Andrew Johnston | 22/09/2015 11:24:57 |
![]() 7061 forum posts 719 photos | Good lord, a JFET! That's a blast from the past. Neil is correct regarding the operation of JFET analog switches. However, I suspect that the majority of switches fabricated using CMOS need to use two MOSFETs per switch. As far as I am aware the drain-source diode is inherent in all MOSFETs due to the way that they are fabricated. So one might as well characterise the diode and use it for something useful. Andrew |
Muzzer | 22/09/2015 13:55:33 |
![]() 2904 forum posts 448 photos | Ah yes, the 2N3819... As Andrew says, you have to put a pair of MOSFETs back to back if you want bidirectional blocking. If you want to keep the losses the same as you would have had with a single unidirectional device, you will need to quadruple the number of devices and hence the cost. But for a motor drive, you don't need to worry about that. You invariably design the motor so that the maximum back EMF is less than the DC bus. You can control the motor in field weakened operation above this point but this requires clever software and you need to have a plan for when you lose control that doesn't involve popping the switches. For best operation, you use a half bridge on each phase of the motor, whether brushed, brushless, 3-phase or whatever. This allows you complete control in all 4 quadrants ie speeds and torques in either direction and in any combination. For easiest and most versatile control of a brushed DC motor, the best solution is a full "H-bridge" controller. This is 2 half bridges and gives 4-quadrant control, which is what the OP was after. You can buy integrated H-bridge driver power devices with all sorts of in-built protection features or doubtless you can get a ropey equivalent from an auction site. Murray Edited By Muzzer on 22/09/2015 13:56:29 |
Muzzer | 22/09/2015 14:01:40 |
![]() 2904 forum posts 448 photos | Something like this would be almost indestructible if you mounted it suitably.Can be controlled with PWM and DIR(ection) inputs. Requires an handful of caps etc to make it complete. |
Tim Stevens | 22/09/2015 17:18:51 |
![]() 1779 forum posts 1 photos | In the old days, cars had dynamos which would act as motors. Connect to a battery as normal (for them) - one side to earth on the casing, and the other side to both Field and Armature connections. The dynamo will then turn at a modest and steady speed, and can be used as a motor. Now drive the 'motor' faster in the same direction and it starts to generate, rather than consume. No fancy switching, no change in connections, as long as the rpm is rather more than motoring speed it generates, and in doing so absorbs power from the driver and charges the battery - ie regenerative braking. All that stops the dynamo acting as a motor at low rpm in 'real life' is the cut-out, which disconnects the output at (about) the cross-over point when output changes to input. Perhaps this will help to understand what is going on? Cheers, Tim |
jason udall | 22/09/2015 18:37:04 |
2032 forum posts 41 photos | I have read ( maybe not understood but) all the above. Heres my question. Agiven motor is charaterised thus. Stall current Off load rpm. Off load current Max rpm ( mechanical.."centigugal"or bearing limits) So far so good. So to drive motor at off load rpm ( which we assume is lower than max rpm) We feed a voltage sufficient to reach this rpm..dc for now. So why is motor sinking so much less than the stall current?. Well at stall we have say 24V and a few tenths of ohms thus we have hundreds of amps.. But at off load rpm we see 5A..? Doing the arithmetic we see that the motor acts as if run on 1-2 V...what gives..well that back emf the motor is generating...say 22V... PUSHING the other way leaving only 1-2V excess .. Now take the point where the motor is loaded and throttled to half rpm... Motor back emf falls and load current rises..... Now apply brakes... Well back emf is about ( pardon my guess ) 1/4 at this rpm so say 6V... Mmm how do we charge a 24V battery from that. . (Yes I know ...inverter of some kind but lets keep this to the..fet/diode/switch level)?.. . |
Muzzer | 22/09/2015 18:50:32 |
![]() 2904 forum posts 448 photos | You probably need to get a book out on the subject or the internet equivalent. You can make a fairly reasonable model of a motor by combining a DC voltage source and an inductance. The voltage of the source is proportional to the rotational speed (scale factor Kv) and the torque generated by the motor is proportional to the current in the stator (the static winding) - scale factor Kt. That pretty much does all you need. When the motor is spinning with no load, the voltage source ("back emf" With wound field rotors (dynamos and alternators), the voltage factor Kv varies with the field current. Higher current gives higher torques but lower speeds. With permanent magnet motors Kv doesn't vary. Murray |
Please login to post a reply.
Want the latest issue of Model Engineer or Model Engineers' Workshop? Use our magazine locator links to find your nearest stockist!
Sign up to our newsletter and get a free digital issue.
You can unsubscribe at anytime. View our privacy policy at www.mortons.co.uk/privacy
You can contact us by phone, mail or email about the magazines including becoming a contributor, submitting reader's letters or making queries about articles. You can also get in touch about this website, advertising or other general issues.
Click THIS LINK for full contact details.
For subscription issues please see THIS LINK.