Relay/transistor question

[Gustav]

This implementation of relays in the NetEQ circuit is confusing to me.

Why are we dropping the voltage to bring it back up with a transistor before reaching the relay? VLR+ = 12V, so why not go direct?

And a bonus question.

I sort of blocked out the LED when looking at the schematic at first, so I spent a little time trying to figure out why there were two values instead of just a single drop resistors, but.

I realized 1K is there to drop the voltage for the LED, but is there any advantage to placing it before the switch?

 

[mdainsd]

"Why are we dropping the voltage to bring it back up with a transistor before reaching the relay? VLR+ = 12V, so why not go direct?"

Not sure I understand what you are asking. The transistor is a switch the grounds one side of the relay (the other side being supplied with 12V). So Im not understanding the "bringing it back up" part.

The two resistors in series? total resistance suitable for base current. But needs to be split to provide LED current limiting.

Does it matter which side of the switch the 1K is on, no.

Or did I totaly mess up understanding your question (s)?

 

[Gustav]

"Why are we dropping the voltage to bring it back up with a transistor before reaching the relay? VLR+ = 12V, so why not go direct?"

Not sure I understand what you are asking. The transistor is a switch the grounds one side of the relay (the other side being supplied with 12V). So Im not understanding the "bringing it back up" part.

The two resistors in series? total resistance suitable for base current. But needs to be split to provide LED current limiting.

Does it matter which side of the switch the 1K is on, no.

Or did I totaly mess up understanding your question (s)?

Thanks a lot - last part clears everything up.

First part - my basic question is "Why a transistor, when we have the 12V on a switch to do the switching already?". I tried to form some idea of what was going on too, and I seem to have rotated the transistor in my attempt.

It just seems more elaborate than it needs to, but I am sure there is a good reason - and that I am failing to see what is actually going on..

Gustav

 

[ruairioflaherty]

Gustav,

Check out this thread - http://groupdiy.com/index.php?topic=43965.msg549113#msg549113

I asked the same question (among others).

Cheers,
Ruairi

 

[PRR]

> seems more elaborate than it needs to,

Agree; but we weren't there when the design came together.

Maybe they didn't want high relay current in the S2 area (though the LED current is not small).

Maybe there is a history of earlier consoles with direct switch, relay working relay, transistor switch, and they lost track of what they had.

Maybe Joe worked on part of it, Fred another part, Sue had to patch-up Joe and Fred's work, and it just came out this way.

 

[PRR]

The 1K gives about (12V-1.6V)/1K 10mA current to LED, reasonable.

The LED at 1.6V and the transistor base at 0.6V, we have 1V in 4.7K, so 0.2mA maximum into transistor base (and only 9.8mA in LED, still fine). Without that, the 0.6V transistor would hog all the 1K's current, the LED would never light.

The 10K gives a for-sure pull-down so the transistor WILL go off when S2 is open. (It probably will, but the LED is not a for-sure pull-down below 1.6V, and Defensive Design encourages a 2-cent resistor just in case S2 leaks.)

The transistor will turn on FAST. We don't need nanosecond turn-on to slap a relay around; OTOH nanosecond switching may get into the audio. The 220nF cap slows transistor turn-on.

 

[mdainsd]

And.....

putting the 1K resistor up stream of the switch protects the power rail (RY+) from a short due to a failed switch or heaven forbid the introduction of an adult beverage into said switch....

Paralleling the two switch contacts increases reliability.

One pole could have been used for base circuit and the other for the LED but that introduces failure modes. Oxidizing of one set of contacts would cause the LED to be on but relay doesnt energize, or vice versa, relay but no LED, both maddening to the operator.

 

[Gustav]

The 1K gives about (12V-1.6V)/1K 10mA current to LED, reasonable.

The LED at 1.6V and the transistor base at 0.6V, we have 1V in 4.7K, so 0.2mA maximum into transistor base (and only 9.8mA in LED, still fine). Without that, the 0.6V transistor would hog all the 1K's current, the LED would never light.

The 10K gives a for-sure pull-down so the transistor WILL go off when S2 is open. (It probably will, but the LED is not a for-sure pull-down below 1.6V, and Defensive Design encourages a 2-cent resistor just in case S2 leaks.)

The transistor will turn on FAST. We don't need nanosecond turn-on to slap a relay around; OTOH nanosecond switching may get into the audio. The 220nF cap slows transistor turn-on.

I think the lights just came on (Still not sure anyones home, though).

Thanks!

Gustav

 

[Gold]

I hesitate a little to respond to this as I'm not anything close to a designer. I think the reason a transistor is used is to isolate the relay in the case of multiple channels. Without some sort of isolation the relays would talk to each other via the ground bus.  I've been told that switching V+ is bad design practice. You want to switch the low side. This can be accomplished by putting a diode in series with the low side of a switch to isolate. My guess is that a transistor was used to manage the loads of the relay and the LED.

 

[Andy Peters]

This implementation of relays in the NetEQ circuit is confusing to me.

Why are we dropping the voltage to bring it back up with a transistor before reaching the relay? VLR+ = 12V, so why not go direct?

I know nothing about the NetEQ but I wonder if this relay was originally under the control of a microprocessor. That circuit is commonly used when one wants to drive a relay coil from a micro pin.

Perhaps a better reason is that the switch is located at the front panel, and rather than route a high-ish current signal with potential noise from spikes up to the front panel and back, they just route the low current control signal to the transistor. I assume that the transistor is located right near the relay.

I sort of blocked out the LED when looking at the schematic at first, so I spent a little time trying to figure out why there were two values instead of just a single drop resistors, but.

I see just one (R103).

I realized 1K is there to drop the voltage for the LED, but is there any advantage to placing it before the switch?

In this case, the 1K is doing two jobs. One, it is limiting the LED current. Two, it's a pullup resistor for the transistor base logic gate. They could have used one of the switch poles for the LED and the other for the transistor, but doing it this way increases reliability as noted above.

When the switch is out, R36 pulls the transistor base to ground, turning it off. The collector pulls to +VRL through the relay coil and the relay is off (in the NC position).  When the switch is in, the base of the transistor is pulled up to +VRL through the two resistors (5.7k total). The LED anode sets the voltage at the switch side of R34, call it 2 V, so the current in R103 is 10 V / 1k = 10 mA, of which 5 mA is the base current and the other 5 mA lights up the LED. 5 mA is enough base drive to saturate the transistor on (plus Vbe = 2 V so the transistor turns on). Transistor on means the collector pulls to ground, turning on the relay coil.