Industrial Control - Experiment #2
mose
Posts: 7
I'm trying to work through experiment #2 in the Industrial Control book Ver1.1 to use PWM to drive a 12VDC brushless fan at variable speed with an LM358 op amp and a 2N3904 transistor.· I measure·the voltage between pin 5 of the LM358 and ground and· can see it change from ~1.1V to ~5V·as I change the (duty) value in the PWM command from 60 to 254, which seems OK.··But the fan voltage measures ~3.5V at a duty of 60, goes to about 7.6V at a duty of 140, and then doesn't increase further as I increase the duty value up to 254.· I'm running off a DC power supply that I know is supplying 9VDC to the Board of Education Vin.· I tried both op amps in the LM358 package with no change.· The voltage measured at the base of the transistor tracks slightly less than a volt higher than the emitter voltage, so I don't think the transistor is bad either.· Plus I swapped out the transistor with same results.
Any suggestions?
Or any other ways to drive a brushless DC fan with variable voltage from 4 to 12 (maybe 14)VDC?
-mose
Post Edited By Moderator (Jon Williams (Parallax)) : 2/6/2006 2:43:09 AM GMT
Any suggestions?
Or any other ways to drive a brushless DC fan with variable voltage from 4 to 12 (maybe 14)VDC?
-mose
Post Edited By Moderator (Jon Williams (Parallax)) : 2/6/2006 2:43:09 AM GMT
Comments
· The apparent/effective/equivalent voltage (pulse pk. ampl. *·DU) is·what you are measuring with your DMM at the op-amp input.· So, with a 9V supply, you'd expect to measure 0.9V with 10% duty, 4.5V with 50%, and 8.1V with 90%.
· Sounds like the output to the fan is from a common-collector (also known as an emitter-follower) amplifier, which is a current amplifier.· Its gain is "1", but in actuality it is <1; the output voltage is VB - VBE.· So, if VB = 8.1V, then: 8.1V - 0.7V = 7.4V
·
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Edit remarks /·follow-up
I read the experiment and you are at "full output" at '140'.· And, as I have confirmed the output amp is a common-collector.· I think that all is in order.· If you want/need more speed/voltage then you'll need to increase the supply voltage.
Post Edited (PJ Allen) : 2/6/2006 5:33:13 PM GMT
·
· If you have a +12V supply, the most you'll see at the transistor emitter (output) is 11.3V.· It seems to me that you could use a 12V supply to power the circuit illustrated.· You could use a higher value of Rf which would increase·the amplification and the output vs. duty value.
· Another thing -- is your power supply 9V Regulated or Unregulated?· An Unregulated supply would peak charge output to 13-14V, but a Regulated supply will stick at 9V (as it should.)
·
Post Edited (PJ Allen) : 2/6/2006 5:40:56 PM GMT
You CAN actually get higher output voltages using a 'charge-pump' circuit configuration -- but that's not the configuration you have here.
What if I supply a separate +14V at Vin instead of the +9V shown, use a separate 9V regulated supply to power the Board of Education, and tie the grounds for the 9V and 12V supply together? Would this allow me to regulate a 4 to 14VDC to the fan without harming the Board of Education?
Or is there another circuit that would allow me to drive 4 to 14VDC to the fan using the PWM (or a different) stamp command?
Thanks for the help.
However, the circuit looks like it will work fine with a 14-volt REGULATED supply, so if that's what you have, you can go for it. The BS2 'Vin' pin drives an on-module regulator, so it can take anything between 5.5 and 21 volts (21 volts might be pushing it, actually).
DO make sure you use that zener the text calls out, to limit the voltage on the BS2 signal pin.
It's a little confusing on the max voltage input for the 2.1mm center-positive plug on the Board of Education:· the writing on the pcboard says 6-9VDC, but the schematic in the BOE manual says 6-30VDC.
·
Imagine, it's almost as though I know anything.
Well, good then.