Voltage follower output protection?

[LeoTech]

Hey EEvblog community,

I have been a viewer of Dave's content for quite some time, and I have been using the forums for browsing a lot lately - and finally decided to join myself, by asking a question.

But first of all, I would like to thank all of you for this - until now - awesome forum for learning and getting inspired!

Chatter aside, let's get down to business!

I am currently working on a current buffer, to improve the output of my signal generator, and I am thinking of using a basic op-amp voltage follower with a transistor for higher current. The basic topology can be seen in the attached image - made with EasyEda. Capacitors and other components are obviously missing. 

So, I was wondering how well the output of this is protected?

The output of the circuit is reverse voltage protected, right? As the transistor already acts like a protection diode, by preventing current flowing in the wrong direction - I might be wrong, so feel free to correct me. But is this even necessary at the output?

Besides that i would like to include protection for:

• Short circuit
• Over currentdraw

I have some ideas, but i doubt that they are the best solution.

Thanks in advance!

Leo

Btw, I am familiar with basic electronics (laws, concepts etc.) and components, so don't have to go easy on me !

[Zero999]

That circuit can only source current, not sink it.

The transistor only protects against the output voltage being 5V above the base voltage, because that's the typical reverse voltage rating of the base-emitter junction.

You need a current booster with a push-pull output stage.

See the following application notes:
http://www.ti.com/lit/an/snoa600b/snoa600b.pdf
http://cds.linear.com/docs/en/application-note/an18f.pdf
http://www.ka-electronics.com/images/pdf/Op_Amp_Booster_Stages_pt1.pdf

[David Hess]

You do not say what your bandwidth requirements are but I would probably do something like that shown below but with current limiting transistors added across the output transistor emitter resistors.

[LeoTech]

That circuit can only source current, not sink it.

The transistor only protects against the output voltage being 5V above the base voltage, because that's the typical reverse voltage rating of the base-emitter junction.

You need a current booster with a push-pull output stage.

See the following application notes:
http://www.ti.com/lit/an/snoa600b/snoa600b.pdf
http://cds.linear.com/docs/en/application-note/an18f.pdf
http://www.ka-electronics.com/images/pdf/Op_Amp_Booster_Stages_pt1.pdf

Thanks for clearing up the reverse voltage protection!

I was indeed aware of the fact that the circuit only sources current and not sinks current. As I only need it for positive voltages - hence only source.

I should have been more clear on that.

But I will have a look at the links anyway - can't hurt!

Leo

[Zero999]

That circuit can only source current, not sink it.

The transistor only protects against the output voltage being 5V above the base voltage, because that's the typical reverse voltage rating of the base-emitter junction.

You need a current booster with a push-pull output stage.

See the following application notes:
http://www.ti.com/lit/an/snoa600b/snoa600b.pdf
http://cds.linear.com/docs/en/application-note/an18f.pdf
http://www.ka-electronics.com/images/pdf/Op_Amp_Booster_Stages_pt1.pdf

Thanks for clearing up the reverse voltage protection!

I was indeed aware of the fact that the circuit only sources current and not sinks current. As I only need it for positive voltages - hence only source.

I should have been more clear on that.

But I will have a look at the links anyway - can't hurt!

Leo

Because it can only source current, the slightest bit of output capacitance will stop it from working and the circuit is only really good for DC, so no good for boosting the output of a signal generator.

One thing you can do to improve the performance is add a resistor from the output to the negative rail. This will provide a path for current flowing in the opposite direction, so output capacitance is less of an issue. A current sink would be ideal, but it's more complex.

The current limit can be implemented, by adding another transistor and sense resistor to turn the output off, when the voltage across a the resistor exceeds the base-emitter threshold voltage.

[f5r5e5d]

you could just use an op amp that has the current output you need with built in protection

CFA op amps for DSL drivers are a class, the THS6012/TPA6120 has been used in many headphone amplifiers, some hobbyist PCB, kits

400 ma per side, dual op amp - parallel the outputs with 25 Ohm series ea and you're mostly limited by supply and heatsinking

[David Hess]

There are also circuits which allow operational amplifiers to be used in parallel increasing their output current and power dissipation limits.

[LeoTech]

There are also circuits which allow operational amplifiers to be used in parallel increasing their output current and power dissipation limits.

you could just use an op amp that has the current output you need with built in protection

CFA op amps for DSL drivers are a class, the THS6012/TPA6120 has been used in many headphone amplifiers, some hobbyist PCB, kits

400 ma per side, dual op amp - parallel the outputs with 25 Ohm series ea and you're mostly limited by supply and heatsinking

Yeah, i thought about that, but i want to use the components i have at hand.

Thanks anyway for the recommendations.

[LeoTech]

That circuit can only source current, not sink it.

The transistor only protects against the output voltage being 5V above the base voltage, because that's the typical reverse voltage rating of the base-emitter junction.

You need a current booster with a push-pull output stage.

See the following application notes:
http://www.ti.com/lit/an/snoa600b/snoa600b.pdf
http://cds.linear.com/docs/en/application-note/an18f.pdf
http://www.ka-electronics.com/images/pdf/Op_Amp_Booster_Stages_pt1.pdf

Thanks for clearing up the reverse voltage protection!

I was indeed aware of the fact that the circuit only sources current and not sinks current. As I only need it for positive voltages - hence only source.

I should have been more clear on that.

But I will have a look at the links anyway - can't hurt!

Leo

Because it can only source current, the slightest bit of output capacitance will stop it from working and the circuit is only really good for DC, so no good for boosting the output of a signal generator.

One thing you can do to improve the performance is add a resistor from the output to the negative rail. This will provide a path for current flowing in the opposite direction, so output capacitance is less of an issue. A current sink would be ideal, but it's more complex.

The current limit can be implemented, by adding another transistor and sense resistor to turn the output off, when the voltage across a the resistor exceeds the base-emitter threshold voltage.

Well, it looks like I have to return to the books and drawing board again!

The part about DC only is by design, I only need the positive part of an arbitrary signal. (I am gonna add that to the original post to avoid further confusion.)

If I were to add R4 to the design wouldn't that create a voltage divider with R3, and thereby change the output and feedback voltage? Or do I just have to increase R4 many times higher than R3?

Thanks,

Leo

[Zero999]

If I were to add R4 to the design wouldn't that create a voltage divider with R3, and thereby change the output and feedback voltage? Or do I just have to increase R4 many times higher than R3?

No, because R3 comes before the feedback point, so the op-amp will increase its output voltage to compensate for any voltage drop across R3.