal. So far so good.
What is the purpose of the 47k resistors? The explanatory text mentions that they are necessary to "add the currents in the diode". What would change if we remove them and just directly connect them to the diode?
resistors modulation am
resistors modulation am
asked 1 hour ago
ahemmetter
239110
asked 1 hour ago
ahemmetter
239110
asked 1 hour ago
ahemmetter
239110
asked 1 hour ago
ahemmetter
239110
asked 1 hour ago
ahemmetter
239110
239110
add a comment |Â
add a comment |Â
1 Answer
1
active
oldest
votes
up vote
7
down vote
accepted
The concept here is that you are adding the signal currents before feeding it to the diode; the resistors are there to convert the voltage sources "NF" and "HF" to current sources.
You'd get exactly the same effect by adding the signal voltages directly — simply connect the "NF" and "HF" boxes in series, without any resistors. The only downside to this is that "NF" and "HF" can't share a common ground, and that's often a desirable feature of a practical system. But some AM transmitters isolate the NF signal with a transformer, which solves that problem.
Note that the circuit as given is not at all practical — you would not want to feed a parallel-tuned circuit, which has a high impedance at resonance, from a current source. Instead, you would use a series-tuned circuit that keeps the diode cathode close to ground potential.
answered 52 mins ago
Dave Tweed♦
112k9136243
1
Two voltage sources fighting one another when connected in parallel could cause intermodulation distortion within one or both generators. One of those distortion products is exactly what you want out of the modulator. But other distortion products are not desired, and "Siebung" won't reject them. The series-connection of sources of Dave's answer is one solution and is far less lossy than those 47k series resistors.
– glen_geek
31 mins ago
@glen_geek Aha! So in a way these resistors also prevent that the signal of one frequency source interferes with the other source before being added, if I understood that correctly.
– ahemmetter
14 mins ago
add a comment |Â
1 Answer
1
active
oldest
votes
1 Answer
1
active
oldest
votes
active
oldest
votes
active
oldest
votes
up vote
7
down vote
accepted
The concept here is that you are adding the signal currents before feeding it to the diode; the resistors are there to convert the voltage sources "NF" and "HF" to current sources.
You'd get exactly the same effect by adding the signal voltages directly — simply connect the "NF" and "HF" boxes in series, without any resistors. The only downside to this is that "NF" and "HF" can't share a common ground, and that's often a desirable feature of a practical system. But some AM transmitters isolate the NF signal with a transformer, which solves that problem.
Note that the circuit as given is not at all practical — you would not want to feed a parallel-tuned circuit, which has a high impedance at resonance, from a current source. Instead, you would use a series-tuned circuit that keeps the diode cathode close to ground potential.
answered 52 mins ago
Dave Tweed♦
112k9136243
1
Two voltage sources fighting one another when connected in parallel could cause intermodulation distortion within one or both generators. One of those distortion products is exactly what you want out of the modulator. But other distortion products are not desired, and "Siebung" won't reject them. The series-connection of sources of Dave's answer is one solution and is far less lossy than those 47k series resistors.
– glen_geek
31 mins ago
@glen_geek Aha! So in a way these resistors also prevent that the signal of one frequency source interferes with the other source before being added, if I understood that correctly.
– ahemmetter
14 mins ago
add a comment |Â
up vote
7
down vote
accepted
The concept here is that you are adding the signal currents before feeding it to the diode; the resistors are there to convert the voltage sources "NF" and "HF" to current sources.
You'd get exactly the same effect by adding the signal voltages directly — simply connect the "NF" and "HF" boxes in series, without any resistors. The only downside to this is that "NF" and "HF" can't share a common ground, and that's often a desirable feature of a practical system. But some AM transmitters isolate the NF signal with a transformer, which solves that problem.
Note that the circuit as given is not at all practical — you would not want to feed a parallel-tuned circuit, which has a high impedance at resonance, from a current source. Instead, you would use a series-tuned circuit that keeps the diode cathode close to ground potential.
answered 52 mins ago
Dave Tweed♦
112k9136243
1
Two voltage sources fighting one another when connected in parallel could cause intermodulation distortion within one or both generators. One of those distortion products is exactly what you want out of the modulator. But other distortion products are not desired, and "Siebung" won't reject them. The series-connection of sources of Dave's answer is one solution and is far less lossy than those 47k series resistors.
– glen_geek
31 mins ago
@glen_geek Aha! So in a way these resistors also prevent that the signal of one frequency source interferes with the other source before being added, if I understood that correctly.
– ahemmetter
14 mins ago
add a comment |Â
up vote
7
down vote
accepted
up vote
7
down vote
accepted
The concept here is that you are adding the signal currents before feeding it to the diode; the resistors are there to convert the voltage sources "NF" and "HF" to current sources.
You'd get exactly the same effect by adding the signal voltages directly — simply connect the "NF" and "HF" boxes in series, without any resistors. The only downside to this is that "NF" and "HF" can't share a common ground, and that's often a desirable feature of a practical system. But some AM transmitters isolate the NF signal with a transformer, which solves that problem.
Note that the circuit as given is not at all practical — you would not want to feed a parallel-tuned circuit, which has a high impedance at resonance, from a current source. Instead, you would use a series-tuned circuit that keeps the diode cathode close to ground potential.
answered 52 mins ago
Dave Tweed♦
112k9136243
The concept here is that you are adding the signal currents before feeding it to the diode; the resistors are there to convert the voltage sources "NF" and "HF" to current sources.
You'd get exactly the same effect by adding the signal voltages directly — simply connect the "NF" and "HF" boxes in series, without any resistors. The only downside to this is that "NF" and "HF" can't share a common ground, and that's often a desirable feature of a practical system. But some AM transmitters isolate the NF signal with a transformer, which solves that problem.
Note that the circuit as given is not at all practical — you would not want to feed a parallel-tuned circuit, which has a high impedance at resonance, from a current source. Instead, you would use a series-tuned circuit that keeps the diode cathode close to ground potential.
answered 52 mins ago
Dave Tweed♦
112k9136243
edited 9 mins ago
answered 52 mins ago
Dave Tweed♦
112k9136243
answered 52 mins ago
Dave Tweed♦
112k9136243
answered 52 mins ago
Dave Tweed♦
112k9136243
112k9136243
1
Two voltage sources fighting one another when connected in parallel could cause intermodulation distortion within one or both generators. One of those distortion products is exactly what you want out of the modulator. But other distortion products are not desired, and "Siebung" won't reject them. The series-connection of sources of Dave's answer is one solution and is far less lossy than those 47k series resistors.
– glen_geek
31 mins ago
@glen_geek Aha! So in a way these resistors also prevent that the signal of one frequency source interferes with the other source before being added, if I understood that correctly.
– ahemmetter
14 mins ago
add a comment |Â
1
Two voltage sources fighting one another when connected in parallel could cause intermodulation distortion within one or both generators. One of those distortion products is exactly what you want out of the modulator. But other distortion products are not desired, and "Siebung" won't reject them. The series-connection of sources of Dave's answer is one solution and is far less lossy than those 47k series resistors.
– glen_geek
31 mins ago
@glen_geek Aha! So in a way these resistors also prevent that the signal of one frequency source interferes with the other source before being added, if I understood that correctly.
– ahemmetter
14 mins ago
1
1
Two voltage sources fighting one another when connected in parallel could cause intermodulation distortion within one or both generators. One of those distortion products is exactly what you want out of the modulator. But other distortion products are not desired, and "Siebung" won't reject them. The series-connection of sources of Dave's answer is one solution and is far less lossy than those 47k series resistors.
– glen_geek
31 mins ago
Two voltage sources fighting one another when connected in parallel could cause intermodulation distortion within one or both generators. One of those distortion products is exactly what you want out of the modulator. But other distortion products are not desired, and "Siebung" won't reject them. The series-connection of sources of Dave's answer is one solution and is far less lossy than those 47k series resistors.
– glen_geek
31 mins ago
@glen_geek Aha! So in a way these resistors also prevent that the signal of one frequency source interferes with the other source before being added, if I understood that correctly.
– ahemmetter
14 mins ago
@glen_geek Aha! So in a way these resistors also prevent that the signal of one frequency source interferes with the other source before being added, if I understood that correctly.
– ahemmetter
14 mins ago
add a comment |Â
Sign up or log in
StackExchange.ready(function ()
StackExchange.helpers.onClickDraftSave('#login-link');
);
Sign up using Google
Sign up using Facebook
Sign up using Email and Password
Post as a guest
StackExchange.ready(
function ()
StackExchange.openid.initPostLogin('.new-post-login', 'https%3a%2f%2felectronics.stackexchange.com%2fquestions%2f402118%2fwhy-are-the-resistors-in-an-am-modulator-necessary%23new-answer', 'question_page');
);
Post as a guest
Sign up or log in
StackExchange.ready(function ()
StackExchange.helpers.onClickDraftSave('#login-link');
);
Sign up using Google
Sign up using Facebook
Sign up using Email and Password
Post as a guest
Sign up or log in
StackExchange.ready(function ()
StackExchange.helpers.onClickDraftSave('#login-link');
);
Sign up using Google
Sign up using Facebook
Sign up using Email and Password
Post as a guest
Sign up or log in
StackExchange.ready(function ()
StackExchange.helpers.onClickDraftSave('#login-link');
);
Sign up using Google
Sign up using Facebook
Sign up using Email and Password
Sign up using Google
Sign up using Facebook
Sign up using Email and Password