4012 NAND driving solenoid

[rnehrboss]

So I have a 4013 NAND trying to pull a solenoid. I think what happened is I pulled too much current from the output? My inexperience with transistors.

I need to pull a 12 volt solenoid with a 28 ohm cable. I think this means I need 12/28=428 ma of current. So I chose a transistor 2N3053 which should give Ic of 700ma. so far so good?

I'm not sure that I understand Hfe gain issue. Can you provide some insight on how to make my NAND gate that only gives about 5 ma drive this solenoid?

Thanks!

 

[btbass]

Hfe can be seen as current gain. The 2n3053 has not got a lot of this. Could be as low as 25.
Take a look a Mosfet's. They have a gate threshold voltage and on resistance.
The gate takes no current, except the gate charge which is typically very low.
You will find loads that can sink 1 Amp.

 

[rnehrboss]

Thanks! I ordered up some IRL520PBF MOSFET's looks like they can sink a lot, but on retrospect, it looks like they only have an input of +-10 volts, which is a problem as I'm running 12VDC through the 4012's. Do you think this will work, or burn out? Do I need a resistor between the ouput of the 4012 and the Gate on the MOSFET?

Thanks again!

 

[IanP]

A gate resistor is recommended in most applications ..

A gate resistor limits the instantaneous current that is drawn when the MOSFET is turned on (charging gate's capacitance) ..
If you are driving a MOSFET directly from a low-current device such as logic gate then gate resistors are recommended and anywhere from 10 to 100 ohms is fine ..

Rgds,
IanP

 

[xaccto]

on the other hand if you have bjt transistor like BD139 on hand, and about 1k resistor btw logic output and BJT base, that would work for your purpose also.
such transistors are still atm cheaper than MOSFETS.........

 

[btbass]

The IRL520 is a logic level device. It will turn on with a gate level of 2V.
The +-10 is the maximum safe level of the gate source voltage before gate punch through and damage. You can use a series gate resistor and a 8v2 zenner diode between gate and source to limit the gate voltage.
Another solution is a potential divider, drive the gate through a 2K2 resistor and have a 2K2 resistor connected between gate and source. This will set 6V on the gate when driven by 12V.
In high frequency applications, the turn on/off time is controlled by the Total gate charge in Coulombs and the gate series resistor.
The Total gate charge for the IRL520 is 12nC
If you drive it at 12V with a 2K2 series gate resistor, the gate charge current will be
12/2200 = 5.45mA
If you divide nC by mA the result is in uS.
So the turn on/off time is 12/5.45 = 2.2uS.

Once the gate is charged and the device is on, it only needs ~250uA to hold it on.

The IRL520 is quite a large device.
There are a lot of Mosfets about now that can sink 1 Amp continuous in a SOT23 package.

 

[rnehrboss]

Thanks guys.

continuing:

I switched to driving the IRL520 MOSFET. This works great. Solenoid engages when all inputs to the 4012 are tied to 12 volts.

Problem is, the NAND gate is burning out (or something?) After a short time, pulling the inputs high (to 12vdc which is what VDD to the chip is) the ouput of the NAND no longer goes to 0 vdc, but 6-7vdc.

I pull the inputs low through a 47k resistor, and pull them high, directly taking them to 12 volts. Any ideas?

 

[btbass]

Make sure you have no open unconnected inputs. Tie unused inputs low.
You could increase the load on the outputs by increasing the value of the mosfet gate resistors. make them 22K nand output to gate, 22K gate to source.
I don't think turn on time is a problem in your application.

 

[rnehrboss]

RE No untied inputs. I assume this is only on one quad of the two quad input 4012 chip. I'm pulling all inputs low through 47k resistors, but one quad input I'm not using at all.

I've disconnected the MOSFET and just measure the output of one of the quad input NAND gates. Doing this, a new chip "breaks" without any other circuitry.

Could the problem be straigt to 12v high? Again, they are tied to ground through the resistors for the low state.

Help!?!

 

[btbass]

You must tie all unused inputs on all gates, any unused floating inputs can oscillate and destroy the chip! You can pull them all down to ground using one resistor, or tie them direct to ground.
Taking an input to supply shouldn't be a problem.
You should also have a decoupling cap, 100nF, connected close to the supply pin.

 

[rnehrboss]

I sure appreciate your quick replies! I'll try pulling them all down.

 

[rnehrboss]

Thanks again. I've attached my circuit diagram. The chips keep breaking, even if I don't have anything hooked to the output of the 4012.



Again, any help is appreciated.

 

[btbass]

Attached is a suggested circuit.
Switches normally closed type.

 

[rnehrboss]

Thanks! It looks like what I've done, except for tying the other NAND to ground, and adding the capacitors...and it looks like the gate resistors.

Which do you think is frying my chips? Or maybe not fried if I tie the unused gate? i'll check tomorrow.

 

[btbass]

The floating unused inputs and the lack of supply decoupling is frying your chips!

They might not be fried, leaving open inputs can make the output go into funny modes, like high speed oscillation?