Dumb question: translating a voltage

[carbonzit]

I say dumb because I oughta know this, but still am having trouble with the implementation details.

Let's say a guy wanted to build this circuit (the famous "LED throbber"):

But let's say that guy also wants it to work a little differently. Instead of a positive-going triangle wave, he wants one that goes both positive and negative, like this:

**broken link removed**

Now, that's got to be pretty easy. I'm just shifting the whole wave downwards, so I just need add (subtract?) a negative DC offset to the wave, right? But now would a guy do that?

The only part of the circuit I'm concerned with here is this:

**broken link removed**

So since I can't "add" a negative voltage to the base, I'd want to make the emitter more positive with respect to the base, which would have the same effect, right? So how would I do that?

Note: Please don't respond by telling me there's a better way to do this. My goal here is not to start a production line of widgets with throbbing LEDs; my aim is to learn an electronic principle. So humor me and let me do it this way, even if this is the most ridiculous possible route, OK?

By the way, even though the author of the circuit gives us a variable "virtual ground", I don't think this will work to bring the triangle wave below zero, will it?

 

[BrownOut]

Well one way would be to use bipolar supplies, so that the emitter would return to some negative V-.

PS I think VR2 in your circuit can be adjusted to translate the voltage the way you want, if you were using bipolar supplies.

 

[carbonzit]

OK, that's one way, but not the way I want to do it. Any way to do it with a single supply? I ASS-U-ME I just need to make the base swing negative with respect to the emitter here.

 

[birdman0_o]

You're making it negative so there's a 50/50 duty cycle?

 

[BrownOut]

OK, that's one way, but not the way I want to do it. Any way to do it with a single supply? I ASS-U-ME I just need to make the base swing negative with respect to the emitter here.

That would cut the transistor off. But you can't do that unless you have bipolar supplies. The emitter returns to ground, and you cant get more negative than that. In fact, you won't even be able to generate a negative going pulse with the single supply. But... you can use a resistive divider to make the emitter return to Vcc/2.

 

[carbonzit]

You're making it negative so there's a 50/50 duty cycle?

Yes, more or less. I want a symmetrical waveform w/respect to the X-axis here.

 

[crutschow]

Aww, you don't want me to tell you how to do it better??

One way to level shift the waveform and get it to go negative without a negative supply is to couple the signal through a series capacitor and resistor to ground. The average value across the resistor will be 0V which will be the half-voltage point on the triangle wave. The output will then be plus and minus 1/2 the pp value of the wave (for a symmetrical wave).

 

[carbonzit]

That would cut the transistor off. But you can't do that unless you have bipolar supplies. The emitter returns to ground, and you cant get more negative than that. In fact, you won't even be able to generate a negative going pulse with the single supply. But... you can use a resistive divider to make the emitter return to Vcc/2.

Doh! I guess that should have been obvious to me: since there's no negative voltage coming in, I can't expect any going out, can I? OK, back to the drawing board ...

 

[carbonzit]

One way to level shift the waveform and get it to go negative without a negative supply is to couple the signal through a series capacitor and resistor to ground. The average value across the resistor will be 0V which will be the half-voltage point on the triangle wave. The output will then be plus and minus 1/2 the pp value of the wave (for a symmetrical wave).

Like this?

**broken link removed**

OK, but now I'm having a hard time understanding how I get a negative-going part of the wave out of this, since my supply rails are + and 0, not + and -.

Isn't this basically just a voltage divider, but one with a reactive component (C)? And why would the average value across the resistor be zero? Wouldn't that depend on the exact values of R and C?

 

[birdman0_o]

I'd like to understand what you are trying to accomplish with the circuit, please elaborate.

 

[carbonzit]

Exactly what I've stated. You don't need any more information. I want a symmetrical (about the voltage axis) triangle wave that goes positive and negative, instead of the zero-to-positive wave generated by the original circuit.

Why I want this is of no consequence here. I just do, and I want to learn how to accomplish this, if possible. (If not possible, that's the end of the lesson here.)

 

[crutschow]

Like this?

**broken link removed**

OK, but now I'm having a hard time understanding how I get a negative-going part of the wave out of this, since my supply rails are + and 0, not + and -.

Isn't this basically just a voltage divider, but one with a reactive component (C)? And why would the average value across the resistor be zero? Wouldn't that depend on the exact values of R and C?

OK. I should have stated that the circuit would have to go at the emitter output of the transistor and whatever load you want to drive is at the output of the capacitor.

When you use a capacitor as a level shifter you make it large enough so that its reactance is very small at the frequencies of interest. Thus virtually all the AC voltage will appear across the load. Since a capacitor can't pass DC the average output of the capacitor will be 0V.

 

[carbonzit]

OK. I should have stated that the circuit would have to go at the emitter output of the transistor and whatever load you want to drive is at the output of the capacitor.

OK, like this?

**broken link removed**

I'm still mystified how one can get a negative-going voltage out of this. Is this because of the capacitor discharging during the bottom part of the waveform?

 

[crutschow]

OK, like this?

**broken link removed**

I'm still mystified how one can get a negative-going voltage out of this. Is this because of the capacitor discharging during the bottom part of the waveform?

You need a transistor bias resistor from the emitter to ground. With the capacitor as you show there is no DC path for the transistor current.

Since the cap output average is zero then any output signal through the cap has to have an average output of zero. For a symmetrical triangle wave the average is 1/2 the pp voltage. See circuit below.

 

[carbonzit]

Thanks. I'll try that.

Then I'll soak my head and try to figure out how the damn thing works!

 

[carbonzit]

It works!

Here's what I ended up with:

**broken link removed**

I know it's going negative because both LEDs pulse (alternately, of course).

Thanks! You're a gentleman and a scholar.

Now, of course, I want tweaks. Never satisfied.

I can tell by the behavior of the LEDs that they're underpowered and are only lighting at the very peaks of the waveform. So I need more volts, man. (Larger peak-to-peak swing.)

How can I get this? Different resistor values? or how about another transistor in there? (Darlington pair?) Can you cook me up a quick circuit fix? (If you're wondering what transistor I used, it's one I grabbed from my junk box. Looks like a Motorola piece, marked "7128/644". Maybe I should use a different xistor?)

Then I promise I'll sit down and study this and figure out how it works.

 

[crutschow]

It would help if we knew the value of V+ and the signal voltage at the transistor base.

Since you have added an extra LED you can remove the 1k resistor in parallel. That may increase the output some.

 

[Jaguarjoe]

Don't forget that the first couple of volts or so that is across the LED's will be lost until Vf is reached.

 

[carbonzit]

It would help if we knew the value of V+ and the signal voltage at the transistor base.

With a supply of 9 volts, I get a base voltage swing of 4.5 to 6.2V.
With a supply of 15 volts, I get about 7 to 9.5V.
I'd like to be able to run this off a 9-volt battery.
Obviously, this transistor stage isn't working too well. I'd like suggestions on how to improve it.

My own thoughts, for what they're worth:

• A higher-gain stage, instead of this emitter follower (A=1, right?)
  
• A Darlington pair
  
• A common-emitter stage followed by an emitter follower
  
• or ???

Since you have added an extra LED you can remove the 1k resistor in parallel. That may increase the output some.

Did, but it didn't make much difference.

 

[crutschow]

What is generating the base voltage? 7 to 9.5V gives you only a 2.5Vpp signal so I'm not surprised you can barely see the LEDs light.

A Darlington won't help since you need to a higher voltage swing and a Darlington follower has a slightly lower gain than a single-stage emitter follower.

Certainly a common-emitter stage driving the emitter follower should help if you bias it properly. Without seeing your circuit I can't help you there.