Probem - circuit for switching a single coil latching relay

[rainman1]

Probem in design?

I built the presented circuit today for switching a single coil latching relay (1W, 5V rated):


I managed to switch the relay by changing the states of V2 and V3 - 0V and 3.3V states.

In switching mode (meaning V2 equals 3.3V, V3 equals 0v), the current that flowed from the 5V PSU was 250mA continuous, but the voltage drop on the relay was less than 1V.

I must say that I measured the relay's voltage after the switching of the relay and not during the switching.

There are two things that i dont understand:
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?

 

[svhb]

Re: Probem in design?

there is indeed a problem with your circuit. A pnp transistor can be seen as two diodes pointing to the base, an npn as two diodes pointing awai from the base.

If you put 0v on bot inputs, you connect your 5V supply with the E-B diodes to R4 and R3. This means indeed around 50mA of current.

If you put 3V3 as input (for example the Q1-Q3 side) you will have a base current for Q3 : (5V - 3V3 - VbeQ3)/190 Ohm, and you will have base current in Q1 : (3V3 - VbeQ1)/190. So basically you made a short in the legt leg of your bridge circuit. The current that will flow depends on the base currents and the HFE of the transistors.

Normally the transistor bridge must be cross coupled, and another transistor is used to control the base currents. See attached drawing.

Stefaan

 

[rainman1]

Re: Probem in design?

Thank you!
I learned so much from your answer.

I'd like to know please, how does T5 (pnp) causes T4 to be open (in cutoff) when IN1 = 3.3V?
I ask that because the problem with my design as you said it that Q1 and Q3 are both close (conduct) when right side has 3.3V.

Thank you!

 

[rainman1]

Probem in design?

I think you have a problem with that design, because T6 wont be able to turn off T1, because in order to do that it needs to have at least 3.8V at input, and input is only 3.3V.

 

[svhb]

Re: Probem in design?

Sorry, T5 and T6 must be npn. Just copied the wrong symbol...

But you discovered the bug, so you are on the right track...

Stefaan

 

[rainman1]

Probem in design?

Thanks.
Can you tell me please on what resistors you're recommending?
I cant find right resistors.
The relay is 5V @ 200mA (1W) rated.
and 5V Vcc can push up to 300mA.

Thank you for your help.

 

[svhb]

Re: Probem in design?

For R7 and R8 :

The voltage on the resistor is VCC - two times the Vbe of a transisor - VceSat of T5 (or T6) = roughly 5-1.4-0.2 = 3.4V
You need enough current through these resistors to make all transistors saturate, so take the less of the HFE of the pnp and the npn (the pnp will be less normally).

Ibase = Icollecor/HFE = Voltage on R8 / R8. This is a good start point for R8. Take the real value at least two times smaller to ensure saturation of the bridge transisors.

The same procedure you follow for R9 and R5, R9 and R5 form a voltage divider, but the voltage on R5 will be always one Vbe+one Vcesat (approx 0.9V). So the base current for T5 is defined by ((Vin-0.9)/R9) - 0.9/R5 . R5 is introduced as a safety factor, it will work without, but if the output that is connected to this input has leakage current, the transistor T5 can start conducting...

If you do the calculations, and add the information about the transistors you will use, I can check the results if you want...