Switching & Powering Relays

[wizard333]

1) When is it and is it not appropriate to use a transistor to switch the coil current for a relay? I.e. switching the relay by referencing the base of a transistor to ground where the collector is connected to the coil, vs. just grounding the coil directly to switch the relay. I see it done both ways.

2) If you look at this layout for a Ceriatone D copy, look at the relay power section. It shows a 6V tap through a standard voltage doubler being regulated by a 7812. Why would this work? Shouldn't the input voltage be 2.5v above the output voltage for the regulator to work, which wouldn't happen by using a voltage doubler on a 6v tap?

:

[Enzo]

Is it better to scratch my back with a regular screwdriver or a phillips screwdriver?

You can do it either way, or other ways, to suit yourself. One way is "better" than another because it fits better into your overall scheme, or makes use of what is already available instead of having to add something. If you already have 12vDC in an amp, I would use 12v relays instead of adding a 24v or 5v supply just for relays of those voltages.

A relay coil is a magnet. COmplete a circuit through it for current and it will energize and throw its contacts. The relay has absolutely no idea whether you used a transistor or a clip wire or an op amp or a triac or a switch.

If I want to control several relays at once, instead of several lines out of the amp, I can run one to control several transistors, which in turn control relays. One nice little circuit is the inverter. Maybe I want a couple relays to alternate. WHen one is on the other is off and vice versa. All I have to do is control the base of one transistor and its relay, then conect the collector of that transistor to the base of the other relay's transistor. Now when I pull up the first transistor base, it pulls the collector down to energize the relay, but it also then pulls the base of transistort #2 down, so it releases its relay. Now when I drop the base of the first guy, it allow the colector to return to full voltage, which also pulls up the base of #2, turning it and its relay on. SO one line to one switch will select which of two trasnsistors and thus two relays is live.

Can you do us a favor? The layout is fine and all, but if yo want to discuss a circuit, it really is better to look at a schematic. My own opinion of course.

Forget for a moment any doublers. If you rectify and filter 6.3VAC, what do you get? About 8.9VDC. Through a doubler then the same principles apply. Remember, that when rectified and filtered, AC charges the supply caps up to the peak voltage, not the RMS voltage.

[loudthud]

Most voltage regulator chips don't need 2.5V input to output differential unless the load current is near maximum. If you are using a 1 amp regulator to supply 100mA, often you only need 1.3 volt or so. You must look at the chip's data sheet. Even if the low part of the ripple voltage falls below the minimum, you only get a brief divit in output voltage which relays ignore. Duncan's PSUD-II has a model for a voltage doubler, run a simulation and see what you get for output voltage. Or just build it.

[wizard333]

Enzo - So other than situations where you want to control more than one transistor/relay with a single switch, there is no reason to use one or the other? Seems like, if you don't need to do that, then not using the transistor sure uses less parts etc. There are no current draw reasons to use a transistor?

As far as schematics, there isn't one for that amp, but the layout pretty much speaks for itself as far as that part of the circuit goes, it is as straight forward as a schematic. Middle left hand side.

If you rectify and filter 6.3VAC, what do you get? About 8.9VDC. Through a doubler then the same principles apply. Remember, that when rectified and filtered, AC charges the supply caps up to the peak voltage, not the RMS voltage.

If you are using a 1 amp regulator to supply 100mA, often you only need 1.3 volt or so.

Right. In this case it's not pulling off the heaters, it's a separate 6-0 tap on the PT, but in any case you don't get more than 12V by using a voltage doubler on a 6VAC potential here, or am I missing something? And you need more than 12V to run a 12V regulator correct?

[loudthud]

Right. In this case it's not pulling off the heaters, it's a separate 6-0 tap on the PT, but in any case you don't get more than 12V by using a voltage doubler on a 6VAC potential here, or am I missing something?

Yes! You don't have a basic understanding of how power supplys work. Enzo tried to explain it above. Look at the pic below. The transformer is 6.3VAC RMS (green trace) but the output is 14.4VDC (red trace).

[wizard333]

I think I have a basic understanding of it. Voltage will vary with load correct? Won't the load change when a relay is charged vs uncharged? Perhaps you can explain it in detail point by point rather than posting the graph? The graph is a simulation of a result, it does not explain the result. Can you?

[wizard333]

No detailed expalanation? I remember from school that the voltage doubler "doubles the voltage at half the current", so I'm still confused as to how you can get more than double under load. That PSUD2 circuit doesn't have regulation; does that isolate the loading on the supply as mutiple relays are charged/uncharged?

[loudthud]

What part of the quote below do you not understand? Did you ever wonder how a 350-0-350 transformer makes a B+ of 500V?

Forget for a moment any doublers. If you rectify and filter 6.3VAC, what do you get? About 8.9VDC. Through a doubler then the same principles apply. Remember, that when rectified and filtered, AC charges the supply caps up to the peak voltage, not the RMS voltage.

Does the square root of 2 ring any bells?

I remember from school that the voltage doubler "doubles the voltage at half the current"

Do you remember how to calculate the peak voltage of 6VAC? I realize electronics education isn't what it used to be but this should have been covered.

It's true that any DC voltage will drop under load. How much it drops depends on the source of voltage and the amount of the load. In the simulation I used 100mA as a load. PSUD2 allows you to use a stepped load so you can see the voltage at two different load currents. How much will your relays draw?

[wizard333]

Does the square root of 2 ring any bells?

You get rectified 8.82v DC nominal off 6.3 VAC if you are using solid state rectification; 1.4*6.3=8.82.

ever wonder how a 350-0-350 transformer makes a B+ of 500V?

No, I haven't. With a full wave rectifier, 350*1.4 = 490.

So if I follow your snotty remarks, 6.3*2*sqrt2= 17.8, at least unloaded. Would have been much easier to just say that without being obnoxious.

[loudthud]

You get rectified 8.82v DC nominal off 6.3 VAC if you are using solid state rectification; 1.4*6.3=8.82.

No, I haven't. With a full wave rectifier, 350*1.4 = 490.

So if I follow your snotty remarks, 6.3*2*sqrt2= 17.8, at least unloaded. Would have been much easier to just say that without being obnoxious.

If you are able to figure all that out, why did you ask this question:

... but in any case you don't get more than 12V by using a voltage doubler on a 6VAC potential here, or am I missing something? And you need more than 12V to run a 12V regulator correct?

You won't get very far in the electronics industry or in amp building until you get the ability to figure things out for yourself. You can't post every question on a forum. Do a search, read a book, run a simulation or just build it. The knowledge you gain will last longer. PSUD2 is a powerfull tool and it is pretty accurate if you take the time to build the models (especially transformers) as close as possible to real world parts. An example is the 6 volt winding on that transformer. Chances are it's more like 6.5VAC into an open circuit and drops to 6VAC under load.

So if I follow your snotty remarks, 6.3*2*sqrt2= 17.8, at least unloaded. Would have been much easier to just say that without being obnoxious.

You were given the answer twice. Enzo and I tried to explain it to you. Enzo explained about peak voltage and I ran a simulation to show that the answer wasn't 12V. The rest is left as an excercise for the student.

[diagrammatiks]

it actually makes very little sense why that ceriatone does it that way or dumble if that's an accurate copy.

6v gives you 8 or so gives you 6 again through a 7806 runs a 5vdc relay.

...

hooray.

[mooreamps]

it actually makes very little sense why that ceriatone does it that way or dumble if that's an accurate copy.

6v gives you 8 or so gives you 6 again through a 7806 runs a 5vdc relay.

...

hooray.

Perhaps you were missing the "other half" of the equation... For example, for a transformer rated at 6.3 volts, the rating could be something like 6.3 volts at 4 amperes. So, with a proper input voltage of 115 VAC, and you put a 4 ampere load on it, then the device is rated to provide 6.3 VAC ; under those conditions...

Having said that, input voltages vary, and actual loads placed on the transformer depends on the part installed.
Unloaded, the output voltage tends to measure a little higher.. Also, if "unloaded", and DC rectified with filter caps charged, then the measured voltage is higher still.

-g

[diagrammatiks]

so what?

is it going to produce between 6.3vdc and 8vdc after rectification?

is it not going to do that? Is is going to make less?

[Enzo]

And a relay rated for say 12v will generally be happy as a clam at 10v or 14v. It is just a magnet coil, not a precision transducer.

[mooreamps]

so what?

is it going to produce between 6.3vdc and 8vdc after rectification?

is it not going to do that? Is is going to make less?

It all depends upon the load. It's a sliding scale.
No load ; 8.8 volts dc
At rated load 6.3 volts dc.

Most common approaches are dc regulators. Me personally, I was looking at an A/C regulator. {less problematic I'm thinking}

-g