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### Network # Need help analyzing a circuit

J

#### James Howe

Jan 1, 1970
0
I'm new to electronics so I'm not great at circuit analysis. I've been
reading some books and I understand some simple circuits, but the circuit
below has me somewhat confused. This circuit is used to convert an input
voltage into a constant current. The capacitor in the circuit is actually
hooked into a timer to periodically discharge it, but I'm not interested
in that part of the circuit at this time.

Basically what I'm trying to figure out is voltage and current at parts of
the circuit. I've tried my best to produce an ASCII version of the
circuit, please excuse my crude attempt. The circuit basically consists
of a handful of resistors, two 2N4126 PNP transistors, a .01uf capacitor
and an LM324 op amp. The positive voltage is a control voltage and can
range from approximately 0v to 5v. The negative voltage is fixed at -12v.

In my reading, I've seen descriptions on how to compute voltage and
current through NPN as well as PNP transitors. However, all the examples
have a positive (or negative for PNP) voltage going into the collector and
the emitter connected to ground. This circuit is different because the
collector is connected to negative voltage through a capacitor and the
emitter is connected to a positive voltage. I'm not sure the best way to
attack this circuit for the purpose of figuring out voltage and current
values. Any help or tips you can provide would be greatly appreciated.

Here is the schematic:

56k 1k
+4v *--/\/\/\--+--/\/\/\----------------------+
| |
| / e
| /
| \ ------| Q1 (PNP)
| -|\ /e \
+------| \ / \ c
| \--/\/\/\-|Q2 (PNP) |
+------| / 47 \ |
| +|/ \c |
| / LM324 | |
0v *----------+ +--------+
|
|
|
|
.01 uf 100 |
-12v *--------]|-------------------/\/\/\-------+

And if the picture is munged, here is a description:

A +4v supply is connected to a 56k and 1k resistor in series. Between the
56k and 1k resistor a connection is made to the LM324 OPAMP on the
inverting input. The non-inverting input is connected to ground. The
output of the OPAMP connects to a 47 ohm resistor and into the base of a
2N4196 PNP Transistor. The emitter is connected to another 2N4146 and the
emitter of the second transistor connects to the other end of the 1k
resistor. The collectors are tied together and are connected to a 100 ohm
resistor which is connected to a .01 uf capacitor and finally to a -12v
power supply. As I said earlier, in the full circuit, the capacitor is
periodically discharged.

Thanks.

J

#### John Popelish

Jan 1, 1970
0
James said:
I'm new to electronics so I'm not great at circuit analysis. I've been
reading some books and I understand some simple circuits, but the circuit
below has me somewhat confused. This circuit is used to convert an input
voltage into a constant current. The capacitor in the circuit is actually
hooked into a timer to periodically discharge it, but I'm not interested
in that part of the circuit at this time.

Basically what I'm trying to figure out is voltage and current at parts of
the circuit. I've tried my best to produce an ASCII version of the
circuit, please excuse my crude attempt. The circuit basically consists
of a handful of resistors, two 2N4126 PNP transistors, a .01uf capacitor
and an LM324 op amp. The positive voltage is a control voltage and can
range from approximately 0v to 5v. The negative voltage is fixed at -12v.

In my reading, I've seen descriptions on how to compute voltage and
current through NPN as well as PNP transitors. However, all the examples
have a positive (or negative for PNP) voltage going into the collector and
the emitter connected to ground. This circuit is different because the
collector is connected to negative voltage through a capacitor and the
emitter is connected to a positive voltage. I'm not sure the best way to
attack this circuit for the purpose of figuring out voltage and current
values. Any help or tips you can provide would be greatly appreciated.

Here is the schematic:

56k 1k
+4v *--/\/\/\--+--/\/\/\----------------------+
| |
| / e
| /
| \ ------| Q1 (PNP)
| -|\ /e \
+------| \ / \ c
| \--/\/\/\-|Q2 (PNP) |
+------| / 47 \ |
| +|/ \c |
| / LM324 | |
0v *----------+ +--------+
|
|
|
|
.01 uf 100 |
-12v *--------]|-------------------/\/\/\-------+

And if the picture is munged, here is a description:

A +4v supply is connected to a 56k and 1k resistor in series. Between the
56k and 1k resistor a connection is made to the LM324 OPAMP on the
inverting input. The non-inverting input is connected to ground. The
output of the OPAMP connects to a 47 ohm resistor and into the base of a
2N4196 PNP Transistor. The emitter is connected to another 2N4146 and the
emitter of the second transistor connects to the other end of the 1k
resistor. The collectors are tied together and are connected to a 100 ohm
resistor which is connected to a .01 uf capacitor and finally to a -12v
power supply. As I said earlier, in the full circuit, the capacitor is
periodically discharged.

Thanks.

The main simplifying assumption to make is that the opamp will use its
output to keep its two inputs at the same voltage. Since the
noninverting input is at 0 volts, the output will use the output
darlington pair of PNP transistors to hold its - (inverting ) input at
zero volts, also.

The second simplifying assumption is that the two inputs of the opamp
draw no current.

So any current that arrives at from the positive control voltage (+4)
through the 56k resistor must be sucked up by the 1k emitter resistor
in such a way that the node between those two resistors stays at zero
volts. The two transistors are connected to multiply the current
gains of the transistors, so that the base current of Q2 is so small
that you can neglect it. So essentially all the current through the
1k resistor also passes through the two collectors to the 100 ohm
resistor. And this current equals the current through the 56k
resistor with the control voltage minus zero volts across it,
regardless of variations on the voltage on the collectors of the two
PNP transistors. So the capacitor charges with a current that equals
the current through the 56k resistor.

R

#### Robert Monsen

Jan 1, 1970
0
James said:
I'm new to electronics so I'm not great at circuit analysis. I've been
reading some books and I understand some simple circuits, but the
circuit below has me somewhat confused. This circuit is used to
convert an input voltage into a constant current. The capacitor in the
circuit is actually hooked into a timer to periodically discharge it,
but I'm not interested in that part of the circuit at this time.

Basically what I'm trying to figure out is voltage and current at parts
of the circuit. I've tried my best to produce an ASCII version of the
circuit, please excuse my crude attempt. The circuit basically
consists of a handful of resistors, two 2N4126 PNP transistors, a .01uf
capacitor and an LM324 op amp. The positive voltage is a control
voltage and can range from approximately 0v to 5v. The negative
voltage is fixed at -12v.

In my reading, I've seen descriptions on how to compute voltage and
current through NPN as well as PNP transitors. However, all the
examples have a positive (or negative for PNP) voltage going into the
collector and the emitter connected to ground. This circuit is
different because the collector is connected to negative voltage
through a capacitor and the emitter is connected to a positive
voltage. I'm not sure the best way to attack this circuit for the
purpose of figuring out voltage and current values. Any help or tips
you can provide would be greatly appreciated.

Here is the schematic:

56k 1k
+4v *--/\/\/\--+--/\/\/\----------------------+
| |
| / e
| /
| \ ------| Q1 (PNP)
| -|\ /e \
+------| \ / \ c
| \--/\/\/\-|Q2 (PNP) |
+------| / 47 \ |
| +|/ \c |
| / LM324 | |
0v *----------+ +--------+
|
|
|
|
.01 uf 100 |
-12v *--------]|-------------------/\/\/\-------+

And if the picture is munged, here is a description:

A +4v supply is connected to a 56k and 1k resistor in series. Between
the 56k and 1k resistor a connection is made to the LM324 OPAMP on the
inverting input. The non-inverting input is connected to ground. The
output of the OPAMP connects to a 47 ohm resistor and into the base of
a 2N4196 PNP Transistor. The emitter is connected to another 2N4146
and the emitter of the second transistor connects to the other end of
the 1k resistor. The collectors are tied together and are connected to
a 100 ohm resistor which is connected to a .01 uf capacitor and finally
to a -12v power supply. As I said earlier, in the full circuit, the
capacitor is periodically discharged.

Thanks.

A capacitor is a device that allows charge to 'pile up' (technical term).

A pile of charge is a voltage.

The formal relation is

charge = capacitance times voltage

This is another way of saying

current = capacitance times 'rate of change of voltage'.

For the above, you know that the opamp will try to keep the junction
between the 56k and 1k resistor at 0V, so the current through the 56k
resistor will be

4V / 56k = 71.429uA

The capacitor is 10nF, so the 'rate of change' of voltage is

71.429uA / 10nF = 71.429e-6/10e-9 = 7.1429e3 volts/sec

If the positive side of the capacitor starts at -12V, then to get to,
say, -2V will take

10/7.1429e3 = 1.4ms

The waveform is constantly increasing (it's linear), so the output,
taken from the positive side of the capacitor, is a 'ramp', which must
be brought back to -12V periorically by shorting across the capacitor,
probably using a transistor.

--
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.

J

#### James Howe

Jan 1, 1970
0
James said:
I'm new to electronics so I'm not great at circuit analysis. I've
been reading some books and I understand some simple circuits, but the
circuit below has me somewhat confused.[...]
Here is the schematic:
56k 1k
+4v *--/\/\/\--+--/\/\/\----------------------+
| |
| / e
| /
| \ ------| Q1 (PNP)
| -|\ /e \
+------| \ / \ c
| \--/\/\/\-|Q2 (PNP) |
+------| / 47 \ |
| +|/ \c |
| / LM324 | |
0v *----------+ +--------+
|
|
|
|
.01 uf 100 |
-12v *--------]|-------------------/\/\/\-------+

[...]

For the above, you know that the opamp will try to keep the junction
between the 56k and 1k resistor at 0V, so the current through the 56k
resistor will be

4V / 56k = 71.429uA

[...]

Thanks for the explanation, but I have one question. I'm familiar with
OPAMP's used with negative feedback and it sounds like this is what you
are describing when you say that the OPAMP will attempt to keep the
junction of th 56k and 1k at 0v. However, what confuses me is that the
output of the OPAMP goes into two PNP transistors where the emitters are
connected to the 1k resistor. In conventional electron flow, wouldn't the
current be coming through the 1k, and through the emitters exiting through
the base at the 47 ohm resistor and out through the collector connected to
the 100 ohm resistor? In this arrangement I don't see how there can be
negative feedback into the inverting pin since it would seem that the
current couldn't flow through the base/emitter junctions back towards the
1k resistor.

I must be misunderstanding something.

Thanks again.

J

#### John Popelish

Jan 1, 1970
0
James said:
James said:
I'm new to electronics so I'm not great at circuit analysis. I've
been reading some books and I understand some simple circuits, but the
circuit below has me somewhat confused.[...]
Here is the schematic:
56k 1k
+4v *--/\/\/\--+--/\/\/\----------------------+
| |
| / e
| /
| \ ------| Q1 (PNP)
| -|\ /e \
+------| \ / \ c
| \--/\/\/\-|Q2 (PNP) |
+------| / 47 \ |
| +|/ \c |
| / LM324 | |
0v *----------+ +--------+
|
|
|
|
.01 uf 100 |
-12v *--------]|-------------------/\/\/\-------+

[...]

For the above, you know that the opamp will try to keep the junction
between the 56k and 1k resistor at 0V, so the current through the 56k
resistor will be

4V / 56k = 71.429uA

[...]

Thanks for the explanation, but I have one question. I'm familiar with
OPAMP's used with negative feedback and it sounds like this is what you
are describing when you say that the OPAMP will attempt to keep the
junction of th 56k and 1k at 0v. However, what confuses me is that the
output of the OPAMP goes into two PNP transistors where the emitters are
connected to the 1k resistor. In conventional electron flow, wouldn't the
current be coming through the 1k, and through the emitters exiting through
the base at the 47 ohm resistor and out through the collector connected to
the 100 ohm resistor? In this arrangement I don't see how there can be
negative feedback into the inverting pin since it would seem that the
current couldn't flow through the base/emitter junctions back towards the
1k resistor.

I must be misunderstanding something.

When the opamp output voltage is two diode drops more negative than
the right end of the 1k resistor, the Q2 begins to conduct, supplying
base current to Q1 which also begins to conduct, and all that current
passes through the 1k resistor, making its right end more negative
than its right. This process continues till the voltage at the
junction of the two resistors balances at zero volts, so that all the
current passing through the 56k resistor also passes through the 1k
resistor, while the voltage across the 56k resistor is 4 volts.

At the same time almost all that current also passes through the 100
ohm resistor. The only part missing is the small base current being
injected by the opamp output into Q2 which becomes part of the emitter
current of Q1 but does not make it to the collectors.

J

#### James Howe

Jan 1, 1970
0
James said:
James Howe wrote:
I'm new to electronics so I'm not great at circuit analysis. I've
been reading some books and I understand some simple circuits, but the
circuit below has me somewhat confused.[...]
Here is the schematic:
56k 1k
+4v *--/\/\/\--+--/\/\/\----------------------+
| |
| / e
| /
| \ ------| Q1 (PNP)
| -|\ /e \
+------| \ / \ c
| \--/\/\/\-|Q2 (PNP) |
+------| / 47 \ |
| +|/ \c |
| / LM324 | |
0v *----------+ +--------+
|
|
|
|
.01 uf 100 |
-12v *--------]|-------------------/\/\/\-------+

[...]

For the above, you know that the opamp will try to keep the junction
between the 56k and 1k resistor at 0V, so the current through the 56k
resistor will be

4V / 56k = 71.429uA

[...]

Thanks for the explanation, but I have one question. I'm familiar with
OPAMP's used with negative feedback and it sounds like this is what you
are describing when you say that the OPAMP will attempt to keep the
junction of th 56k and 1k at 0v. [...]

I must be misunderstanding something.

When the opamp output voltage is two diode drops more negative than
the right end of the 1k resistor, the Q2 begins to conduct, supplying
base current to Q1 which also begins to conduct, and all that current
passes through the 1k resistor, making its right end more negative
than its right. This process continues till the voltage at the
junction of the two resistors balances at zero volts, so that all the
current passing through the 56k resistor also passes through the 1k
resistor, while the voltage across the 56k resistor is 4 volts.

[...]

Ok, that makes sense. Thanks for the explanation.

R

#### Robert Monsen

Jan 1, 1970
0
James said:
James said:
James Howe wrote:
I'm new to electronics so I'm not great at circuit analysis. I've
been reading some books and I understand some simple circuits,
but the
circuit below has me somewhat confused.[...]
Here is the schematic:
56k 1k
+4v *--/\/\/\--+--/\/\/\----------------------+
| |
| / e
| /
| \ ------| Q1 (PNP)
| -|\ /e \
+------| \ / \ c
| \--/\/\/\-|Q2 (PNP) |
+------| / 47 \ |
| +|/ \c |
| / LM324 | |
0v *----------+ +--------+
|
|
|
|
.01 uf 100 |
-12v *--------]|-------------------/\/\/\-------+

[...]

For the above, you know that the opamp will try to keep the junction
between the 56k and 1k resistor at 0V, so the current through the 56k
resistor will be

4V / 56k = 71.429uA

[...]

Thanks for the explanation, but I have one question. I'm familiar with
OPAMP's used with negative feedback and it sounds like this is what you
are describing when you say that the OPAMP will attempt to keep the
junction of th 56k and 1k at 0v. [...]

I must be misunderstanding something.

When the opamp output voltage is two diode drops more negative than
the right end of the 1k resistor, the Q2 begins to conduct, supplying
base current to Q1 which also begins to conduct, and all that current
passes through the 1k resistor, making its right end more negative
than its right. This process continues till the voltage at the
junction of the two resistors balances at zero volts, so that all the
current passing through the 56k resistor also passes through the 1k
resistor, while the voltage across the 56k resistor is 4 volts.

[...]

Ok, that makes sense. Thanks for the explanation.

One more thing. One nice way to get to play around with a circuit like
this is to either build it and see what happens (not really possible
unless you have a scope), or simulate it. Terry Pinnell has a great page
on simulators (free and not free) that you can use. It's here

I like circuitmaker student for simple stuff, and pspice student for
temperature dependent analysis. There is a also a free one which people
here tend to post with, called LTSpice, at the linear site.

--
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.

J

#### James Howe

Jan 1, 1970
0
One more thing. One nice way to get to play around with a circuit like
this is to either build it and see what happens (not really possible
unless you have a scope), or simulate it. Terry Pinnell has a great page
on simulators (free and not free) that you can use. It's here

I like circuitmaker student for simple stuff, and pspice student for
temperature dependent analysis. There is a also a free one which people
here tend to post with, called LTSpice, at the linear site.

Funny you should mention building the circuit to just play around with as
I was planning to do just that! In fact, I just picked up a used scope
that I'm hoping to use for this sort of thing. I'll also look into the
simulators. I had downloaded an evaluation copy of one and played around
with it, but I like the idea of working with the real circuit.

R

#### Robert Monsen

Jan 1, 1970
0
James said:
Funny you should mention building the circuit to just play around with
as I was planning to do just that! In fact, I just picked up a used
scope that I'm hoping to use for this sort of thing. I'll also look
into the simulators. I had downloaded an evaluation copy of one and
played around with it, but I like the idea of working with the real
circuit.

The problem with simulators is they tend to lie if you screw up the
parameters, or if the circuit contains things that are difficult to
model, or for any number of reasons. Also, you often get into situations
where the mathematical models they use don't converge, and you end up
trying to figure out ways to fix the circuit just to get it to converge.

However, they let you 'see' things that are fairly difficult to see with
a scope, like currents and power through circuit items. You can also
show any number of varying voltages at once, thus overcoming the
two-trace limitations of many oscilloscopes.

--
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.

J

#### James Howe

Jan 1, 1970
0
James said:
I'm new to electronics so I'm not great at circuit analysis. I've
been reading some books and I understand some simple circuits, but the
circuit below has me somewhat confused. This circuit is used to
convert an input voltage into a constant current.
[...]
56k 1k
+4v *--/\/\/\--+--/\/\/\----------------------+
| |
| / e
| /
| \ ------| Q1 (PNP)
| -|\ /e \
+------| \ / \ c
| \--/\/\/\-|Q2 (PNP) |
+------| / 47 \ |
| +|/ \c |
| / LM324 | |
0v *----------+ +--------+
|
|
|
|
.01 uf 100 |
-12v *--------]|-------------------/\/\/\-------+

[...]

For the above, you know that the opamp will try to keep the junction
between the 56k and 1k resistor at 0V, so the current through the 56k
resistor will be

4V / 56k = 71.429uA

Ok, I've got some really dumb followup questions. If the OPAMP is going
to keep the junction between the 56k and 1k resistor at 0V, what is the
purpose of the 47 ohm resistor and the Darlington pair? If the current
through the 56k resistor basically is used to charge the capacitor, what
role do the other components play? Why the 1k resistor (why that
particular size)? Why the 100 ohm resistor?

Thanks.

J

#### John Popelish

Jan 1, 1970
0
James said:
James said:
I'm new to electronics so I'm not great at circuit analysis. I've
been reading some books and I understand some simple circuits, but the
circuit below has me somewhat confused. This circuit is used to
convert an input voltage into a constant current.
[...]
56k 1k
+4v *--/\/\/\--+--/\/\/\----------------------+
| |
| / e
| /
| \ ------| Q1 (PNP)
| -|\ /e \
+------| \ / \ c
| \--/\/\/\-|Q2 (PNP) |
+------| / 47 \ |
| +|/ \c |
| / LM324 | |
0v *----------+ +--------+
|
|
|
|
.01 uf 100 |
-12v *--------]|-------------------/\/\/\-------+

[...]

Ok, I've got some really dumb followup questions. If the OPAMP is going
to keep the junction between the 56k and 1k resistor at 0V, what is the
purpose of the 47 ohm resistor and the Darlington pair? If the current
through the 56k resistor basically is used to charge the capacitor, what
role do the other components play? Why the 1k resistor (why that
particular size)? Why the 100 ohm resistor?

I'm not sure I would have included those components. The 47 ohm
resistor looks like a current limit resistor that comes into play if
the collectors of the transistors have no current path (like if the
capacitor is allowed to charge up till the first collector to base
junction swings to forward bias). The 100 ohm resistor is more
problematic, unless the discharge switch you mentioned earlier
connects between it and the collectors, so that the peak discharge
current is limited by the 100 ohms. You have to think about all the
strange and unusual operating conditions the circuit might experience
to guess what the designer had in mind for some components.

J

#### James Howe

Jan 1, 1970
0
James said:
James Howe wrote:
I'm new to electronics so I'm not great at circuit analysis. I've
been reading some books and I understand some simple circuits, but
the circuit below has me somewhat confused.>> [...]
56k 1k
+4v *--/\/\/\--+--/\/\/\----------------------+
| |
| / e
| /
| \ ------| Q1 (PNP)
| -|\ /e \
+------| \ / \ c
| \--/\/\/\-|Q2 (PNP) |
+------| / 47 \ |
| +|/ \c |
| / LM324 | |
0v *----------+ +--------+
|
|
|
|
.01 uf 100 |
-12v *--------]|-------------------/\/\/\-------+

[...]
Ok, I've got some really dumb followup questions. If the OPAMP is going
to keep the junction between the 56k and 1k resistor at 0V, what is the
purpose of the 47 ohm resistor and the Darlington pair? If the current
through the 56k resistor basically is used to charge the capacitor, what
role do the other components play? Why the 1k resistor (why that
particular size)? Why the 100 ohm resistor?

[...] The 100 ohm resistor is more
problematic, unless the discharge switch you mentioned earlier
connects between it and the collectors, so that the peak discharge
current is limited by the 100 ohms. [...]

In the full circuit, there is a connection between the .01uf cap and the
100 ohm resistor. The connection goes one way and connects to the
non-inverting pin of another LM324 op amp, and the other direction
connects to the discharge pin on a 555. The buffered output from the
second op amp is fed through a 33k resistor and connected to the trigger
and threshold pins on the 555. So the current goes through the 100 ohm
resistor when the cap is charging, but the discharge does not. This
particular circuit is used to create a voltage controlled oscillator.

R

#### Robert Monsen

Jan 1, 1970
0
James said:
James said:
I'm new to electronics so I'm not great at circuit analysis. I've
been reading some books and I understand some simple circuits, but
the circuit below has me somewhat confused. This circuit is used
to convert an input voltage into a constant current.
[...]
56k 1k
+4v *--/\/\/\--+--/\/\/\----------------------+
| |
| / e
| /
| \ ------| Q1 (PNP)
| -|\ /e \
+------| \ / \ c
| \--/\/\/\-|Q2 (PNP) |
+------| / 47 \ |
| +|/ \c |
| / LM324 | |
0v *----------+ +--------+
|
|
|
|
.01 uf 100 |
-12v *--------]|-------------------/\/\/\-------+

[...]

For the above, you know that the opamp will try to keep the junction
between the 56k and 1k resistor at 0V, so the current through the 56k
resistor will be

4V / 56k = 71.429uA

Ok, I've got some really dumb followup questions. If the OPAMP is
going to keep the junction between the 56k and 1k resistor at 0V, what
is the purpose of the 47 ohm resistor and the Darlington pair? If the
current through the 56k resistor basically is used to charge the
capacitor, what role do the other components play? Why the 1k resistor
(why that particular size)? Why the 100 ohm resistor?

Thanks.

The darlington minimizes current flow into the opamp for control. With a
normal PNP, you have a 1% error. With a darlington, you'll have a 0.01%
error.

You could minimize error even more using a p-channel JFET, which has
essentially no leakage when the cap is charging, but also allows the
current to be diverted when the cap reaches 0V. In order for this to
matter, you need to use a good capacitor, such as a teflon cap.

--
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.

J

#### John Popelish

Jan 1, 1970
0
James said:
James said:
James Howe wrote:
I'm new to electronics so I'm not great at circuit analysis. I've
been reading some books and I understand some simple circuits, but
the circuit below has me somewhat confused.>> [...]
56k 1k
+4v *--/\/\/\--+--/\/\/\----------------------+
| |
| / e
| /
| \ ------| Q1 (PNP)
| -|\ /e \
+------| \ / \ c
| \--/\/\/\-|Q2 (PNP) |
+------| / 47 \ |
| +|/ \c |
| / LM324 | |
0v *----------+ +--------+
|
|
|
|
.01 uf 100 |
-12v *--------]|-------------------/\/\/\-------+

[...]
Ok, I've got some really dumb followup questions. If the OPAMP is going
to keep the junction between the 56k and 1k resistor at 0V, what is the
purpose of the 47 ohm resistor and the Darlington pair? If the current
through the 56k resistor basically is used to charge the capacitor, what
role do the other components play? Why the 1k resistor (why that
particular size)? Why the 100 ohm resistor?

[...] The 100 ohm resistor is more
problematic, unless the discharge switch you mentioned earlier
connects between it and the collectors, so that the peak discharge
current is limited by the 100 ohms. [...]

In the full circuit, there is a connection between the .01uf cap and the
100 ohm resistor. The connection goes one way and connects to the
non-inverting pin of another LM324 op amp, and the other direction
connects to the discharge pin on a 555. The buffered output from the
second op amp is fed through a 33k resistor and connected to the trigger
and threshold pins on the 555. So the current goes through the 100 ohm
resistor when the cap is charging, but the discharge does not. This
particular circuit is used to create a voltage controlled oscillator.

In that case, I see no point to having the 100 ohm resistor. The
transistors represent a limited source of current, whether charging
the cap or dumping into the discharge pin of a 555.

J

#### James Howe

Jan 1, 1970
0
James said:
]
Ok, I've got some really dumb followup questions. If the OPAMP is
going to keep the junction between the 56k and 1k resistor at 0V, what
is the purpose of the 47 ohm resistor and the Darlington pair? If the
current through the 56k resistor basically is used to charge the
capacitor, what role do the other components play? Why the 1k
resistor (why that particular size)? Why the 100 ohm resistor?
Thanks.

The darlington minimizes current flow into the opamp for control. With a
normal PNP, you have a 1% error. With a darlington, you'll have a 0.01%
error.
[...]

That's interesting, could you explain that a little more? Is this
something that is talked about in the data sheets for the 734 OPAMP? I
very low voltage levels. Does this configuration help the op amp produce
accurate lower voltages. Is that what you meant by the 0.01% error?
Also, would this configuration help with making the current converter more
resistant to temperature issues? This particular circuit is part of an
analog music synthesizer so they would want a particular voltage to
produce a particular frequency with consistency.

Thanks

R

#### Robert Monsen

Jan 1, 1970
0
James said:
James said:
]
Ok, I've got some really dumb followup questions. If the OPAMP is
going to keep the junction between the 56k and 1k resistor at 0V,
what is the purpose of the 47 ohm resistor and the Darlington
pair? If the current through the 56k resistor basically is used to
charge the capacitor, what role do the other components play? Why
the 1k resistor (why that particular size)? Why the 100 ohm resistor?
Thanks.

The darlington minimizes current flow into the opamp for control. With
a normal PNP, you have a 1% error. With a darlington, you'll have a
0.01% error.
[...]

That's interesting, could you explain that a little more? Is this
something that is talked about in the data sheets for the 734 OPAMP? I
at very low voltage levels. Does this configuration help the op amp
produce accurate lower voltages. Is that what you meant by the 0.01%
error? Also, would this configuration help with making the current
converter more resistant to temperature issues? This particular
circuit is part of an analog music synthesizer so they would want a
particular voltage to produce a particular frequency with consistency.

Thanks

No, the issue I was referring to is current leakage for control. The way
a bipolar transistor works is that some of the current is taken by the
diode formed by the emitter-base (for PNP). That current flow allows a
much larger current to flow from emitter to collector. For a normal
transistor, the ratio of collector current to base current is called
'beta'. It's usually between 15 and 300.

Thus, if you are depending on the fact that the 56k resistor is
delivering exactly 1/14000 A, some of that current will be diverted to
the base rather than delivered to the capacitor. A ballpark figure is
1/100, and this is where I got my 1% figure.

For a darlington pair, however, the control current for the 'first'
transistor is again diverted to the output, using 1/100 of *that*
current. Thus, the real control current which flows into the opamp is
1/100 of 1/100, or 1/10000 of the current that flows into the capacitor.
Thus, the error required for control is 1/10000, or 0.01%.

--
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.

R

#### Rich Grise

Jan 1, 1970
0
The problem with simulators is they tend to lie if you screw up the
parameters, or if the circuit contains things that are difficult to
model, or for any number of reasons. Also, you often get into situations
where the mathematical models they use don't converge, and you end up
trying to figure out ways to fix the circuit just to get it to converge.

However, they let you 'see' things that are fairly difficult to see with
a scope, like currents and power through circuit items. You can also
show any number of varying voltages at once, thus overcoming the
two-trace limitations of many oscilloscopes.

I once built an 8-input MUX for a scope, but this was in the days of 1 MHz
clocks and stuff, i.e., I knew better than to use it for any kind of fast,
or critical, stuff, but it did give a pretty good indication of what the
circuit was doing! I think I was using it to watch the data bus on a
micro I was debugging. Sort of a beer-budget logic analyzer. Cheers!
Rich

R

#### Rich Grise

Jan 1, 1970
0
James said:
James Howe wrote:

James Howe wrote:
I'm new to electronics so I'm not great at circuit analysis. I've
been reading some books and I understand some simple circuits, but
the circuit below has me somewhat confused.>> [...]
56k 1k
+4v *--/\/\/\--+--/\/\/\----------------------+
| |
| / e
| /
| \ ------| Q1 (PNP)
| -|\ /e \
+------| \ / \ c
| \--/\/\/\-|Q2 (PNP) |
+------| / 47 \ |
| +|/ \c |
| / LM324 | |
0v *----------+ +--------+
|
|
|
|
.01 uf 100 |
-12v *--------]|-------------------/\/\/\-------+

[...]

Ok, I've got some really dumb followup questions. If the OPAMP is going
to keep the junction between the 56k and 1k resistor at 0V, what is the
purpose of the 47 ohm resistor and the Darlington pair? If the current
through the 56k resistor basically is used to charge the capacitor, what
role do the other components play? Why the 1k resistor (why that
particular size)? Why the 100 ohm resistor?

[...] The 100 ohm resistor is more
problematic, unless the discharge switch you mentioned earlier
connects between it and the collectors, so that the peak discharge
current is limited by the 100 ohms. [...]

In the full circuit, there is a connection between the .01uf cap and the
100 ohm resistor. The connection goes one way and connects to the
non-inverting pin of another LM324 op amp, and the other direction
connects to the discharge pin on a 555. The buffered output from the
second op amp is fed through a 33k resistor and connected to the trigger
and threshold pins on the 555. So the current goes through the 100 ohm
resistor when the cap is charging, but the discharge does not. This
particular circuit is used to create a voltage controlled oscillator.

In that case, I see no point to having the 100 ohm resistor. The
transistors represent a limited source of current, whether charging
the cap or dumping into the discharge pin of a 555.

I could see where it offloads some of the dissipation in the current
source. Other than that, I have no clue.

Thanks!
Rich

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