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5volt 50ma input, need to output 12 up to 5 amp

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Hello,
Have a circuit already built but its output is not enough to activate a solenoid. Power supply comes from 12vdc car battery amperage a plenty. There are so many transistors out there and i've been googling for a day or two and they all try to get the voltage from the input source. I'm just trying to get the 5 volt 50ma outputs of my circuit to turn on a transistor that will output 12 volts at up to 5 amps. Solenoid is 25 ohms.


Thanks so much.
 
You can use a logic-level Mosfet to perform your task at hand.

That is if I am reading your problem description correctly, which is to drive a solenoid powered by a 12v battery, with a logic level signal.

Googling "logic level Mosfet" provides hundreds of thousands of results, including this:
https://www.sparkfun.com/products/10213

Don't forget the freewheeling diode in parallel to the solenoid coil.
 
Yes basically the input to transistor or turn on will be logic 5v 50ma and supply voltage will be 12vdc automotive battery and i wish to power 12 volt 5 amps to a solenoid when input is on. Thank you for telling me key words to search for.
 
So i'm learning I need to use a P channel mosfet to output 12 volt supply to load (if coil is used, include flyback diode. I've learned that wiring up race cars with 20 + relays and getting shocked when they all turn off)

And to use N channel if I want to output ground to a component.

Is all this correct. This really is going to change the way I wire race cars, and relays will be a thing of the past for me.
 
Is one end of the solenoid tied to +12V or to Gnd? That determines how you drive it.

Dont forget that if using a PFET as a high-side switch, it is the source that gets tied to +12V, and that it is turned on by making its gate more negative than its source, which means you need a level shifter if the switching signal originates from a microprocessor or other logic that is operated between Gnd and +5V
 
Is one end of the solenoid tied to +12V or to Gnd? That determines how you drive it.

Dont forget that if using a PFET as a high-side switch, it is the source that gets tied to +12V, and that it is turned on by making its gate more negative than its source, which means you need a level shifter if the switching signal originates from a microprocessor or other logic that is operated between Gnd and +5V
in the transmission the one leg of the solenoid solenoid is grounded(i dont have access to that easily). Inside the car to activate the solenoid i have to send 12vdc.

I'm learning now that P and N differs in the voltages that "turn on" which i think is what you just said.

My circuit activating the mosfets are 0 volts "intended off" and 5 volts "intended on" configuration that I need to amplify to 12vdc with at least 5 amp capacity.


I'm doing ALOT of googling about this so bear with me.
 
A 25 Ohm solenoid draws at most about 1/2A from a 12V battery, so a switching transistor (bjt or fet) that can handle 5A gives plenty of margin.
 
This is what you need to drive a grounded load from a 5V logic signal:
**broken link removed**
 
D1= 1n4001, 1n4002, 1n4003.......
Q2=mps2222 or 2n2222a
Q1=IRF9510 (maybe too big but a place to start)
 
after looking all day, and seeing all the options, could you suggest some model numbers for q1 and q2
My favorite for Q2 is a 2n3904 (mostly because I have about 250 of them).

For Q1, **broken link removed**. Take your pick. I've sorted to the ones that would work for you...
 
Here is an ltspice simulation of your circuit if you want to get a quick way to understand it
ltspice is a free simulator from linear.com, just load the file, and hit the running man icon.
 

Attachments

  • millenium.asc
    2.2 KB · Views: 136
Mr Flyback,

re your LTSpice file where you said:

Beware of inrush
into a solenoid, -it
may be bad if the coil resistance
is low. Some solenoids are almost
just an air cored inductor when first energised,
so there is little inductance to fight inrush
current


Please explain how there can be any inrush current into a solenoid with a finite inductance. I took your .asc file, modified it to closer reflect reality, and show both the turn-on and turn-off behavior of the solenoid driver. Note that I used much more realistic values for the solenoid inductance (~10mH, and its DC resistance = 3Ω. This makes the final current just under 4A, which is typical of an automotive door lock solenoid, for example.

I show the plot of current through the solenoid. Note that the initial current (right after the PFET turns on) is zero. The current approaches its steady-state value of ~4A, and it takes ~3 time-constants = L1*R3 to get there. Note that the settling time is consistent with what you would actually measure in a solenoid of this type...

When the PFET turns off at time=30ms, note that because of the snubber diode, the decaying current through L1 is well behaved, with the same time-constant. Note how D1 clamps the drain voltage V(drain) at one diode drop below ground (-0.7V) until the solenoid current stops flowing.

mil.gif
 
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i bought 4 of these.https://www.digikey.com/product-detail/en/SPP15P10PL H/SPP15P10PL H-ND/2783608

a couple days ago i was at radio shack they didnt have anything logic level. SO i got some TIP42's(for no reason other than pnp transistor) and played with those, R2 got hot. Then i assumed there wasnt a diode in the TIP42 like what looks to be one built into the logic level stuff. I played some more but nothing really worked. Waiting for the stuff i ordered to come in and play some more. Burned up a 2n2222 on accident, dunno how but i have atleast a couple more and radio shack stocks them.

having fun with "no idea what i'm doing"
 
You are connecting the source to +12V?
 
yes but i wasnt using a logic level at the time, was using a pnp tip42
...
A TIP42 is not suitable for this application. First, its minimum current gain is only 15, which means that to switch a 4A load, its base current would have to be about 300mA (400mA is better). There is no way to get the 2n3904 to switch that much current. No wonder the resistor between the base of the TIP42 and the collector of the 2N3904 is getting hot, because it would need to have ~10V across it at 0.4A, so it would be dissipating 4W!

A PNP Darlington would sort of work, but would have to be bolted to a fairly large heatsink. The minimum Vce for the Darlington is about 1.4V, so at 4A, it would be dissipating >5W, which takes a big heatsink.

The PFET is the only way to make this circuit work efficiently (without producing a whole bunch of heat). Since the gate drive to the PFET is ~-10V, there is no need for it to be a "logic-level" type.
 
Please explain how there can be any inrush current into a solenoid with a finite inductance
..with many solenoids, the core is the "contactor", and it contacts by getting pulled into the windings, when the windings are energised with current........at first the "plunger" is outside the windings, and so the windings have just air inside them, its only when the (high permeability) plunger pulls into the windings that the permeability of the plunger is then in the windings to give that bigger inductance.

I think a lot of solonoids are purposely built with too-high resistance just to mitigate inrush, and this then means that they waste loads of power..its a great shame when you think all that's needed is a bit of inrush protection..take solenoid valves in electric showers.....they dissipate something like 10 watts just in the coil resistance...totally unnecessary....I can hear the regulators moving in...blue angel etc
 
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Dang Mike, your still awesome. Long live the Saturn III. :)
 
..with many solenoids, the core is the "contactor", and it contacts by getting pulled into the windings, when the windings are energised with current........at first the "plunger" is outside the windings, and so the windings have just air inside them, its only when the (high permeability) plunger pulls into the windings that the permeability of the plunger is then in the windings to give that bigger inductance.

That is what happens to inductance vs plunger position, but your conclusion about what effect that has on the current in the coil is wrong. The position of the plunger effects the inductance of the coil, in a highly non-linear way, with the initial inductance low, and the final (gap closed) inductance several times higher, but the resistance of the winding does not change. The inductance would be proportional to 1/(remaining gap).

If you assume that the initial inductance is zero, then the coil current would would be entirely determined by the coil resistance. If you wait a while, the Ldi/dt term (whatever L is) goes to zero, and the current is again entirely determined by the coil resistance (with no change between initial current and final current). There is no "in-rush"!
 
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