Timeline for Amplify voltage between transistors in a current mirror setup?
Current License: CC BY-SA 4.0
21 events
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| Oct 21, 2019 at 18:15 | comment | added | remcycles | I don't see the impedance measurements you mention, but after some quick LTSpice simulations (with a diode connected NMOS in parallel with 300kOhm like in my example schematic), I'd agree with @Hearth that an instrumentation amp would be a good idea. It really depends on the actual device characteristics and your accuracy requirements though. | |
| Oct 21, 2019 at 18:02 | comment | added | remcycles | There's a joke that "all sensors are temperature sensors, but some are better than others". | |
| Oct 21, 2019 at 17:07 | comment | added | remcycles | That assumes that the radiation response is independent of absolute temperature, and I don't know if that's a safe assumption. | |
| Oct 19, 2019 at 7:35 | comment | added | Leo | remicles2, Interesting point the one about temperature, I would need to do some tests. One of the reasons for getting the voltage difference between 2 transistors is that if I expose one to radiation, and one not, then their difference should just be due to radiation, eliminating Tdiff, etc. | |
| Oct 19, 2019 at 7:33 | comment | added | Leo | @Hearth I've added the impedence measurement of the TFT, would this still work with the opamp setup ? | |
| Oct 18, 2019 at 20:15 | comment | added | remcycles | Certainly, but I don't really know what the input impedance requirements are, so I didn't mention it. @leoelectrics Look into instrumentation amplifiers if you find you need a high performance differential amplifier. en.wikipedia.org/wiki/Instrumentation_amplifier | |
| Oct 18, 2019 at 19:52 | comment | added | Hearth | An in-amp might be a better idea than a plain op-amp here, since that way you can avoid loading the circuit being measured. | |
| Oct 18, 2019 at 19:18 | vote | accept | Leo | ||
| Oct 18, 2019 at 19:18 | comment | added | Leo | Ok, thanks, by my understanding this setup works! | |
| Oct 18, 2019 at 19:07 | history | edited | remcycles | CC BY-SA 4.0 |
Add information about related PTAT temperature sensors
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| Oct 18, 2019 at 17:34 | comment | added | remcycles | I guess it's a habit from my circuits textbook, which has lots of diagrams like the one in this post that include an ideal current source in series with a voltage source: electronics.stackexchange.com/questions/377804/… | |
| Oct 18, 2019 at 17:29 | comment | added | remcycles | Good catch. You are right, it is redundant. You can replace the ideal voltage source with a wire to get the same effect. I put it in there just because most real current sources need a source of power, but of course, an ideal current source is its own source of power. | |
| Oct 17, 2019 at 19:33 | comment | added | Leo | ok thanks, last thing I don't understand, why did you connect and ideal voltage source before the 2 ideal current sources, what is the use of it? Can't the with only the ideal current source | |
| Oct 17, 2019 at 17:37 | comment | added | remcycles | @leoelectrics I don't have any experience using CircuitLab for simulation, but you should be able to change transistor properties and see a difference between VDS1 and VDS2. | |
| Oct 17, 2019 at 17:36 | comment | added | remcycles | @leoelectrics In general, schematics are easier to understand if you draw them with higher voltages on top, and lower voltages at the bottom of the drawing, so that current flows from top to bottom (towards "ground" to use that metaphor). It's also good to put inputs to the circuit on the left, and draw the schematics so outputs are on the right. Following this advice, your NMOSFET is upside down, as is the ground symbol on it's gate. You can also flip the MOSFET so the gate is on the left. Doing that will make it easier to connect the voltmeter symbol correctly. | |
| Oct 17, 2019 at 17:30 | comment | added | remcycles | @leoelectrics The voltmeter is connected to the correct terminals of the MOSFET, but the polarity is backwards. You want to measure between the Source and the Gate to get \$V_{GS}\$ which is equal to \$V_{DS}\$ in this circuit. In your updated schematic, the negative lead of the meter is connected to the Gate, but should be connected to the Source. That's not a huge problem, but you should know that a positive \$V_{GS}\$ will look like a negative voltage on that meter as it is currently connected. | |
| Oct 17, 2019 at 6:09 | comment | added | Leo | @remicles, following your advice, I have updated circuit1, I am not sure I am connecting the voltmeter correctly, does it have to be connected between VGD or VGS and VGD just shorted? | |
| Oct 17, 2019 at 5:21 | comment | added | Leo | @ElliotAlderson As I didn't setup the experiment, I probably remember it wrong, my bad. I think remicles2's solution with the drain and Gate shorted and the source set to ground is probably what they did. Can this circuit be simulated now with 2 different transistors to see that it actually is reading the VDS1-VDS2? | |
| Oct 16, 2019 at 23:10 | comment | added | Elliot Alderson | What makes you think that the OP intended to connect the transistor gates to their drains? In both of the OP's schematics the gates are not connected to the drain, or to any source of current. | |
| Oct 16, 2019 at 23:02 | history | edited | remcycles | CC BY-SA 4.0 |
added 65 characters in body, s/source/drain/
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| Oct 16, 2019 at 22:55 | history | answered | remcycles | CC BY-SA 4.0 |