Cumulus Support

Although not as well done as Gina's, a string of outdoor Christmas lights wrapped around the gauge might do it.

You might experiment with a cup of water to see of it works near your dwelling before running power to your station.

I use the lights woven through my citrus trees and pineapple plants to get past the few very cold days we have.

Randy

Orion wrote:Hi Gina
Wondered if your thermostatically controlled heater circuit could be used to control one of these running off an old 12v car battery
Maplin 12v In-Car Ceramic Heater £7.49
http://www.maplin.co.uk/Module.aspx?ModuleNo=48775

The Maplin heater spec doesn't say how much current it uses or the power but quotes a 15A fuse so I would think it takes more current than the power darlington transistor can take (5A). There are higher power devices available though.

PS have you done any wind vane mods yet

No, not recently. I tried extending the vane some time back but it was worse. I think the best form of damping would be magnetic induction and I will be looking into this. But the problem with the F.O. unit is interfering with the reed switches. Two possible directions I have in mind - 1. Keep the F.O. vane and replace the read switches with reflective optical couplers and a Gray encoded optical disc, or 2. making a complete new wind vane, probably using 1-wire to collect the data.

If OTOH you mean heating the unit to prevent freezing up... Not yet but hopefully soon. I have components on order.

Charlie wrote:Just in case somebody not familiar with resistors tries to duplicate what you are doing, I believe you connected the 33 ohm resistors in series (not parallel) to achieve 68 ohms (66, but close enough). Or were you doing something different? I can't figure out a series or parallel combo that gives 0.5 A from 24 V or even 12 V? (maybe 16 V?)

The 33 ohm resistors are in parallel giving 16.5 ohms. Some volt drop occurs in the bell wire I'm using to connect heaters to power supply so I used the ammeter on the PSU to find out how much power the heater resistors were getting. Using Ohms law - P = I squared times R. Current was 0.5A squaring that gives 0.25 and multiplying by 16.5 gives 0.25x16.5 = 4.125W

Also, are you going to use the darlington as a switch, or is it actually being operated in it's (mostly) linear region?

It's used as a switch. The positive feedback on the op-amp ensures this.

Orion wrote:I=V/R = 24v/68R = 0.35294amps
P = I 2 x R = 8.47053 watts

Only 20volts available from bench supply so I put the resistors in parallel. But the connecting wire resistance was more than I thought - I didn't bother measuring it - just quickly strung some bell wire from rain gauge, along the fence, through the trees and into the shed ASAP coz' it was b cold! I might have done better to have put the resistors in series, but went by guesswork. Yes, 24v and 66 or 68 ohms gives around 8W. The bell wire resistance works out at 8.7 ohms. So...
Series :- 66 + 8.7 = approx 75 ohms. 20v will give 20/75=0.27A In 66 ohms that's 4.7W. (2.3W each resistor)
Parallel :- 16.5 + 8.7 = 25 ohm. 20v will give 20/25 = 0.8A. In 16.5 ohms that's just over 10W (5W each resistor).

So I was right that parallel was better.