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Figuratively speaking, Olin Lathrop's answer is a wonderful story about the unequal "tug of war" between a "pulling up" resistor and "pulling down" switch. And, when the switch is 'off', the "strug...
Answer
#8: Post edited
by
Circuit fantasist
·
2020-10-18T19:06:02Z (about 4 years ago)
Minor edit
- Figuratively speaking, Olin Lathrop's answer is a wonderful story about the unequal "tug of war" between a "pulling up" resistor and "pulling down" switch. And, when the switch is 'off', the "struggle" is between the resistor and input leakages… and then the resistor must win.
- To be more precise, the positive terminal of the power supply is what "pulls up" the common node through the resistor and the negative terminal "pulls down" it (either through the switch or leakages).
- This arrangement is the same "voltage divider configuration", widely used in simple common-source logic gates with resistor load… and the considerations are the same.
In my teaching practice, I have observed that students find it difficult to understand that the voltage after the resistor (open circuit) should be almost equal to the supply voltage. When I ask how much this voltage is, I receiv all sorts of answers - 0, 1/2 power supply and, most rarely, the power supply. That is why, tomorrow morning I will start the lab on Semiconductor devices dedicated to diodes with my favorite experiment - I will make my students connect a resistor in series to an "ideal" voltage source and then measure the voltage after the resistor with both an "ideal" and real voltmeter.- Then I will suggest to them to close the circuit by another resistor (thus "inventing" the voltage divider)... and finally, by a diode. Then I will ask them why we still need the resistor... and to see why, we may short the resistor:) Fortunately, the power supply is 12V/1A (protected) and the diode is 1N4007.
- Figuratively speaking, Olin Lathrop's answer is a wonderful story about the unequal "tug of war" between a "pulling up" resistor and "pulling down" switch. And, when the switch is 'off', the "struggle" is between the resistor and input leakages… and then the resistor must win.
- To be more precise, the positive terminal of the power supply is what "pulls up" the common node through the resistor and the negative terminal "pulls down" it (either through the switch or leakages).
- This arrangement is the same "voltage divider configuration", widely used in simple common-source logic gates with resistor load… and the considerations are the same.
- In my teaching practice, I have observed that students find it difficult to understand that the voltage after the resistor (open circuit) should be almost equal to the supply voltage. When I ask how much this voltage is, I receive all sorts of answers - 0, 1/2 power supply and, most rarely, the power supply. That is why, tomorrow morning I will start the lab on Semiconductor devices dedicated to diodes with my favorite experiment - I will make my students connect a resistor in series to an "ideal" voltage source and then measure the voltage after the resistor with both an "ideal" and real voltmeter.
- Then I will suggest to them to close the circuit by another resistor (thus "inventing" the voltage divider)... and finally, by a diode. Then I will ask them why we still need the resistor... and to see why, we may short the resistor:) Fortunately, the power supply is 12V/1A (protected) and the diode is 1N4007.
#7: Post edited
by
Circuit fantasist
·
2020-10-18T19:05:23Z (about 4 years ago)
Minor edit
- Figuratively speaking, Olin Lathrop's answer is a wonderful story about the unequal "tug of war" between a "pulling up" resistor and "pulling down" switch. And, when the switch is 'off', the "struggle" is between the resistor and input leakages… and then the resistor must win.
- To be more precise, the positive terminal of the power supply is what "pulls up" the common node through the resistor and the negative terminal "pulls down" it (either through the switch or leakages).
- This arrangement is the same "voltage divider configuration", widely used in simple common-source logic gates with resistor load… and the considerations are the same.
In my teaching practice, I have observed that students find it difficult to understand that the voltage after the resistor (open circuit) should be almost equal to the supply voltage. When I asked how much this voltage is, I received all sorts of answers - 0, 1/2 power supply and, most rarely, the power supply. That is why, tomorrow morning I will start the lab on Semiconductor devices dedicated to diodes with my favorite experiment - I will make my students connect a resistor in series to an "ideal" voltage source and then measure the voltage after the resistor with both an "ideal" and real voltmeter.- Then I will suggest to them to close the circuit by another resistor (thus "inventing" the voltage divider)... and finally, by a diode. Then I will ask them why we still need the resistor... and to see why, we may short the resistor:) Fortunately, the power supply is 12V/1A (protected) and the diode is 1N4007.
- Figuratively speaking, Olin Lathrop's answer is a wonderful story about the unequal "tug of war" between a "pulling up" resistor and "pulling down" switch. And, when the switch is 'off', the "struggle" is between the resistor and input leakages… and then the resistor must win.
- To be more precise, the positive terminal of the power supply is what "pulls up" the common node through the resistor and the negative terminal "pulls down" it (either through the switch or leakages).
- This arrangement is the same "voltage divider configuration", widely used in simple common-source logic gates with resistor load… and the considerations are the same.
- In my teaching practice, I have observed that students find it difficult to understand that the voltage after the resistor (open circuit) should be almost equal to the supply voltage. When I ask how much this voltage is, I receiv all sorts of answers - 0, 1/2 power supply and, most rarely, the power supply. That is why, tomorrow morning I will start the lab on Semiconductor devices dedicated to diodes with my favorite experiment - I will make my students connect a resistor in series to an "ideal" voltage source and then measure the voltage after the resistor with both an "ideal" and real voltmeter.
- Then I will suggest to them to close the circuit by another resistor (thus "inventing" the voltage divider)... and finally, by a diode. Then I will ask them why we still need the resistor... and to see why, we may short the resistor:) Fortunately, the power supply is 12V/1A (protected) and the diode is 1N4007.
#6: Post edited
by
Circuit fantasist
·
2020-10-18T19:03:04Z (about 4 years ago)
About the setup
- Figuratively speaking, Olin Lathrop's answer is a wonderful story about the unequal "tug of war" between a "pulling up" resistor and "pulling down" switch. And, when the switch is 'off', the "struggle" is between the resistor and input leakages… and then the resistor must win.
- To be more precise, the positive terminal of the power supply is what "pulls up" the common node through the resistor and the negative terminal "pulls down" it (either through the switch or leakages).
- This arrangement is the same "voltage divider configuration", widely used in simple common-source logic gates with resistor load… and the considerations are the same.
- In my teaching practice, I have observed that students find it difficult to understand that the voltage after the resistor (open circuit) should be almost equal to the supply voltage. When I asked how much this voltage is, I received all sorts of answers - 0, 1/2 power supply and, most rarely, the power supply. That is why, tomorrow morning I will start the lab on Semiconductor devices dedicated to diodes with my favorite experiment - I will make my students connect a resistor in series to an "ideal" voltage source and then measure the voltage after the resistor with both an "ideal" and real voltmeter.
Then I will suggest to them to close the circuit by another resistor (thus "inventing" the voltage divider)... and finally, by a diode. Then I will ask them why we still need the resistor... and to see why, we may short the resistor:)
- Figuratively speaking, Olin Lathrop's answer is a wonderful story about the unequal "tug of war" between a "pulling up" resistor and "pulling down" switch. And, when the switch is 'off', the "struggle" is between the resistor and input leakages… and then the resistor must win.
- To be more precise, the positive terminal of the power supply is what "pulls up" the common node through the resistor and the negative terminal "pulls down" it (either through the switch or leakages).
- This arrangement is the same "voltage divider configuration", widely used in simple common-source logic gates with resistor load… and the considerations are the same.
- In my teaching practice, I have observed that students find it difficult to understand that the voltage after the resistor (open circuit) should be almost equal to the supply voltage. When I asked how much this voltage is, I received all sorts of answers - 0, 1/2 power supply and, most rarely, the power supply. That is why, tomorrow morning I will start the lab on Semiconductor devices dedicated to diodes with my favorite experiment - I will make my students connect a resistor in series to an "ideal" voltage source and then measure the voltage after the resistor with both an "ideal" and real voltmeter.
- Then I will suggest to them to close the circuit by another resistor (thus "inventing" the voltage divider)... and finally, by a diode. Then I will ask them why we still need the resistor... and to see why, we may short the resistor:) Fortunately, the power supply is 12V/1A (protected) and the diode is 1N4007.
#5: Post edited
by
Circuit fantasist
·
2020-10-18T18:58:41Z (about 4 years ago)
Minor edit
- Figuratively speaking, Olin Lathrop's answer is a wonderful story about the unequal "tug of war" between a "pulling up" resistor and "pulling down" switch. And, when the switch is 'off', the "struggle" is between the resistor and input leakages… and then the resistor must win.
- To be more precise, the positive terminal of the power supply is what "pulls up" the common node through the resistor and the negative terminal "pulls down" it (either through the switch or leakages).
- This arrangement is the same "voltage divider configuration", widely used in simple common-source logic gates with resistor load… and the considerations are the same.
In my teaching practice, I have observed that students find it difficult to understand that the voltage after the resistor (open circuit) should be almost equal to the supply voltage. When I asked how much this voltage is, I received all sorts of answers - 0, 1/2 the power supply and, most rarely, the power supply. That is why, tomorrow morning I will start the lab on Semiconductor devices dedicated to diodes with my favorite experiment - I will make my students connect a resistor in series to an "ideal" voltage source and then measure the voltage after the resistor with both an "ideal" and real voltmeter.- Then I will suggest to them to close the circuit by another resistor (thus "inventing" the voltage divider)... and finally, by a diode. Then I will ask them why we still need the resistor... and to see why, we may short the resistor:)
- Figuratively speaking, Olin Lathrop's answer is a wonderful story about the unequal "tug of war" between a "pulling up" resistor and "pulling down" switch. And, when the switch is 'off', the "struggle" is between the resistor and input leakages… and then the resistor must win.
- To be more precise, the positive terminal of the power supply is what "pulls up" the common node through the resistor and the negative terminal "pulls down" it (either through the switch or leakages).
- This arrangement is the same "voltage divider configuration", widely used in simple common-source logic gates with resistor load… and the considerations are the same.
- In my teaching practice, I have observed that students find it difficult to understand that the voltage after the resistor (open circuit) should be almost equal to the supply voltage. When I asked how much this voltage is, I received all sorts of answers - 0, 1/2 power supply and, most rarely, the power supply. That is why, tomorrow morning I will start the lab on Semiconductor devices dedicated to diodes with my favorite experiment - I will make my students connect a resistor in series to an "ideal" voltage source and then measure the voltage after the resistor with both an "ideal" and real voltmeter.
- Then I will suggest to them to close the circuit by another resistor (thus "inventing" the voltage divider)... and finally, by a diode. Then I will ask them why we still need the resistor... and to see why, we may short the resistor:)
#4: Post edited
by
Circuit fantasist
·
2020-10-18T18:58:00Z (about 4 years ago)
More about the voltage after R
- Figuratively speaking, Olin Lathrop's answer is a wonderful story about the unequal "tug of war" between a "pulling up" resistor and "pulling down" switch. And, when the switch is 'off', the "struggle" is between the resistor and input leakages… and then the resistor must win.
- To be more precise, the positive terminal of the power supply is what "pulls up" the common node through the resistor and the negative terminal "pulls down" it (either through the switch or leakages).
- This arrangement is the same "voltage divider configuration", widely used in simple common-source logic gates with resistor load… and the considerations are the same.
In my teaching practice, I have observed that students find it difficult to understand that the voltage after the resistor (open circuit) should be almost equal to the supply voltage. That is why, tomorrow morning I will start the lab on Semiconductor devices dedicated to diodes with my favorite experiment - I will make my students connect a resistor in series to an "ideal" voltage source and then measure the voltage after the resistor with both an "ideal" and real voltmeter.- Then I will suggest to them to close the circuit by another resistor (thus "inventing" the voltage divider)... and finally, by a diode. Then I will ask them why we still need the resistor... and to see why, we may short the resistor:)
- Figuratively speaking, Olin Lathrop's answer is a wonderful story about the unequal "tug of war" between a "pulling up" resistor and "pulling down" switch. And, when the switch is 'off', the "struggle" is between the resistor and input leakages… and then the resistor must win.
- To be more precise, the positive terminal of the power supply is what "pulls up" the common node through the resistor and the negative terminal "pulls down" it (either through the switch or leakages).
- This arrangement is the same "voltage divider configuration", widely used in simple common-source logic gates with resistor load… and the considerations are the same.
- In my teaching practice, I have observed that students find it difficult to understand that the voltage after the resistor (open circuit) should be almost equal to the supply voltage. When I asked how much this voltage is, I received all sorts of answers - 0, 1/2 the power supply and, most rarely, the power supply. That is why, tomorrow morning I will start the lab on Semiconductor devices dedicated to diodes with my favorite experiment - I will make my students connect a resistor in series to an "ideal" voltage source and then measure the voltage after the resistor with both an "ideal" and real voltmeter.
- Then I will suggest to them to close the circuit by another resistor (thus "inventing" the voltage divider)... and finally, by a diode. Then I will ask them why we still need the resistor... and to see why, we may short the resistor:)
#3: Post edited
by
Circuit fantasist
·
2020-10-18T18:43:14Z (about 4 years ago)
Minor edit
Figuratively speaking, Olin Lantrop's answer is a wonderful story about the unequal "tug of war" between a "pulling up" resistor and "pulling down" switch. And, when the switch is 'off', the "struggle" is between the resistor and input leakages… and then the resistor must win.- To be more precise, the positive terminal of the power supply is what "pulls up" the common node through the resistor and the negative terminal "pulls down" it (either through the switch or leakages).
- This arrangement is the same "voltage divider configuration", widely used in simple common-source logic gates with resistor load… and the considerations are the same.
- In my teaching practice, I have observed that students find it difficult to understand that the voltage after the resistor (open circuit) should be almost equal to the supply voltage. That is why, tomorrow morning I will start the lab on Semiconductor devices dedicated to diodes with my favorite experiment - I will make my students connect a resistor in series to an "ideal" voltage source and then measure the voltage after the resistor with both an "ideal" and real voltmeter.
- Then I will suggest to them to close the circuit by another resistor (thus "inventing" the voltage divider)... and finally, by a diode. Then I will ask them why we still need the resistor... and to see why, we may short the resistor:)
- Figuratively speaking, Olin Lathrop's answer is a wonderful story about the unequal "tug of war" between a "pulling up" resistor and "pulling down" switch. And, when the switch is 'off', the "struggle" is between the resistor and input leakages… and then the resistor must win.
- To be more precise, the positive terminal of the power supply is what "pulls up" the common node through the resistor and the negative terminal "pulls down" it (either through the switch or leakages).
- This arrangement is the same "voltage divider configuration", widely used in simple common-source logic gates with resistor load… and the considerations are the same.
- In my teaching practice, I have observed that students find it difficult to understand that the voltage after the resistor (open circuit) should be almost equal to the supply voltage. That is why, tomorrow morning I will start the lab on Semiconductor devices dedicated to diodes with my favorite experiment - I will make my students connect a resistor in series to an "ideal" voltage source and then measure the voltage after the resistor with both an "ideal" and real voltmeter.
- Then I will suggest to them to close the circuit by another resistor (thus "inventing" the voltage divider)... and finally, by a diode. Then I will ask them why we still need the resistor... and to see why, we may short the resistor:)
#2: Post edited
by
Circuit fantasist
·
2020-10-18T18:26:43Z (about 4 years ago)
Minor edit
- Figuratively speaking, Olin Lantrop's answer is a wonderful story about the unequal "tug of war" between a "pulling up" resistor and "pulling down" switch. And, when the switch is 'off', the "struggle" is between the resistor and input leakages… and then the resistor must win.
- To be more precise, the positive terminal of the power supply is what "pulls up" the common node through the resistor and the negative terminal "pulls down" it (either through the switch or leakages).
- This arrangement is the same "voltage divider configuration", widely used in simple common-source logic gates with resistor load… and the considerations are the same.
In my teaching practice, I have observed that students find it difficult to understand that the voltage after the resistor (open circuit) should be almost equal to the supply voltage. That is why, tomorrow morning I will start the lab on Semiconductor devices dedicated to diodes with a favorite experiment - I will make my students connect a resistor in series to an "ideal" voltage source and then measure the voltage after the resistor with both an "ideal" and real voltmeter.- Then I will suggest to them to close the circuit by another resistor (thus "inventing" the voltage divider)... and finally, by a diode. Then I will ask them why we still need the resistor... and to see why, we may short the resistor:)
- Figuratively speaking, Olin Lantrop's answer is a wonderful story about the unequal "tug of war" between a "pulling up" resistor and "pulling down" switch. And, when the switch is 'off', the "struggle" is between the resistor and input leakages… and then the resistor must win.
- To be more precise, the positive terminal of the power supply is what "pulls up" the common node through the resistor and the negative terminal "pulls down" it (either through the switch or leakages).
- This arrangement is the same "voltage divider configuration", widely used in simple common-source logic gates with resistor load… and the considerations are the same.
- In my teaching practice, I have observed that students find it difficult to understand that the voltage after the resistor (open circuit) should be almost equal to the supply voltage. That is why, tomorrow morning I will start the lab on Semiconductor devices dedicated to diodes with my favorite experiment - I will make my students connect a resistor in series to an "ideal" voltage source and then measure the voltage after the resistor with both an "ideal" and real voltmeter.
- Then I will suggest to them to close the circuit by another resistor (thus "inventing" the voltage divider)... and finally, by a diode. Then I will ask them why we still need the resistor... and to see why, we may short the resistor:)
#1: Initial revision
by
Circuit fantasist
·
2020-10-18T18:22:04Z (about 4 years ago)
Figuratively speaking, Olin Lantrop's answer is a wonderful story about the unequal "tug of war" between a "pulling up" resistor and "pulling down" switch. And, when the switch is 'off', the "struggle" is between the resistor and input leakages… and then the resistor must win. To be more precise, the positive terminal of the power supply is what "pulls up" the common node through the resistor and the negative terminal "pulls down" it (either through the switch or leakages). This arrangement is the same "voltage divider configuration", widely used in simple common-source logic gates with resistor load… and the considerations are the same. In my teaching practice, I have observed that students find it difficult to understand that the voltage after the resistor (open circuit) should be almost equal to the supply voltage. That is why, tomorrow morning I will start the lab on Semiconductor devices dedicated to diodes with a favorite experiment - I will make my students connect a resistor in series to an "ideal" voltage source and then measure the voltage after the resistor with both an "ideal" and real voltmeter. Then I will suggest to them to close the circuit by another resistor (thus "inventing" the voltage divider)... and finally, by a diode. Then I will ask them why we still need the resistor... and to see why, we may short the resistor:)