Buffer Amplifier's Questions

Dear all,

I'm reading the chapter on "Voltage Buffer". It's stated that it helps
to isolate the input from the load.

From my understanding, it helps to prevent "Too high currents" in the
input the provide voltage to the load. Am i correct?

I'm connecting a signal from a test point from an IC in a DVD+RW drive,
namely point A. This point A's signal is further connected to 4
Op-Amps. It's discovered that the IC is burnt after i connected them
for a while. I suspect that it's burnt because of "Too High Current" is
needed to send to the 4 Op-Amps. In this case, can connecting a voltage
buffer to the test point A before connecting to the 4 op-Amps help?

Pls correct me if i wrote anything wrong above. Also, pls help to
explain my question.

Thanks.

Regards,
Albert

 

Thanks Robert.

The grounds of the Op-Amp and the IC are the same.

I'm doing a project which does the calculation of addition and
subtraction of voltages which involve Vref. For the circuit's
configurations, pls refer to

http://www.us.oup.com/us/pdf/microcircuits/students/amps/ina105-burr.pdf

After adding the buffer amplifier to the Vref, the IC's still burnt.
(Note: Vref is one of the point in the IC).
Any suggestions? Or any experience to share?

 

[Robert Monsen]

albertleng@gmail.com wrote:

Dear all,

I'm reading the chapter on "Voltage Buffer". It's stated that it helps
to isolate the input from the load.

From my understanding, it helps to prevent "Too high currents" in the
input the provide voltage to the load. Am i correct?

I'm connecting a signal from a test point from an IC in a DVD+RW drive,
namely point A. This point A's signal is further connected to 4
Op-Amps. It's discovered that the IC is burnt after i connected them
for a while. I suspect that it's burnt because of "Too High Current" is
needed to send to the 4 Op-Amps. In this case, can connecting a voltage
buffer to the test point A before connecting to the 4 op-Amps help?

Pls correct me if i wrote anything wrong above. Also, pls help to
explain my question.

Thanks.

Regards,
Albert

Is the ground for the opamps the same as for the IC? They must be the
same for any of this to work.

How are the opamps connected? The inputs of opamps are usually very high
impedance, meaning they will source or sink only tiny amounts of
current. However, if you hook them up in certain feedback
configurations, the feedback can source or sink current, possibly
causing the problem. If this is your problem, then just using another
opamp as a buffer seems like it may work.

--
Regards,
Robert Monsen

"Your Highness, I have no need of this hypothesis."
- Pierre Laplace (1749-1827), to Napoleon,
on why his works on celestial mechanics make no mention of God.

 

[Andrew Holme]

albertleng@gmail.com wrote:

Thanks Robert.

The grounds of the Op-Amp and the IC are the same.

I'm doing a project which does the calculation of addition and
subtraction of voltages which involve Vref. For the circuit's
configurations, pls refer to


http://www.us.oup.com/us/pdf/microcircuits/students/amps/ina105-burr.pdf

After adding the buffer amplifier to the Vref, the IC's still burnt.
(Note: Vref is one of the point in the IC).
Any suggestions? Or any experience to share?

Have you replaced the burnt IC? When you say "burnt" - do you mean it
gets hot when your external circuitry is attached - or, is it
burnt-out? Have you replaced it?

Does it get hot now even without the external circuiry, or only when it
is connected?

Are you sure you are only connecting outputs to inputs? Could it be
output-to-output i.e. outputs fighting each other?

Have you observed the test point on an oscilloscope or multimeter?
Does the voltage / signal level at the test point change when you
connect the external circuitry? That would be a sure sign of a
short-circuit.

Can you post a schematic of your circuitry? Can you give more details
of the circuit and/or what you are trying to achieve?

 

Thanks.

The "burnt" i meant is the IC is damaged. I could see some "sparks"
come out. After the IC is being replaced, the drive becomes okay again?

Yup... the Vref voltage that's supposed to be ~1.945V would change to
~10V or above when we connected it to the external circuit.

I attach the link to my circuit diagram and also the drive's IC TZA1047
as below:

http://uploads.savefile.com/users/uploads/beam_landing_circuit.jpg

http://uploads.savefile.com/users/uploads/tza1047.jpg


My design will take 5 voltage signals from the IC TZA1047, which are
Vref, A, B, C and D. My circuit will first perform (A - Vref), (B -
Vref), (C - Vref) and (D - Vref) (Note: Let A*= (A-Vref), B* =
(B-Vref) and so on.)

Then, A*, B*, C* and D* will go into a circuit which does

Vout= [(A*+D*)-(B*+C*)]/(A*+B*+C*+D*)

The addition and subtraction part will be done by INA105kp while the
division part will be done by mpy634.

The following are the links of the datasheets for INA105 and MPY634:

http://www.us.oup.com/us/pdf/microcircuits/students/amps/ina105-burr.pdf

http://www.jcmicon.com/PDF/mpy634.pdf

(Note: TZA1047 is an IC of the DVD+RW drive.)

Please give any suggestions. Sorry that my scanned circuit diagram may
not be very clear.

Thanks.

 

Thanks.

The "burnt" i meant is the IC is damaged. I could see some "sparks"
come out. After the IC is being replaced, the drive becomes okay again?

Yup... the Vref voltage that's supposed to be ~1.945V would change to
~10V or above when we connected it to the external circuit.

I attach the link to my circuit diagram and also the drive's IC TZA1047
as below:

http://uploads.savefile.com/users/uploads/beam_landing_circuit.jpg

http://uploads.savefile.com/users/uploads/tza1047.jpg


My design will take 5 voltage signals from the IC TZA1047, which are
Vref, A, B, C and D. My circuit will first perform (A - Vref), (B -
Vref), (C - Vref) and (D - Vref) (Note: Let A*= (A-Vref), B* =
(B-Vref) and so on.)

Then, A*, B*, C* and D* will go into a circuit which does

Vout= [(A*+D*)-(B*+C*)]/(A*+B*+C*+D*)

The addition and subtraction part will be done by INA105kp while the
division part will be done by mpy634.

(Note: TZA1047 is an IC of the DVD+RW drive.)

Please give any suggestions. Sorry that my scanned circuit diagram may
not be very clear.

Thanks.

 

[Andrew Holme]

albertleng@gmail.com wrote:

Thanks.

The "burnt" i meant is the IC is damaged. I could see some "sparks"
come out. After the IC is being replaced, the drive becomes okay
again?

Yup... the Vref voltage that's supposed to be ~1.945V would change to
~10V or above when we connected it to the external circuit.

Are you missing a ground connection? What about the INA105? Did you
replace that as well? Maybe its inputs are blown.

Why don't you make a test harness consisting of resistive potential
dividers, powered from the DVD. Connect your inputs to them (or it?)
one at a time and make sure the voltage at the centre of the divider
doesn't change. If it does, maybe that input has gone short circuit.
This will also prove you have a sound ground reference.

I attach the link to my circuit diagram and also the drive's IC
TZA1047
as below:

http://uploads.savefile.com/users/uploads/beam_landing_circuit.jpg

http://uploads.savefile.com/users/uploads/tza1047.jpg

Those links don't work for me; it redirects to the savefile home page.

My design will take 5 voltage signals from the IC TZA1047, which are
Vref, A, B, C and D. My circuit will first perform (A - Vref), (B -
Vref), (C - Vref) and (D - Vref) (Note: Let A*= (A-Vref), B* =
(B-Vref) and so on.)

Then, A*, B*, C* and D* will go into a circuit which does

Vout= [(A*+D*)-(B*+C*)]/(A*+B*+C*+D*)

The addition and subtraction part will be done by INA105kp while the
division part will be done by mpy634.

The following are the links of the datasheets for INA105 and MPY634:


http://www.us.oup.com/us/pdf/microcircuits/students/amps/ina105-burr.pdf

http://www.jcmicon.com/PDF/mpy634.pdf

(Note: TZA1047 is an IC of the DVD+RW drive.)

Please give any suggestions. Sorry that my scanned circuit diagram
may
not be very clear.

Thanks.