Great way of demonstrating the function of the base by put a switch to the collector and an to the base.i have the first edition book ,that idea isn't in it.thanks.
I have the second edition of the book, and most of those circuits are not in my book, either. I wonder if all those circuits are indeed in Laurence's third edition, or if Laurence is showing us his own experimental circuits in addition to those in the book?
I can assure you that I'm building just the circuits from the third edition of the book (I can't speak for the first or second editions), however, I'm often having to deviate from the breadboard layouts when I have components that are of a different size from those shown. I promise if I do ever show something not in the book, I'll make that very clear.
@@M7GNP , thanks for confirming that the experiments are indeed in the book, even if you need to tweak them. I find it impressive that the third edition of the book has more experiments, and better experiments, than my second edition. It's a credit to the author that the revisions are not just errata, but are significant expansions.
You suggested that a "problem" with a relay is that it's electromechanical. Rather than being a problem, I believe that's actually the main "benefit" of a relay, because it introduces electrical isolation between the control side (the coil) and the load side (the contacts). This electrical isolation means the control side can be a low-power DC circuit, while the load side can be a high-power DC or AC circuit. Anyway, just passing along a clarification which ChatGPT provided to me.
@@SomeNewKid Yes, 'problem' was not a good choice of word - 'limitations' would have been better. My main point was to make a rather clumsy segue into transistors by pointing out that because they are purely electronic there are no mechanical parts to go wrong and they can achieve much better switching speeds than an electromechanical relay. However you are quite right to point out that where there is a need to electrically isolate the switching circuit from the circuit being switched that relays are ideal for that application.
I need your help with the following issue. I am working on experiment #10-17 from the book, as demonstrated at 14:10 in your video. I made a mistake while assembling the circuit: I connected the capacitor to the positive rail on the breadboard instead of the negative rail. This led to the following observations: 1. The circuit worked, but the LED lights up immediately without any delay (see pic1 and pic2). 2. When I connect a multimeter, it shows -0.44V when the circuit is open and 8.04V when the circuit is closed. I don’t understand why there is a voltage reading when the slide switch is in the off position (see pic3 and pic4). Although I’ve identified my mistake, I would like to understand why the multimeter shows 0.44V when the switch is off. Pictures uploaded as a video in my profile section. ua-cam.com/video/jwAS02GsW_k/v-deo.html Thank you!
@@johnminty8243 if both sides of the capacitor were connected to the positive bus then that would be why you didn't get a delay as the potential difference across the capacitor would have been non-existent or very small so there would be almost no charging delay before sufficient current flowed to the base to turn the transistor on. The small negative voltage you are getting when the switch is open could just be the discharge current from the capacitor. It's a little difficult to tell from your photos as it looks like your meter leads are just connected across the buses.
Great video,i really enjoyed it.
Sam.
Great way of demonstrating the function of the base by put a switch to the collector and an to the base.i have the first edition book ,that idea isn't in it.thanks.
I have the second edition of the book, and most of those circuits are not in my book, either. I wonder if all those circuits are indeed in Laurence's third edition, or if Laurence is showing us his own experimental circuits in addition to those in the book?
I can assure you that I'm building just the circuits from the third edition of the book (I can't speak for the first or second editions), however, I'm often having to deviate from the breadboard layouts when I have components that are of a different size from those shown. I promise if I do ever show something not in the book, I'll make that very clear.
@@M7GNP , thanks for confirming that the experiments are indeed in the book, even if you need to tweak them. I find it impressive that the third edition of the book has more experiments, and better experiments, than my second edition. It's a credit to the author that the revisions are not just errata, but are significant expansions.
You suggested that a "problem" with a relay is that it's electromechanical. Rather than being a problem, I believe that's actually the main "benefit" of a relay, because it introduces electrical isolation between the control side (the coil) and the load side (the contacts). This electrical isolation means the control side can be a low-power DC circuit, while the load side can be a high-power DC or AC circuit. Anyway, just passing along a clarification which ChatGPT provided to me.
@@SomeNewKid Yes, 'problem' was not a good choice of word - 'limitations' would have been better. My main point was to make a rather clumsy segue into transistors by pointing out that because they are purely electronic there are no mechanical parts to go wrong and they can achieve much better switching speeds than an electromechanical relay. However you are quite right to point out that where there is a need to electrically isolate the switching circuit from the circuit being switched that relays are ideal for that application.
I need your help with the following issue. I am working on experiment #10-17 from the book, as demonstrated at 14:10 in your video. I made a mistake while assembling the circuit: I connected the capacitor to the positive rail on the breadboard instead of the negative rail. This led to the following observations:
1. The circuit worked, but the LED lights up immediately without any delay (see pic1 and pic2).
2. When I connect a multimeter, it shows -0.44V when the circuit is open and 8.04V when the circuit is closed. I don’t understand why there is a voltage reading when the slide switch is in the off position (see pic3 and pic4).
Although I’ve identified my mistake, I would like to understand why the multimeter shows 0.44V when the switch is off.
Pictures uploaded as a video in my profile section.
ua-cam.com/video/jwAS02GsW_k/v-deo.html
Thank you!
@@johnminty8243 if both sides of the capacitor were connected to the positive bus then that would be why you didn't get a delay as the potential difference across the capacitor would have been non-existent or very small so there would be almost no charging delay before sufficient current flowed to the base to turn the transistor on. The small negative voltage you are getting when the switch is open could just be the discharge current from the capacitor. It's a little difficult to tell from your photos as it looks like your meter leads are just connected across the buses.
@@M7GNPThanks! I will create a video today and post it tonight, along with a few questions.