Repeatable protection on the relay - problem. - Page 1
Forum section: electronics / beginners

author: kacper10714
author_rank: Newbie
timestamp: October 29, 2017, 09:30:08 am
content: Welcome.I apologise for my weak English:)I made a chip on 3 transistors. The T1 transistor acts as a current sensor. T2 turns on the relay by disconnecting the conductivity. T3 is designed to lock the safety catch to reset it manually.After the first modification 2 resistors and a blocking diode on the relay coil (black elements) were added.The circuit now switches on during the current flow and disconnects it as if I was pushing a button, not as before that it is running all the time.The problem is that the protection is activated at 0.3V - 0.4V Ube T1, and not as I would expect 0.65V - 0.7V.How to solve this problem?

author: capt bullshot
author_rank: Super Contributor
timestamp: October 29, 2017, 09:39:04 am
content: This is normal, as the transistor starts to conduct at lower base voltages than your "textbook 0.7V". Either make the sensing resistor R1 smaller to compensate for the lower voltage, or use a resistive voltage divider from the right hand side pin of R1 to the base of T1.Anyway, you should add a small series resistor (100R ... 1k) into the base of T1 for protection of the transistor. If you have a short circuit at the output, without the resistor a quite high current can flow through the base of T1, destroying it.

author: kacper10714
author_rank: Newbie
timestamp: October 29, 2017, 10:23:08 am
content: "resistive voltage divider from the right hand side pin of R1 to the base of T1"The resistor based on T1 will act as a T1 protection, but also as a divider?Calculate this resistor?

author: capt bullshot
author_rank: Super Contributor
timestamp: October 29, 2017, 11:48:19 am
content: The new resistor is OK, acts as protection for T1. Value should be in the range 100R ... 1k for the protection. As part of a voltage divider, the value can be the same. For example, use two 1k resistors to set the trip point to about 0.7 ... 0.8V voltage drop at R1.Vbe of a transistor will give a quite coarse threshold for the trip current, if you need exact values, you should better use some kind of comparator.Maybe useful will be a capacitor across B and E of T1 to slow down the circuit, otherwise short peaks (e.g. from a charging capacitor at the output) may trip the detector. C2 in your circuit does a similar job.

author: danadak
author_rank: Super Contributor
timestamp: October 29, 2017, 12:21:36 pm
content: Add a series diode in base of current sensing transistor, then it will"turn on" in ~ 2 x Vbe.Of course recalculate Rsense.Regards, Dana.

author: kacper10714
author_rank: Newbie
timestamp: October 29, 2017, 12:28:21 pm
content: C2 is to cut out the current spike.I do not understand how to draw this divider? where should I give my second divider resistor? One resistor should be on the gateway as it is and the other resistor connected to the emitter?Adding a diode in series with the gate will not cause me any more problems, more power loss on the measuring resistor?I forgot to add that the protection is to be disconnected at 3.5A - 4.5A. Up to 3A should not disconnect anything.

author: capt bullshot
author_rank: Super Contributor
timestamp: October 29, 2017, 06:43:16 pm
content: Yes, one resistor is from the right hand side of R1 to base of T1, as drawn. The other is added from base of T1 to emitter of T1 / left hand side of R1.The resistors are connected in series across R1, and the tap is fed to the base of T1.By using the voltage divider formula on this you can adjust the trip point for the circuit. At 3A, there's 0.45V drop across R1 that should not turn on T1, but it does. Using a divider by 2 (two resistors of 470R), 0.225 is applied to T1's base. At 4A, there's 0.6V drop, resulting in 0.3V for T1, probably still to low to trigger the circuit. So go for a lower ratio, maybe 1.5, easy to make out of a 1k and a 470R resistor. This gives 0.4V for T1 at 4A, and 0.3V at 3A. There's still quite a large uncertainty where the trip point will be and it'll vary over temperature.As I said, defining a precise trip with with this kind of circuit is difficult. You'll have to design with a large margin between largest current that doesn't trip and smallest current that must trip.Connecting a diode in series results in approximately doubling the trip point. You may then have to use a larger value for R1, which will result in higher power loss.

author: kacper10714
author_rank: Newbie
timestamp: October 29, 2017, 07:03:32 pm
content: It is important that it works above 3A