How to find biasing resistor values if LC circuit is present at collector instead of a resistor?

Thread Starter

rockgro

Joined Sep 7, 2021
13
This is a circuit for Transistor Based AM modulation.
Here, How can I find the values of the biasing resistors (R1 and R2) to set the Q point in mid??
Pls, explain all the calculations to do while designing this circuit!

1651125679301.png
 

crutschow

Joined Mar 14, 2008
29,799
For that circuit you can't set the Q to the midpoint since no collector resistance is shown to cause a voltage drop.
The collector average voltage will always be at the supply voltage.
You select a collector current that gives the desired output signal voltage based upon the transistor transconductance, the carrier signal frequency. and the LC tank impedance.
 

Ramussons

Joined May 3, 2013
1,264
That transistor will be in a Class C mode of operation for the Carrier of 11 KHz.
The right way to Amplitude Modulate this carrier is to inject the Message signal between the +Vcc and Collector LC tank.

Yes, you can skip R1 and R2
 

Ramussons

Joined May 3, 2013
1,264
Don't understand that.
How will the transistor be biased without those resistors?
In the present setup, what happens to the transistor "bias" on the -ve part of the carrier wave? It will cutoff. (Hopefully, that Base-Emitter junction will survive the almost -15 Volts reverse voltage). That transistor will not operate much on the Linear region.
Putting it simply, that transistor will conduct only on the Positive part of the carrier input, it will be cut off on the Negative part of the cycle. And it will do that whether R1 and R2 is present or not.

As I said in the begining of my first post, that transistor is a Class C amplifier for the Carrier, and AM is acheived by varying the Collector voltage. The R1-R2 bias has a meaning ONLY if the Carrier amplitude is Less that the bias voltage. Else it will be Class B or C operation.
 
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crutschow

Joined Mar 14, 2008
29,799
In the present setup, what happens to the transistor "bias" on the -ve part of the carrier wave? It will cutoff. (Hopefully, that Base-Emitter junction will survive the almost -15 Volts reverse voltage). That transistor will not operate much on the Linear region.
Putting it simply, that transistor will conduct only on the Positive part of the carrier input, it will be cut off on the Negative part of the cycle. And it will do that whether R1 and R2 is present or not.

As I said in the begining of my first post, that transistor is a Class C amplifier for the Carrier, and AM is acheived by varying the Collector voltage. The R1-R2 bias has a meaning ONLY if the Carrier amplitude is Less that the bias voltage. Else it will be Class B or C operation.
How do you know it's Class C without knowing the signal levels?
 

BobTPH

Joined Jun 5, 2013
4,907
It class C is only without bias. With proper bias, there will always be a current, which can change with the input in both directions. The voltage at the collector will be centered at V+ and go above and below V+, following the input waveform but inverted.

Bob
 

sparky 1

Joined Nov 3, 2018
717
It is important to set a goal and choose a good model to build and understand. Abbreviation can lead to confusion.
From what I can guess you have a frequency mixer drawn in a way that makes it very difficult to grasp
the concepts. Why not make a double balanced mixer for AM that is well documented and discussed?

There really should be more discussion on AM modulation. It is common to see confusion with this subject and the confusion extends to amateur radio experimenters on youtube some of which have been derived from half baked modulation models along with poorly drawn schematics.
The details have been omitted by way of abbreviated shorthand. Here is a post where one builder asks and suggestions given.
https://forum.allaboutcircuits.com/threads/modulation.32019/post-199879

There were a few that took an interest and built an AM modulator along with Charles Wenzel. We want to give Charles credit for his efforts.
Since that link is copyrighted (openly published on techlib.com), The original page link is here.
AM band Transmitter (techlib.com)
 
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Thread Starter

rockgro

Joined Sep 7, 2021
13
The circuit which I referred to at the beginning of the post is taken from the "Lab Manual" of my "Analog Communication system" course!
I have a few questions now:
  1. Is this circuit (in its original form) even a suitable circuit for AM modulation? If yes, pls explain all the biasing calculations!
  2. What suitable modification can I do to make it more suitable for AM modulation?
  3. Which kind of circuit can be a better transistor-based AM modulator? I can think of Class A based circuit! Correct me if I am wrong!
  4. Should I challenge the "course" authorities to make changes to the existing circuit?
 

Ramussons

Joined May 3, 2013
1,264
How do you know it's Class C without knowing the signal levels?
Bias Resistors are known.
Signal Levels are known.
The input at the Base of BC107 will be between +12 and -12 Volts. That transistor will be in cutoff for a little more than half the carrier input cycle.
That's class C no?
 

DickCappels

Joined Aug 21, 2008
8,710
  1. Is this circuit (in its original form) even a suitable circuit for AM modulation? If yes, pls explain all the biasing calculations!
  2. What suitable modification can I do to make it more suitable for AM modulation?
  3. Which kind of circuit can be a better transistor-based AM modulator? I can think of Class A based circuit! Correct me if I am wrong!
  4. Should I challenge the "course" authorities to make changes to the existing circuit?
• It is a good example of an AM transmitter.
• The schematic below shows the details of a similar transmitter. Notice that the resistor driven by a signal source has been replaced by a modulated current source. The differential pair and the current source form a two quadrant multiplier.
* "Better" may be subjective. In my book, the circuit below is fine, though linearity of the modulation might be improved a little bit by using a transistor as a modulated current source.
• This circuit can be class A class C depending upon the amplitude of the carrier injected into the differential pair and how the bias is set up.
• You should not challenge anybody until you better understand what you are challenging.
1651665452804.png

This was designed by Charles Wenzel. There is also an improved version on this website: http://techlib.com/electronics/amxmit.htm
Scroll down to near the bottom of the page until you get to the photo of the modulation envelope this transmitter produces.
 

BobTPH

Joined Jun 5, 2013
4,907
Notice also that the carrier is capacitive coupled to the modulator in this case.

The original circuit is coupled by a resistor so low that the biasing is completely overridden.

Bob
 

Papabravo

Joined Feb 24, 2006
18,426
The basic methodology of DC analysis:
  1. Set all AC sources to 0
  2. Replace all capacitors with an open circuit
  3. Replace all inductors with a short circuit
What you are left with is the DC sources, resistors, and active devices.
 

MrAl

Joined Jun 17, 2014
8,992
This is a circuit for Transistor Based AM modulation.
Here, How can I find the values of the biasing resistors (R1 and R2) to set the Q point in mid??
Pls, explain all the calculations to do while designing this circuit!

View attachment 265992

Hello there,

Is this circuit missing an input series capacitor? It looks like it is missing that.
I say that because with no signal the 1k input resistor is in parallel with R2 so it makes R2 redundant. The normal thing to do is add series capacitor to the input, unless of course the input source already has that.

The DC bias point would be based on collector current where the max would be Vcc/10000 so half of that would probably be good assuming the message signal source has zero output resistance. That also leads to about Vcc/2 volts at the emitter so rather than use the collector voltage you can use the emitter voltage as a design point.

The short answer to the bias setup is to assume some Beta for the transistor, then choose the bias resistor(s) to create that half max current through the collector. You can get an estimate really quick that way. If you estimate the base emitter diode voltage to be around 0.7 volts then the base voltage will be about 0.7 volts higher than the emitter voltage, and the emitter voltage will be around Vcc/2 because that causes about half max current to flow through the emitter resistor which means about half through the collector too.
The two bias resistors R1, R2, set the bias but also part of the input impedance so you have to adjust them based on those two factors. That's of course with the input capacitively coupled as mentioned above. If the input is not capacitively coupled then the 1k resistor is in parallel with R2 for the DC bias setup.

Another point is that the 15v input Ac signal may be too much for the circuit with that 1k input resistor. That sounds too high or something else is wrong. This combined with the above notes leads one to ask where this circuit came from it may not work as is.

Let us know what you come up with and if you need any more help.
 
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DickCappels

Joined Aug 21, 2008
8,710
With that amount of overdrive the transistor would certainly be off a good deal of the time. If off 50% or more of the time it would be called a class C amplifier (per @Ramussons in post #13). It would still transmit fine, and in fact the transistor would be a little cooler because of this more efficient mode (than class A).
 

Thread Starter

rockgro

Joined Sep 7, 2021
13
I want to add some more clarification here:
  • The circuit was directly given in the lab manual without any calculations or design equations
  • The Frequency and the voltage of the message and carrier signal given in the circuit gave no proper modulation. So we need to tinker and figure out on our own which voltage and frequency combination works!! This does not seem engineering!
  • It turns out that we need to set the Voltage of both the signals greater than the supply voltage which does not make sense to me
  • Also, our teacher told us to tune the LC frequency equal to carrier frequency but this also was incorrect. After experimentation, I found out that the LC frequency should be set to the difference in frequencies of the 2 signals to get the best results.
  • I was getting much better results if I directly use a class A configuration. Ofc the class A has bad efficiency but that's not important if your objective is to learn AM modulation!

This is the result I got with the original circuit after a lot of tinkering
1652168720230.png 1652168796528.png1652168819161.png

And This is the result of my simple class A based modulator which I designed:
1652168942905.png 1652168969683.png1652169173130.png
 

Attachments

DickCappels

Joined Aug 21, 2008
8,710
It is very unusual to see the emitter of an NPN transistor being fed from a positive power supply. If the power supply is really connected that way and you are getting any emitter current the transistor is in avalanche mode, acting like a Zener diode. The 2N3904 (real ones) avalanche when the emitter is about six volts more positive than the base. That might be part of the reason you get unexpected results.
 
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