How does this Bipolar RF Amplifier + Output Transformer circuit work?

In summary: It's better to keep the full model of the transformer in mind, IMO. That is correct.The load impedance is transformed back to appear in parallel with the magnetizing inductance Lm of the primary winding, with the leakage inductance Lk in series with all of that....
  • #1
Boltzman Oscillation
233
26
Homework Statement
How does the transformer work in this circuit?
Relevant Equations
KVL, KCL, BJT equations, etc.
phyfor0.jpg

I am trying to create a receiver for a personal project I am working on. This is the RF amplifier that I was given by a book I am following. How does this circuit work? Usually for a BJT amplifier I find the DC bias point and then use the small signal model after I bias the circuit? I understand that if I wanted to find the DC bias point then all the capacitors would become open but what would happen to that transformer? Will it just become an open like an inductor? Am I not supposed to do DC biasing for this circuit? I would also appreciate any sources on RF amplifiers. I want to be able to do the actual circuit analysis of them, the book I am reading does not analyze them but instead just gives me the circuits required.
 
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  • #2
Boltzman Oscillation said:
Homework Statement:: How does the transformer work in this circuit?
Relevant Equations:: KVL, KCL, BJT equations, etc.

Am I not supposed to do DC biasing for this circuit
Looks like it is using a voltage divider dc bias. It is formed by Vcc-2 resistors connected to base terminal -ground.
Antenna feeds the input signal through the two coupling capacitors. The capacitors "block" dc component in the input.
 
  • #3
cnh1995 said:
Looks like it is using a voltage divider dc bias. It is formed by Vcc-2 resistors connected to base terminal -ground.
Antenna feeds the input signal through the two coupling capacitors. The capacitors "block" dc component in the input.
When we look at just the DC signal of the circuit then the capacitor in parallel with the transformer will become an open. What happens to the transformer?
 
  • #4
Boltzman Oscillation said:
What happens to the transformer?
I don't know RF amplifiers in detail, but the LC loop looks like a tank circuit tuned to a certain frequency fcertain. This tank circuit will pick only the component of input signal with frequency fcertain.

In presence of only dc bias, I don't think the capacitor will be open circuited. The tank circuit will still resonate at its natural frequency.

Anyways, I am not an expert in RF (or any other) amplifiers. I request experts to weigh in here (and correct me too, if my response is incorrect).
 
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  • #5
Yup, @cnh1995 got it right. The variable capacitor in parallel with the inductor is adjusted to resonant at the frequency of the radio station you want to receive.

Since it is a parallel resonant circuit, it acts as a high impedance Collector load at its resonant frequency, and as a lower impedance load at other frequencies. The end result is a signal at the resonant frequency is amplified more than other frequencies.

Now to get the amplified signal to the next stage in the receiver you have a couple choices:
  • you could use capacitor coupling by connecting a small capacitor from that Collector to the Base of the next stage
  • you could wrap a few turns of wire around the inductor to make a transformer and connect that secondary winding to the next stage
A couple of advantages of using the transformer approach is it can be a little more effecient in signal transfer by better matching the input impedance of the next stage, and if you choose you can also tune the secondary winding to increase the selectivity (rejection of nearby frequencies.) (Also you can make transformers at home, capacitors are a little tougher)

In this case, for DC biasing you can usually treat the Collector inductor as a short circuit, and bias for perhaps 1mA Collector current, actual value is not critical.

Let us know how it turns out!

Cheers,
Tom
 
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  • #6
Tom.G said:
In this case, for DC biasing you can usually treat the Collector inductor as a short circuit
So when the amplifier is just powered on without antenna input, does the LC circuit act as a damped oscillator, which settles down with capacitor shorted out by the inductor?
 
  • #7
cnh1995 said:
I don't know RF amplifiers in detail, but the LC loop looks like a tank circuit tuned to a certain frequency fcertain. This tank circuit will pick only the component of input signal with frequency fcertain.

In presence of only dc bias, I don't think the capacitor will be open circuited. The tank circuit will still resonate at its natural frequency.

Anyways, I am not an expert in RF (or any other) amplifiers. I request experts to weigh in here (and correct me too, if my response is incorrect).
So the primary winding of the transformer can be treated as an inductor?
 
  • #8
Boltzman Oscillation said:
So the primary winding of the transformer can be treated as an inductor?
For dc, yes.
 
  • #9
Boltzman Oscillation said:
So the primary winding of the transformer can be treated as an inductor?
It's better to keep the full model of the transformer in mind, IMO. The load impedance is transformed back to appear in parallel with the magnetizing inductance Lm of the primary winding, with the leakage inductance Lk in series with all of that. Depending on the application, the full model (including the capacitances) may be important. If the transformer output is lightly loaded and the ration of Lm/Lk is high, you can just treat it as an inductor.

https://en.wikipedia.org/wiki/Transformer

1592951426753.png
 
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  • #10
cnh1995 said:
So when the amplifier is just powered on without antenna input, does the LC circuit act as a damped oscillator, which settles down with capacitor shorted out by the inductor?
Well, I suppose you could say that. If you replace "damped oscillator" with "damped tuned circuit" I think the description would be a bit more understandable.

edit:
So the primary winding of the transformer can be treated as an inductor?
For dc, yes.

And for AC as an simple inductor IF the secondary is lightly loaded, or loosely coupled to the primary; as @berkeman pointed out.
end edit;

Cheers,
Tom
 
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  • #11
Tom.G said:
Well, I suppose you could say that. If you replace "damped oscillator" with "damped tuned circuit" I think the description would be a bit more understandable.

edit:
And for AC as an simple inductor IF the secondary is lightly loaded, or loosely coupled to the primary; as @berkeman pointed out.
end edit;

Cheers,
Tom

In this case the secondary has no load, and thus the primary is treated as an inductor.
 
  • #12
Boltzman Oscillation said:
In this case the secondary has no load, and thus the primary is treated as an inductor.
What good is an input amp with no output load? One hand clapping? :wink:
 
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  • #13
berkeman said:
What good is an input amp with no output load? One hand clapping? :wink:

Eh I am learning the different stages of a receiver and an RF amplifier is the first stage so I just want to test it out first. Will LTSpice be able to simulate this? I am just trying to make sure i understand the concept and not blow up a component in the process. The ultimate goal is to make a device that tracks the location of an RF signal source.
 
  • #14
Boltzman Oscillation said:
Will LTSpice be able to simulate this?
Sure, although it may not appreciate the open circuit at the output of the coupling transformer. Start with a 1MEG resistor on the output coil to keep the simulation from throwing an error.
 
  • #15
In regard to the question of whether the tank circuit will self oscillate, the answer is no. To experience sustained oscillation an electronic circuit has to have a feedback from the output (collector) to the input (base) in the proper phase. The output signal is 180 degrees out of phase from the input so feedback applied directly would suppress oscillations, negative feedback. Audio and RF amplifier designs frequently employ negative feedback to suppress any tendency for an amplifier circuit design to oscillate. This circuit employs no feedback. If it employed positive feedback (a sampling of the output signal is directed in phase to the input) you have an oscillator. The ARRL has some excellent literature on rf receiver, amplifier and transmitter circuit design of a practical nature. Texts should be available in your library.
 
  • #16
John Kovach said:
In regard to the question of whether the tank circuit will self oscillate, the answer is no. To experience sustained oscillation an electronic circuit has to have a feedback from the output (collector) to the input (base) in the proper phase. The output signal is 180 degrees out of phase from the input so feedback applied directly would suppress oscillations, negative feedback. Audio and RF amplifier designs frequently employ negative feedback to suppress any tendency for an amplifier circuit design to oscillate. This circuit employs no feedback. If it employed positive feedback (a sampling of the output signal is directed in phase to the input) you have an oscillator. The ARRL has some excellent literature on rf receiver, amplifier and transmitter circuit design of a practical nature. Texts should be available in your library.
Thank you, i will check them out.
 
  • #17
It would help if you labelled the components in the diagram

The purpose of the capacitor across the emitter resistance is to increase the ac, or signal frequency, gain. Its time constant, T = CR, is set to be much longer than the signal frequency.

The voltage divider sets the base voltage to Vbe * R2/(R1+R2) by assuming base current is negligible (ie Ic/Ib gain is high).

The emitter resistance sets the emitter current, and hence the collector current, as Ie = (Vb - junction voltage) / Remitter.

If the emitter capacitor is omitted, and the input rises by 1V, the emitter and collector current increases by 1V divided by R emitter. To get a high gain you need R emitter to be small but you cannot because of the bias requirement.

So, the capacitor is placed across the emitter resistance in order to hold the emitter voltage steady (or, if you prefer, to reduce the ac impedance at signal frequency to a low value). A 1V increase in base voltage at signal frequency will now create a huge increase in collector current and, as the change in output voltage is equal to the collector load times the change in collector current, the voltage gain is high.
 
  • #18
Boltzman Oscillation said:
Homework Statement:: How does the transformer work in this circuit?
Relevant Equations:: KVL, KCL, BJT equations, etc.

View attachment 265046
I am trying to create a receiver for a personal project I am working on. This is the RF amplifier that I was given by a book I am following. How does this circuit work? Usually for a BJT amplifier I find the DC bias point and then use the small signal model after I bias the circuit? I understand that if I wanted to find the DC bias point then all the capacitors would become open but what would happen to that transformer? Will it just become an open like an inductor? Am I not supposed to do DC biasing for this circuit? I would also appreciate any sources on RF amplifiers. I want to be able to do the actual circuit analysis of them, the book I am reading does not analyze them but instead just gives me the circuits required.
 
  • #19
This is an amplifier having a tuned transformer output. A load resistor of some form (maybe the input of another stage) is connected across the right hand winding, which is the secondary, and creates a different value of resistance across the primary as a consequence of the turns ratio. At the resonant frequency, the transistor sees just this resistance, because the parallel LC circuit has very high impedance. At frequencies away from resonance the LC circuit has lower impedance and shunts current away from the resistive load, giving filter action. The sharpness of the filter is decided by the ratio of the resistance divided by the reactance of either L or C (they are the same) at the resonant frequency. This ratio is Q, the Quality Factor of the resonant circuit. The bandwidth is equal to the resonant frequency divided by Q (measured to the half power points).
In calculating gain, the transistor is working into the resistance coupled into the primary. You can often consider the transformer to be perfect because leakage inductance is removed by the resonant action - it just shifts the resonant frequency a bit.
Regarding DC operating conditions, the Cs are open circuits, but the L provides very low DC resistance, so the collector is at supply voltage but has a resistive load as described above.
 

Related to How does this Bipolar RF Amplifier + Output Transformer circuit work?

1. How does a Bipolar RF Amplifier work?

A Bipolar RF Amplifier is a type of amplifier that uses bipolar junction transistors (BJTs) to amplify radio frequency (RF) signals. BJTs are semiconductor devices that can amplify signals by controlling the flow of current through the device. In a Bipolar RF Amplifier, the input signal is first amplified by a low-power transistor, and then further amplified by a high-power transistor. The amplified signal is then sent to the output transformer for further processing.

2. What is the purpose of an Output Transformer in this circuit?

The Output Transformer in this circuit serves two main purposes. Firstly, it matches the impedance of the amplifier to the impedance of the load (e.g. an antenna), ensuring maximum power transfer. Secondly, it isolates the amplifier from the load, protecting it from any potential damage caused by mismatched impedance or reflected power.

3. How does the Output Transformer step up the voltage in this circuit?

The Output Transformer in this circuit has a higher number of turns on the secondary coil compared to the primary coil. This creates a step-up transformer, where the voltage on the secondary coil is higher than the voltage on the primary coil. This step-up voltage is necessary to drive high-power loads, such as antennas, which require a higher voltage to operate.

4. What is the purpose of the capacitors in this circuit?

The capacitors in this circuit serve two main purposes. Firstly, they act as coupling capacitors, separating the DC and AC components of the signal. This ensures that only the AC component is amplified, while the DC component is blocked. Secondly, they act as bypass capacitors, providing a low-impedance path for high-frequency signals to ground, improving the amplifier's stability and performance.

5. How does the feedback loop in this circuit work?

The feedback loop in this circuit is used to stabilize the amplifier's gain and ensure it operates in a linear region. It works by taking a portion of the output signal and feeding it back to the input of the amplifier through a resistor. This creates a negative feedback loop, which reduces the gain of the amplifier. As the output signal increases, the feedback signal also increases, causing the amplifier to reduce its gain and maintain a stable output signal.

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