help understanding BUFfer 634

Thread Starter

justtrying

Joined Mar 9, 2011
439
So, my summer PID job has buffers. I know why buffers are used, but like everything theory and practice are two different things.

Like... I thought buffer would provide a stable current regardless of what happens on the input side (I now this is horrible oversimplification, but that is my basic understanding to the extend of one hour lecture that covered buffers). In the circuit that I am working on, buffers can apparently provide anywhere between 0 to 250 mA of current each. I do not understand this at all. I have read the spec sheets for BUF 634, and it has not made it any clearer.

If someone could make any suggestions... the circuit attached is not exactly the same that I am working on, but is its predecessor and operates on the same principles.

p.s. it will supply current to a laser, or so I am told.
 

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praondevou

Joined Jul 9, 2011
2,942
I had a look at it's datasheet. It says:

"Internal circuitry is protected with a diode clamp connected
from the input to output of the BUF634—see Figure 1. If the
output is unable to follow the input within approximately 3V

(such as with an output short-circuit), the input will conduct
increased current from the input source. This is limited by
the internal 200Ω resistor. If the input source can be dam-
aged by this increase in load current, an additional resistor
can be connected in series with the input."

So from what I understand (I may be wrong though) the output voltage follows the input voltage because it has unity gain. However, the output is able to provide up to 250mA. So it's just a normal (analog) buffer that can provide much more current than the previous stage.

In the circuit you posted the current to the laser would be measured by the differential amplifier whose output provides the negative feedback to the loop.
 

SgtWookie

Joined Jul 17, 2007
22,230
<snip>
Like... I thought buffer would provide a stable current regardless of what happens on the input side <snip>
Nope - the buffer attempts to mimic the input voltage at the output by sinking or sourcing current from/to the output, up to it's 250mA limit.

The buffer's input impedance is very high, and its' output impedance is very low.

Another name for a unity-gain buffer is a "voltage follower". Google will show you lots of examples of using an opamp as a voltage follower or unity-gain buffer.
 

MrChips

Joined Oct 2, 2009
30,708
Nope. Buffers are used to reduce loading effect, that is, to relay a signal without affecting its amplitude. It can do this because it has a very high input impedance.

It can also be considered a current booster. If the previous stage has high output impedance, it cannot supply much current to the load. The buffer has low output impedance and may be able to drive low impedance load.
 

Thread Starter

justtrying

Joined Mar 9, 2011
439
I think I am starting to get it. I knew that buffers were used as interface between components and to control loading effects, but was unclear about how they functioned.

I have another question. Attached is the actual circuit I am trying to work out. It is designed to be multipurpose. Previous circuit is one of the possible options that could be implmented. Another few appear to involve inputing current through J3 or J5 and J6 (using sense resistor). What is not specified is whether or not buffers remain within the circuit. As you can see there are means to short them out (via R51 and R55).

If buffers are unity gain, then it is fine to keep them in the circuit, and since the control signal (going to the diode) is still taken at that point, it is probably good to have them there.

V
 

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Thread Starter

justtrying

Joined Mar 9, 2011
439
Right, what is the question... the question is whether the buffers should be included when the circuit is implemented so that external current input is available. (I guess I ask this in 2nd paragraph) Or did I answer it myself? Confirmation never hurts.
 
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