Problem with 2N3904

[harshthegreat89]

Hello everyone.....

I'm making a electronic combinational lock using D-flip-flop IC 4013 and transistor 2N3904..... I made the circuit on general purpose board.... the circuit is working but the problem is that i'm getting very low current for K1 load (Plz see the circuit diagram attached).... I'm able to glow a LED attached to K1 load, but i'm not able to operate a 6V RELAY.... I want to use a RELAY bcoz i want to operate a electromagnetic striker or something like that to make a lock mechanism which will operate on fairly high voltage & current....

Plz help me out.....

https://img261.imageshack.us/my.php?image=electroniccombinationloeg5.gif

 

[ericgibbs]

hi,
The 18K in the transistor base is much to high a value.

eg: (9-0.7)/18K = 460uA, with a transistor gain of say 50 this would give a collector current of 50 * 350 =23mA.

Reduce the resistor value to about 2K2

eg: (9-0.7)/2200 = 3.7mA

What current does the relay require.?

Is this clear.?

 

[KMoffett]

What's the state of ICb pin 13 to unlock...HI or LO? Seems a little odd to use a 2N3904 as an emitter follower output.

Ken

 

[Roff]

What's the state of ICb pin 13 to unlock...HI or LO? Seems a little odd to use a 2N3904 as an emitter follower output.

Ken

The logic is shifting a zero from left to right. It looks to me like the transistor should be a 2N3906, or other PNP.
As Eric said, the required base resistor is dependent on the relay. It should be about 15 times the resistance of the relay coil.
The LED may be overstressed with the full relay current passing through it. It should probably have a resistor in parallel with it.

 

[audioguru]

I think the 2N3904 is trying to be an emitter-follower but its emitter and collector are connected backwards. So the poor CD4013 is driving the relay and LED through the 18k resistor in series with the base-collector diode of the transistor. Less than 1mA.

 

[Roff]

I think the 2N3904 is trying to be an emitter-follower but its emitter and collector are connected backwards. So the poor CD4013 is driving the relay and LED through the 18k resistor in series with the base-collector diode of the transistor. Less than 1mA.

Could be, but the logic calls for a PNP. If he wants to use an NPN emitter follower, he needs to tie the leftmost D input to Vcc instead of GND, and tie all the "illegal" parallel switches to the Set input, instead of Reset, as he now has it.

 

[harshthegreat89]

The resistance of the relay coil is 97hm: ..........

So, as far as i know according to V=IR current drawn by relay would be (6/97) = 61.85 mA ......... Am i right ?? Or may be i'm wrong.....

Anyway the resistance of the coil is 97hm: ...... so what should be the value of the resistor connected to the base instead of 18K ??

"ericgibbs" said 2K2hm: but i'm afraid it will blow-off the transistor..... or is it SAFE ??

 

[ericgibbs]

The resistance of the relay coil is 97hm: ..........

So, as far as i know according to V=IR current drawn by relay would be (6/97) = 61.85 mA ......... Am i right ?? Or may be i'm wrong.....

Anyway the resistance of the coil is 97hm: ...... so what should be the value of the resistor connected to the base instead of 18K ??

"ericgibbs" said 2K2hm: but i'm afraid it will blow-off the transistor..... or is it SAFE ??

hi,
If its a 2N3904 its a NPN transistor, your drawing shows a PNP.

If its a NPN then its an emitter follower with the relay in the emitter, a 2K2 would still be OK.

IF its a PNP then the 2K2 is still OK.

Which is it supposed to be NPN or PNP..

Is this clear.?

 

[audioguru]

Like this.

 

[blueroomelectronics]

The poor LED below the relay will live a short and unhapply life.

 

[on1aag]

Like this.

 

[Roff]

If he wants to use a 2N3904, he can do it like this.

 

[Dean Huster]

I don't understand why one would want the LED and the relay coil in series in the first place as the currents through both may not be compatible for good operation of both. Why not just put the LED with a current-limiting resistor in parallel with the coil? Or hang it off either the Q or Q-not output of the last chip so that the transistor does't have to deal with the LED current at all.

Dean

 

[audioguru]

Yeah, the 9V battery will quickly drop to the 6V rating of the relay coil anyway.

With the 6V relay coil in series with the 2V red LED then the relay will stop working soon when the battery voltage drops.

 

[wingerr]

If you need it to be more secure, you might want to diode isolate each of the reset switch inputs to prevent someone with a DMM from just sweeping the contacts to find your switch pattern, if they're able to get access to the contacts.
Might want to check out the LS7220 digital lock IC is an alternative to the circuit if you just need to make this for the function as opposed to the project experience, it has a few extra amenities like status save.
http://www.datasheetcatalog.com/datasheets_pdf/L/S/7/2/LS7220.shtml

 

[wingerr]

Didn't know the attachments option was good for png files also, that makes it easier- I'll try it out:

 

[chemelec]

This is a Circuit I Designed about 4 years ago.

The Article is here:
**broken link removed**

The Schematic is here:
**broken link removed**

It works very good.

 

[wingerr]

The touch pad would be a neat idea but it'd probably be more important to diode isolate your resets. Otherwise a DMM would reveal all the reset inputs by a quick continuity check, and then all it would take is to touch all the remaining inputs-
Might also add some clamps to protect the inputs from ESD if there's going to be direct contact with a charged up finger-