As a radio ham this was particularly interesting for me. I'm sure there are many like me who would be glad of more RF related topics. Yes, I can find them on UA-cam but not as well explained as you do. Thanks for this video.
Class C makes an excellent AM modulator all you have to do is vary the collector voltage with the modulation input..one trick i used is to basically use a class B audio amp output with direct dc coupling to act as the voltage supply for the class C stage, then the audio will cause AM modulation of the output
Very interesting and great that you show the actual effects and all the steps you take, that makes it super easy to follow and understand for me. You're a great guy FesZ Electronics to share all this knowledge online, keep up the good work. Best regards, Ricardo Penders
Hello! Good video! In my oppinion- it`s better practice to use a wideband long line (twisted pair) transformer on BJT input to decrease Miller effect and therefore increasing stability of the amplifier. For example an input transformer 1:4 decrease input impedance from 50 to 12.5Ohm.
AM is done by modulating the collector power of the class C ampl - also called high level modulation. Yes, it wont be able to further amplify already modulated signal
Excellent video. Still one querry, in Frequency modulation , class C amplifier is used . Whether full sinusoildal Wave is the output of Class C amplifier or a limited angle ( less < 2 pi ) is used ? Thanks
yes this is all very well done and all relevant, great.... maybe you want to talk about the FCC license??? I obtained one a long time ago and and the universal mystery question answer was always: grid leak bias. It was almost comical...And one question that NOBODY (except old fart me) knew the answer: Why must a battery always be placed on a felt surface? Because batteries are made from glass (right?) and one little sand grain could cause it to crack. So .. your oscillator circuitry discussions could indeed be very helpful to those who need to obtain the FCC license here in the US. Unless the US no longer requires it for certain working fields, it would not surprise me if the entire topic is moot.
Dear @FesZ Electronics, special thank for your nice video. but I have a question about Class c amplifiers without tank and only an RF choke. Are calculations the same? how does that choke be selected?
Great tutorial, I find your tutorials very interesting, I have a transmitter transmitting at 110mhz, the final output is -15db I have tried to amplify it to 1watts but it remains the same, my output impedance is matched to 50 ohms Can you advise on how best to amplify a -15db signal
Great video! You covered the topic very clearly, and I like how you explained the concepts from the ground up. Would it be possible to cover the class A amplifier with the same characteristics, e.g. operating at 1 MHz with a 60 dB gain to a ~5 W output?
Many thanks for the great videos. I have a question, I built this class C amplifier with a 2N3904 BJT, 10V Power Supply and a 1,6MHZ Collpitts Oscillator at the input. The problem is that the PA is not able to provide the 20Vpp output signal as expected. I tryed also, with a Q = 10 LC tank and Q= 5 but no change. In your video you didn't discuss about the Resistor and Capacitor at the input, how there are defined ? ... it would be very nice if you could help me .. Thanks in avance Cyril / France
Hello Cyril. What was the input signal amplitude? and also what is the output signal amplitude? - exactly 20Vpp will not be possible since you have losses in the LC circuit and the transistor will always have some finite CE voltage drop. Also, make sure the peak collector current is within the possibilities of the transistor - this is a low power transistor. Anyway, I think there is a transistor model for this component by default in LTspice, you should be able to simulate the circuit and see what signal amplitude the simulator is predicting.
Hello Fes, thank you so much for your quick response. Yes the LTSpice Simulation was my first step, same as you, and the expected output voltage should be 20Vpp according LTSpice. I have just 14Vpp with the Q= 10 or Q= 5 LC Tank but a great Sinus signal. The input signal amplitude is more than 0,7V (when the oscillator is connected to the PA). You see 7Volts are missing .. I d'on't why
If you have a series resistor in place in-between the signal generator and transistor base, you might be loosing some voltage there. The "resistor" might be the output impedance of the signal generator. If you have the possibility, I would recommend to increase the input signal amplitude.
I simulated this circuit with no load and Vce goes above 50V, output impedance is ~0 with an occasional spike to > kohm at the voltage spike. Is there any way to make this circuit more resilient to open circuit operation? Would replacing the LC matching network with a transformer help, or does one have to resort to active measures?
Maybe you can try a varistor or TVS type diode - something that will start conducting only when voltage exceeds a specific threshold. This will add extra capacitance on the output though that might need to be considered.
@@FesZElectronics I was thinking of having a second bjt that would pull q1 base to ground when q1vce > 30, maybe feed it from a voltage divider that would map 30v to 0.7V when q2 will start conducting. Although I'm not sure if it would reach stedy state.
Now I crave to see what a push pull class C amplifier would be capable to provide Also interestingly in the past of 1960s 50s AM transmitters used tubes in a class C operation in AM modulated transmitters. They way they modulated the output was by having a transformer in the cathode of the tube (or the emitter of the transistor) and fed the AF to it. This allowed the bias of the tube to be mosulated on the on the power side not on the inout side. Not sure how this would perform. It is also notable that such modulation is inefficient as fuck and lossy..it usually required a 20-30W audio amplifier to drive the cathode transformer of such class C output stage. If we can disregard efficiency altogether and would want to modulate AM perfectly then a differential pair modulator is probably the best. Or if you want a totally cursed AM transmitter look no further but get a ZVS driver tune the LC to the desired frequency and current modulate it 🤣
One thing I have a hard time understanding about a class C is the signal on the base, you will see it "drop" and then it does a strange "spikeup" . Almost like it's getting reverse voltage breakdown back against the input signal from the collector. It's hard to describe and understand. Doesn't make full sense to me
If you can make the LC network components large enough, any frequency is possible. Usually it will be more cost effective to make some other type of amplifier at such a low frequency though(like an audio class D)
If you only need 15KHz, this is in the audio range - any audio class D amplifier should be good depending on your power need of course. A decent class D should be able to output up to 20KHz. The nice thing about an audio class D is that you can easily change the output frequency, add amplitude modulation and other things - which are not possible with class C.
@@FesZElectronics would like to see this as well, its hard to figure because when the transistor turns on its gonna be base current...25mv over Ie or whatever times beta..kind of the inverse of the impedance. Its gonna be pretty low i think but its a challenge to calculate overall for my tiny brain
I did this experiment myself. I can't calculate the input and output impedance of a class C amplifier, which is the biggest I've lived here.At 6:06 , you wrote in blue that the load is 100ohm.Doesn't the load being 100 ohms mean the output impedance is 100 ohms?How did you calculate the load is 100ohm? I looked at your other videos, but I couldn't find the answer to my question?
At around 2:00 I calculate the necessary load that the base amplifier needs to have, considering the supply voltage, so that after lossless impedance matching it can drive 0.5W into 50R. The base amplifier (transistor+LC circuit in the collector) does have an 100R output impedance, but after impedance matching this is taken down to 50R; so the total output impedance of the circuit is 50R
@@FesZElectronics Got it, thanks. But why didn't you do the impedance matching at the input part of the circuit?You got the AC signal from a signal generator. The output of the signal generators is usually 50ohm. I guess the input impedance is 100ohm//Rpi (Rpi=resistance between base and ground in transistor AC analysis). Is there any particular reason why you don't do this?Aren't the input and output resistor formulas used in a class A amplifier not used in a class C amplifier? I'm wondering about this.Also, in my experiment, when I placed a coil with a suitable value between the base and ground instead of a resistance, the output amplitude was almost twice the DC voltage. When a coil is used instead of a 100ohm resistor, the output amplitude is much higher.
I tried looking for a good explanation on how to find the input impedance, but I was unable to find an easy way to obtain it - some papers just suggest to just vary the network parameters until it works most effectively and them measure what you set and from there backtrack to figure out the transistor equivalent impedance. Since the transistor is operated in a highly non-linear fashion, you cannot use small signal analysis - you need to perform a large signal analysis somehow. Regarding output signal, the point is not just to get high amplitude, its to obtain good efficiency (you can get more amplitude but efficiency will go down); I did observe other schematics and recommendations to use inductors instead of pulldown resistors, but at least in the simulator this did not give me better results - maybe there was an issue in the transistor model or something.
@@FesZElectronics The resistor value that halves the output amplitude of the amplifier is determined. Can this value not be accepted as the output resistance? Vs: Output amplitude, Ro=output resistance, Rx=variable resistance.If Ro=Rx , Output amplitude =Vs.(Rx/(Rs+Rx) from here, Output amplitude =Vs/2. So Rx is the output resistance of the amplifier. Is it wrong to determine the output resistance in this way?
😗Nice..... Best place for high voltage transformer like a 240v in to 6000 out doing a booster one to up one to 6to 12 max 20,000 up voltage off pulse transformer on ebay done a little transformer works but really need those wind charts for a ion EMP unit system! Tesla coil ladder rotate.
I guess from a writing point of view simply writing ".x" is not as obvious as "0.x" and when you read it out, you read what you see. Anyway, at least in my country(Romania) its never spoken as ".x" - I think its only an English thing.
@@FesZElectronics Thank you and I apologize. For some reason it has become a pet peeve. Rest assured, almost all English speakers ALSO say "0.x" ... to me the zero is unneeded. Thank you for the response and sorry to be such a grump about it!
Don't worry, I do understand why you are saying that the zero is not needed; I guess that a good compromise though is saying "O point" (the letter o) - it takes less time than saying zero :D
As a radio ham this was particularly interesting for me. I'm sure there are many like me who would be glad of more RF related topics. Yes, I can find them on UA-cam but not as well explained as you do. Thanks for this video.
Class C makes an excellent AM modulator all you have to do is vary the collector voltage with the modulation input..one trick i used is to basically use a class B audio amp output with direct dc coupling to act as the voltage supply for the class C stage, then the audio will cause AM modulation of the output
Very interesting and great that you show the actual effects and all the steps you take, that makes it super easy to follow and understand for me.
You're a great guy FesZ Electronics to share all this knowledge online, keep up the good work.
Best regards,
Ricardo Penders
Testiculo
Very rare video to find. Circuit descriptions are awesome!
Thanks! Your videos are amazing. Theory and practice help to understand better.
Hello! Good video! In my oppinion- it`s better practice to use a wideband long line (twisted pair) transformer on BJT input to decrease Miller effect and therefore increasing stability of the amplifier. For example an input transformer 1:4 decrease input impedance from 50 to 12.5Ohm.
Thank you! I am happy to see pen and paper again! It reminds me of your good old videos. Thanks!
I think I will be using the "pen and paper" a bit - I think its a bit better than the whiteboard segments.. what do you think?
Awesome! More rf amplifier videos please!
AM is done by modulating the collector power of the class C ampl - also called high level modulation. Yes, it wont be able to further amplify already modulated signal
Excellent video. Thanks!
Very nice presentation
What exactly is the purpose of that 100ohm resistor at the input? I understand its an RC filter, but what is it's use?
Excellent video.
Still one querry, in Frequency modulation , class C amplifier is used . Whether full sinusoildal Wave is the output of Class C amplifier or a limited angle ( less < 2 pi ) is used ?
Thanks
yes this is all very well done and all relevant, great.... maybe you want to talk about the FCC license??? I obtained one a long time ago and and the universal mystery question answer was always: grid leak bias. It was almost comical...And one question that NOBODY (except old fart me) knew the answer: Why must a battery always be placed on a felt surface? Because batteries are made from glass (right?) and one little sand grain could cause it to crack. So .. your oscillator circuitry discussions could indeed be very helpful to those who need to obtain the FCC license here in the US. Unless the US no longer requires it for certain working fields, it would not surprise me if the entire topic is moot.
Wonderful!
Dear @FesZ Electronics, special thank for your nice video. but I have a question about Class c amplifiers without tank and only an RF choke. Are calculations the same? how does that choke be selected?
Hello, and welcome back!!
Great video!
Nice video, thanks for sharing :)
Great tutorial, I find your tutorials very interesting,
I have a transmitter transmitting at 110mhz, the final output is -15db I have tried to amplify it to 1watts but it remains the same, my output impedance is matched to 50 ohms
Can you advise on how best to amplify a -15db signal
Great video! You covered the topic very clearly, and I like how you explained the concepts from the ground up. Would it be possible to cover the class A amplifier with the same characteristics, e.g. operating at 1 MHz with a 60 dB gain to a ~5 W output?
Many thanks for the great videos. I have a question, I built this class C amplifier with a 2N3904 BJT, 10V Power Supply and a 1,6MHZ Collpitts Oscillator at the input. The problem is that the PA is not able to provide the 20Vpp output signal as expected. I tryed also, with a Q = 10 LC tank and Q= 5 but no change. In your video you didn't discuss about the Resistor and Capacitor at the input, how there are defined ? ... it would be very nice if you could help me .. Thanks in avance Cyril / France
Hello Cyril. What was the input signal amplitude? and also what is the output signal amplitude? - exactly 20Vpp will not be possible since you have losses in the LC circuit and the transistor will always have some finite CE voltage drop. Also, make sure the peak collector current is within the possibilities of the transistor - this is a low power transistor. Anyway, I think there is a transistor model for this component by default in LTspice, you should be able to simulate the circuit and see what signal amplitude the simulator is predicting.
Hello Fes, thank you so much for your quick response. Yes the LTSpice Simulation was my first step, same as you, and the expected output voltage should be 20Vpp according LTSpice. I have just 14Vpp with the Q= 10 or Q= 5 LC Tank but a great Sinus signal. The input signal amplitude is more than 0,7V (when the oscillator is connected to the PA). You see 7Volts are missing .. I d'on't why
If you have a series resistor in place in-between the signal generator and transistor base, you might be loosing some voltage there. The "resistor" might be the output impedance of the signal generator. If you have the possibility, I would recommend to increase the input signal amplitude.
@@FesZElectronics ok than i have to put an amplifier between the Colpitts and the PA .. I will try ... Thanks
I simulated this circuit with no load and Vce goes above 50V, output impedance is ~0 with an occasional spike to > kohm at the voltage spike. Is there any way to make this circuit more resilient to open circuit operation? Would replacing the LC matching network with a transformer help, or does one have to resort to active measures?
Maybe you can try a varistor or TVS type diode - something that will start conducting only when voltage exceeds a specific threshold. This will add extra capacitance on the output though that might need to be considered.
@@FesZElectronics I was thinking of having a second bjt that would pull q1 base to ground when q1vce > 30, maybe feed it from a voltage divider that would map 30v to 0.7V when q2 will start conducting. Although I'm not sure if it would reach stedy state.
Now I crave to see what a push pull class C amplifier would be capable to provide
Also interestingly in the past of 1960s 50s AM transmitters used tubes in a class C operation in AM modulated transmitters. They way they modulated the output was by having a transformer in the cathode of the tube (or the emitter of the transistor) and fed the AF to it. This allowed the bias of the tube to be mosulated on the on the power side not on the inout side. Not sure how this would perform. It is also notable that such modulation is inefficient as fuck and lossy..it usually required a 20-30W audio amplifier to drive the cathode transformer of such class C output stage.
If we can disregard efficiency altogether and would want to modulate AM perfectly then a differential pair modulator is probably the best. Or if you want a totally cursed AM transmitter look no further but get a ZVS driver tune the LC to the desired frequency and current modulate it 🤣
about, 15:40 , cant one protect the junction by adding a diode in series to the base ?
Good ind informative tutorials.
I m watching your tutorials.
I want to communicate with you Sir.
Try to reach him in the lower 80 meter band
One thing I have a hard time understanding about a class C is the signal on the base, you will see it "drop" and then it does a strange "spikeup" . Almost like it's getting reverse voltage breakdown back against the input signal from the collector. It's hard to describe and understand. Doesn't make full sense to me
is it possible to use such an amplifier at a frequency of 15 kHz?
If you can make the LC network components large enough, any frequency is possible. Usually it will be more cost effective to make some other type of amplifier at such a low frequency though(like an audio class D)
It's for rf helmonds coil, What would be the best amplifier for this?
If you only need 15KHz, this is in the audio range - any audio class D amplifier should be good depending on your power need of course. A decent class D should be able to output up to 20KHz. The nice thing about an audio class D is that you can easily change the output frequency, add amplitude modulation and other things - which are not possible with class C.
Thanks for interesting tutorial. One question is left - input impedance of amplifier. How to calculate or measure?
Good question :D I'm still researching this and I'll try to do a video on it at some point.
@@FesZElectronics would like to see this as well, its hard to figure because when the transistor turns on its gonna be base current...25mv over Ie or whatever times beta..kind of the inverse of the impedance. Its gonna be pretty low i think but its a challenge to calculate overall for my tiny brain
🔥👍
cool
Nice, can you tell anything about colpitts oscillator (calculations, etc)?
I did this experiment myself. I can't calculate the input and output impedance of a class C amplifier, which is the biggest I've lived here.At 6:06 , you wrote in blue that the load is 100ohm.Doesn't the load being 100 ohms mean the output impedance is 100 ohms?How did you calculate the load is 100ohm? I looked at your other videos, but I couldn't find the answer to my question?
At around 2:00 I calculate the necessary load that the base amplifier needs to have, considering the supply voltage, so that after lossless impedance matching it can drive 0.5W into 50R.
The base amplifier (transistor+LC circuit in the collector) does have an 100R output impedance, but after impedance matching this is taken down to 50R; so the total output impedance of the circuit is 50R
@@FesZElectronics Got it, thanks. But why didn't you do the impedance matching at the input part of the circuit?You got the AC signal from a signal generator. The output of the signal generators is usually 50ohm. I guess the input impedance is 100ohm//Rpi (Rpi=resistance between base and ground in transistor AC analysis). Is there any particular reason why you don't do this?Aren't the input and output resistor formulas used in a class A amplifier not used in a class C amplifier? I'm wondering about this.Also, in my experiment, when I placed a coil with a suitable value between the base and ground instead of a resistance, the output amplitude was almost twice the DC voltage. When a coil is used instead of a 100ohm resistor, the output amplitude is much higher.
I tried looking for a good explanation on how to find the input impedance, but I was unable to find an easy way to obtain it - some papers just suggest to just vary the network parameters until it works most effectively and them measure what you set and from there backtrack to figure out the transistor equivalent impedance. Since the transistor is operated in a highly non-linear fashion, you cannot use small signal analysis - you need to perform a large signal analysis somehow.
Regarding output signal, the point is not just to get high amplitude, its to obtain good efficiency (you can get more amplitude but efficiency will go down); I did observe other schematics and recommendations to use inductors instead of pulldown resistors, but at least in the simulator this did not give me better results - maybe there was an issue in the transistor model or something.
@@FesZElectronics The resistor value that halves the output amplitude of the amplifier is determined. Can this value not be accepted as the output resistance? Vs: Output amplitude, Ro=output resistance, Rx=variable resistance.If Ro=Rx , Output amplitude =Vs.(Rx/(Rs+Rx) from here, Output amplitude =Vs/2. So Rx is the output resistance of the amplifier. Is it wrong to determine the output resistance in this way?
@@FesZElectronics if you use a mosfet you can set the input resistance with a resistor only, easier to reason about :)
hi
please put schematic and pcb file links
😗Nice..... Best place for high voltage transformer like a 240v in to 6000 out doing a booster one to up one to 6to 12 max 20,000 up voltage off pulse transformer on ebay done a little transformer works but really need those wind charts for a ion EMP unit system! Tesla coil ladder rotate.
چیلامپه سیلامپه
سیلاتو یاراپه
what nationality are you ?
why do people always say "zero point x" instead of simply "point x"? the zero is unnecessary!
I guess from a writing point of view simply writing ".x" is not as obvious as "0.x" and when you read it out, you read what you see. Anyway, at least in my country(Romania) its never spoken as ".x" - I think its only an English thing.
@@FesZElectronics Thank you and I apologize. For some reason it has become a pet peeve. Rest assured, almost all English speakers ALSO say "0.x" ... to me the zero is unneeded. Thank you for the response and sorry to be such a grump about it!
Don't worry, I do understand why you are saying that the zero is not needed; I guess that a good compromise though is saying "O point" (the letter o) - it takes less time than saying zero :D
@@FesZElectronics haha :)