LF generator for swept field electromagnet - Page 1
Forum section: electronics / projects

author: ChristofferB
author_rank: Frequent Contributor
timestamp: January 28, 2018, 04:06:16 pm
content: Hi! I'm still messing around with NMR and EPR spectroscopy, getting together the magnet currently.I need to be able to sweep the field of a permanent magnet, so I have a coil around one of the neodymium magnets in a U-yoke.I need an LF supply to this, this will be both sweeping the field and providing X input for an oscilloscope, ideally a ramp, so i have a linear sweep, but i've considered that it might not be terrible if it's a sine. I've seen that used before, atleast. Would a mains->12V transformer, and then a ceramic potentiometer in series with the coil as current limiter do?I guess the slightly more advanced route would be a separate saw generator and an LF amp, but i'm not certain that's really needed. i'll ad a phasing network on the scope port so I can get the entire sweep on screen.thanks in advance!

author: dmills
author_rank: Super Contributor
timestamp: January 28, 2018, 08:44:47 pm
content: X output from the 'scope (usually around the back) wired into an audio amp?You generally find this on a BNC or sometimes a 4mm banana socket on the back of the scope. Nice linear sawtooth that tracks the X axis defection somewhat linearly.Regards, Dan.

author: danadak
author_rank: Super Contributor
timestamp: January 28, 2018, 11:28:31 pm
content: Use a PSOC to generate the saw and a synch trigger to start the sweep.Much more capability in it, could add a control loop for current or voltageas well. Pipe out via USB or UART to a Python app to display system info.Attached basic example.Regards, Dana.

author: ChristofferB
author_rank: Frequent Contributor
timestamp: January 29, 2018, 01:57:26 am
content: Quote from: dmills on January 28, 2018, 08:44:47 pmThat's a great idea! maybe a transistor buffer would be enough, I really don't need to change the field much. but else, yeah, one of those 2-5W SIP audio amps might be good if the scope output's not enough. Maybe just feed it out though an emitter follower with a 2n3055 for beefiness. I guess i could do away with the whole phase network too!

author: ChristofferB
author_rank: Frequent Contributor
timestamp: January 29, 2018, 02:38:04 am
content: prelim. design for reference:

author: dmills
author_rank: Super Contributor
timestamp: January 29, 2018, 12:26:16 pm
content: Improve your buffer by making the emitter load a resistor and placing the coil (plus a catch diode) in the collector circuit.This makes the transistor drive a current in the coil proportional to voltage and means that as long as you have sufficient supply voltage, changing the sweep rate should not change the strength of the gradient.Note that scope X axis is often a signal that goes negative for the left part of the screen, so you may been to play some level shifty shenanigans.Regards, Dan.

author: danadak
author_rank: Super Contributor
timestamp: January 29, 2018, 01:02:14 pm
content: For the scopes that dont have SAW out, they may have a gate out thatyou can integrate. You will have to build a integrator reset circuit upondetecting end of sweep.Many analog scopes had SAW out, many DSOs do not. Some DSOs havechannel out, you could use math functions to develop sweep waveform.Regards, Dana.

author: ChristofferB
author_rank: Frequent Contributor
timestamp: January 29, 2018, 02:14:13 pm
content: Quote from: danadak on January 29, 2018, 01:02:14 pmyou're right, making a triggerable ramp gen digitally is probably the modern approach, but luck has it I have an analog scope in excess i can dedicate to the project, and i'm fairly certain it has the ramp out signal.here's an improved schematic. I haven't sized the load resistor yet, but I think a 24 ohm power resistor for half an amp current would be sufficient

author: dmills
author_rank: Super Contributor
timestamp: January 29, 2018, 02:27:24 pm
content: Ah that load resistor would more usually be a few ohms or so depending on what voltage your ramp generator outputs, most of the voltage is dropped across the transistor, the resistor just makes the thing into a current output amplifier by providing current feedback to the emitter of the transistor.If your ramp peaks at say 5V then the peak emitter voltage will be 5-0.7ish so say about 4.3V, for half an amp, you would then use about an 8 ohm resistor, the higher you make the supply voltage the more power you burn in the transistor, but the faster you can sweep because the voltage across the coil due to V=L.di/dt can be higher.Regards, Dan.

author: ChristofferB
author_rank: Frequent Contributor
timestamp: January 29, 2018, 02:34:17 pm
content: Gotcha, i'll try with a ~8ohm. Maybe with a low range power potentiometer so i can vary it a bit.