Bistable relay as monostable - why 2 transistors ?

[@xi@g@me]

Hello everyone !
I'm still working on my 8 to 1 home made SCART switch, and found I needed to use a relay for the "slow switch" signal, because it can go up to +12V, and the 4051 I use are limited to +6V input.

Anyway, I eventually found this page :
https://www.electroschematics.com/6546/low-current-relay/

I basically understood the circuit and how it works, however I wonder why this one uses 2 transistors ? Isn't it possible to make it work using only one PNP transistor, as we usually do with a capacitor charge / discharge circuit ?
One colleague of mine stated that the 2 transistors are here to limit the current increase speed when removing the power, to avoid a extremely high voltage at the ends of the relay coil, which may damage the capacitor and / or the semiconductors, but I don't understand how and why it would work.

Can someone show me the light on this ?

 

[BobK]

The two transistor circuit gives you positive feedback, which makes the transitions fast and sharp. A single trasistor with a charging capacitor would turn on and off slowly.

Bob

 

[@xi@g@me]

The two transistor circuit gives you positive feedback, which makes the transitions fast and sharp. A single trasistor with a charging capacitor would turn on and off slowly.

Bob

Thanks for your quick answer BobK ! This is really interesting ! Can you explain further ? Or do you know a place where I could read more about this ?

 

[AnalogKid]

The circuit in post #1 is not a standard 2-transistor bistable latch. There is no latching except what is done electromechanically inside the relay. Both transistors are either on or off at the same time.

The transistors are two saturated switches in series. The only reason for two of them is that the base current through the 100K resistor is so low that a single transistor does not have enough gain to sink the relay coil current. So the first transistor turns on the second transistor, and the second transistor turns on the relay coil. This is a common situation when a circuit is optimized for very low static current consumption.

A problem with the schematic is that there is no base current limiting for T2. The circuit assumes that the limited gain of T1 will limit the T2 base current. The 2SA499 has a max gain of 200, for a T2 base current of over 20 mA. That's a lot, especially considering that T2 has a collector current rating of only 150 mA. This is poor design.

ak

 

[BobK]

The two transistor circuit gives you positive feedback, which makes the transitions fast and sharp. A single trasistor with a charging capacitor would turn on and off slowly.

Bob

Silly me, I did not look at the cicuit but assumed from title that it was a monostable multivibrator he was talking about.

So my previous post is not relevant.

Bob

 

[@xi@g@me]

Thanks you two!
I started to read things about positive and negative feedback, and was still searching the link between them and this schematic. I understand better now! so BobK it is because you did the mistake that I discovered news things I didn't even know about
AnalogKid, thanks for the explanation, I get it better now. I'll try to fix that design according to what you said

 

[@xi@g@me]

I just did some calculations, and here's what I found (I put the image here again, so no need to look for the link at the top of this topic) :

I plan to use this using a +5V voltage, and +3V bistable relay.
I also consider here that the capacitor is full and that the power is cut.

T1 :
Icb should be (5 / 100k) = 0.05mA.
Ice may be up to 10mA if we consider a gain of 200.
(here I don't see the purpose of the 10kOhms resistor here)
I guess that all the current goes to the base of T2.

T2 :
Ibe = 10.05mA.
Ice can go up to 2.01A if we still consider a gain of 200.

BUT : the capacitor to base of T2 circuit has a 47Ohms resistor, thus limiting the Ice current to (5 / 47) = ~106mA, which is well under the said maximum of 150mA for T2.

So... I think this circuit will work right, won't it ? (but consuming more than the 50mA they say in the linked page)
I also did not take into account the small voltage drop of the transistors on purpose. I'm not sure this would change a lot of things here, and the relay coil only needs 3V to operate.

 

[@xi@g@me]

... up ?

 

[AnalogKid]

Your image did not come through. The purpose of the 10K resistor is to make sure T2 turns off rather than leave the base floating when T1 is off.

ak

 

[@xi@g@me]

About the picture, that's weird... I thought it was because of the fact I was new here and it needed validation. The image displayed correctly in the post edition form.

Anyway, so I guess the 10k resistor is used as a pull-down... makes sense.

(tried again, but the image fails to display on the forum, even if I can see it here in the edit form... but whatever, this is the same image which is in the link I posted with the topic)