Emitter follower differ GND potential

So, the DC level is about 3V?

of a common

If your follower is NPN Darlington, you need to translate the DC level from 3V to (about) 10V (the Darlington will drop it to nine or so volts). If the +12 and +6 aren't floating, they DO have a common ground, of course.

One easy translator would be an NPN transistor; pull the collector up with a resistor to +12 (the collector is the output). Connect the base to +6. Connect the emitter through a resistor to your sinewave source.

Are you concerned with lack of Earth ground or are you concerned because you have 0V, 6V, and 12V? Need more info.

Bob

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of a common

Thank you. I will try this on the bench. So I understand fully can you please advise if the following is correct?

The 20Hz 5Vpp signal from my op amp buffer goes to the NPN emitter, though say a 470R resistor (no coupling cap). The base is connected to the +6V rail. The translated signal then appears at the collector side of the resistor between the collector and +12V.

How is the value of the resistor to +12V determined?

Any further relevant details would be much appreciated.

Jerry Norris

The end result is intended to be a symmetrical output swing between the +12 and +6 volt rails of the darlington.

What I need to know is how to input to it a 5V signal with its 0V ground reference.

IOW the input 0V needs to be shifted up to +6V to match the higher "ground" potential in the darlington.

Jerry Norris

Correct

It's determined by the Darlington and the load, because if you try for too much gain (voltage gain = 1 might be too much) the output will clip. With the emitter and collector resistors equal, you get voltage gain of 1.

OK, thanks. That's clear. Very handy circuit.

One last point please. Does the ground of the original signal get conected in any way to the translator you provided, or the darlington? If not, there appears to be only 1 wire carrying the signal to them.

IOW What happens to the signal 0V ground in relation to the +6 virtual ground? I feel as though I am mssing something.

Jerry Norris

To be clear, I drew up the circuit as it is now. It can be viewed here:

Many thanks.

Jerry

Actually, this is the latest version.

Thanks again,

Jerry

You're putting a lot of reverse current through that single 6V battery, which may not be such a good thing to do.

On Friday, June 10, 2011 12:21:55 AM UTC-7, Jerry Norris wrote: [about a sinewave translated to the (6V, 12V) range to drive a Darlington follower]

The 12V battery presumably has a ground return connection, though.

This scheme necessarily depends on the 6V and 12V rails being stable as the ground is; for highest precision, other translation schemes (using op amps) might be preferable.

The circuit looks like a battery charger with superimposed ripple? So, is the intent to determine the battery charge by measuring its AC impedance?

There's really no need to talk about "ground" in this type of circuit. A lot of people speak of some particular supply in a circuit as being ground, although it's really just a circuit "common"

In your circuit, the net you've labeled as GND is just a power supply node that happens to be the most negative voltage in your circuit, so this "darlington ground" you speak of does not exist. The collectors of the darlingtons simply connecto to +12V and the 2.2K ohm resistor just connects to +6V - with respect to what you're calling GND.

The circuit you're showing should get you close to what you want - depending on the load you need to drive. What is the load and what is it referenced to (i.e. GND, +6V, or +12V)?

Bob

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you

A 'symmetrical' output swing between 12V and 6V would be centered at

1/2 x (12 +6)=3D9V. So you want to shift your 0V signal reference to 9V. Then 5VPP implies +/-2.5V about 9V or IOW a maximum of 2.5+9=3D11.5V and a minimum of -2.5+9=3D6.5V. You're obviously not going to make the 11.5V on the load with a darlington.

Thanks for the pointer. Here is a circuit like what you suggest above that I designed, with a little help. I will build it up today.

No, it's just an amp with an offset signal for an experiment.

Jerry Norris

What kind of a problem is it in practical terms? For example, will the batteries tend to equalize, or create a black hole?

And if so, is there anyway around it while maintaining the orignal intent?

BTW here is the latest circuit, using an op amp.

Jerry Norris

Is that all that battery is there for? To provide a potential for the resistor load 6V down from the 12V? It does not supply current to any other loads? As your circuit stands now, it is supplying an average of (9-6)/

100=3D30mA continuous charging current. This will reduce battery capacity and eventually destroy it, the time depending on its AH. If all you need is a potential, then it can be done electronically, for a lot less, and last forever. I am assuming it is a lead-acid and you're not doing something stupid like using alkaline.

You seem to have noticed my project is a little eccentric. I have probably managed to confuse just about everyone since it's not standard practice.

The point of the exercise is that the ouput swings symmetrically between the +12V and +6V rails that are provided _solely_ by the 3 batteries. Being two in series and one in parallel providing a potential of +6V.

IOW it's an absolute potential with respect to the batteries, not floating above ground. This is physically different than establishing a +6V offset from a common used by both the signal source and the amp.

I would consider fitting an optoisolator between the load and the amp for this reason.

Sounded simple enough at the beginning, but I am still wrestling with it two weeks later. Any further insights along these lines would be much appreciated.

Jerry Norris

Doesn't this get you the same thing??? Please view in a fixed-width font such as Courier.

. . . . . . . ---------+--------+-----+--- . | | | | | . | 3.3k | | | . 6V --- | --|\| c | . - | | | >----b c . | >-||-+---|--|/ e--b . | | | | e . | | | | | . | | ----|---------+ . | | | | . | | | [100] . | | | | . +---------------------------- . | | | . | | | . | 10k | . 6V --- | | . - | | . | | | . ---------+-------- . | . --- . /// . . . . . . . .

Oh, but I do not want the ground reference to come into play.

Please have a look at this diagram which uses audio frequency with a coupling transformer, 12V audio amp module, and 3 SLA's.

Yes I know it is unconventional, but I have experiment in mind to demonstrarte the difference between relative and absolute charge.

The circuit you have kindly provided is the relative version typical to expected electronic design.

BTW is there anyway around the charging problem on the you mentioned in your previous post, without substantially changing the configuration shown.

Many thanks,

Jerry Norris

The ground reference comes into play through the grounded 6V battery whichever of those configurations you use. The three battery configuration is a waste.