Nice test, Happy New Year! 1) The huge resistance in your ladder matches the input impedance of the meter (parallel with the resistor you are measuring) dropping the impedance to 1/2 (5M). The ladder is essentially 15M so the drop across 5M would be 1/3 of 30V = 10V. 2) Disconnecting the power to an OPAMP will stop all amplification and inversion functions. This will leave the two passive resistors in the circuit. The output will be essentially identical to the input signal. 3) You mentioned this was an audio control, so it is safe to assume this isn't a linear potentiometer (log). The log of 50% would be right about 3VDC. 4) There may be several components that could react in this manner. I feel a low-voltage unidirectional TVS ESD diode could be used. As long as this voltage exceeds the "Diode Check" voltage your multimeter outputs.
I downloaded the audio file using mediahuman and detected a complex waveform, looks like compound sine waves. After doing FFT of the sound at 7:49 I see peaks at 2400 Hz, 4200 Hz, 8300 Hz, and 13100 Hz, the highest at 4200. So I suspect the mystery component is exciting the circuit around 4.2 kHz, perhaps some sort of active oscillator or passive piezo speaker (both wouldn't measure on a multimeter). The remainder of the peak frequencies probably have something to do with the frequency response of the buzzer (2400 Hz). If you look at the datasheet for a typical passive speaker, the frequency curve looks like a roller coaster ride, which would explain why it looks like it contains many harmonics. I ran the experiment using an active speaker and the passive one created a waveform with decreasing amplitude spikes at 4, 8, 12, and 16 kHz, nothing at 2 kHz, but I have different speakers than Richard. I think the 10uF capacitor is just there to maintain a constant DC level, the RC circuit with a 1K resistor is only 16 Hz. Removing it decreased the sound level and slightly altered the frequency in my setup.
Hi Richard and Happy New Year. Thanks for another year of entertaining and informative videos and for an amusing test to distract me from the miserable weather here in the UK! Regarding your test questions I believe the answers to be: 1 - Probably about 10v as the input impedance of the fluke (about 10M) will be in parallel with the lower of the two 10M resistors resulting in an effective resistance of about 5M so the junction of the two resistors will be at about 1/3 of supply. 2 - With power off then the I/P signal will be apparent, non-inverted, at the O/P pin as a result of the i/p and feedback resistors in series. 3 - Potentiometer voltage at slider pin will depend upon the potentiometer law or taper and if not linear then which way around it is connected. In the video we can see the pot is marked as "type A", in Asia and the US that indicates audio or logarithmic, but in Europe it can mean linear! If it were linear then the voltage would be close to half the supply or 5v regardless of which way the pot is connected. However, if as I suspect it's an Asian "type A" then it will be logarithmic and the mid position would be approximately 10% resistance from one end and consequently 90% from the other, in which case depending upon which way around you connected the pot the voltage will be either 1v or 9v. 4 - Trickier one this but given the behavior and the multimeter readings it must be some kind of directional and active component. I suspect the mystery component is an SCR with a floating gate allowing it to trigger on noise, discharge the capacitor through the speaker until the current drops sufficiently for the SCR to turn off allowing the capacitor to recharge via the resistor until another noise spike re-triggers the SCR again.
The potentiometer terminals are marked on bottom and polarized location 11’ 22’ 33’. The spec sheets show the voltage vs position for the 1-2 terminal pair when voltage is applied to 1-3. This should help determine which voltage to choose for the slider.
Q1: 10V The resistive divider with two equal resistors divides the voltage by two and the voltage should be 15V. But as your DVM has a similar input resistance of 10 megohm, the resistance of the part were the DVM is connected becomes only 5 megohm. Hence the voltage is divided 1:2, 10V on the measured part and 20V on the other. Q2: the same as the input signal Without a supply voltage the LM358 shoult no longer influence the circuit behaviour. What remains are two resistors which connect the input to the output. 1K+3K will not significantly influence the signal as the scope input resistance is at least 1 megohm. Q3: 1V or 9V dependent on the direction of the potentiometer Given the use as a volume control in audio circuits and the letter A in the part number I assume this is a logarithmic potentiometer. Therefore in the center position the resitance is divided 1:9 approximately. Depending on which side you measure the voltage it is around 1V or 9V. Q4: Thyristor This is probably not the right solution as a thyristor will usually not measure open circuit in both directions. However, the circuit can work with a thyristor. You have to connect gate and anode to the positive end of the capacitor and the cathode to the speaker. This will give you a relaxation oscillator. The thyristor triggers and discharges the capacitor until the current drops below the holding current. Than the capacitor charges again until the trigger voltage is reached. If you flip the connections the thyristor will not trigger as the gate is always at zero volts via the speaker. I tested the curcuit with several thyristors and not all worked as holding currents and trigger voltages differ and may not fit this circuit. With some fine tuning of the resistor you get more thyristors to work.
G,day Sir and Happy New Year. Quiz questions. I submit the following answers and MY reasonings. Q1. Two resistors (value 10.6 mega OMS each in series. 10 volts of DC current would s 0:01 till flow through the two. (I'm getting a diode and the resistor confused, so my guess is 10 volts). Q2. The measurement at the op amp with nil current (and no stored energy e.g caps) is 0 voltage. (Reasoning- there's no source of other DC supply). Q3.Potentiometer slide. Is only to measure or monitor displacement of volts. Q4. Bread board with speaker circuit. The mistry component is a relay or gate. 🧲🤨💡
#1: half of supply, #2: input waveform if instrument doesn't load circuit. #3: if linear pot half of supply, if log pot 30% of supply. #4: bimetallic flasher
@@a1fliema1fie well this electrolytic capacitor and it doesn't block DC in some cases - trust me. And cannot be diode because it tests as OL in both directions
@@a1fliema1fie before a capacitor 'blocks dc' it will have to charge... but one capacitor will charge faster than the other because one of them is in series with the buzzer (ie load) causing a kind of perpertual imbalance where the capacitors charge and discharge sending singals through the buzzer
Yoloooo 1) not half the input volts , because of the input impedence of the multimeter the answer will vary if the multumeter is 10Mohm or 1Mohm input impedence 2) i've no idea i think i have to watch the all you have to know series of op amp hahah , but i guess output will be same 3) around half the input so around 5v 4) DIAC ? ... no reading on multimeter on diode mode because DMM is testing at around 2v and the diac is more than that , zener would show up at diode mode , the capacitor charges reaches the opening voltage of diac and discharges through the speaker diac closes capacitor charges so and so.... Hope i answered 1 sensei hahah Kudos all the way from Africa, Ethiopia
Dear Richard, Congratulations for your hard work, you never stop, it is marvellous. I do not know if I will succeed the test but I wish you a Happy New Year and keep on with these most interesting videos. Marc from Belgium
Thanks for all your interesting videos, hope you keep doing that in 2024. Wish you a great 2024. 1: less than 15V, probably close to 10 Volt as the multimeter has also an input resistents of 10Mohm. 2: output will be the same as input, between input and output are only the 2 resistors, as the input impedance of the scope is high there will be almost no loss . 3: if it is a linear slider the output will close to 5V 4: maybe a crystal?
1.)10 V 2.) 2,5Vss or 1,77Veff,,,,,, Ve=Vo 3.) 2V if the poti log otherweise 4,9936V lin 4.) One zener Diode and one normal diode both anodes together. And... Happy NEW YEAR dear Richard and all other nice people🥳🥳🥳🥳🥳😇😇😇😇😇 Live long and prosper 🖖🖖🖖
Heya, nice quiz here are mine ideas:, Hope you wil have an nice new year with all good goodies and a good health. a) 10V b) notthing C 1,5V or 8,5V D) capacitor
Hello and A Happy New Year, 1. Something between 10 to 15 v, 2. Same as the input, 3. That's depends if the potentiometre is logaritmic or liniar, I'm not shure, 4. Thyristor, triac 🎉🥂🍾🍻
Without looking at anyone else's answers, I got: 1. 10v (assuming a 10MΩ dmm input impedance) 2. gotta be 0v, hasn't it? 3. 1, 5, or 9v depending on linear/log and which way round it is if log 4. can't think of a single component, but it could be a diode and capacitor in series Happy new year!
I thought the last question was going to be what melted the center of your breadboard. I wonder if the mystery component is an active buzzer or active oscillator. Since the potentiometer is for audio it likely uses a logarithmic resistance profile which is usually about 15% voltage output at the middle position, assuming it is volume control and you connected the voltmeter to terminals 1' and 2' and the voltage was connected to 3'. I'm going to guess 1.5V. Gotta love those confusing 5-color resistor bands. Oh, and if you switch the power off, nothing will be seen because the studio will go dark. If only the op amp is powered off, you'll see observe the input signal in it's original amplitude and phase.
FYI, I tried my suggestion by using a active piezo speaker which does not have a measurable resistance either direction. When hooked up as the mystery "component", both the active speaker and the passive one made sound, so this could be one answer, although it's not truly a "component" nor is it as small as what was shown. I would suppose if the piezo is taped quiet, or the driver disconnected, it would work the same since the circuit is still active? I didn't dissemble the active speaker to find out (it's epoxied). I tried some of the other suggestions: transistors, diac, triac, zener, varactor, capacitor, and only the active speaker worked. Unless I used the wrong value or type? Really had hopes for the varactor! Another possibility is a relay wired to self-interrupt itself, but that would read continuity on a multimeter and would need to work at very low current. Perhaps a solid state relay would work.
Here's my guesses. 1. Approximately 10 volts. 2. Output will be the same as the input. 3. Approximately 5 Volts. 4. NPN BJT in avalanche breakdown mode (Collector to RC junction, Emitter to speaker, Base is not connected). (edit) oops ... I got the leads reversed for #4. It should be Emitter to RC junction and Collector to speaker. Tried it on the bench. Went through about 5 different types of NPN until finally a BC337 worked for me. (edit 2) Revisiting Q3: I was assuming the slide pot is linear. if the pot is logarithmic, then my answer is approximately 1 volt.
@@pault6533 I did finally try this on my bench. My initial comment had C & E reversed. Got it to work with a BC337-40 when using a tiny inductive speaker. It did not work with a bare piezo transducer, nor with a 1k resistor across the piezo. Looking at 7:09 it's not clear just what type of transducer Richard is using (due to the multiple solder points on the rear), so it's possible that mine is not the correct answer.
Q1: 10V, you wanted me to say 15V but i won't because inside the multimeter theres an 10M resistor Q2: it would look the same as the input signal Q3: 9V or 1V because audio pots is in logarithmic scale Q4: idk some diode?
Question 1: We have to consider the input impedance of the multimeter. Note that I haven't been able to locate the manual for the Fluke 79 Series, as shown in the video. I've been using the value stated for the Fluke 79 Series II instead, which states > 10 Megaohm. I'd say a typical value for a modern multimeter, but as you use an old model, the value might differ, Using a value of 10 Megaohm. This gives a 10 Megaohm resistor parallel to the 10 Megaohm resistor in the circuit, thus equivalent to 5 Megaohm. So we have 10 Megaohm is series with 5 megaohm, thus a total of 15 megaohm, As we supply 30 volts, the voltage drop over the resistor and multimeter will be 10 volts. Question 2: Output voltage cannot exceed supply voltage. However, when the supply voltage lines are floating, could they float in a way that allows some output voltage? I still expect no output. Question 3: Note the part number is V50KAX2. The unit is a 50K potmeter. The A determines the characteristic. Now, we need to know whether this is an European American or Asian part, as the meaning of the A and B in this position are swapped. For European parts, A is linear and B is logarithmic. For American and Asian parts, A is logarithmic and B is linear. For a linear part, it would be at half the voltage, thus 5 volts. For a logarithmic part, it's about a quarter, thus 2.5 volts Question 4: I am not sure but it is probably a zener diode with a zener voltage that exceeds the measuring capabilities of the multimeter. Note that the supply voltage is 12 volts, and thus, the capacitor could charge up to 12 volts. Also note this speaker is likely to be an inductive type. Happy New Year!
@@pault6533 That's right... hmmm... what about a TVS diode, having the high voltage in both directions. Nah, that won't work either as the polarity matters in circuit.
For Question 2: I think that leaving the power lines of an inverting opamp floating will result in the input voltage showing at its output. That is due to the fact that the gain setting resistors are in series between the input and the output.
Q1. There is indeed a disagreement in the potential of several points shown in this diagram, if I kerred enuf chuff about the solution I could write a law about it.
I don't know about anyone else, but I HATE the 5 band colour code for resistors, and I think the way it is implemented is all wrong. Back in the day, I could look at a resistor and instantly know its value without having to work it out, when you've done it thousands of times, your brain learns. NOW! Not only do I have to work it out, I often have to use a multimeter on many resistors because they have a brown band at each end. 0:22 These resistors could also be 150 Ohms, depending on which brown band is the tolerance. 🤷
Nice test, Happy New Year!
1) The huge resistance in your ladder matches the input impedance of the meter (parallel with the resistor you are measuring) dropping the impedance to 1/2 (5M). The ladder is essentially 15M so the drop across 5M would be 1/3 of 30V = 10V.
2) Disconnecting the power to an OPAMP will stop all amplification and inversion functions. This will leave the two passive resistors in the circuit. The output will be essentially identical to the input signal.
3) You mentioned this was an audio control, so it is safe to assume this isn't a linear potentiometer (log). The log of 50% would be right about 3VDC.
4) There may be several components that could react in this manner. I feel a low-voltage unidirectional TVS ESD diode could be used. As long as this voltage exceeds the "Diode Check" voltage your multimeter outputs.
Thanks Richard for a video laden 2023. Can't wait to see what 2024 has in store for us. Keep up the good work.
I downloaded the audio file using mediahuman and detected a complex waveform, looks like compound sine waves. After doing FFT of the sound at 7:49 I see peaks at 2400 Hz, 4200 Hz, 8300 Hz, and 13100 Hz, the highest at 4200. So I suspect the mystery component is exciting the circuit around 4.2 kHz, perhaps some sort of active oscillator or passive piezo speaker (both wouldn't measure on a multimeter). The remainder of the peak frequencies probably have something to do with the frequency response of the buzzer (2400 Hz). If you look at the datasheet for a typical passive speaker, the frequency curve looks like a roller coaster ride, which would explain why it looks like it contains many harmonics. I ran the experiment using an active speaker and the passive one created a waveform with decreasing amplitude spikes at 4, 8, 12, and 16 kHz, nothing at 2 kHz, but I have different speakers than Richard. I think the 10uF capacitor is just there to maintain a constant DC level, the RC circuit with a 1K resistor is only 16 Hz. Removing it decreased the sound level and slightly altered the frequency in my setup.
Over-achiever! 😂
Wow I admire your enthusiasm 😁
Hi Richard and Happy New Year. Thanks for another year of entertaining and informative videos and for an amusing test to distract me from the miserable weather here in the UK!
Regarding your test questions I believe the answers to be:
1 - Probably about 10v as the input impedance of the fluke (about 10M) will be in parallel with the lower of the two 10M resistors resulting in an effective resistance of about 5M so the junction of the two resistors will be at about 1/3 of supply.
2 - With power off then the I/P signal will be apparent, non-inverted, at the O/P pin as a result of the i/p and feedback resistors in series.
3 - Potentiometer voltage at slider pin will depend upon the potentiometer law or taper and if not linear then which way around it is connected. In the video we can see the pot is marked as "type A", in Asia and the US that indicates audio or logarithmic, but in Europe it can mean linear!
If it were linear then the voltage would be close to half the supply or 5v regardless of which way the pot is connected.
However, if as I suspect it's an Asian "type A" then it will be logarithmic and the mid position would be approximately 10% resistance from one end and consequently 90% from the other, in which case depending upon which way around you connected the pot the voltage will be either 1v or 9v.
4 - Trickier one this but given the behavior and the multimeter readings it must be some kind of directional and active component.
I suspect the mystery component is an SCR with a floating gate allowing it to trigger on noise, discharge the capacitor through the speaker until the current drops sufficiently for the SCR to turn off allowing the capacitor to recharge via the resistor until another noise spike re-triggers the SCR again.
#4 might also be achieved using a low voltage DIAC such as the ST4.
@@viccurtis7448 Yes, a simple DIAC sawtooth oscillator ..
The potentiometer terminals are marked on bottom and polarized location 11’ 22’ 33’. The spec sheets show the voltage vs position for the 1-2 terminal pair when voltage is applied to 1-3. This should help determine which voltage to choose for the slider.
Q1: 10V
The resistive divider with two equal resistors divides the voltage by two and the voltage should be 15V. But as your DVM has a similar input resistance of 10 megohm, the resistance of the part were the DVM is connected becomes only 5 megohm. Hence the voltage is divided 1:2, 10V on the measured part and 20V on the other.
Q2: the same as the input signal
Without a supply voltage the LM358 shoult no longer influence the circuit behaviour. What remains are two resistors which connect the input to the output. 1K+3K will not significantly influence the signal as the scope input resistance is at least 1 megohm.
Q3: 1V or 9V dependent on the direction of the potentiometer
Given the use as a volume control in audio circuits and the letter A in the part number I assume this is a logarithmic potentiometer. Therefore in the center position the resitance is divided 1:9 approximately. Depending on which side you measure the voltage it is around 1V or 9V.
Q4: Thyristor
This is probably not the right solution as a thyristor will usually not measure open circuit in both directions. However, the circuit can work with a thyristor. You have to connect gate and anode to the positive end of the capacitor and the cathode to the speaker. This will give you a relaxation oscillator. The thyristor triggers and discharges the capacitor until the current drops below the holding current. Than the capacitor charges again until the trigger voltage is reached. If you flip the connections the thyristor will not trigger as the gate is always at zero volts via the speaker. I tested the curcuit with several thyristors and not all worked as holding currents and trigger voltages differ and may not fit this circuit. With some fine tuning of the resistor you get more thyristors to work.
G,day Sir and Happy New Year.
Quiz questions.
I submit the following answers and MY reasonings.
Q1. Two resistors (value 10.6 mega OMS each in series. 10 volts of DC current would s 0:01 till flow through the two.
(I'm getting a diode and the resistor confused, so my guess is 10 volts).
Q2. The measurement at the op amp with nil current (and no stored energy e.g caps) is 0 voltage.
(Reasoning- there's no source of other DC supply).
Q3.Potentiometer slide. Is only to measure or monitor displacement of volts.
Q4. Bread board with speaker circuit. The mistry component is a relay or gate.
🧲🤨💡
#1: half of supply, #2: input waveform if instrument doesn't load circuit. #3: if linear pot half of supply, if log pot 30% of supply. #4: bimetallic flasher
Q1: 15v; Q2: 2.5v; Q3: 5V; Q4: 100 pF electrolytic capacitor or less; Happy new Year everyone. Rick, special thanks to you for the job you are doing.
Q4 cant be a cap, it would block DC. My guess a 6v-12v Zener
I have much faith in your set of answers
@@a1fliema1fie well this electrolytic capacitor and it doesn't block DC in some cases - trust me. And cannot be diode because it tests as OL in both directions
@@a1fliema1fie before a capacitor 'blocks dc' it will have to charge... but one capacitor will charge faster than the other because one of them is in series with the buzzer (ie load) causing a kind of perpertual imbalance where the capacitors charge and discharge sending singals through the buzzer
Are electrolytic capacitors made at 100 pF?
Yoloooo
1) not half the input volts , because of the input impedence of the multimeter the answer will vary if the multumeter is 10Mohm or 1Mohm input impedence
2) i've no idea i think i have to watch the all you have to know series of op amp hahah , but i guess output will be same
3) around half the input so around 5v
4) DIAC ? ... no reading on multimeter on diode mode because DMM is testing at around 2v and the diac is more than that , zener would show up at diode mode , the capacitor charges reaches the opening voltage of diac and discharges through the speaker diac closes capacitor charges so and so....
Hope i answered 1 sensei hahah
Kudos all the way from Africa, Ethiopia
Dear Richard,
Congratulations for your hard work, you never stop, it is marvellous.
I do not know if I will succeed the test but I wish you a Happy New Year and keep on with these most interesting videos. Marc from Belgium
Thanks for all your interesting videos, hope you keep doing that in 2024. Wish you a great 2024.
1: less than 15V, probably close to 10 Volt as the multimeter has also an input resistents of 10Mohm.
2: output will be the same as input, between input and output are only the 2 resistors, as the input impedance of the scope is high there will be almost no loss .
3: if it is a linear slider the output will close to 5V
4: maybe a crystal?
Probably a log taper (volume control x2) You can see it's marked as V50kAx2 in the video.
1.)10 V
2.) 2,5Vss or 1,77Veff,,,,,, Ve=Vo
3.) 2V if the poti log otherweise 4,9936V lin
4.) One zener Diode and one normal diode both anodes together.
And... Happy NEW YEAR dear Richard and all other nice people🥳🥳🥳🥳🥳😇😇😇😇😇
Live long and prosper 🖖🖖🖖
Hi Richard.
Happy New Year. Great job on the videos and see you in 2024.
Heya, nice quiz here are mine ideas:, Hope you wil have an nice new year with all good goodies and a good health.
a) 10V
b) notthing
C 1,5V or 8,5V
D) capacitor
Hello and A Happy New Year,
1. Something between 10 to 15 v,
2. Same as the input,
3. That's depends if the potentiometre is logaritmic or liniar, I'm not shure,
4. Thyristor, triac
🎉🥂🍾🍻
Without looking at anyone else's answers, I got:
1. 10v (assuming a 10MΩ dmm input impedance)
2. gotta be 0v, hasn't it?
3. 1, 5, or 9v depending on linear/log and which way round it is if log
4. can't think of a single component, but it could be a diode and capacitor in series
Happy new year!
No 4 only an educated guess, no one has mentioned this, could it be a DIAC.
My thought exactly. Question 4a: what voltage is the diac breakdown at? :)
Except diacs are bidirectional so it would work the same either polarity, so nevermind.
Q1 15V
Q2 same as input
Q3 2.2V
Q4 transistor
You wicked man,
Happy New Year🎉🎂🍾🥂
I thought the last question was going to be what melted the center of your breadboard. I wonder if the mystery component is an active buzzer or active oscillator. Since the potentiometer is for audio it likely uses a logarithmic resistance profile which is usually about 15% voltage output at the middle position, assuming it is volume control and you connected the voltmeter to terminals 1' and 2' and the voltage was connected to 3'. I'm going to guess 1.5V. Gotta love those confusing 5-color resistor bands. Oh, and if you switch the power off, nothing will be seen because the studio will go dark. If only the op amp is powered off, you'll see observe the input signal in it's original amplitude and phase.
FYI, I tried my suggestion by using a active piezo speaker which does not have a measurable resistance either direction. When hooked up as the mystery "component", both the active speaker and the passive one made sound, so this could be one answer, although it's not truly a "component" nor is it as small as what was shown. I would suppose if the piezo is taped quiet, or the driver disconnected, it would work the same since the circuit is still active? I didn't dissemble the active speaker to find out (it's epoxied). I tried some of the other suggestions: transistors, diac, triac, zener, varactor, capacitor, and only the active speaker worked. Unless I used the wrong value or type? Really had hopes for the varactor! Another possibility is a relay wired to self-interrupt itself, but that would read continuity on a multimeter and would need to work at very low current. Perhaps a solid state relay would work.
#4 unknown part could be a NTC or PTC Thermistor.or a Zener Diode.
Q1 in the ideal world it be half so 15 volts as it is a divider and as long the values are the same it be half the input voltage
Q3.. voltage will be close to 5v since u ve created a voltage divider using variable resistor
We don't know if It's a linear or logarithmic pot. That would be true if it were linear. But that would be too easy.
A1: Ur2=U1*R2/(R1+R2) it is a voltage divider
Here's my guesses.
1. Approximately 10 volts.
2. Output will be the same as the input.
3. Approximately 5 Volts.
4. NPN BJT in avalanche breakdown mode (Collector to RC junction, Emitter to speaker, Base is not connected).
(edit) oops ... I got the leads reversed for #4. It should be Emitter to RC junction and Collector to speaker. Tried it on the bench. Went through about 5 different types of NPN until finally a BC337 worked for me.
(edit 2) Revisiting Q3: I was assuming the slide pot is linear. if the pot is logarithmic, then my answer is approximately 1 volt.
I tried your NPN suggestion and it didn't work. Nor did a PNP. Wondered if you tried this? Or a Mosfet?
@@pault6533 I did finally try this on my bench. My initial comment had C & E reversed.
Got it to work with a BC337-40 when using a tiny inductive speaker.
It did not work with a bare piezo transducer, nor with a 1k resistor across the piezo.
Looking at 7:09 it's not clear just what type of transducer Richard is using (due to the multiple solder points on the rear), so it's possible that mine is not the correct answer.
@@BusyElectrons Thanks for checking! We shall have an answer soon.
You are right about all the questions, nice Job
Happy New Year.
Beautiful
Q3 actually same is Q1 in the middle both ends should have even resistence so it becomes a divider half of 10 volts in the ideal world be 5 volts
Happy new year rich to you and yours 2024 is the year of white smoke 😅👍
Q4 I think its a christal as it oscilates and if i am not mistaken they are polarity sensitive
Q1: 10V, you wanted me to say 15V but i won't because inside the multimeter theres an 10M resistor
Q2: it would look the same as the input signal
Q3: 9V or 1V because audio pots is in logarithmic scale
Q4: idk some diode?
You could say that the mystery component is well endowed and swings a different way.....
Question 1:
We have to consider the input impedance of the multimeter.
Note that I haven't been able to locate the manual for the Fluke 79 Series,
as shown in the video. I've been using the value stated for the Fluke 79
Series II instead, which states > 10 Megaohm.
I'd say a typical value for a modern multimeter, but as you use an old model,
the value might differ, Using a value of 10 Megaohm.
This gives a 10 Megaohm resistor parallel to the 10 Megaohm resistor in the
circuit, thus equivalent to 5 Megaohm. So we have 10 Megaohm is series with 5
megaohm, thus a total of 15 megaohm, As we supply 30 volts, the voltage drop
over the resistor and multimeter will be 10 volts.
Question 2:
Output voltage cannot exceed supply voltage.
However, when the supply voltage lines are floating, could they float in a
way that allows some output voltage? I still expect no output.
Question 3:
Note the part number is V50KAX2. The unit is a 50K potmeter. The A determines
the characteristic. Now, we need to know whether this is an European American
or Asian part, as the meaning of the A and B in this position are swapped.
For European parts, A is linear and B is logarithmic. For American and Asian
parts, A is logarithmic and B is linear.
For a linear part, it would be at half the voltage, thus 5 volts.
For a logarithmic part, it's about a quarter, thus 2.5 volts
Question 4:
I am not sure but it is probably a zener diode with a zener voltage
that exceeds the measuring capabilities of the multimeter.
Note that the supply voltage is 12 volts, and thus, the capacitor could charge
up to 12 volts. Also note this speaker is likely to be an inductive type.
Happy New Year!
A case where the measurement method affects the result! Wouldn't the Zener be measurable forward biased?
@@pault6533 That's right... hmmm... what about a TVS diode, having the high voltage in both directions. Nah, that won't work either as the polarity matters in circuit.
@@AndreDeLimburger See if you have any varactors to try. I have only a few small ones and they didn't work, might need to be a different spec.
@@AndreDeLimburgerI was thinking maybe tunnel diode or Gunn diode, something with a negative resistance that would make it oscillate.
For Question 2: I think that leaving the power lines of an inverting opamp floating will result in the input voltage showing at its output.
That is due to the fact that the gain setting resistors are in series between the input and the output.
Happy new toy to you all
My Answers:
Q1) 15 Volt
Q2) 2.5 Volt
Q3) 5 Volt
Q4) MOSFET
...q3, as the letter codes are not standardized internationally, can I ask if it's a volume or a balance stereo pot?
Q1. There is indeed a disagreement in the potential of several points shown in this diagram, if I kerred enuf chuff about the solution I could write a law about it.
Happy New Year and Three Kings.
Si Feliz Anos Nuevo y Tres Rienos
Q1: can we assume you using the Fluke 79?
yes
Happy new Year everybody!
Q1: 15V
Q2: no idea i think a stairway down :)
Q3: 5 V
Q4: Schottky Diode
Q2 when you shut the power off then you have nothing as it has no voltage to use to amplify
I don't know about anyone else, but I HATE the 5 band colour code for resistors, and I think the way it is implemented is all wrong.
Back in the day, I could look at a resistor and instantly know its value without having to work it out, when you've done it thousands of times, your brain learns.
NOW! Not only do I have to work it out, I often have to use a multimeter on many resistors because they have a brown band at each end.
0:22 These resistors could also be 150 Ohms, depending on which brown band is the tolerance. 🤷