# Could someone please explain how this works?

E

#### Eric R Snow

Jan 1, 1970
0
Greetings All,
is the schematic for a thermostat kit I built. Even though it works I
don't understand how it works. The high temperature limit is too low
for what I want to use it for. But I think that using less resistance
for R5 will raise the upper temperature limit. Am I right? And if I
am, why?
Thank You,
Eric

J

#### Jamie

Jan 1, 1970
0
Eric said:
Greetings All,
is the schematic for a thermostat kit I built. Even though it works I
don't understand how it works. The high temperature limit is too low
for what I want to use it for. But I think that using less resistance
for R5 will raise the upper temperature limit. Am I right? And if I
am, why?
Thank You,
Eric
Yes, Use less R on R5. You need to increase the set point range
so that the NTC will be allowed to decrease in R to deliver more
input to the - input which when it over shoots the + input will
switch the output off.
in that circuit, the RELAY remains on until the - input exceeds the
+ input of the op-amp. Kind of a safety shut down i guess.
Beware that NTC have a max temp range before they start to lose the
dynamic range.

J

#### Jamie

Jan 1, 1970
0
Eric said:
Greetings All,
is the schematic for a thermostat kit I built. Even though it works I
don't understand how it works. The high temperature limit is too low
for what I want to use it for. But I think that using less resistance
for R5 will raise the upper temperature limit. Am I right? And if I
am, why?
Thank You,
Eric
To add to the description of the circuit. the NTC decreases in
resistance as it gets hotter.
NTC = negative thermo coefficient (reduces in R as temp goes up)
PTC = Positive xxxxxxxxxxxxxxxxx (increases in R as temp goes up).

J

#### John Popelish

Jan 1, 1970
0
Eric said:
Greetings All,
is the schematic for a thermostat kit I built. Even though it works I
don't understand how it works. The high temperature limit is too low
for what I want to use it for. But I think that using less resistance
for R5 will raise the upper temperature limit. Am I right? And if I
am, why?
Thank You,

This circuit compares the voltages from two resistive
voltage dividers, one made with 3 fixed resistors, one of
which has a variable tap, and one made with a fixed resistor
and a temperature dependent resistor.

As temperature falls, the resistance of the thermistor
rises, lowering the voltage from its divider. When that
voltage falls below that from the adjustable tap, the
comparator output swings positive, turning on the relay
driver transistor.

A limitation of the circuit is that the comparator does not
work when its two inputs are closer than 1.5 volts to its
positive supply, but since the dividers are driven by only a
fraction of that supply (regulated by a zener diode) I doubt
this limitation will be a factor. But rather than change
R5, I would lower R4 to raise the temperature range where
the thermistor divider passes through the tap choices. This
will also raise the lower end of the adjustment range,
though, keeping the total temperature span about the same.

J

#### John Fields

Jan 1, 1970
0
Greetings All,
is the schematic for a thermostat kit I built. Even though it works I
don't understand how it works. The high temperature limit is too low
for what I want to use it for. But I think that using less resistance
for R5 will raise the upper temperature limit. Am I right?
---
Yes.
---

And if I am, why?
Thank You,
Eric

---
Why...?

IC1 is an LM324, which is four low power operational amplifiers, one
of which (IC1C) is being used as a non-inverting buffer for the 5.1V
reference being generated by ZD1 and R1, and another (IC1D) as a
voltage comparator. IC1B and IC1C aren't being used and are wired
to force all inputs and outputs to zero volts.

SK1 is used to connect the DC supply to the circuit, and D2 is used
to provide protection in case the supply is connected backwards, in
which case D2 will be reverse biased, cutting off the supply of
current to the circuit and preventing IC1 from being destroyed.

C4 is used as a reservoir capacitor and serves to help remove ripple
from the input DC.

The LM324 draws about 2mA, worst case, per opamp, for a total of
8mA.

The relay has a 360 ohm coil so, subtracting the drop across D2 and
VCE(sat) of T1 from the 12V supply it draws about:

E 12V - (0.7V + 0.1V)
I = --- = -------------------- ~ 31mA
R 360R

D1 draws:

(+V) - Vzk 11.3V - 5.1V
I = ------------ = ------------ ~ 6mA
R1 1000R

(+V) - 2V 11.3V - 2V
I = ----------- = ------------ ~ 1mA
R3 10kR

The total current needed, then, would be about 46mA with the relay
closed and the ripple across C4 would be:

I dt 4.6E-2A * 8.3E-3s
V = ------ = ------------------- ~ 3.82 volts
C 1.0E-4F

That's assuming a 12V peak, 60Hz full-wave rectified sine wave into
D2, so a previously filtered 12VDC supply would make the ripple
essentially nonexistent.

Looking at the divider comprising R5 RV1 R6, you have two limits:

A. When RV1 is at zero ohms, the reference voltage into IC1-13 will
be:

+V E1
|
[170kR] R5
|
+---->Vref E2
|
[0R] RV1
|
[56kR] R6
|
GND

E1 * (RV1 + R6) 11.3V * (0R + 5.6E4R)
E2 = ----------------- = ---------------------- = 2.8V, and:
R5 + RV1 + R6 1.7E5R + 0R + 5.6E4R

B. When RV1 is set to the maximum resistance, the reference voltage
into IC1-13 will be:

+V E1
|
[170kR] R5
|
+---->Vref E2
|
[100kR] RV1
|
[56kR] R6
|
GND

and E2 will be:

E1 * (RV1 + R6) 11.3V * (1.0E5R + 5.6E4R)
E2 = ----------------- = --------------------------- = 5.41V
R5 + RV1 + R6 1.7E5R + 1.0E5R + 5.6E4R

Since the circuit is using an NTC (Negative Temperature Coefficient)
thermistor as a sensor, the thermistor's resistance will decrease as
its temperature increases, and since it's connected like this:

5.1V
|
[NTC]
|
+--->Vout
|
[10kR]
|
GND

Vout will _increase_ as temperature increases.

Now, since the output of the thermistor is connected to the - input
of the comparator, the output of the comparator will go high and the
relay will be energized whenever the voltage corresponding to the
temperature the thermistor is sensing drops _below_ the voltage on
the comparator.

Conversely, When the output voltage from the sensor rises to a
voltage higher than the voltage on the + input of the comparator the
comparator's output will go low and the relay will be de-energized.

If you need the upper temp trip of your circuit to be increased,
then you need to either make the voltage from the sensor lower or
the voltage from the reference divider into the comparator's + input
higher at that temperature.

In order to lower the voltage from the sensor, the value of R4 could
be decreased. However, that's not a good idea because it might
increase the self-heating of the thermistor, leading to erroneous
switching of the comparator.

That leaves fiddling with the reference divider, and in order to
increase the voltage into the + input of the comparator R5 can be
made smaller or R6 can be made larger, both at the expense of
raising the lower temp limit.

A better way would be to determine what you wanted the upper and
lower temp limits to be and then to choose R5 and R6 (while keeping
the 100k pot in place) in order to meet those criteria.

If you have the R/T specs for the thermistor or a manufacturer and
part number and can get the data, then you can do it like this:

Assume your temp limits are 0C and 100C and you want to use a YSI
P/N 44016:

http://www.ysitemperature.com/techdocs/R_vs_T std part.xls

with a 10k ohm resistance at 25C, you'll have 678.5 ohms at 100C and
32660 ohms at 0C, so for 100C you'll have:

5.1V E1
|
[678.5R] R1
|
+---> E2
|
[10kR] R2
|
GND

Then, solving for E2:

E1 R2 5.1V * 10kR
E2 = --------- = --------------- = 4.776V
R1 + R2 678.5R + 10kR

Likewise, for 0C:

5.1V E1
|
[32660R] R1
|
+---> E2
|
[10kR] R2
|
GND

Then, solving for E2:

E1 R2 5.1V * 10kR
E2 = --------- = --------------- = 1.195V
R1 + R2 32660R + 10kR

Now what you have to do is to set up your reference divider so that
with the pot cranked to one end the voltage into IC1-12 is 1.195V,
and with it cranked to the other end the voltage is 4.776V, like
this:

5.1V 5.1V
| |
[R5] [RT1]
| |
+--4.776V E1 | +--[6M8]--+
| | | |
[100kR]<----------|-+-|+\ | C
| | | >----+--[10kR]---B
+--1.195V E2 +---|-/ E
| |
[R6] [10k] R4
| |
GND GND

In order to do that, you can calculate the current through the pot:

E1 - E2 4.776V - 1.195V
I = --------- = ----------------- ~ 3.58E-5A ~ 36µA
R 1E5R

Since the current in a series circuit is everywhere the same, and
the drop across R6 needs to be 1.195V, then:

E2 1.195V
R6 = ---- = --------- = 33194.4 ohms
It 3.6E-5A

The closest standard 1% resistor smaller than 33194 is 32400 ohms,
so let's use that to start off with.

Now, since the same current is in R5, we can say:

5.1V - 4.776V
R5 --------------- = 9000 ohms
3.6E-5A

The closest standard 1% resistor smaller than 9000 ohms is 8870
ohms, so our divider now looks like this:

5.1V
|
[8870R] R5
|
+----E1
|
[100kR] RV1
|
+----E2
|
[32400R] R6
|
GND

Checking:

E 5.1V
It = ---- = ------------------------ = 36.1µA
Rt 8870R + 100kR + 32400R

So:

E2 = It * R6 = 3.61E-5A * 3.24E4R = 1.17 volts

and:

E1 = It * (R6 + RV1) = 4.78 volts

E1 is off, but not by a lot and it can be fixed by diddling with the
values of R5 and R6, but there's another issue which hasn't been
dealt with yet, and that's the resistance tolerance of the pot.

Most pots are +/- 10%, so in order to make sure that the desired
voltage range can be accommodated by cranking the pot between its
limits, the calculations should be made with RV1 equal to 90k. That
way, if it's actually more than that the voltage range will always
be within the range of the pot if enough leeway is given by the end
resistors the rest of the circuit tolerances are paid attention to.

E

#### Eric R Snow

Jan 1, 1970
0
Greetings All,
is the schematic for a thermostat kit I built. Even though it works I
don't understand how it works. The high temperature limit is too low
for what I want to use it for. But I think that using less resistance
for R5 will raise the upper temperature limit. Am I right?
---
Yes.
---

And if I am, why?
Thank You,
Eric

---
Why...?

IC1 is an LM324, which is four low power operational amplifiers, one
of which (IC1C) is being used as a non-inverting buffer for the 5.1V
reference being generated by ZD1 and R1, and another (IC1D) as a
voltage comparator. IC1B and IC1C aren't being used and are wired
to force all inputs and outputs to zero volts.

SK1 is used to connect the DC supply to the circuit, and D2 is used
to provide protection in case the supply is connected backwards, in
which case D2 will be reverse biased, cutting off the supply of
current to the circuit and preventing IC1 from being destroyed.

C4 is used as a reservoir capacitor and serves to help remove ripple
from the input DC.

The LM324 draws about 2mA, worst case, per opamp, for a total of
8mA.

The relay has a 360 ohm coil so, subtracting the drop across D2 and
VCE(sat) of T1 from the 12V supply it draws about:

E 12V - (0.7V + 0.1V)
I = --- = -------------------- ~ 31mA
R 360R

D1 draws:

(+V) - Vzk 11.3V - 5.1V
I = ------------ = ------------ ~ 6mA
R1 1000R

(+V) - 2V 11.3V - 2V
I = ----------- = ------------ ~ 1mA
R3 10kR

The total current needed, then, would be about 46mA with the relay
closed and the ripple across C4 would be:

I dt 4.6E-2A * 8.3E-3s
V = ------ = ------------------- ~ 3.82 volts
C 1.0E-4F

That's assuming a 12V peak, 60Hz full-wave rectified sine wave into
D2, so a previously filtered 12VDC supply would make the ripple
essentially nonexistent.

Looking at the divider comprising R5 RV1 R6, you have two limits:

A. When RV1 is at zero ohms, the reference voltage into IC1-13 will
be:

+V E1
|
[170kR] R5
|
+---->Vref E2
|
[0R] RV1
|
[56kR] R6
|
GND

E1 * (RV1 + R6) 11.3V * (0R + 5.6E4R)
E2 = ----------------- = ---------------------- = 2.8V, and:
R5 + RV1 + R6 1.7E5R + 0R + 5.6E4R

B. When RV1 is set to the maximum resistance, the reference voltage
into IC1-13 will be:

+V E1
|
[170kR] R5
|
+---->Vref E2
|
[100kR] RV1
|
[56kR] R6
|
GND

and E2 will be:

E1 * (RV1 + R6) 11.3V * (1.0E5R + 5.6E4R)
E2 = ----------------- = --------------------------- = 5.41V
R5 + RV1 + R6 1.7E5R + 1.0E5R + 5.6E4R

Since the circuit is using an NTC (Negative Temperature Coefficient)
thermistor as a sensor, the thermistor's resistance will decrease as
its temperature increases, and since it's connected like this:

5.1V
|
[NTC]
|
+--->Vout
|
[10kR]
|
GND

Vout will _increase_ as temperature increases.

Now, since the output of the thermistor is connected to the - input
of the comparator, the output of the comparator will go high and the
relay will be energized whenever the voltage corresponding to the
temperature the thermistor is sensing drops _below_ the voltage on
the comparator.

Conversely, When the output voltage from the sensor rises to a
voltage higher than the voltage on the + input of the comparator the
comparator's output will go low and the relay will be de-energized.

If you need the upper temp trip of your circuit to be increased,
then you need to either make the voltage from the sensor lower or
the voltage from the reference divider into the comparator's + input
higher at that temperature.

In order to lower the voltage from the sensor, the value of R4 could
be decreased. However, that's not a good idea because it might
increase the self-heating of the thermistor, leading to erroneous
switching of the comparator.

That leaves fiddling with the reference divider, and in order to
increase the voltage into the + input of the comparator R5 can be
made smaller or R6 can be made larger, both at the expense of
raising the lower temp limit.

A better way would be to determine what you wanted the upper and
lower temp limits to be and then to choose R5 and R6 (while keeping
the 100k pot in place) in order to meet those criteria.

If you have the R/T specs for the thermistor or a manufacturer and
part number and can get the data, then you can do it like this:

Assume your temp limits are 0C and 100C and you want to use a YSI
P/N 44016:

http://www.ysitemperature.com/techdocs/R_vs_T std part.xls

with a 10k ohm resistance at 25C, you'll have 678.5 ohms at 100C and
32660 ohms at 0C, so for 100C you'll have:

5.1V E1
|
[678.5R] R1
|
+---> E2
|
[10kR] R2
|
GND

Then, solving for E2:

E1 R2 5.1V * 10kR
E2 = --------- = --------------- = 4.776V
R1 + R2 678.5R + 10kR

Likewise, for 0C:

5.1V E1
|
[32660R] R1
|
+---> E2
|
[10kR] R2
|
GND

Then, solving for E2:

E1 R2 5.1V * 10kR
E2 = --------- = --------------- = 1.195V
R1 + R2 32660R + 10kR

Now what you have to do is to set up your reference divider so that
with the pot cranked to one end the voltage into IC1-12 is 1.195V,
and with it cranked to the other end the voltage is 4.776V, like
this:

5.1V 5.1V
| |
[R5] [RT1]
| |
+--4.776V E1 | +--[6M8]--+
| | | |
[100kR]<----------|-+-|+\ | C
| | | >----+--[10kR]---B
+--1.195V E2 +---|-/ E
| |
[R6] [10k] R4
| |
GND GND

In order to do that, you can calculate the current through the pot:

E1 - E2 4.776V - 1.195V
I = --------- = ----------------- ~ 3.58E-5A ~ 36µA
R 1E5R

Since the current in a series circuit is everywhere the same, and
the drop across R6 needs to be 1.195V, then:

E2 1.195V
R6 = ---- = --------- = 33194.4 ohms
It 3.6E-5A

The closest standard 1% resistor smaller than 33194 is 32400 ohms,
so let's use that to start off with.

Now, since the same current is in R5, we can say:

5.1V - 4.776V
R5 --------------- = 9000 ohms
3.6E-5A

The closest standard 1% resistor smaller than 9000 ohms is 8870
ohms, so our divider now looks like this:

5.1V
|
[8870R] R5
|
+----E1
|
[100kR] RV1
|
+----E2
|
[32400R] R6
|
GND

Checking:

E 5.1V
It = ---- = ------------------------ = 36.1µA
Rt 8870R + 100kR + 32400R

So:

E2 = It * R6 = 3.61E-5A * 3.24E4R = 1.17 volts

and:

E1 = It * (R6 + RV1) = 4.78 volts

E1 is off, but not by a lot and it can be fixed by diddling with the
values of R5 and R6, but there's another issue which hasn't been
dealt with yet, and that's the resistance tolerance of the pot.

Most pots are +/- 10%, so in order to make sure that the desired
voltage range can be accommodated by cranking the pot between its
limits, the calculations should be made with RV1 equal to 90k. That
way, if it's actually more than that the voltage range will always
be within the range of the pot if enough leeway is given by the end
resistors the rest of the circuit tolerances are paid attention to.
Thank You Jamie, John and John,
All of your answers helped me to understand what's going on. And John
Field's lengthy explanation was really helpful. Thanks especially for
taking the time to write it all.
Cheers,
Eric R Snow

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