I want to act on a valve that a pulse of 30 ms to open and the same to close it, at least once a day, with the H_Bridge design in the image included, but I have doubts if the battery will be exhausted quickly or not. Can someone help me to solve the doubt?
In the image, Rl = 9 Ohm resistance is equivalent to the solenoid valve. The Zener diodes at the base of the transistors allows me to maintain the tension in Rl about 5 volts. A voltage for 30 ms opens the valve, and other pulse in B close it.
Argh! You need to draw an H-bridge with +ve rail at the top, -ve rail at the
bottom, otherwise you'll cause enormous confusion. People talk about top-side
PWM and such all the time.
Circuit diagrams are the language of electronics, follow the conventions!
As for power consumption everything in the system may be important - in particular
what's driving the H-bridge (an Arduino? which Arduino?) are you using sleep mode,
what is the battery capacity, etc etc.
And make sure the components in the diagram can be seen on a 200dpi screen, yours
are tiny...
I'm not electronic, that is why I ask advice. Apologies for the previous attached drawing.
The theme of the Arduino that i use is not important, since I'm using an atinny85 powered independently.
What I simply want is to know if with 9V, 500 or 600mAH battery, and opening and closing the valve once a day the circuit will work for a long time. I'm not and expertized in this matter and I need some help.
I don't know about that H-bridge schematic in particular, but H-bridges in general should consume no power* in the 'coast' state.
A generic full H-bridge can have several states: Forward, high braking, low braking, reverse, and coast. (Technically, there are also shoot-through states, but we will ignore those.)
In the attached example of a half-bridge, no power is wasted in the coast state (when the two inputs are logic-LOW). Similarly, in a full H-bridge, no power is wasted in 'coast' when all four inputs are logic-LOW.
However your valve may not work right when the H-bridge is in the coast state. It may require you to keep it in forward or reverse, or I dunno, it might leak or something. In that case, you will be wasting however mA it takes to keep your transistors biased, which is likely to amount to several mA which will drain a 600mAh battery in a few hours (you'll never get the full 600mAh out of a 600mAh battery). So -- it depends.
So, why do you want to use the 'pulse' feature of the schematic you posted? If you need to keep the H-bridge in a certain state while the ATTiny sleeps, you can use an ordinary H-bridge. Whatever state (HIGH or LOW) you put an ATTiny output pin in when you go to sleep, it will stay in that state. No external 'latching' circuit is necessary.
Yeah, ok, transistors may leak a negligible few picoamps of current.
In the OP's circuit, I'm not convinced the top Darlington can ever be fully turned on. I've never seen an H-Bridge that didn't use PNP's on the +rail side. If he can only turn it partially on with a 5V pulse from the Arduino, I think you will have 7V across the top transistor set. If you look at the 2nd poster's schematic, that one is the more traditional circuit.
Yes, isn't a traditional H-Bridge as I saw searching in the web. What I want is turn forward or backward only for 30ms and only supply 5V (in reality this circuit supply 5.4V when battery is full and, I suposed, when batery supply 7v it will work correcly, but, I don't know)
to my valve. The reason is try to supply only the current necesary for open and close the valve.
Yes, rmetzner49, the transistor that you say isn't be fully turned on. This two darlington with zener diode at the top are a primitive voltage regulator.
In reality, my schematic isn't a H-Bridge. For open or close valve, 500mA for 30ms is only what I need.
The initial idea that I thought was to feed my circuit with a 9 volt battery and create a voltage regulator that provide me 5 volts which are what valve need. If I do so independent of the driver, I can package the valve independently with its control circuit. A 9 volt battery seemed more comfortable, but I always have me wrong in the concept.
But in reality, I wanted to make independent the driver that I thought that the drop of tension produced by valve could cause me problems.
The issue of all is to make it work for long periods of time, and this is the part that's going to consume more.
aglio:
Yes, isn't a traditional H-Bridge as I saw searching in the web. What I want is turn forward or backward only for 30ms
A standard H-bridge will do that, though.
and only supply 5V ... to my valve. The reason is try to supply only the current necesary for open and close the valve.
This is usually not the way to go about it. A solenoid is a coil of wire that creates a magnetic field. If you reduce the current or voltage, you reduce the strength of the magnetic field, which means your solenoid isn't as "strong". Now, if you're trying to run a 5V solenoid off a stiff 9V supply, that may be exactly what you want. But a square PP3 9V battery is the exact opposite of stiff: under a 9? load, you will probably find that measured voltage is closer to 5V than it is to 9V. Alkaline cells of that size can't source much current, so voltage drops dramatically under load.
In reality, my schematic isn't a H-Bridge. For open or close valve, 500mA for 30ms is only what I need.
A standard H-bridge will do that, though.
I thought that the drop of tension produced by valve could cause me problems.
If you mean the inductive kickback from the solenoid coil's collapsing magnetic field, yes, that is a problem, but can be dealt with by a flyback diode (and a capacitor for good measure).
The issue of all is to make it work for long periods of time, and this is the part that's going to consume more.
I don't see how your circuit is any improvement over a standard H-bridge, though.
JohnLincoln:
But if you are using the same battery to power the Arduino, then the current drawn by the Arduino flatten the battery much sooner.
aglio:
I din't know if powered coil directly could cause any damage.
Leaving a 5V coil powered at something above 5V for long periods of time could cause it to overheat. But for 30 milliseconds with a 1/2880000 duty cycle? I feel pretty confident it will be okay.
If it makes you feel better, add a 20? resistor in series with the solenoid coil. Your 9? @ 5V coil wants to draw 550mA or 2.7 Watts. If you have a stiff 9V source, adding a 20? resistor to a 9? coil brings you back in line with 2.7W of power consumption (Ohm's Law). But like I said, a PP3 isn't stiff:
A fresh PP3 can pack a little punch at first, but as you can see from the discharge curves, spends much of it's life producing about 6V with a 500mA load. You'd need to do some empirical testing, but an additional 20? series resistor means your PP3 is too weak to drive the solenoid towards the end of the curve. You might consider using a middle-of-the-road 10? series resistor to tame the bite from a fresh battery, yet still allow the solenoid to operate when the battery starts to run out of juice.
By the way, are you sure you need an H-bridge to apply reverse current to the solenoid? If all you need is to power and not-power the coil, a single solitary transistor is all that is needed.
aglio:
I din't know if powered coil directly could cause any damage.
If it makes you feel better, add a 20? resistor in series with the solenoid coil. Your 9? @ 5V coil wants to draw 550mA or 2.7 Watts. If you have a stiff 9V source, adding a 20? resistor to a 9? coil brings you back in line with 2.7W of power consumption (Ohm's Law).
I do not know how you calculate.
The coil with 9 ohm needs 5 volt the resistor needs to burn 9-5= 4 volts. 4volts / 550mA (countin in my head) about 7 ohms.
A 9 volt battery and an big electrolytic capacitor can deliver a high current for 30 ms
Alkaline batteries, in general, shouldn't be taxed at more than a 3 hour rate. You shouldn't be asking for more than 180ma from your 600maH battery. And to be precise, that 600maH rating refers to a 10 hour discharge time; you won't have that full 600maH if you discharge it in a timespan shorter than 10 hours.
AA cells, which have a higher capacity, or LiPo cells, which are capable of high discharge rates, would be a much better choice.
I'm surprised you don't put some more thought into an alternative energy source. With such little demand you could probably derive enough energy from a cpu fan being driven by a light breeze to keep the batteries topped up.
Yes, that's a good little circuit, you can adapt it for different voltages and currents by
choosing the right output transistors for the current, and adjusting the 220 resistors to
provide enough base current from a given supply voltage. Watch out for the power
dissipation figures for those resistors too.
I used Ohms Law Calculator to plug in 2.7W and 9V. I didn't verify it by hand but it says 30?. Since the solenoid coil is 9?, that leaves 21? extra series resistance to get 2.7W.
But I think I see my mistake. That's 2.7W for the solenoid and the resistor, which is wrong: we want 2.7W just in the solenoid. Good catch Pelleplutt.
