Not yet complete
Working on a project with an OP-AMP
requires a lot of skill and understanding.
SERVICING OP-AMP CIRCUITS
The input impedance of an OP-AMP is very high and probing either
input with a multimeter or CRO will change the voltage on the input and
alter the state of the output.
The reason is this: The voltage on either input is extremely critical. It
only has to change by 1/10th of a millivolt and the output will change a
considerable amount. The actual change will depend on the gain of the
OP-AMP and this is determined by the value of the components surrounding
If the gain is not controlled, it can be as high as 10,000 to 100,000 but
most OP-AMP have surrounding components that limit the gain to between 2
It is also impossible to
measure the difference in potential between the inverting input and
Thus the normal method of probing and testing an OP-AMP with a multimeter
or CRO DOES NOT WORK! You have to use a new method of
testing. It's simple and quite brilliant.
It's a 47k on jumper leads with
a short length of tinned wire in the second alligator clip to act as a
The lead can be connected between the pin under investigation and either
the positive, 0v rail or negative rail, while monitoring (measuring ) the output
of the OP-AMP.
Don't be tricked by a CRO. It puts a load on the OP-AMP and if the line
is HIGH IMPEDANCE, the CRO will affect the amplitude of the
The amplitude on the display will be reduced (attenuated) as the
frequency increases. For instance, for a 10MHz CRO, the amplitude will be
50% of the real value when the signal is 10MHz. The load from the leads
of the probe on the CRO will attenuate (reduce) the signal and if the
circuit is operating at say 100MHz, the load of the CRO can quite often
"kill" the operation of the circuit.
The first circuit we will discuss is an example of BAD DESIGN.
Before you start, you need to know the basics. We have covered these in the previous
pages of this topic and the most important point to remember is the
voltage on the "+" input must be slightly higher than the "–" input for
the output to be HIGH. This is the way an OP-AMP naturally sits when it
is connected as shown below. The "–" input rises (normally due to the
voltage from the output) until it is just lower than the "+" input and
this makes the output nearly equal to the voltage on the input. If the
"+" input changes by say 1mV, the biasing of the two other pins will
adjust by approx the same value. This makes the circuit inherently fairly
However in the following circuit this is not the case.
The main fault is the connection of the two inputs to
the 5v rail.
The correct way to bias an OP-AMP is to allow the circuit to easily
produce the voltage on the inverting input as shown in the diagram below:
In the first circuit, the output must fall by 100mV if the "+" input
falls 1mV, to maintain the bias conditions.
As the 1u gets older, its leakage current will increase and this will
change the bias on the output of the OP-AMP. For this reason, the 1u must
be a low leakage device such as a TANTALUM.
A project containing this circuit was sent in for repair and the solution
was to replace the 1u with a tantalum.
To find out if a circuit is critical, an easy approach is to press you
finger across each of the electrolytics, on the underside of the board.
The slightest press across a faulty electrolytic will turn the
ifficult for the OP-AMP to produce a
lower voltage on the inverting input. For each millivolt lower than the
"+" input, the output must be 100millivolts lower than 5v. With a supply
of 5v, the "+" input can only drop 50mV or less for the OP-AMP to hold
its self-bias conditions.
An improved design is to place the 10k on the 0V rail.
The circuit above includes the circuitry connected to the "+" input
as this is a very important part when servicing the stage. You cannot
work on a OP-AMP stage if you don't know how it is being driven as the
input line is very sensitive to the slightest change in voltage.
When the circuit turns on, the 1u electrolytic will be uncharged and the
output will be LOW. It will charge via the two 100k resistors and after
about 5 seconds the "+" input will be higher than the "–" input and the
output will go HIGH. The actual voltage on the output will be lower than
5v so that the "–" input is a fraction of a millivolt below the "+"
This how the OP-AMP sits. When the signal on the output of the first
OP-AMP is above 0.7v, it will pass through D2, via the 4u7 electrolytic and charge the 2u2
electrolytic. Diode D1 discharges the 4u7 when the signal goes low so
the 4u7 remains discharged so it can pass its signal to D2.
Any slight voltage on the 2u2 will be passed to the non-inverting input
of the OP-AMP and cause the output to rise.
If the circuit is not operating correctly, the only point you can monitor
(read) is the output of the second OP-AMP. It should be about 5v.
The rest of the circuit is classified as HIGH IMPEDANCE and any
probing with a multimeter will upset the conditions.
The most critical component in the circuit is the 1u electrolytic. It
must be a low leakage type to allow the voltage on non-inverting input to
rise above the inverting input.
To see if the OP-AMP is sitting correctly, place the 47k (on test leads)
between the non-inverting input and the 2v rail, while monitoring the
output. The output voltage should rise. Placing the resistor between the
"+" input and 0v rail, will make the output go LOW.
Place the jumper lead between pint "A" and 0v rail - in other words place
a "short" between point "A" and ground to see if a signal from the first
OP-AMP is being detected by the second OP-AMP. You will have to wait a
few seconds for the circuit to settle down before taking a reading as the
1u will have to charge via the two 100k resistors before taking a
If the "charge-pump" section is generating a voltage but no input signal
is being delivered to the project, the fault may lie in "hum" being
generated by a previous stage or some form of self-oscillation.
You can also get an effect called "motor-boating." This is a low
frequency feedback through the 5v rail. It occurs like this: If a
large load is placed on the power rail, a slight voltage drop will occur
and this will be passed to the 5v rail. We have already seen that the
voltage on the 5v rail is very critical and any slight change can alter
that state of a stage. This effect will be passed through the circuit to
create a repeat "hiccup."
To see if the "charge-pump" section is capable of feeding the OP-AMP
amplifier, place the 47k between point "A" and the 12v rail. This will
charge the 2u2 and feed the OP-AMP. You can also repeat the test at
the join of the two diodes, to make sure they are around the correct way.
The OP-AMP circuit above is very similar to the voltage-follower
arrangement described on a previous page of this topic.
The output rises slowly to the same level as that on the "+" input due to
the effect of the 10u electrolytic charging slowing via the 1M resistor.
It will sit (quiescent condition) with the output at approx 5v.
When a signal is being processed, the electrolytic will not have time to
charge or discharge and thus it can be considered to have zero
resistance between its terminals. The gain of the stage will be the ratio
of the two voltage-viding resistors on the output:
1,000,000/10,000 = 100.
If the output is not 5v, place the 47k on jumper leads between the "+" input and 12v and
see if the output rises. If it does not rise, the fault may lie in a
fault OP-AMP or something on the output preventing it from rising. Make
sure the 1M is the correct value. Make sure the 5v rail is the correct
voltage. You will not be able to accurately measure the voltage on the
non-inverting input and that's why the 47k on jumper leads is needed to
check the operation of the stage.
This circuit may work but the output is NEGATIVE! The circuit is designed
for audio, but how many audio circuits require a negative voltage from
the output of a stage?
Before building a circuit you have to be aware of its operation as one
would expect the output of a stage to be POSITIVE.
Fig 4: AUDIO MIXER
When probing any type of circuit there is always a desire to
"short between two points" to "see what will happen."
This is always dangerous as many chips and transistors etc will not
accept an overload, even for a fraction of a second.
That's why we suggest using a 47k on jumper leads. It's a fairly good
value to use for high impedance circuits - especially OP-AMPs, to get a
reading on the output pin of the chip.
There are, however, some places where you need to "short" as the
voltage being tested is very small and a 47k resistor will not produce
Before you use the test lead as a "short," make sure the section is
Impedance. In other words, make sure the components you are probing
are not directly connected to the power rail or via a low-value resistor.
If you are not sure about the applying a "short," you should start with a
10k resistor, then a 1k and finally a 100R resistor on jumper leads.
A 100R will normally provide the same results as a "short" while
protecting the components from damage.
Another very good "component" for testing the action of a circuit is the
resistance of your body.
By holding the metal part of the alligator clips with your fingers, you
can adjust the effective resistance by squeezing the clips.
This is a very simple way to get a variable resistance and many circuits
can be activated in this way.