Why is the square wave signal so distorted at the output of a push-pull pair?
I have this example circuit on my workbench:
simulate this circuit – Schematic created using CircuitLab
If I remove the RLOAD=1K resistor, then the output signal gets distorted.
Yellow is Node1 and blue is node2. With R load:
Without R load:
It is not easy to read from the picture but the lower part of the square signal starts at 600mV (one junction) and slowly goes down to about 50mV. Here is it zoomed:
The same thing happens at node1 too, but I'm less concerned about that. I wonder why there is distortion when there is no load?
bjt push-pull
asked Dec 23 at 10:21
nagylzs
212110
|
show 1 more comment
I have this example circuit on my workbench:
simulate this circuit – Schematic created using CircuitLab
If I remove the RLOAD=1K resistor, then the output signal gets distorted.
Yellow is Node1 and blue is node2. With R load:
Without R load:
It is not easy to read from the picture but the lower part of the square signal starts at 600mV (one junction) and slowly goes down to about 50mV. Here is it zoomed:
The same thing happens at node1 too, but I'm less concerned about that. I wonder why there is distortion when there is no load?
bjt push-pull
asked Dec 23 at 10:21
nagylzs
212110
Is your probe compensated?
– TemeV
Dec 23 at 10:33
@TemeV Why would that have anything to do with the problem? Unless he uncompensates the probe after the first reading.
– pipe
Dec 23 at 10:50
It just looks like uncompensated probe, so wanted to check that. Maybe using different probe or something.
– TemeV
Dec 23 at 11:40
But the first waveform looks square. This this is load dependant, current to feed the bjt's. If this was probe compensation then all the waveforms would show that characteristic
– JonRB
Dec 23 at 12:41
Yes of course. It is unlikely, but the probe could have been changed between the measurements or something. It is easy to forget the compensation, and that is why I always ask it first when I see that kind of waveform.
– TemeV
Dec 23 at 13:06
|
show 1 more comment
I have this example circuit on my workbench:
simulate this circuit – Schematic created using CircuitLab
If I remove the RLOAD=1K resistor, then the output signal gets distorted.
Yellow is Node1 and blue is node2. With R load:
Without R load:
It is not easy to read from the picture but the lower part of the square signal starts at 600mV (one junction) and slowly goes down to about 50mV. Here is it zoomed:
The same thing happens at node1 too, but I'm less concerned about that. I wonder why there is distortion when there is no load?
bjt push-pull
asked Dec 23 at 10:21
nagylzs
212110
I have this example circuit on my workbench:
simulate this circuit – Schematic created using CircuitLab
If I remove the RLOAD=1K resistor, then the output signal gets distorted.
Yellow is Node1 and blue is node2. With R load:
Without R load:
It is not easy to read from the picture but the lower part of the square signal starts at 600mV (one junction) and slowly goes down to about 50mV. Here is it zoomed:
The same thing happens at node1 too, but I'm less concerned about that. I wonder why there is distortion when there is no load?
bjt push-pull
bjt push-pull
asked Dec 23 at 10:21
nagylzs
212110
asked Dec 23 at 10:21
nagylzs
212110
asked Dec 23 at 10:21
nagylzs
212110
asked Dec 23 at 10:21
nagylzs
212110
asked Dec 23 at 10:21
nagylzs
212110
212110
Is your probe compensated?
– TemeV
Dec 23 at 10:33
@TemeV Why would that have anything to do with the problem? Unless he uncompensates the probe after the first reading.
– pipe
Dec 23 at 10:50
It just looks like uncompensated probe, so wanted to check that. Maybe using different probe or something.
– TemeV
Dec 23 at 11:40
But the first waveform looks square. This this is load dependant, current to feed the bjt's. If this was probe compensation then all the waveforms would show that characteristic
– JonRB
Dec 23 at 12:41
Yes of course. It is unlikely, but the probe could have been changed between the measurements or something. It is easy to forget the compensation, and that is why I always ask it first when I see that kind of waveform.
– TemeV
Dec 23 at 13:06
|
show 1 more comment
Is your probe compensated?
– TemeV
Dec 23 at 10:33
@TemeV Why would that have anything to do with the problem? Unless he uncompensates the probe after the first reading.
– pipe
Dec 23 at 10:50
It just looks like uncompensated probe, so wanted to check that. Maybe using different probe or something.
– TemeV
Dec 23 at 11:40
But the first waveform looks square. This this is load dependant, current to feed the bjt's. If this was probe compensation then all the waveforms would show that characteristic
– JonRB
Dec 23 at 12:41
Yes of course. It is unlikely, but the probe could have been changed between the measurements or something. It is easy to forget the compensation, and that is why I always ask it first when I see that kind of waveform.
– TemeV
Dec 23 at 13:06
Is your probe compensated?
– TemeV
Dec 23 at 10:33
Is your probe compensated?
– TemeV
Dec 23 at 10:33
@TemeV Why would that have anything to do with the problem? Unless he uncompensates the probe after the first reading.
– pipe
Dec 23 at 10:50
@TemeV Why would that have anything to do with the problem? Unless he uncompensates the probe after the first reading.
– pipe
Dec 23 at 10:50
It just looks like uncompensated probe, so wanted to check that. Maybe using different probe or something.
– TemeV
Dec 23 at 11:40
It just looks like uncompensated probe, so wanted to check that. Maybe using different probe or something.
– TemeV
Dec 23 at 11:40
But the first waveform looks square. This this is load dependant, current to feed the bjt's. If this was probe compensation then all the waveforms would show that characteristic
– JonRB
Dec 23 at 12:41
But the first waveform looks square. This this is load dependant, current to feed the bjt's. If this was probe compensation then all the waveforms would show that characteristic
– JonRB
Dec 23 at 12:41
Yes of course. It is unlikely, but the probe could have been changed between the measurements or something. It is easy to forget the compensation, and that is why I always ask it first when I see that kind of waveform.
– TemeV
Dec 23 at 13:06
Yes of course. It is unlikely, but the probe could have been changed between the measurements or something. It is easy to forget the compensation, and that is why I always ask it first when I see that kind of waveform.
– TemeV
Dec 23 at 13:06
|
show 1 more comment
1 Answer
1
active
oldest
votes
Think about what the load resistor is doing. Without that resistor, how are you going to put base current into either transistor and turn Q2 and Q3 on properly. With the resistor, Q2 can be effectively turned on and that same resistor acts as a decent pull-down when Q3 is supposedly being activated. Without proper base biasing you won't have a decent push pull stage. Try using a 10 kohm in parallel with collector/emitter on each transistor to see what happens. Or, alternatively try biasing the bases as per how a class AB stage operates.
answered Dec 23 at 10:59
Andy aka
239k10176407
This will be used for a charge pump, so I don't want to operate it in AB mode. Okay, so let me se if I understand correctly. When Q2 is opened and Q3 is closed, then there will be 0.6V (one junction) between node2 and ground. When the signal switches, then Q2 is closed first, leaving some charge at node2. Only after that Q3 is opened, but it cannot discharge node2 below 0.6V (because of Q3 junction). Probably the scope probe is discharging it slowly, and that is what I see. Am I interpreting this correctly?
– nagylzs
Dec 23 at 16:49
add a comment |
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1 Answer
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1 Answer
1
active
oldest
votes
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oldest
votes
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oldest
votes
Think about what the load resistor is doing. Without that resistor, how are you going to put base current into either transistor and turn Q2 and Q3 on properly. With the resistor, Q2 can be effectively turned on and that same resistor acts as a decent pull-down when Q3 is supposedly being activated. Without proper base biasing you won't have a decent push pull stage. Try using a 10 kohm in parallel with collector/emitter on each transistor to see what happens. Or, alternatively try biasing the bases as per how a class AB stage operates.
answered Dec 23 at 10:59
Andy aka
239k10176407
This will be used for a charge pump, so I don't want to operate it in AB mode. Okay, so let me se if I understand correctly. When Q2 is opened and Q3 is closed, then there will be 0.6V (one junction) between node2 and ground. When the signal switches, then Q2 is closed first, leaving some charge at node2. Only after that Q3 is opened, but it cannot discharge node2 below 0.6V (because of Q3 junction). Probably the scope probe is discharging it slowly, and that is what I see. Am I interpreting this correctly?
– nagylzs
Dec 23 at 16:49
add a comment |
Think about what the load resistor is doing. Without that resistor, how are you going to put base current into either transistor and turn Q2 and Q3 on properly. With the resistor, Q2 can be effectively turned on and that same resistor acts as a decent pull-down when Q3 is supposedly being activated. Without proper base biasing you won't have a decent push pull stage. Try using a 10 kohm in parallel with collector/emitter on each transistor to see what happens. Or, alternatively try biasing the bases as per how a class AB stage operates.
answered Dec 23 at 10:59
Andy aka
239k10176407
This will be used for a charge pump, so I don't want to operate it in AB mode. Okay, so let me se if I understand correctly. When Q2 is opened and Q3 is closed, then there will be 0.6V (one junction) between node2 and ground. When the signal switches, then Q2 is closed first, leaving some charge at node2. Only after that Q3 is opened, but it cannot discharge node2 below 0.6V (because of Q3 junction). Probably the scope probe is discharging it slowly, and that is what I see. Am I interpreting this correctly?
– nagylzs
Dec 23 at 16:49
add a comment |
Think about what the load resistor is doing. Without that resistor, how are you going to put base current into either transistor and turn Q2 and Q3 on properly. With the resistor, Q2 can be effectively turned on and that same resistor acts as a decent pull-down when Q3 is supposedly being activated. Without proper base biasing you won't have a decent push pull stage. Try using a 10 kohm in parallel with collector/emitter on each transistor to see what happens. Or, alternatively try biasing the bases as per how a class AB stage operates.
answered Dec 23 at 10:59
Andy aka
239k10176407
Think about what the load resistor is doing. Without that resistor, how are you going to put base current into either transistor and turn Q2 and Q3 on properly. With the resistor, Q2 can be effectively turned on and that same resistor acts as a decent pull-down when Q3 is supposedly being activated. Without proper base biasing you won't have a decent push pull stage. Try using a 10 kohm in parallel with collector/emitter on each transistor to see what happens. Or, alternatively try biasing the bases as per how a class AB stage operates.
answered Dec 23 at 10:59
Andy aka
239k10176407
answered Dec 23 at 10:59
Andy aka
239k10176407
answered Dec 23 at 10:59
Andy aka
239k10176407
answered Dec 23 at 10:59
Andy aka
239k10176407
239k10176407
This will be used for a charge pump, so I don't want to operate it in AB mode. Okay, so let me se if I understand correctly. When Q2 is opened and Q3 is closed, then there will be 0.6V (one junction) between node2 and ground. When the signal switches, then Q2 is closed first, leaving some charge at node2. Only after that Q3 is opened, but it cannot discharge node2 below 0.6V (because of Q3 junction). Probably the scope probe is discharging it slowly, and that is what I see. Am I interpreting this correctly?
– nagylzs
Dec 23 at 16:49
add a comment |
This will be used for a charge pump, so I don't want to operate it in AB mode. Okay, so let me se if I understand correctly. When Q2 is opened and Q3 is closed, then there will be 0.6V (one junction) between node2 and ground. When the signal switches, then Q2 is closed first, leaving some charge at node2. Only after that Q3 is opened, but it cannot discharge node2 below 0.6V (because of Q3 junction). Probably the scope probe is discharging it slowly, and that is what I see. Am I interpreting this correctly?
– nagylzs
Dec 23 at 16:49
This will be used for a charge pump, so I don't want to operate it in AB mode. Okay, so let me se if I understand correctly. When Q2 is opened and Q3 is closed, then there will be 0.6V (one junction) between node2 and ground. When the signal switches, then Q2 is closed first, leaving some charge at node2. Only after that Q3 is opened, but it cannot discharge node2 below 0.6V (because of Q3 junction). Probably the scope probe is discharging it slowly, and that is what I see. Am I interpreting this correctly?
– nagylzs
Dec 23 at 16:49
This will be used for a charge pump, so I don't want to operate it in AB mode. Okay, so let me se if I understand correctly. When Q2 is opened and Q3 is closed, then there will be 0.6V (one junction) between node2 and ground. When the signal switches, then Q2 is closed first, leaving some charge at node2. Only after that Q3 is opened, but it cannot discharge node2 below 0.6V (because of Q3 junction). Probably the scope probe is discharging it slowly, and that is what I see. Am I interpreting this correctly?
– nagylzs
Dec 23 at 16:49
add a comment |
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Is your probe compensated?
– TemeV
Dec 23 at 10:33
@TemeV Why would that have anything to do with the problem? Unless he uncompensates the probe after the first reading.
– pipe
Dec 23 at 10:50
It just looks like uncompensated probe, so wanted to check that. Maybe using different probe or something.
– TemeV
Dec 23 at 11:40
But the first waveform looks square. This this is load dependant, current to feed the bjt's. If this was probe compensation then all the waveforms would show that characteristic
– JonRB
Dec 23 at 12:41
Yes of course. It is unlikely, but the probe could have been changed between the measurements or something. It is easy to forget the compensation, and that is why I always ask it first when I see that kind of waveform.
– TemeV
Dec 23 at 13:06