A Digital Display Adapter For Your Curve Tracer Or Oscilloscope.
🕑 Added 2017-08-01 15:32:33 +0000 UTCBuild this simple circuit too give your curve tracer or analog oscilloscope a digital display.
Quiz questions included.
Patreon NEW LIST of Videos: https://www.patreon.com/posts/8239565
This is video #20, include this when requesting attachments.
Please see below attachment for schematic.
Comments
Steven Sadler
Hi Paul I. Have 2 questions for you 1st is would this circuit be ok with a quad opamp packages. Second would there be any benefit in having switchable frequency ranges ie 10khz 100khz for testing higher frequency devices caps etc
Donny Bazinet
Hello Mr Carlson. I'm a big noob but I love your videos, its a whole new world. Could a version of this circuit work with a frequency counter? I found an old one made by General Radio which can also measure period and ratio between its inputs, but only goes to 300kilocycles! The five digits are displayed with edge-lit acrylic sheets using small lamps and look beautiful.
LoSaYa
Thank you
Harpan Budi Santoso
I have only old oscilloscope, and this is truly helped me alot.
Bruce Dow
Mr Carlson, you do TIG welding to boot?? Now I *know* I am in the right place! Haha
Mr Carlson's Lab
Hi Robert. The bulb is part of the oscillator circuit in a different (previous) video, this circuit is to give that project an external display.
Fried Mule
I feel a bit stupid for asking but I am a totally noob. In the end you talk about the bulb, I do not understand where it comes in, could not find it on circuit?
Mr Carlson's Lab
A clean sine wave with low distortion will yield the best results in this solid state circuit. If you were to build an "octopus" then distortion on the supplied source signal isn't as important. If you were to use a triangle wave, your signal would have to be split evenly above and below the line.
Mr Carlson's Lab
Hi Dennis, if you attach the peak detection circuit to the buffer input or output, when you put a load across the probes, the reading will altered by the device under test. You want the ability to manually sweep the current applied to the DUT so you can see the knee Voltage properly.
Dennis Cabell
And the second question (fumbled fingers made this 2 posts) You said to attach the peak detection circuit to the "high" side of the shunt resistor (directly to the oscillator output). I would have attached it to the "low" side (same point as the buffer to the scope) so that it would record the voltage that the device would see, and the same voltage as the peak of the wave on the scope. Why attach it where you indicate?
Dennis Cabell
I am working though these catching up, and at this point I have a couple questions I hope you can answer. I have been working up my own curve tracer before finding this. I was using a triangle wave instead of a sine wave. Are there advantages/disadvantages to one or the other?
Mr Carlson's Lab
Glad your enjoying Peter!
peter jeffs
Another fantastic video, thank you. It's great that you have challenged us to answer questions about it to - another great learning aid. Once again, your clear instructions and detailed explanations ensured a right first time build. I'm really enjoying this curve tracer project and can't wait to put all the building blocks together to get the finished product.
Mr Carlson's Lab
Hello Dr. Ramin. A great versatile IC is the old LM723. These are still commonly used in the Astron power supplies, and many other "linear" type supplies.
Ramin Azizli
Hi Paul. I need your advice. I'm building an adjustable dual rail linear power supply at about 10A for each channel with voltage and current regulation at the output. For this purpose I got a transformer with 8V AC for powering voltmeters and ammeter and etc, 15V AC for powering fan and finally two 30V line with center tap to get +30V and -30V AC. My question is what would be your advice about the best choice of the controller chip for positive and negative rail that I can choose to build a voltage and current regulation with over-current and over-voltage protection? I was looking at LT1083 or LM337 but still cannot decide. Maybe you can offer better option? Thanks in advance.
Toni Pomar
It's raining a lot here too. Nasty weather, indeed.
Psient
Mr Carlson's Lab
Thanks for your input Mark!
Mr Carlson's Lab
Thanks for your understanding Mark. It's 1:47AM, and I just finished cutting and welding another bracket for the new camera... It seems there isn't enough time in the day.... or night. A few hours of sleep, and I'll be back at it again.
Mark Bosse
Technical question. I was a child of the digital age. I was in the first crop of Home-brew Computer Club Members. I was there the night of the Gates vs. Jobs fight. I'm not dumb. But the thing that drove me repeatedly out of analog was a complete ignorance of coax. Sounds bad but I still don't get it. Maybe something for your main channel, but a tutorial about this weird stuff from a science point of view would be great. YouTube is full of "do it this way" stuff but never why. Just another idea for the pile
Mark Bosse
Mr. Carlson, don't worry about life's impact on production schedule. I have been at this electronics thing for 47 years. Never my career, just a passion. I can wait a week or two or there. In my career, I took the same attitude. My clients always got the best work product, not the fastest. In the end, my company had gone from $5M market cap to $100B. My billing rate was $5K/hr. Quality always is why we are your cultists.
Toni Pomar
Indeed, the circuit is a peak detector and non-inverting amplifier. When the jumper is on, the amplification is just 1 (voltage follower), and when the jumper is off, the 4.7k resistor defines the amplification with the 470k resistor (in this case its 1+(4.7k/470k) = 1.01). About replacing the 4.7k resistor and jumper, maybe it could be replaced by a 470k potentiometer. At 0% it would be like the jumper, so the output is the peak voltage of the wave. At 100% the amplification would be x2, giving then the peak to peak voltage at the output. Anyway, this last part is a bit of a long shot, mainly because with just a single component there I can't quite see another option right now.
Diabolical Artificer
I think it's a peak detector circuit. The diode half wave rectifies the OP from the opamp, this charges the 22u cap . The 100k R in parallel the the 22u cap acts as a bleed resistor, keeping the charge on the cap constant, no bumps. The 470k and 4.7k R's form a 100,000:1 potential divider dictating what FB goes from OP to the inverting IP . But I'm struggling to suss out why shorting out the 4.7k R changes the least significant digit, apart from it changes the FB ratio. As others have said I'd replace the jumper with a pot, but that's based more on intuition than solid theory. I havn't built this circuit yet, (been busy fixing RF sig gen and finishing work on my 100w valve amp. If interested see here - <a href="https://www.youtube.com/watch?v=MZTTpBLvRCw" rel="nofollow noopener" target="_blank">https://www.youtube.com/watch?v=MZTTpBLvRCw</a> ) so am still in the dark a bit about what does what where
nj Phil
Paul... You had quite a week! Thanks for the new video. These little circuits are awesome. Quick builds and cool functionality!
Eddie Bray
I think it's a comparator circuit.
Andre Gopee
I think the jumper and resistor should be replace with a 5K Pot in order to calibrate the feedback so as to essentially calibrate the output voltage readout. I maybe wrong... just thought I'd give it a shot.
Steve Crisp
I think the component you would change to is a VR say a 503 with a jumper to the wiper from one end.
Rod Smallwood
Rectifier charges capacitor. Switch discharges
Jon Fredrik Våle
Nice! Belive you made a VU meter.:)
jasper_fracture
Peak detector?
Reb Elba
Works very good. The DMM follows very smooth. I like that we should think and explain the circuit itself. Many thanks Paul!
Eddie Bray
That's the true value of a breadboard, you can mix and match components to discover different things.
Erik G
Mmmmm.......my version, using the +/- 9V (rather than the +/-15V) doesn't give the correct voltage for the zener (3.9V) in circuit... the voltage shows a little less than half what it should be at the knee. Do you think it would be possible to alter the resistor values to make this work?