Circuit to drive a 5V relay

Thread Starter

woon_h88

Joined Mar 25, 2009
49
Hi,

Im trying to create a circuit which when a short-circuit is found at load (RL), The sensing resistor (RS) will sense the change and amp using the LM324.
But the output of LM324 only able to supply about 3.3V which not enough to trigger the 5V relay to break the circuit from damage. The 5V is supply by another power supply circuit which can varies from 3~16V (Fixed at 5V for now). Is there some way i can do to get an output of around 4.5V with the use of only discrete component? Tried to put a darlington transistor using 2x 2N3904 but its double all the output of the LM324 to 4.9V.

upload_2015-12-20_0-53-49.png


Thank.
 

Thread Starter

woon_h88

Joined Mar 25, 2009
49
Hi,

Hmm, I should replace the whole darlington transistor to something like this?

upload_2015-12-20_2-5-55.png

I see the datasheet the 2N7000 need at at least 0.8V to the gate. So i have to re-adjust the gain for the LM 324 to fit the spec i set. Am i right?

Thank
 

GopherT

Joined Nov 23, 2012
8,012
Hi,

Im trying to create a circuit which when a short-circuit is found at load (RL), The sensing resistor (RS) will sense the change and amp using the LM324.
But the output of LM324 only able to supply about 3.3V which not enough to trigger the 5V relay to break the circuit from damage. The 5V is supply by another power supply circuit which can varies from 3~16V (Fixed at 5V for now). Is there some way i can do to get an output of around 4.5V with the use of only discrete component? Tried to put a darlington transistor using 2x 2N3904 but its double all the output of the LM324 to 4.9V.

View attachment 96859


Thank.

If your relay needs less than 20 to 30 mA, you can try this. If you don't have sharp on/off transitions, you can add some positive feedback for hysteresis (not negative feedback like you had)

image.jpg
 

AnalogKid

Joined Aug 1, 2013
8,485
In the drawing in post #4 you want to reverse the + and - inputs into the opamp or comparator. Still, the LM324 would have a hard time making 30 mA, even if it were sinking the current to ground. A better way is a small driver transistor as in post #3, either a 2N7000 FET or something like a bipolar 2N4401, 2222, or 3904. The bipolars would need a 1K base resistor. One end of the relay goes to the +5V and the other end goes to the collector. The emitter goes to GND.

ak
 

hp1729

Joined Nov 23, 2015
2,304
Hi,

Im trying to create a circuit which when a short-circuit is found at load (RL), The sensing resistor (RS) will sense the change and amp using the LM324.
But the output of LM324 only able to supply about 3.3V which not enough to trigger the 5V relay to break the circuit from damage. The 5V is supply by another power supply circuit which can varies from 3~16V (Fixed at 5V for now). Is there some way i can do to get an output of around 4.5V with the use of only discrete component? Tried to put a darlington transistor using 2x 2N3904 but its double all the output of the LM324 to 4.9V.

View attachment 96859

I am missing where this short you want to detect should be in respect to this circuit.

Thank.
 

GopherT

Joined Nov 23, 2012
8,012
In the drawing in post #4 you want to reverse the + and - inputs into the opamp or comparator.

ak
Since we need it to cut out when the voltage across the sense resistor gets too high. And, the relay is connected to Vcc, the polarity in my drawing works as simulated and as shown.
 

GopherT

Joined Nov 23, 2012
8,012
Hi,

The relay i got was: http://sg.rs-online.com/web/p/non-latching-relays/1966260/
The coil resistance is 168 ohm, So the current is about 29mA. Will give it a tried once i received the item.

Thank
Here is a circuit that can handle the load if your relay coil. Any NPN transistor suggested by analogkid (above) will work. Almost any NPN transistor with gain over 100 and can handle 50 mA or more will work, analog kid listed the most common types.

Adjust the pot to have the relay open at the required voltage on the current-sensing resistor.

NOTE: the op amp inputs are inverted from the earlier schematic because the transistor essentially inverts the logic.


image.jpg image.jpg
 

hp1729

Joined Nov 23, 2015
2,304
Here is a circuit that can handle the load if your relay coil. Any NPN transistor suggested by analogkid (above) will work. Almost any NPN transistor with gain over 100 and can handle 50 mA or more will work, analog kid listed the most common types.

Adjust the pot to have the relay open at the required voltage on the current-sensing resistor.

NOTE: the op amp inputs are inverted from the earlier schematic because the transistor essentially inverts the logic.


View attachment 96968 View attachment 96969
Your math may be correct. Is the objective to measure the motor current? Move that 1.3 ohm resistor over between the motor and ground.
 

GopherT

Joined Nov 23, 2012
8,012
Correction per @hp1729

Also, your relay will start chattering when you hit your current limi because the op amp will turn the transistor on as soon as the relay is cut out. You should maybe use
- a current limiter instead of relay
- a latching relay (so it stays open once current limit is reached

Even with hysteresis, you will have chattering because the relay is either open or closed.

Good luck

image.jpg image.jpg
 

Thread Starter

woon_h88

Joined Mar 25, 2009
49
Hi,

Thank you..Just tested the circuit and the relay is working!..The RL is the load (will varies). Depend on the load, the relay will trigger to stop supplying to the load.

Thank again..
 
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