02-03-2018, 12:50 PM
Perhaps it's time to take a step back and go back to basics here...
Forget about how it's actually made for a moment - a current source is a 1 port device, so no "input" and "output" as such (an amplifier would be a 2-port device, with an input and an output).
If we were talking about voltage sources, this would be much more intuitive. But the current source is very similar indeed, and thinking applicable to a voltage source is transferable to the current source. So, perhaps we should briefly recap voltage sources...
We know that an "ideal" voltage source maintains a constant voltage between its terminals, whatever current flows from it. We also know that a "real-world" voltage source isn't quite so ideal - that's because it has some resistance in series with the output, and so as the current increases, the voltage drops at the terminals because of this internal resistance. And it's very easy to calculate using nothing more than Ohm's law.
In short, a "real" voltage source can be modelled with a perfect voltage source and a resistor. Even if the actual circuit is rather more complex than this, it can be represented by this model. You've no-doubt read all about Thévenin, so I'll leave it at that. Just wanted to set the scene - not teach it all from new...
Moving on to current sources, there's no doubt that people find these much less intuitive. I'm not entirely sure why, but over the years I've seen it a lot. You're in good company!
Firstly, an ideal current source just causes a constant current to flow, whatever the voltage across its 2 terminals. Because we spend so much time talking about voltages in circuits, that's actually a harder concept to grasp - as I say, current seems rather more abstract than voltages to many. Probably because it's easier to measure voltage than current.
In the same way a real-world voltage source has series resistance, a real-world current source has some resistance too. But this resistance is not in series (think about it - if it were, would it change the current flowing?). Instead, this unwanted resistance is in parallel with the current source. Ideally, we want this resistance to be as high as possible, so that the minimum of current goes through this unwanted resistance (note that's the opposite to the voltage source, where we want the resistance to be as low as possible).
Again, I'm not trying to teach this anew - just establishing the basics...
So what we're trying to measure here is the value of this parallel resistance. It's not a real resistor, of course - rather, it's an apparent resistance caused by the imperfections of the circuit. It might well vary under certain conditions, so all we can do is arrive at an approximate value, ideally under the circumstances the current source will see under normal operation.
What is the effect of this parallel resistance? It means that the current delivered from the current source will vary with applied voltage. This is getting to the nub of the question.
Let's start with DC, as that's easiest...
Suppose you measure 20V across your current source, and note that 5mA exactly flows. Then, you increase the voltage across the voltage source to 30V, and note that 5.1mA is now flowing. The change in voltage was 10V, and that caused a change in current of 0.1mA. That implies 100k of parallel resistance (and also implies that the true value of the "ideal" current sink in our Norton model would be 4.8mA, as 0.2mA of that was flowing in the 100k resistance at 20V).
That might take a bit of thinking about, but once you're happy, we can turn this into an AC problem.
As before, we need to put a voltage across the current source and vary it. We also need to be able to measure the current.
For this, we need to put a DC voltage source in series with an AC voltage source. The 5mA current will flow through each (and this shouldn't be a problem in practice - there should be no need for DC blocking capacitors, etc - but I'll take a look at your HP 3770 in a moment).
In order to observe the current, we need a current probe for your oscilloscope. But you probably don't have one - I don't. But we can make one, thanks to the magic of Ohm's law - we simply insert a low-value resistor in series with our circuit and monitor the voltage across it. A 1k resistor with a constant 5mA flowing through it will have a constant 5V across it (i.e. 1V per milliamp), and in the context of this particular project, that's probably OK. If you were dealing with lower voltages, perhaps you'd choose a lower value resistance.
Alternatively, if your DMM has a good frequency response at the frequencies of interest, just use that.
So that's all the pieces in place. On paper, draw a dotted line around your current source, partitioning off the internal workings from the test setup. Then draw a series circuit of the DC voltage source, the AC voltage source, and the 1k sense resistor. And once - and only once - you understand all that, then you can think about the practicalities of building it for real. What happens next depends on whether your HP has a floating output, but I want to leave that as an exercise for you to work out.
Across the resistance, there will be a DC voltage with a (hopefully!) small AC component. Switch the 'scope to AC input, and measure that. Let's say it's 0.1V peak to peak, which represents a current change of 0.1mA. Let's assume the AC voltage source was set to 10V peak to peak. That implies an internal resistance of 100k, at that particular frequency (you might like to try a few).
If you did use a DMM to measure the AC current, then remember that was an RMS measurement, not a peak-to-peak. In which case, make sure you've measured the RMS value of the AC voltage source. Or convert I RMS to I peak-to-peak if you've got a peak-to-peak voltage measurement. Compare apples with applies
All of this would be simpler if we had a whiteboard in front of us. Unfortunately, I'm a bit too busy to produce diagrams, but in a way, that's good because it forces you to think it through and that'll make it "stick" better than just following a diagram.
Good luck,
Mark
Forget about how it's actually made for a moment - a current source is a 1 port device, so no "input" and "output" as such (an amplifier would be a 2-port device, with an input and an output).
If we were talking about voltage sources, this would be much more intuitive. But the current source is very similar indeed, and thinking applicable to a voltage source is transferable to the current source. So, perhaps we should briefly recap voltage sources...
We know that an "ideal" voltage source maintains a constant voltage between its terminals, whatever current flows from it. We also know that a "real-world" voltage source isn't quite so ideal - that's because it has some resistance in series with the output, and so as the current increases, the voltage drops at the terminals because of this internal resistance. And it's very easy to calculate using nothing more than Ohm's law.
In short, a "real" voltage source can be modelled with a perfect voltage source and a resistor. Even if the actual circuit is rather more complex than this, it can be represented by this model. You've no-doubt read all about Thévenin, so I'll leave it at that. Just wanted to set the scene - not teach it all from new...
Moving on to current sources, there's no doubt that people find these much less intuitive. I'm not entirely sure why, but over the years I've seen it a lot. You're in good company!
Firstly, an ideal current source just causes a constant current to flow, whatever the voltage across its 2 terminals. Because we spend so much time talking about voltages in circuits, that's actually a harder concept to grasp - as I say, current seems rather more abstract than voltages to many. Probably because it's easier to measure voltage than current.
In the same way a real-world voltage source has series resistance, a real-world current source has some resistance too. But this resistance is not in series (think about it - if it were, would it change the current flowing?). Instead, this unwanted resistance is in parallel with the current source. Ideally, we want this resistance to be as high as possible, so that the minimum of current goes through this unwanted resistance (note that's the opposite to the voltage source, where we want the resistance to be as low as possible).
Again, I'm not trying to teach this anew - just establishing the basics...
So what we're trying to measure here is the value of this parallel resistance. It's not a real resistor, of course - rather, it's an apparent resistance caused by the imperfections of the circuit. It might well vary under certain conditions, so all we can do is arrive at an approximate value, ideally under the circumstances the current source will see under normal operation.
What is the effect of this parallel resistance? It means that the current delivered from the current source will vary with applied voltage. This is getting to the nub of the question.
Let's start with DC, as that's easiest...
Suppose you measure 20V across your current source, and note that 5mA exactly flows. Then, you increase the voltage across the voltage source to 30V, and note that 5.1mA is now flowing. The change in voltage was 10V, and that caused a change in current of 0.1mA. That implies 100k of parallel resistance (and also implies that the true value of the "ideal" current sink in our Norton model would be 4.8mA, as 0.2mA of that was flowing in the 100k resistance at 20V).
That might take a bit of thinking about, but once you're happy, we can turn this into an AC problem.
As before, we need to put a voltage across the current source and vary it. We also need to be able to measure the current.
For this, we need to put a DC voltage source in series with an AC voltage source. The 5mA current will flow through each (and this shouldn't be a problem in practice - there should be no need for DC blocking capacitors, etc - but I'll take a look at your HP 3770 in a moment).
In order to observe the current, we need a current probe for your oscilloscope. But you probably don't have one - I don't. But we can make one, thanks to the magic of Ohm's law - we simply insert a low-value resistor in series with our circuit and monitor the voltage across it. A 1k resistor with a constant 5mA flowing through it will have a constant 5V across it (i.e. 1V per milliamp), and in the context of this particular project, that's probably OK. If you were dealing with lower voltages, perhaps you'd choose a lower value resistance.
Alternatively, if your DMM has a good frequency response at the frequencies of interest, just use that.
So that's all the pieces in place. On paper, draw a dotted line around your current source, partitioning off the internal workings from the test setup. Then draw a series circuit of the DC voltage source, the AC voltage source, and the 1k sense resistor. And once - and only once - you understand all that, then you can think about the practicalities of building it for real. What happens next depends on whether your HP has a floating output, but I want to leave that as an exercise for you to work out.
Across the resistance, there will be a DC voltage with a (hopefully!) small AC component. Switch the 'scope to AC input, and measure that. Let's say it's 0.1V peak to peak, which represents a current change of 0.1mA. Let's assume the AC voltage source was set to 10V peak to peak. That implies an internal resistance of 100k, at that particular frequency (you might like to try a few).
If you did use a DMM to measure the AC current, then remember that was an RMS measurement, not a peak-to-peak. In which case, make sure you've measured the RMS value of the AC voltage source. Or convert I RMS to I peak-to-peak if you've got a peak-to-peak voltage measurement. Compare apples with applies
All of this would be simpler if we had a whiteboard in front of us. Unfortunately, I'm a bit too busy to produce diagrams, but in a way, that's good because it forces you to think it through and that'll make it "stick" better than just following a diagram.
Good luck,
Mark
