DC Offset Output of Emitter Follower

The circuit linked below is intended to drive a resistive load which is represented by a 47R resistor.

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It works well as is, but I would like to add an extra feature and do not know how. The maximum output swing is 8Vpp.

I would like to be able to offset the DC level of the output so the lower excursion of the signal floats above DC.

IOW so it is pushed toward the positive rail. I am willing to cut back the gain if it starts to clip.

Can someone please tell me how to do this, and make it variable via a pot so the offset can be adjusted from 0 to 4 volts.

All suggestions appreciated.

Thank you,

Mike Towner

Reply to
Mike Towner
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The usual suggestion is that you should tell us what you are trying to do. Telling us why you are trying to do it can often provide useful extra information, and can sometimes can allow us to suggest rather better approaches.

The somewhat dubious circuit you've revealed via drop-box suggests that you need more help than you've asked for.

The LM358 is a very useful device if you want something cheap and can live with rotten performance, but there are lots of better op amps around, some of which will be able to do whatever it is you are trying to do a whole better and with fewer components.

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Bill Sloman, Sydney
Reply to
Bill Sloman

Thank you for your reply.

The amp is intended to activate an electromechanical buzzer, but the speial need is that the drive signal is offest positively from DC as I described. It relates to the sound quality.

The circuit as shown produces a 8Vpp swing. I believed it to be fairly standard, apart from the 103 cap which is there to eliminate distortion.

You mention the circuit and choice of IC is not idea. If so, can you please suggest a better alternative that works off a single supply?

I will then redraw it and post the link here.

Mike Towner

Reply to
Mike Towner

A few questions:

What is the value of C1? What is the DC collector current of Q1? Is the source a short for DC?

And is the circuit really working? Assuming that the source is indeed a short for DC, the non-inverting input of the opamp seems to be at zero volt, while the inverting one is at a certain factor of the Ve and consequently would drive the transistor completely in the off-state, except for when the input voltage is big enough to have it conduct an give a distorted output signal.

joe

Reply to
joe hey

As I stated, the circuit works fine as is. I just want to ad the offet feature to the output as described in my OP.

Anyone?

Mike Towner

Reply to
Mike Towner

Resistor from node 2 of R1 to +12V will shift the input up and thus the output. It'll reduce the signal amplitude a bit, but you can compensate for that with your gain control. The amount of offset will depend on the gain.

Is your signal source bipolar, or does it go between 0V and 1V?

So 1V p-p goes to 8V p-p so a gain of 8. Shifting this to centre around

6V means adding 2V to the output which means adding 2/8 or 250mV at the input.

So lets say we want to allow double this and shift the input by 0.5V max. That means a 230k resistor to 12V - call it 220k. Put a 10k pot between 0V and 12V, connect the wiper to your 220k, the other end of which goes to R1 node 2.

The 220k is much bigger than the 10k input resistor so won't affec the amplitude by much. The 10k pot is much smaller than the 220k so won't be loaded much.

Approximate enough to play with.

Cheers

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Syd
Reply to
Syd Rumpo

I think Joe is trying to get you to understand that your circuit does not bias Q1 independent of the input signal DC level. If that is what you wanted, fine. But it is hard to explain to you how to adjust the DC bias of Q1 without understanding the DC bias of the input signal.

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Rick
Reply to
rickman

How about adding the DC offset to the input signal, the lm358 is a dual, can you use the extra opamp as a summer?

George H.

Reply to
George Herold

Here is my new circuit, which I have modified according to my interpretation of the suggestions received.

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I look forward to bench testing it tomorrow.

Mike Towner

Reply to
Mike Towner

You might want to block the DC from "seeing" the load of the input signal and R2 by using a DC blocking cap in series with R2. I'm not sure what the purpose of R6 is since the opamp will drive whatever current is required to close the loop. Does R6 minimize some internal issue with the opamp?

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Rick
Reply to
rickman

Keep in mind that Slowman is a clueless egotistical pompous ass. Killfile him and save yourself some indigestion.

However you do have some DC bias issues.

Where/what range do you want the output quiescent point? ...Jim Thompson

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| James E.Thompson                                 |    mens     | 
| Analog Innovations                               |     et      | 
| Analog/Mixed-Signal ASIC's and Discrete Systems  |    manus    | 
| San Tan Valley, AZ 85142     Skype: skypeanalog  |             | 
| Voice:(480)460-2350  Fax: Available upon request |  Brass Rat  | 
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I love to cook with wine.     Sometimes I even put it in the food.
Reply to
Jim Thompson

Not me after this answer. :)

joe

Reply to
joe hey

I'm still not sure of what you're trying to achieve or if you're approaching it the best way.

Do you really need to generate DC with an impressed AC signal on it to the "buzzer", or do you need to offset the signal away from ground to keep the opamp happy?

My first thought for doing what you say you need would be to make the amplifier circuit with a DC blocking cap on the input and having it idle at about half the supply voltage.

My second thought, if it didn't win out and become my first thought, would be to use an audio amplifier chip, of which there are tons out there.

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Tim Wescott 
Wescott Design Services 
http://www.wescottdesign.com
Reply to
Tim Wescott

This morning I built the abve circuit which incorporates Syd's recommendations and it works perfectly.

Perhaps I should have mentioned the input signal is full wave recitified, no negative-going swing.

I do like to be positive.

Thanks to all for their comments.

Mike Towner

Reply to
Mike Towner

k

do. Telling us why you are trying to do it can often provide useful extra i nformation, and can sometimes can allow us to suggest rather better approac hes.

you need more help than you've asked for.

ive with rotten performance, but there are lots of better op amps around, s ome of which will be able to do whatever it is you are trying to do a whole better and with fewer components.

Keep in mind that Jim-out-of-touch-with-reality-Thompson is an egotistical pompous ass, and kill-filing him wouldn't cut you off from all that much us eful information. He's been designing integrated circuits since the 1970s. He's too egotistical to make his ICs easy to use (or at least not the one's I've had the misfortune of having had to get to work) but they do work wel l enough to get him repeat business, and he has been known to say useful th ings about them here, not much more useful than what has been said in his p art's application notes, but still useful.

Which have been addressed in more detail by other posters

Somebody with a better grasp of reality than Jim would opt for the half-way between the rails point about 6V.

This wastes a lot of dissipation in your output transistor - a class A/B ou tput would have been better, but it's more components, and tricky-ish to bi as neatly (not that your application seems to need neat).

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Bill Sloman, Sydney
Reply to
Bill Sloman

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