I am attempting to open a cash draw which is fitted with a 24v 1.5 amp solenoid.
I am planning on using an ArduinoUNO with Ethernet Shield linking to our till.
The following circuit appears to be ideal
The electrical diagram is:
However it appears I do not have enough amps to fire the solenoid.
I am using a 32v 600ma transformer to convert to DC and a L7824Cv voltage regulator wired in before the Solenoid Power V+ to drop the voltage down to the required 24v This is working fine and the vold meter is reading 24v
I think I need to put a capacitor in on the circuit to hold a charge which will be released when the arduino input is fired.
Please can somebody advise where I need to put the capicitor and which one I need?
(The circuit is being used to operate a till draw, closing the till draw manuall resets the solenoid back to the starting position.)
Ok, this is a little physics lesson - I don't have all the figures I need so you may have to redo this with your knowledge.
So we need to operate the solenoid for a time, t. I'll assume this is about 1 second for this answer.
We know the solenoid wants 24V 1.5A.
But we can be fairly sure the solenoid will stay operating a lot lower than 24V once it has moved - this means we can afford the voltage to droop during the time t. Lets assume it will work down to 16V (again a guess).
So in time t the voltage shouldn't droop more than 8V. The solenoid current is 1.5A (a little less as the voltage falls, but we'll assume a constant 1.5A for now).
The equation for a capacitor is dQ = C dV (change in charge is C times change in voltage). Thus dQ/dt = C dV/dt (current is C times rate-of-change of voltage).
We have 8V in 1s, so dV/dt = 8. We know the current is about 1A (the other 0.5A the supply can deliver remember), thus 1 = C x 8
Thus C = 0.125F, or 125,000uF.
That's pretty huge - v large and expensive. Hopefully the solenoid doesn't need to operate for that long. Say t was actually 1/10 second, then C would be 12,500uF. A 25V 15,000uF capacitor isn't hugely expensive, but is large, something like 30x50mm
A solenoid is like the coil of a motor - it has some stall current to get past to get things moving from dead stop, and then the hold current can be less.
Got a spec for your solenoid?
Get yourself a better supply: http://www.mpja.com/24VDC-18A-POWER-SUPPLY/productinfo/17434+PS/
There are several things wrong with your diagram. First it NEEDS a 4700uF cap right where the solenoid connects to + power... This is essential. The calculations presented in the precious post's are correct but I designed those types of driver circuits for 12 years for Bi-Polar solenoids of s similar voltage and current and I drove mine from a capacitor of that value... with a 120R 1/4W resistor in series with Vsupply to the solenoid. I was controlling the solenoid with the charge stored in the capacitor and all of the several hundred I did at that voltage worked well and would be capable of operating every 5 seconds or so... From a 4v (Gates cell battery) to 24V .5A switcher. There is another error in the reference drawing for the driver and that is that the Vcesat is (from the data sheet)
Collector–Emitter Saturation Voltage Vcesat
(IC = 3.0 Adc, IB = 12 mAdc) 2.0 V A "Little" high.
(IC = 5.0 Adc, IB = 20 mAdc) 4.0 V A "LOT" higher and with the heavy load imposed by an inductive load, possible.
Although a Darlington has the required current gain it cannot supply full power to the load because it is a switch driven by an emitter follower. The curves later in the sheet indicate that it might work well BUT the absolute Max ratings seem to cast a little doubt. I didn't use darlington's because of that effect. In my early designs I used a "Super Pair" an NPN Driving a PNP as a "high side switch" (the switch is in the supply rather than the ground side) and later I used an NPN driving a PNP driving a grounded source Mosfet. The NPN was an open collector switch that drove a "High Side" PNP switch that supplied 12v to the gate of the mosfet.
Doc - I think I understand the first paragraph about inserting the capacitor. - I have attached an amended diagram. - Am I correct in thinking I can use a smaller dc transformer but the capicitor would take longer to charge?
N.B I just noticed I have put the resistor the wrong side of the capicitor in this diagram
With regards to replacing the darlington transistor. are you able to recommend an PNP to use instead? Sorry but I only studied electrics at college X years ago and I didnt understand it then!
