Date
1 - 18 of 18
7CT1N - Vertical Traces Truncated?
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n4buq
I've been pondering asking this question for I know this is likely to be a "new to curve tracing" question but here goes.
For reference, pictures here: https://groups.io/g/TekScopes/album?id=252850 For the traces shown in the first two photographs in that album, I have set up the 7CT1N and a timebase in chop mode. This lets me see the signals being sent to the 7A26. I'm testing a 2N3904. In the first image, the VERTICAL AMPERES/DIV is set to 0.2mA and the STEP AMPL is set to 2uA. With those settings, all the curves are shown and the output from the 7CT1N to the 7A26 shows the voltage peaks with smaller, "flatter" tops that I presume indicates the tracer is getting CE current and, thus, the trace extends through the curve and then horizontally for a way. In the second image, the only thing that changed is the STEP AMPL and it's now set to 5uA. In that configuration, all of the peaks of the tracer's output do not have the "flatter" portions and the curves are essentially truncated for those steps. If I understand the way the tracer works when testing junctions transistors, the Step Amplitude is supplying a stepped current to B. If the current is increased in the second image, then I don't understand why the current through the CE junction (which, again as I understand it, is presented as a voltage to the 7A26) is being limited. That doesn't quite make sense to me. It's probably obvious that I'm thinking of this incorrectly but I'm just not seeing where I'm going wrong. Anyone want to tackle an explanation for me? Thanks, Barry - N4BUQ |
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Clark Foley
Barry,
Consider the influence of a series resistance between the voltage source and the collector of the DUT. The more collector current, the lower the collector voltage. As you select a different Amp/div, a different series resistance is also selected. |
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n4buq
Hi Clark,
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I'm afraid I'm still not following. Are you saying that when the STEP AMPL is changed, that also changes the series resistance in the collector current circuit? Thanks, Barry - N4BUQ ----- Original Message -----
From: "Clark Foley" <clarkfoley@...> Barry, |
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Clark Foley
The series resistance is changed when you select the vertical scale factor Amp/div. This is not the step size/STEP AMPL. Refer to the schematic.
Remember that the bipolar transistor is a current-controlled current source. As you step the base current (apply a constant base current for the duration of one voltage cycle), you get a proportional collector current. That is why the vertical deflection is nearly a horizontal line segment i.e. constant current. Current multiplied by the series resistance results in a voltage drop between the voltage source and the collector voltage. The collector voltage, not the source voltage, drives the horizontal deflection. This results in a shorter horizontal deflection for each step in collector current. |
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Clark Foley
Another way to look at it is to note that as a higher constant current is desired, the series resistance requires a higher source voltage to avoid current limit i. e. Insufficient headroom. For a given maximum sweep voltage, a higher constant current will be realized for shorter duration.
When the step generator is driving a very low collector current, the voltage source can supply this at a relatively low voltage and continue for nearly the full cycle with plenty of headroom. This gives a long duration of constant collector current. When the step generator attempts to drive a relatively high constant collector current, the voltage source cannot supply it until it has enough headroom given the same source resistance. This might occur near the peak/maximum if the sweep voltage. This gives a very short duration at constant collector current. |
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n4buq
I'm not changing the AMPERES/DIV switch - just the STEP AMPL. Of course I can get it to truncate the traces by adjusting both but in this instance, I was only increasing the base current step amplitudes.
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From your follow-on email, though, I think I get what you're saying. By increasing the base current, the increase in collector current is causing an increase in the current through the collect supply and, thus, more voltage drop across the collector supply's resistance. I think that makes more sense to me now and I understand it a bit better now. Thanks, Barry - N4BUQ ----- Original Message -----
From: "Clark Foley" <clarkfoley@...> The series resistance is changed when you select the vertical scale factor |
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On Mon, May 23, 2022 at 06:13 PM, n4buq wrote:
If I understand correctly what you're saying, what you're seeing is the effect of the (constant) power line; an imaginary line through all the end points of the curves. The line gets steeper when the series resistor becomes lower. On a curve tracer that allows you to set the "max. power", like a 576 or 577, this is clearly visible. I don't think a 7CT1 allows separately setting the current resistor. If I have the time, I'll post two pictures of an ordinary NPN, both with with 10uA base current steps and 8 steps total, later tonight (it's 22:30h over here). One will have the series R set to 140 Ohm, the other with it set to 650 Ohm. All other settings untouched. Raymond |
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On Mon, May 23, 2022 at 10:34 PM, Raymond Domp Frank wrote:
See https://groups.io/g/TekScopes/album?id=275321 Raymond |
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n4buq
Hi Raymond,
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That does look similar and it makes sense that the horizontal portions are truncated the higher the steps go when there's a larger series resistance. In my case, not only are the horizontal portions truncated but the vertical parts of the trace are truncated as well. If you were to increase the series resistance beyond 650 ohms, would you begin to lose the vertical parts of the traces (the ones for which there would be no collector current)? Thanks, Barry - N4BUQ ----- Original Message -----
From: "Raymond Domp Frank" <hewpatek@...> On Mon, May 23, 2022 at 10:34 PM, Raymond Domp Frank wrote:See |
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On Tue, May 24, 2022 at 02:28 AM, n4buq wrote:
Yes, definitely, that's what will happen. In fact, the top curve @ 650 Ohm almost does that already. A curve above that wouldn't be visible. Extend the imaginary constant power line through where it crosses the line Vce=0. Raymond |
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On Tue, May 24, 2022 at 02:36 AM, Raymond Domp Frank wrote:
I added an image with the same settings, 650 Ohm resistor and Vce reduced. Raymond |
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n4buq
Thank you! I think it makes sense to me now.
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Thanks, Barry - N4BUQ ----- Original Message -----
From: "Raymond Domp Frank" <hewpatek@...> On Tue, May 24, 2022 at 02:36 AM, Raymond Domp Frank wrote:I added an image with the same settings, 650 Ohm resistor and Vce reduced. |
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Harvey White
You're feeding the base with a constant current for any step.
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That means that the collector current is limited (by the transistor) to beta * Ib From 0 voltage, and for any collector resistor (current sense), as you increase voltage the current will rise with a slope of Beta. The higher the beta, the more vertical the beginning portion of the curve will be. The lower the beta, the more slanted it will be (assuming NPN), to the right. Once the collector voltage reaches the point where Ic = Beta * Ib, the collector current cannot increase, and the curve flattens out. That's saturation. The transistor behaves like a constant current source (within limits) so that with increasing collector voltage, the current (vertical) remains constant. It might be interesting to note that the length of all the curves from 0 to end is the same. Harvey On 5/23/2022 8:27 PM, n4buq wrote:
Hi Raymond, |
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On Tue, May 24, 2022 at 03:35 AM, Harvey White wrote:
Hi Harvey, You mean from the origin to the end of each curve? That'd be "interesting" but it isn't so. Have a look at the 140 Ohm curves in my images. Raymond |
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Harvey White
I think it is true when the collector voltage sweep is high enough. But looking at the curves in reality, I suspect that the vertical part is modified by the beta.
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I do think that they take the same physical time, though. Harvey On 5/23/2022 9:43 PM, Raymond Domp Frank wrote:
On Tue, May 24, 2022 at 03:35 AM, Harvey White wrote:It might be interesting to note that the length of all the curves from 0 toHi Harvey, |
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n4buq
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----- Original Message -----
From: "Harvey White" <madyn@...> You're feeding the base with a constant current for any step.I thought the step circuit produced a stepped current for transistor mode and stepped voltage for FET mode. Do you mean that for any one step in transistor mode, the current is constant for the duration of that step but not all steps are the same current? If not, then I'm misunderstanding the step circuit. Thanks, Barry - N4BUQ |
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Harvey White
On this, I always thought that the idea was to feed a constant base current, sweep the collector voltage, then feed another step of base current.
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this is on a "per step" basis. The steps for base current should be a constant increment. Ditto with the FET voltages for the gate. Sorry for the confusion. Harvey On 5/23/2022 10:50 PM, n4buq wrote:
----- Original Message -----From: "Harvey White" <madyn@...>I thought the step circuit produced a stepped current for transistor mode and stepped voltage for FET mode. Do you mean that for any one step in transistor mode, the current is constant for the duration of that step but not all steps are the same current? If not, then I'm misunderstanding the step circuit. |
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n4buq
Yes, that's my understanding as well. I think I simply misconstrued your reply.
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Thanks, Barry - N4BUQ ----- Original Message -----
From: "Harvey White" <madyn@...> On this, I always thought that the idea was to feed a constant base |
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