PS re-read the BSR15 datasheet. It's 1.6V drop at -500ma
Interesting. Seems high for a non-darlington transistor.
Ken
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PS re-read the BSR15 datasheet. It's 1.6V drop at -500ma
Did you turn your PNP transistors over? To turn the PNP transistors OFF, you need to drive their bases higher than V1-Vbe...+5V-0.6V...~+4.5V. +3.3V will turn the NPN transistors ON, but not turn the PNP's OFF. You need to do that the get 5V across the coil.
Not sure where your priorities are if you are talking mass production. A 2-coil latching relay, 2 resistors, and 2 transistors, would seem cheaper. Are you stuck on this design for some reason...like...no,I won't go there.
Ken
The only way to create a new design that may be less complex is for you to fully explain what you want to do and provide all the parameters.
What is the relay part number?
1. The relay must consume at least 75% of 5V in order to change its state, how come that its coil's voltage (that shouldn't change after switching) was less than 1V?
2. Another problem that occured was that when both inputs V1,V2 were 0V, the current from the 5V PSU was 50mA.
It means that the current were flowing from the PNP collector to the base? Should it happen?
Assuming your second schematic is what you built and the right side has +3.3V applied and the left side 0V, Q4 and Q2 are both "turned on" which is wasting power by tending to short out the +5V supply. The voltage at the right side of the relay will be somewhere between 1.6V and 3.4V, depending greatly on the particular transistors personalities (characteristics), as the 5V is voltage divided between them. The left side of the relay will have somewhere around 4V because Q3 is also turned on. With 4 volts on the left side and 1.6-3.4V on the right side, you can see why the relay coil only has about 1 volt on it.
With both inputs at 0V, Q3 and Q4 are turned on, the base current of each of those transistors being 23 mA, for a total of 46 mA. (5V-0.6V)/190 ohm= 23 mA. 0.6V is the base-emitter voltage.
I looked over this design, are you seeing any problem with it?
Do you know about a design where this problem is fixed?
To turn a PNP off, the base voltage needs to be at least as positive as the emitter. You only have 3.3V on the base of Q4, while the emitter is at 5V. Under this condition, Q2 and Q4 will both be on, and Q4 will most likely will die due to excess power dissipation. In your simulation, try probing the currents in Q2 and Q4.That could be a great idea.
Before we go to the next level (adding parts which make the design to cost more - we wanna go for mass prodution), i really want to understand what is wrong with this design, why the relay's voltage wasn't around 5V, and why was the stand-by current 50mA and not less?
Using the same number and type of components as in your latest design, the attached circuit will work. You should optimize the resistor values for the greatest margin and performance with your transistors.
The circuit is very similar to the Collin55 motor circuit with a few tweaks for two inputs instead of one so you can turn all transistors off.
I would like to switch up to 16Arms loads.
I would like that the relay would consume 1W when switched, and minimum voltage drop of 3.75V @ 267mA, or maximum voltage drop of 5V @ 200mA.
I would like the relay driver to be as low cost as possible (demand of head of crew).
None of your requirments make any sense.
You have got to stick with my line of thought.
What have you provided in your first discussion is an "H-bridge" An H-bridge is designed to deliver a voltage (and current) to a load in one direction (and then in the opposite direction). It is mainly designed to drive a MOTOR in forward and reverse direction. This is not what you want.
What do you want to do?
I assume you want to operate a relay from the output of a microcontroller.
This is a very simple requirement.
How many times per second do you want the relay to be activated?
What else do you want to do?
A potential problem with this design is it raises the voltage on the I/O pins of the microcontroller above the chip's Vcc. See **broken link removed**.Using the same number and type of components as in your latest design, the attached circuit will work. You should optimize the resistor values for the greatest margin and performance with your transistors.
The circuit is very similar to the Collin55 motor circuit with a few tweaks for two inputs instead of one so you can turn all transistors off.