I wanted to convert 0-1v, 2ns pulse to 0-7v, 2ns pulse
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I wanted to generate a 2ns pulse width signal in particular 2ns ON time and 6ns OFF time signal. I can generate a 0-1v signal having 2ns pulse width using FPGA(by any means we cannot have more than 3.3v from FPGA). And now I wanted to use the 1v, 2ns pulse width signal to generate 7v signal of same pulse width. I have tried using BJT as a switch with Vcc 7v in different configurations but could not amplify the signal. I would like to have some ideas to solve this problem out.
:> Is there any way to generate a 2ns pulse signal just like using 555timer.
bjt pulse
asked Aug 22 at 6:49
Martin
193
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up vote
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I wanted to generate a 2ns pulse width signal in particular 2ns ON time and 6ns OFF time signal. I can generate a 0-1v signal having 2ns pulse width using FPGA(by any means we cannot have more than 3.3v from FPGA). And now I wanted to use the 1v, 2ns pulse width signal to generate 7v signal of same pulse width. I have tried using BJT as a switch with Vcc 7v in different configurations but could not amplify the signal. I would like to have some ideas to solve this problem out.
:> Is there any way to generate a 2ns pulse signal just like using 555timer.
bjt pulse
asked Aug 22 at 6:49
Martin
193
1
Avalanche mode in a BJT might achieve 2 ns rise time and perhaps 10 ns FWHM. Probably have to hand-select the BJTs. And probably something like a 2N2369A which is intended for high-speed saturated switching and high-frequency applications, generally. Dead-bug the circuit. I'd not consider this easy with discrete parts. Might get to your specs if you select well and keep spacing tight.
– jonk
Aug 22 at 6:57
I was considering BFR93AW an RF bjt for the simulation which has 6GHz trasition frequency, so I thought it should work but I could not see any amplification happening.
– Martin
Aug 22 at 7:16
2
Do you really need to amplify the pulse and keep its waveform, or can you use your fpga pulse as a trigger to generate another pulse?
– PlasmaHH
Aug 22 at 9:03
2
Your task is under-specified, you need to define rise and fall times for the pulse. Also you need to understand the difference between small-signal bandwidth of an amplifier (transistor) and large amplitude capabilities (aka slew rate).
– Ale..chenski
Aug 22 at 14:28
add a comment |Â
up vote
3
down vote
favorite
up vote
3
down vote
favorite
I wanted to generate a 2ns pulse width signal in particular 2ns ON time and 6ns OFF time signal. I can generate a 0-1v signal having 2ns pulse width using FPGA(by any means we cannot have more than 3.3v from FPGA). And now I wanted to use the 1v, 2ns pulse width signal to generate 7v signal of same pulse width. I have tried using BJT as a switch with Vcc 7v in different configurations but could not amplify the signal. I would like to have some ideas to solve this problem out.
:> Is there any way to generate a 2ns pulse signal just like using 555timer.
bjt pulse
asked Aug 22 at 6:49
Martin
193
I wanted to generate a 2ns pulse width signal in particular 2ns ON time and 6ns OFF time signal. I can generate a 0-1v signal having 2ns pulse width using FPGA(by any means we cannot have more than 3.3v from FPGA). And now I wanted to use the 1v, 2ns pulse width signal to generate 7v signal of same pulse width. I have tried using BJT as a switch with Vcc 7v in different configurations but could not amplify the signal. I would like to have some ideas to solve this problem out.
:> Is there any way to generate a 2ns pulse signal just like using 555timer.
bjt pulse
asked Aug 22 at 6:49
Martin
193
edited Aug 22 at 7:20
asked Aug 22 at 6:49
Martin
193
asked Aug 22 at 6:49
Martin
193
asked Aug 22 at 6:49
Martin
193
193
1
Avalanche mode in a BJT might achieve 2 ns rise time and perhaps 10 ns FWHM. Probably have to hand-select the BJTs. And probably something like a 2N2369A which is intended for high-speed saturated switching and high-frequency applications, generally. Dead-bug the circuit. I'd not consider this easy with discrete parts. Might get to your specs if you select well and keep spacing tight.
– jonk
Aug 22 at 6:57
I was considering BFR93AW an RF bjt for the simulation which has 6GHz trasition frequency, so I thought it should work but I could not see any amplification happening.
– Martin
Aug 22 at 7:16
2
Do you really need to amplify the pulse and keep its waveform, or can you use your fpga pulse as a trigger to generate another pulse?
– PlasmaHH
Aug 22 at 9:03
2
Your task is under-specified, you need to define rise and fall times for the pulse. Also you need to understand the difference between small-signal bandwidth of an amplifier (transistor) and large amplitude capabilities (aka slew rate).
– Ale..chenski
Aug 22 at 14:28
add a comment |Â
1
Avalanche mode in a BJT might achieve 2 ns rise time and perhaps 10 ns FWHM. Probably have to hand-select the BJTs. And probably something like a 2N2369A which is intended for high-speed saturated switching and high-frequency applications, generally. Dead-bug the circuit. I'd not consider this easy with discrete parts. Might get to your specs if you select well and keep spacing tight.
– jonk
Aug 22 at 6:57
I was considering BFR93AW an RF bjt for the simulation which has 6GHz trasition frequency, so I thought it should work but I could not see any amplification happening.
– Martin
Aug 22 at 7:16
2
Do you really need to amplify the pulse and keep its waveform, or can you use your fpga pulse as a trigger to generate another pulse?
– PlasmaHH
Aug 22 at 9:03
2
Your task is under-specified, you need to define rise and fall times for the pulse. Also you need to understand the difference between small-signal bandwidth of an amplifier (transistor) and large amplitude capabilities (aka slew rate).
– Ale..chenski
Aug 22 at 14:28
1
1
Avalanche mode in a BJT might achieve 2 ns rise time and perhaps 10 ns FWHM. Probably have to hand-select the BJTs. And probably something like a 2N2369A which is intended for high-speed saturated switching and high-frequency applications, generally. Dead-bug the circuit. I'd not consider this easy with discrete parts. Might get to your specs if you select well and keep spacing tight.
– jonk
Aug 22 at 6:57
Avalanche mode in a BJT might achieve 2 ns rise time and perhaps 10 ns FWHM. Probably have to hand-select the BJTs. And probably something like a 2N2369A which is intended for high-speed saturated switching and high-frequency applications, generally. Dead-bug the circuit. I'd not consider this easy with discrete parts. Might get to your specs if you select well and keep spacing tight.
– jonk
Aug 22 at 6:57
I was considering BFR93AW an RF bjt for the simulation which has 6GHz trasition frequency, so I thought it should work but I could not see any amplification happening.
– Martin
Aug 22 at 7:16
I was considering BFR93AW an RF bjt for the simulation which has 6GHz trasition frequency, so I thought it should work but I could not see any amplification happening.
– Martin
Aug 22 at 7:16
2
2
Do you really need to amplify the pulse and keep its waveform, or can you use your fpga pulse as a trigger to generate another pulse?
– PlasmaHH
Aug 22 at 9:03
Do you really need to amplify the pulse and keep its waveform, or can you use your fpga pulse as a trigger to generate another pulse?
– PlasmaHH
Aug 22 at 9:03
2
2
Your task is under-specified, you need to define rise and fall times for the pulse. Also you need to understand the difference between small-signal bandwidth of an amplifier (transistor) and large amplitude capabilities (aka slew rate).
– Ale..chenski
Aug 22 at 14:28
Your task is under-specified, you need to define rise and fall times for the pulse. Also you need to understand the difference between small-signal bandwidth of an amplifier (transistor) and large amplitude capabilities (aka slew rate).
– Ale..chenski
Aug 22 at 14:28
add a comment |Â
1 Answer
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To the inexperienced it sound like a "trivial" task to amplify the signal from 1 V to 7 V, right?
The problem is the speed you need for this. Your signal is 2 ns on, 6 ns off so that's 8 ns total, corresponding to 1 / 8ns = 125 MHz.
125 MHz is getting close to "RF" (radio frequency) signals, this means you have to consider parasitic capacitances and inductances of wires and traces on a PCB. Also even very small capacitances present in a transistor will ruin your day even at 125 MHz.
But it gets worse.
Since this is a pulse signal (not a sinewave) it is not just a 125 MHz signal, it will also contain harmonics of 125 MHz. Even for a half-decent looking pulse/square you need at least the 10th harmonic to be present as well. So 125 MHz then jumps to 1250 MHz = 1.25 GHz Oops! That won't even give you a decent rise time, as the rise time is still limited to 1 / 1250 MHz = 0.8 ns Oops again. How much risetime can you allow for ?
Sure amplifiers that amplify 7x with a 5 GHz bandwidth do exist but they require experience to use. You cannot just buy one if you don't know how to use it as the amplifier will only work when it has the right impedance at its input and output. Some impedance matching might need to be done. Yes, work for experts, not something anyone can do without the right experience. Many engineers who know what is required to make such a setup work call it "black magic". The magic wears off a bit when you understand it though.
answered Aug 22 at 7:10
Bimpelrekkie
41.3k23689
1
Terms for OP to google for: "Gibb's phenomenon" is what we signal processing folks call what happens to a step under restricted bandwidth (and where the 10. harmonic comes from). Real world is more complicated, as things get time-variant pretty quickly at these scales.
– Marcus Müller
Aug 22 at 7:14
1
The BFR93AW has an Ft of 6 GHz so at 6 GHz you would get a gain of 1 (no amplification). For 7x gain the BW drops to less than 1 GHz and that's under ideal circumstances.
– Bimpelrekkie
Aug 22 at 7:30
2
But my input bandwidth is not 6GHz You stating that means that you didn't understand the "Gibb's phenomenon" mentioned by Marcus. Go read: en.wikipedia.org/wiki/Gibbs_phenomenon Then you should learn that the bandwidth of your signal is much larger than 6 GHz. The point is that we're not dealing with a sine wave, for a 125 MHz sinewave, a 1 GHz BW is more than enough. But for a 125 MHz pulse, a 1 GHz BW is by far not enough.
– Bimpelrekkie
Aug 22 at 8:29
4
I should see at least some gain at the expense of bandwidth. Why? Your input signal has more bandwidth than the amplifier can handle, it will not amplify the harmonics (much). Since your signal has a duty cycle of 2ns/8ns = 25 % there is (relatively, compared to a 50% duty cycle signal) less power in the fundamental frequency and more in the harmonics. The harmonics aren't amplified (much) and will also be phase shifted due to group delays that can result in the "almost no gain" that you see. So: what you see is to be expected and can be explained theoretically.
– Bimpelrekkie
Aug 22 at 8:57
3
I wish you wouldn't bring up Willard Gibbs' stuff in this context. It's not really directly relevant. How much bandwidth to get a reasonable approximation of a square wave is more relevant than the mathematical curiosity of imperfection, even with non-causal representation and unlimited terms. Low pass filtering and slew rates (like finite current into a finite load capacitance) are more relevant, in my opinion.
– Spehro Pefhany
Aug 22 at 10:24
 |Â
show 10 more comments
1 Answer
1
active
oldest
votes
1 Answer
1
active
oldest
votes
active
oldest
votes
active
oldest
votes
up vote
12
down vote
To the inexperienced it sound like a "trivial" task to amplify the signal from 1 V to 7 V, right?
The problem is the speed you need for this. Your signal is 2 ns on, 6 ns off so that's 8 ns total, corresponding to 1 / 8ns = 125 MHz.
125 MHz is getting close to "RF" (radio frequency) signals, this means you have to consider parasitic capacitances and inductances of wires and traces on a PCB. Also even very small capacitances present in a transistor will ruin your day even at 125 MHz.
But it gets worse.
Since this is a pulse signal (not a sinewave) it is not just a 125 MHz signal, it will also contain harmonics of 125 MHz. Even for a half-decent looking pulse/square you need at least the 10th harmonic to be present as well. So 125 MHz then jumps to 1250 MHz = 1.25 GHz Oops! That won't even give you a decent rise time, as the rise time is still limited to 1 / 1250 MHz = 0.8 ns Oops again. How much risetime can you allow for ?
Sure amplifiers that amplify 7x with a 5 GHz bandwidth do exist but they require experience to use. You cannot just buy one if you don't know how to use it as the amplifier will only work when it has the right impedance at its input and output. Some impedance matching might need to be done. Yes, work for experts, not something anyone can do without the right experience. Many engineers who know what is required to make such a setup work call it "black magic". The magic wears off a bit when you understand it though.
answered Aug 22 at 7:10
Bimpelrekkie
41.3k23689
1
Terms for OP to google for: "Gibb's phenomenon" is what we signal processing folks call what happens to a step under restricted bandwidth (and where the 10. harmonic comes from). Real world is more complicated, as things get time-variant pretty quickly at these scales.
– Marcus Müller
Aug 22 at 7:14
1
The BFR93AW has an Ft of 6 GHz so at 6 GHz you would get a gain of 1 (no amplification). For 7x gain the BW drops to less than 1 GHz and that's under ideal circumstances.
– Bimpelrekkie
Aug 22 at 7:30
2
But my input bandwidth is not 6GHz You stating that means that you didn't understand the "Gibb's phenomenon" mentioned by Marcus. Go read: en.wikipedia.org/wiki/Gibbs_phenomenon Then you should learn that the bandwidth of your signal is much larger than 6 GHz. The point is that we're not dealing with a sine wave, for a 125 MHz sinewave, a 1 GHz BW is more than enough. But for a 125 MHz pulse, a 1 GHz BW is by far not enough.
– Bimpelrekkie
Aug 22 at 8:29
4
I should see at least some gain at the expense of bandwidth. Why? Your input signal has more bandwidth than the amplifier can handle, it will not amplify the harmonics (much). Since your signal has a duty cycle of 2ns/8ns = 25 % there is (relatively, compared to a 50% duty cycle signal) less power in the fundamental frequency and more in the harmonics. The harmonics aren't amplified (much) and will also be phase shifted due to group delays that can result in the "almost no gain" that you see. So: what you see is to be expected and can be explained theoretically.
– Bimpelrekkie
Aug 22 at 8:57
3
I wish you wouldn't bring up Willard Gibbs' stuff in this context. It's not really directly relevant. How much bandwidth to get a reasonable approximation of a square wave is more relevant than the mathematical curiosity of imperfection, even with non-causal representation and unlimited terms. Low pass filtering and slew rates (like finite current into a finite load capacitance) are more relevant, in my opinion.
– Spehro Pefhany
Aug 22 at 10:24
 |Â
show 10 more comments
up vote
12
down vote
To the inexperienced it sound like a "trivial" task to amplify the signal from 1 V to 7 V, right?
The problem is the speed you need for this. Your signal is 2 ns on, 6 ns off so that's 8 ns total, corresponding to 1 / 8ns = 125 MHz.
125 MHz is getting close to "RF" (radio frequency) signals, this means you have to consider parasitic capacitances and inductances of wires and traces on a PCB. Also even very small capacitances present in a transistor will ruin your day even at 125 MHz.
But it gets worse.
Since this is a pulse signal (not a sinewave) it is not just a 125 MHz signal, it will also contain harmonics of 125 MHz. Even for a half-decent looking pulse/square you need at least the 10th harmonic to be present as well. So 125 MHz then jumps to 1250 MHz = 1.25 GHz Oops! That won't even give you a decent rise time, as the rise time is still limited to 1 / 1250 MHz = 0.8 ns Oops again. How much risetime can you allow for ?
Sure amplifiers that amplify 7x with a 5 GHz bandwidth do exist but they require experience to use. You cannot just buy one if you don't know how to use it as the amplifier will only work when it has the right impedance at its input and output. Some impedance matching might need to be done. Yes, work for experts, not something anyone can do without the right experience. Many engineers who know what is required to make such a setup work call it "black magic". The magic wears off a bit when you understand it though.
answered Aug 22 at 7:10
Bimpelrekkie
41.3k23689
1
Terms for OP to google for: "Gibb's phenomenon" is what we signal processing folks call what happens to a step under restricted bandwidth (and where the 10. harmonic comes from). Real world is more complicated, as things get time-variant pretty quickly at these scales.
– Marcus Müller
Aug 22 at 7:14
1
The BFR93AW has an Ft of 6 GHz so at 6 GHz you would get a gain of 1 (no amplification). For 7x gain the BW drops to less than 1 GHz and that's under ideal circumstances.
– Bimpelrekkie
Aug 22 at 7:30
2
But my input bandwidth is not 6GHz You stating that means that you didn't understand the "Gibb's phenomenon" mentioned by Marcus. Go read: en.wikipedia.org/wiki/Gibbs_phenomenon Then you should learn that the bandwidth of your signal is much larger than 6 GHz. The point is that we're not dealing with a sine wave, for a 125 MHz sinewave, a 1 GHz BW is more than enough. But for a 125 MHz pulse, a 1 GHz BW is by far not enough.
– Bimpelrekkie
Aug 22 at 8:29
4
I should see at least some gain at the expense of bandwidth. Why? Your input signal has more bandwidth than the amplifier can handle, it will not amplify the harmonics (much). Since your signal has a duty cycle of 2ns/8ns = 25 % there is (relatively, compared to a 50% duty cycle signal) less power in the fundamental frequency and more in the harmonics. The harmonics aren't amplified (much) and will also be phase shifted due to group delays that can result in the "almost no gain" that you see. So: what you see is to be expected and can be explained theoretically.
– Bimpelrekkie
Aug 22 at 8:57
3
I wish you wouldn't bring up Willard Gibbs' stuff in this context. It's not really directly relevant. How much bandwidth to get a reasonable approximation of a square wave is more relevant than the mathematical curiosity of imperfection, even with non-causal representation and unlimited terms. Low pass filtering and slew rates (like finite current into a finite load capacitance) are more relevant, in my opinion.
– Spehro Pefhany
Aug 22 at 10:24
 |Â
show 10 more comments
up vote
12
down vote
up vote
12
down vote
To the inexperienced it sound like a "trivial" task to amplify the signal from 1 V to 7 V, right?
The problem is the speed you need for this. Your signal is 2 ns on, 6 ns off so that's 8 ns total, corresponding to 1 / 8ns = 125 MHz.
125 MHz is getting close to "RF" (radio frequency) signals, this means you have to consider parasitic capacitances and inductances of wires and traces on a PCB. Also even very small capacitances present in a transistor will ruin your day even at 125 MHz.
But it gets worse.
Since this is a pulse signal (not a sinewave) it is not just a 125 MHz signal, it will also contain harmonics of 125 MHz. Even for a half-decent looking pulse/square you need at least the 10th harmonic to be present as well. So 125 MHz then jumps to 1250 MHz = 1.25 GHz Oops! That won't even give you a decent rise time, as the rise time is still limited to 1 / 1250 MHz = 0.8 ns Oops again. How much risetime can you allow for ?
Sure amplifiers that amplify 7x with a 5 GHz bandwidth do exist but they require experience to use. You cannot just buy one if you don't know how to use it as the amplifier will only work when it has the right impedance at its input and output. Some impedance matching might need to be done. Yes, work for experts, not something anyone can do without the right experience. Many engineers who know what is required to make such a setup work call it "black magic". The magic wears off a bit when you understand it though.
answered Aug 22 at 7:10
Bimpelrekkie
41.3k23689
To the inexperienced it sound like a "trivial" task to amplify the signal from 1 V to 7 V, right?
The problem is the speed you need for this. Your signal is 2 ns on, 6 ns off so that's 8 ns total, corresponding to 1 / 8ns = 125 MHz.
125 MHz is getting close to "RF" (radio frequency) signals, this means you have to consider parasitic capacitances and inductances of wires and traces on a PCB. Also even very small capacitances present in a transistor will ruin your day even at 125 MHz.
But it gets worse.
Since this is a pulse signal (not a sinewave) it is not just a 125 MHz signal, it will also contain harmonics of 125 MHz. Even for a half-decent looking pulse/square you need at least the 10th harmonic to be present as well. So 125 MHz then jumps to 1250 MHz = 1.25 GHz Oops! That won't even give you a decent rise time, as the rise time is still limited to 1 / 1250 MHz = 0.8 ns Oops again. How much risetime can you allow for ?
Sure amplifiers that amplify 7x with a 5 GHz bandwidth do exist but they require experience to use. You cannot just buy one if you don't know how to use it as the amplifier will only work when it has the right impedance at its input and output. Some impedance matching might need to be done. Yes, work for experts, not something anyone can do without the right experience. Many engineers who know what is required to make such a setup work call it "black magic". The magic wears off a bit when you understand it though.
answered Aug 22 at 7:10
Bimpelrekkie
41.3k23689
answered Aug 22 at 7:10
Bimpelrekkie
41.3k23689
answered Aug 22 at 7:10
Bimpelrekkie
41.3k23689
answered Aug 22 at 7:10
Bimpelrekkie
41.3k23689
41.3k23689
1
Terms for OP to google for: "Gibb's phenomenon" is what we signal processing folks call what happens to a step under restricted bandwidth (and where the 10. harmonic comes from). Real world is more complicated, as things get time-variant pretty quickly at these scales.
– Marcus Müller
Aug 22 at 7:14
1
The BFR93AW has an Ft of 6 GHz so at 6 GHz you would get a gain of 1 (no amplification). For 7x gain the BW drops to less than 1 GHz and that's under ideal circumstances.
– Bimpelrekkie
Aug 22 at 7:30
2
But my input bandwidth is not 6GHz You stating that means that you didn't understand the "Gibb's phenomenon" mentioned by Marcus. Go read: en.wikipedia.org/wiki/Gibbs_phenomenon Then you should learn that the bandwidth of your signal is much larger than 6 GHz. The point is that we're not dealing with a sine wave, for a 125 MHz sinewave, a 1 GHz BW is more than enough. But for a 125 MHz pulse, a 1 GHz BW is by far not enough.
– Bimpelrekkie
Aug 22 at 8:29
4
I should see at least some gain at the expense of bandwidth. Why? Your input signal has more bandwidth than the amplifier can handle, it will not amplify the harmonics (much). Since your signal has a duty cycle of 2ns/8ns = 25 % there is (relatively, compared to a 50% duty cycle signal) less power in the fundamental frequency and more in the harmonics. The harmonics aren't amplified (much) and will also be phase shifted due to group delays that can result in the "almost no gain" that you see. So: what you see is to be expected and can be explained theoretically.
– Bimpelrekkie
Aug 22 at 8:57
3
I wish you wouldn't bring up Willard Gibbs' stuff in this context. It's not really directly relevant. How much bandwidth to get a reasonable approximation of a square wave is more relevant than the mathematical curiosity of imperfection, even with non-causal representation and unlimited terms. Low pass filtering and slew rates (like finite current into a finite load capacitance) are more relevant, in my opinion.
– Spehro Pefhany
Aug 22 at 10:24
 |Â
show 10 more comments
1
Terms for OP to google for: "Gibb's phenomenon" is what we signal processing folks call what happens to a step under restricted bandwidth (and where the 10. harmonic comes from). Real world is more complicated, as things get time-variant pretty quickly at these scales.
– Marcus Müller
Aug 22 at 7:14
1
The BFR93AW has an Ft of 6 GHz so at 6 GHz you would get a gain of 1 (no amplification). For 7x gain the BW drops to less than 1 GHz and that's under ideal circumstances.
– Bimpelrekkie
Aug 22 at 7:30
2
But my input bandwidth is not 6GHz You stating that means that you didn't understand the "Gibb's phenomenon" mentioned by Marcus. Go read: en.wikipedia.org/wiki/Gibbs_phenomenon Then you should learn that the bandwidth of your signal is much larger than 6 GHz. The point is that we're not dealing with a sine wave, for a 125 MHz sinewave, a 1 GHz BW is more than enough. But for a 125 MHz pulse, a 1 GHz BW is by far not enough.
– Bimpelrekkie
Aug 22 at 8:29
4
I should see at least some gain at the expense of bandwidth. Why? Your input signal has more bandwidth than the amplifier can handle, it will not amplify the harmonics (much). Since your signal has a duty cycle of 2ns/8ns = 25 % there is (relatively, compared to a 50% duty cycle signal) less power in the fundamental frequency and more in the harmonics. The harmonics aren't amplified (much) and will also be phase shifted due to group delays that can result in the "almost no gain" that you see. So: what you see is to be expected and can be explained theoretically.
– Bimpelrekkie
Aug 22 at 8:57
3
I wish you wouldn't bring up Willard Gibbs' stuff in this context. It's not really directly relevant. How much bandwidth to get a reasonable approximation of a square wave is more relevant than the mathematical curiosity of imperfection, even with non-causal representation and unlimited terms. Low pass filtering and slew rates (like finite current into a finite load capacitance) are more relevant, in my opinion.
– Spehro Pefhany
Aug 22 at 10:24
1
1
Terms for OP to google for: "Gibb's phenomenon" is what we signal processing folks call what happens to a step under restricted bandwidth (and where the 10. harmonic comes from). Real world is more complicated, as things get time-variant pretty quickly at these scales.
– Marcus Müller
Aug 22 at 7:14
Terms for OP to google for: "Gibb's phenomenon" is what we signal processing folks call what happens to a step under restricted bandwidth (and where the 10. harmonic comes from). Real world is more complicated, as things get time-variant pretty quickly at these scales.
– Marcus Müller
Aug 22 at 7:14
1
1
The BFR93AW has an Ft of 6 GHz so at 6 GHz you would get a gain of 1 (no amplification). For 7x gain the BW drops to less than 1 GHz and that's under ideal circumstances.
– Bimpelrekkie
Aug 22 at 7:30
The BFR93AW has an Ft of 6 GHz so at 6 GHz you would get a gain of 1 (no amplification). For 7x gain the BW drops to less than 1 GHz and that's under ideal circumstances.
– Bimpelrekkie
Aug 22 at 7:30
2
2
But my input bandwidth is not 6GHz You stating that means that you didn't understand the "Gibb's phenomenon" mentioned by Marcus. Go read: en.wikipedia.org/wiki/Gibbs_phenomenon Then you should learn that the bandwidth of your signal is much larger than 6 GHz. The point is that we're not dealing with a sine wave, for a 125 MHz sinewave, a 1 GHz BW is more than enough. But for a 125 MHz pulse, a 1 GHz BW is by far not enough.
– Bimpelrekkie
Aug 22 at 8:29
But my input bandwidth is not 6GHz You stating that means that you didn't understand the "Gibb's phenomenon" mentioned by Marcus. Go read: en.wikipedia.org/wiki/Gibbs_phenomenon Then you should learn that the bandwidth of your signal is much larger than 6 GHz. The point is that we're not dealing with a sine wave, for a 125 MHz sinewave, a 1 GHz BW is more than enough. But for a 125 MHz pulse, a 1 GHz BW is by far not enough.
– Bimpelrekkie
Aug 22 at 8:29
4
4
I should see at least some gain at the expense of bandwidth. Why? Your input signal has more bandwidth than the amplifier can handle, it will not amplify the harmonics (much). Since your signal has a duty cycle of 2ns/8ns = 25 % there is (relatively, compared to a 50% duty cycle signal) less power in the fundamental frequency and more in the harmonics. The harmonics aren't amplified (much) and will also be phase shifted due to group delays that can result in the "almost no gain" that you see. So: what you see is to be expected and can be explained theoretically.
– Bimpelrekkie
Aug 22 at 8:57
I should see at least some gain at the expense of bandwidth. Why? Your input signal has more bandwidth than the amplifier can handle, it will not amplify the harmonics (much). Since your signal has a duty cycle of 2ns/8ns = 25 % there is (relatively, compared to a 50% duty cycle signal) less power in the fundamental frequency and more in the harmonics. The harmonics aren't amplified (much) and will also be phase shifted due to group delays that can result in the "almost no gain" that you see. So: what you see is to be expected and can be explained theoretically.
– Bimpelrekkie
Aug 22 at 8:57
3
3
I wish you wouldn't bring up Willard Gibbs' stuff in this context. It's not really directly relevant. How much bandwidth to get a reasonable approximation of a square wave is more relevant than the mathematical curiosity of imperfection, even with non-causal representation and unlimited terms. Low pass filtering and slew rates (like finite current into a finite load capacitance) are more relevant, in my opinion.
– Spehro Pefhany
Aug 22 at 10:24
I wish you wouldn't bring up Willard Gibbs' stuff in this context. It's not really directly relevant. How much bandwidth to get a reasonable approximation of a square wave is more relevant than the mathematical curiosity of imperfection, even with non-causal representation and unlimited terms. Low pass filtering and slew rates (like finite current into a finite load capacitance) are more relevant, in my opinion.
– Spehro Pefhany
Aug 22 at 10:24
 |Â
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1
Avalanche mode in a BJT might achieve 2 ns rise time and perhaps 10 ns FWHM. Probably have to hand-select the BJTs. And probably something like a 2N2369A which is intended for high-speed saturated switching and high-frequency applications, generally. Dead-bug the circuit. I'd not consider this easy with discrete parts. Might get to your specs if you select well and keep spacing tight.
– jonk
Aug 22 at 6:57
I was considering BFR93AW an RF bjt for the simulation which has 6GHz trasition frequency, so I thought it should work but I could not see any amplification happening.
– Martin
Aug 22 at 7:16
2
Do you really need to amplify the pulse and keep its waveform, or can you use your fpga pulse as a trigger to generate another pulse?
– PlasmaHH
Aug 22 at 9:03
2
Your task is under-specified, you need to define rise and fall times for the pulse. Also you need to understand the difference between small-signal bandwidth of an amplifier (transistor) and large amplitude capabilities (aka slew rate).
– Ale..chenski
Aug 22 at 14:28