Assistance on connections to Power Supply

[zexclo]

Hi,

I would like to seek some assistance on connecting my PID controller to my DC power supply.

I am currently making a temperature regulator using

1. WATLOW controller model ( 982C-25CC-KRGG)
2. Lightbulbs (Philip Halogen 50W, 12V lightbulb)
3. GW DC power supply (Model GPS-3030)

I am encountering some difficulties connecting the controller to the DC power supply.
From my learning,I understand that the controller is a PID controller, which sends pulses to the DC supply which in turn will power up the light bulb should the temperature drops what is needed to maintain.

However, I am having difficulty understanding and connecting the PID controller. As to should the setting of the power supply be set to master output? and how should the circuitry be connected.

For any kind assistance!

 

[Tony Stewart]

Lightbulbs are nonlinear (PTC) resistive devices which your linear control may not be able to handle , such as the surge currents. Cold tungsten takes 8x to 10x the current as hot so it looks like a 400~500W load if pulsed cold. Do you have the specs of PSU and COntroller?

 

[zexclo]

I am putting this into a small box to regulate temperature, currently testing out on the small light bulb and might shift to small a few LEDs light bulbs later upon getting the system up.


I am not very sure, as I think the PSU can only take in 0 to 10V.

Have attached the links for both PSU and controller below.
Kindly pardon me for the flooding of the specs.
Thank you for your help!!

The specs for the PSU as I obtained from the website is (http://www.gwinstek.com/en-global/p...Channel_DC_Power_Supplies/GPS-Series)attached below.

GPS-3030
Display Meter - Analog
Output Watts (W) - 90W
Output Volts (V)- 0 ~ 30V
Output Amps (A) -0 ~3A
Weight (kg)- 5KG

CONSTANT VOLTAGE OPERATION
Regulation Line regulation≦0.01% + 3mV
Load regulation≦0.01% + 3mV (rating current≦3A)
Load regulation≦0.01% + 5mV (rating current > 3A)
Ripple & Noise ≦0.5mVrms 5Hz ~ 1MHz (rating current≦3A)
≦1mVrms 5Hz ~ 1MHz (rating current > 3A)
Recovery Time ≦100μS ( 50% load change, minimum load 0.5A )
Temp. Coefficient ≦300ppm/°C
Output range 0 to rating voltage continuously adjustable
CONSTANT CURRENT OPERATION

Regulation Line regulation≦0.2% + 3mA
Load regulation≦0.2% + 3mA
Ripple Current ≦3mArms
Output Range 0 to rating current continuously adjustable (Hi/Lo range switchable)
METER
Analog

For the controller the specs is as attached (**broken link removed**)

Control Mode • Dual input, quad output, optional retransmit of set point or process variable. • Programmable direct and reverse acting control outputs. • One-step auto-tuning. Operator Interface • Local/remote set point capability. • Dual, 4-digit LED displays: upper, 0.4” (10mm); lower, 0.3” (8mm). • Mode, Auto/Man, Display, Up and Down keys. Input • Contact input for software function select. • Type J, K, T, N, C(W5)2, D(W3)2, E, R, S, B, Pt 22 thermocouple, 1° or 0.1° RTD. • 0-50mVÎ (dc), 0-20mA, 4-20mA, 0-5VÎ (dc), 1-5VÎ (dc), 0-10VÎ (dc) process. • Slidewire, digital event input or heater current options. • Sensor break protection de-energizes control output to protect system or selectable bumpless transfer to manual operation. Latching or non-latching. • °F or °C display or process units, user selectable

Output Options • Solid-state relay, 0.5A @ 24VÅ (ac) min., 253VÅ (ac) max., opto-isolated, burst fire. With or without contact suppression. • Open collector: Max. voltage 42VÎ (dc), max. current 1A. • Switched dc signal: Min. turn-on voltage of 3VÎ (dc) into min. 500Ω load; max. On voltage not greater than 32VÎ (dc) into an infinite load, isolated. • Electromechanical relay1, Form C, 5A @ 120/240VÅ (ac), 6A @ 28VÎ (dc), 1/8 hp. @ 120VÅ (ac) or 125VA @ 120VÅ (ac). With or without contact suppression. Off-state output impedance with RC suppression is 20kΩ.


Do let me know if you need any other info. Thanks again!

 

[Les Jones]

Hi zexclo,
Your power supply will not supply your 12 volt 50 watt bulb as it requires 50/12 = 4.17 amps. Your power supply is only rated at 3 amps. (This is before even considering the switch on surge.) You would be better off using a combination of wire wound resistors for heating. I would suggest having the power supply set to about 24 volts so that you can 50 watts within the current rating of the power supply. So you would need a resistor of (24 x 24)/50 ohms = 11.52 ohms If you prefer to use the lamp then a 12 volt lighting transformer would be a cheaper solution than your variable bench power supply The link you show to the PID controller does not cover the model 982C that you have. If the output on you 982C is the same as the 988 series then you will also need a solid state relay to switch the power.

Les.

 

[zexclo]

Hi Les,
Thank you for your valuable input and help! am looking to change the light bulb to another heating element, am looking at the possibility of using the heating element of hair dryer or a hair dryer and wire wound resistor.
For the use of Solid state relay(SSR), am I removing the need for the bench power supply?
I am quite new to electronics, not sure on the connection between the power supply, solid state relay and the PID.

Correct me if I am wrong, the scenario should be...
the PID, power supply, solid state relay and the heating element are all connected.
when the temperature drops below Set temperature,
the PID will send a signal to the SSR which will close the circuit and the heating element will heat up.
when the temperature is reached,
PID will send a signal to the SSR to open the circuit.

thanks all for the guidance!

Zexclo

 

[Les Jones]

Hi zexclo,
You are not quite correct. The PID controller must be powered up all of the time. It has a low power dc output which controls the solid state relay. When the solid state relay is energised by the PID controller it allows power to flow to the heating element. (You can think of a solid state relay in the same way as a normal magnetic relay. They use a triac to do the switching instead of contacts and probably use an opto isolator to provide isolation. I have seen some that use a small relay to provide isolation.) The way the PID controller works is not as simple as just turning on the power until the set temperature is reached. It uses very low frequency PWM (Pulse width modulation) to control the power to the heater. If you want to understand how PID control (Proportional integral derivative) then Goggle it.) You will probably find that your PID controller is self tuning so you will not have to set the gain for the proportional, integral and derivative components. The solid state relay does not necessarily remove the need for the power supply. If power the heating element directly from the mains that removes the need for a power supply. The power supply is only needed to provide the correct voltage to the heating element. (As in the original choice of a 12 volt 50W bulb.) NOTE that if you still only want 50 watts of heating power using a hair dryer element you would either have to have several in series connected to mains voltage or use a power supply or transformer to provide a suitable voltage. Assuming that the hair drier is 1000 watt rating then (assuming 240 volt mains) then you would need 240 x the square root (50/1000) This would be 50/1000 = 0.05, root 0.05 = 0.2236, 240 x 0.2236 = 53.7 volts

Les.

 

[zexclo]

Hi Les,

Thanks for the explanation, and have roughly understood the operation of PID. It has to be powered at all time and with the signal from the thermocouple(or temperature sensor) it will regulate the pulse of DC to the solid state relay, which in turn control the power to the heater ensuring the power does not go beyond the set band width. The PID is continuously adjusting the pulse, until it approaches the set value where it will pulse and try to get the optimal response.

Say, I am using the power supply of the heater element to provide the voltage, how do I connect the SSR to the heating element?

For a normal hair dyer, the power will be directly connected to the dryer, hence for the temperature regulator case. Am I right to say that I need to place the SSR between the power supply and the hair dyer? If so, is there any wire connection between PID and the hair dryer?

Thank you for your patience in explaining!

Zexclo

 

[Les Jones]

Hi zexclo,
If by "power supply" you mean the mains power (110 or 240 V AC depending which country you are in.) then there will be a connection as the mains neutral will supply the PID controller and also the hair dryer. If by "power supply" you mean the output of a DC power supply or transformer then there will be no connection between the PID and the hair dryer. Bear in mind that if you are using the complete hair drier (Rather than just the element.) then the motor will run slower if the voltage to it is lower than normal. You will also have to make sure that the SSR is suitably rated for voltage, current and weather it is switching AC or DC. I would not advise using the SSR to switch the mains input to a DC power supply or an electronic lighting transformer. It should be OK if you are using a normal transformer.

Les.

 

[zexclo]

Hi Les,
In my country, the power supply is 220/240V ac.

To avoid confusion, from here on, I shall refer the power supply as mains power, and I am also removing the DC power supply from my setup.

Some queries.
1. You mentioned that, if I am using the mains power, the main power neutral will be in connection to the PID and the hair dryer.
Can I understand this as, the hair dryer will be sharing the same neutral wire with the PID when the PID power plug is plug into the main?
And how will the SSR be connected to the hair dryer? Are they connected via ground wiring?

2. May I know why would you not recommend the SSR to switch the main input to an electronic lighting transformer?
Will it be recommended to get a heatsink for the SSR?

Thank you!

Zexclo

 

[KeepItSimpleStupid]

I've done this sort of thing and it gets MESSY and expensive. Really, it straightforward with the RIGHT EQUIPMENT. What we had was messy in a different way.

The controller generally needs a voltage like 0-10 V out 0-5 V out or 0-20 mA out. ISOLATED here is preferred. Isolated 0-20 mA has been easy to get. That can be converted to a remote 0-5V voltage with a resistor EASILY.

The power supply needs to be controlled with a voltage like 0-5V. Isolated control has been pricey so put the isolation in the controller.

Hook together and everyone is happy.

I used Eurotherm controllers with Xantrex or Kikisui power supplies. The Xantrex was 1 RU unit with AC 120/240.

Cake+Expensive.

There are all kinds of control modes for the end device ad some are better than others. 0-100% of something is the best of all.

I also did low voltage variac control with triacs, BUT the power unit MUST have current limiting ad Phase angle fired.

The Caveot is that a typical control input to a DC power supply is a voltage from the positive terminal to some point inside the supply. When you isolate this it becomes cake. Te modules to isolate the I and V were like $500. The controllers did it for free. You could not buy non isolated.

 

[zexclo]

Yeah, its quite confusing. Am glad to have come to this forum and so great that you guys are willing to help and explain.
Alot of the terms are quite foreign to me and I aim to understand the components, and from there purchase the right equipment.

Thanks for sharing your setup, but I am quite confused onto the power supplies and how to do about the setup.

Hence for now,I aim to get a power source from the main.
I was reading online and came across this article at https://makezine.com/projects/universal-temperature-controller-for-70/
Their set up does not have a power supply and it seems that the PID controller and the hair dryer( heating element) shares the same neutral and one of the points (of the dryer) is connected to SSR. Just wondering for their setup, does that mean that their PID controller and heating element uses the same voltage? Or that there is a voltage divider or something which they missed out. As it seems that their PID and heating element uses the same voltage.

Sorry for keep asking questions as I am really new to electronics.

 

[KeepItSimpleStupid]

Let me try to explain a few output control modes.
ON/OFF control; prone to swings unless it's like a tank of water
Slow Cycle - Varies the On and off time; e.g. 5 AC cycles on 10 off; Can be relay; can be triac
Phase angle firing - the device usually outputs to an input proportional to V^2; Call it k*Watts, but k is unknown and R is assumed constant
Phase angle firing with current limit - Used for transformers.
True power control - Rare

 

[Les Jones]

Hi zexclo,
Here are my answers to the questions in post #9
1 If you are plugging different parts of your circuit into different sockets then the neutrals will be connected together via the sockets. (This is assuming the the SSR is switching the live feed to the heater.) If you are switching the neutral with the SSR then it would be the live that would be connected together.

2 The reason I suggest avoiding switching the mains to the input of an electronic transformer is that I suspect they use inrush current protection in the form of a thermistor. As the PID will be switching on and off quite quickly it will not allow the thermistor to cool down to ambient temperature so it would not provide inrush protection. You would need to check the design of the particular item to see if this assumption applied. The answer to weather the SSR requires a heat sink depends on the current it is switching which is unknown until you decide on the heater power and voltage.
Some idea about the item you are controlling may help with suggested solutions.

Les.

 

[zexclo]

Hey thanks keepitsimplestupid for the explanation on the kinds of output control out there.
From the explanation, can I safely assume the PID is similar to the close angle firing control system? As its always on and that its constantly checking the system, with the small deviation K unknown and the big R assumed constant.

What do you meant by isolated controller?

Hi Les,

Thanks for the clarifications. Currently still deciding on my heating element, and also deciding the SSR unit. I believe during my testing and experimenting process, I might have blown the fuse of my PID controller. The voltage reading from DC out is 0.005v to 0.006v from previously 27V.
Hence, I should be purchasing a new unit. Will proceed to get a compatible SSR and decide a heating element from there.

Earlier, you suggested on wire-wound resistor, would you recommend it as a cheaper and safer, more efficient unit as compared to a hair dryer? As the space which I am try to control is only 120cm(length), 40cm(height) 40cm(width). The surrounding temperature is around 22 degree Celsius, and my targeted temperature is 25 degree Celsius.

Zexclo

 

[KeepItSimpleStupid]

KISS works.
Simple terms with PID the setpoint and measured value can agree.
P = is a proportional constant P*e
I= is constant that is multipled by the integral of the error. (May be value Reset in repeats/min)
D = is the derivative constant (prevents overshoot); the derivative of the error (May be called rate)
e= MV-PV or Measured Value-Process Value; (heat/cool) may come into play here.
p= 0-100% is the output
The constants of P,I and D are found using process upsets. A self-tuning controller will find them automagically.
So, basically out of PID, you get a 0 to 100% number is we assume (heating)
I've made non auto-tuning controllers from scratch.

==

The physical output has to turn that 0-100% into something. It can be a voltage, current,% ON/OFF signal etc,

Generally we have another device which would take that 0-100% signal and do something with it. Letting out a piece of the AC waveform as a function of V^2 is one of the better methods, because V^2 is proportional to power and AC is easy to control. Special designs can operate into transformers or tungsten loads (lamps).

Other methods might use 10 sec on 90 sec off for 10% out, others might use exact 1/2 cycles of the AC waveform.

==

Isolated is a term where the output is not physically connected to anything else. Put it simply, the output looks like a battery. You can put it's terminals anywhere. SO, if the output of a power supply is grounded, the isolated output does not interact with that ground. If the control voltage reference is the output of the power supply which varies, it's perfectly content. The way most voltage controlled (e.g 0-5 V = 1-12 V) supplies are designed, it just so happens that the voltage reference is the output of the supply and for a lot of reasons, the negatives are grounded.

 

[Les Jones]

Hi zexclo,
I would suggest using a few wire wound resistors mounted in front of a computer type fan so that the air is circulated in the container to minimise variations in temperature at different points. I would think 50 - 100 watts would be more than enough to heat that space. I would suggest running the heater on 12 or 24 volts from a normal transformer with the SSR switching the low voltage side. Electronic transformers would not be suitable as they normally need a minimum load. (They would have zero load when the SSR was in the open state.) You could supply the computer fan with a bridge rectifier, a capacitor and voltage regulator IC powered from the unswitched 12 or 24 volts AC from the transformer.

Les.

 

[zexclo]

Nice name there, KISS.. hahaha.
Believe that I am extremely long way from building my own non-auto tuning controllers.
And I am also taking quite awhile to understand the concept and how this things work.
Guess I am better off first trying to put things together, before trying to fix something up from scratch.

===

Hi Les,

Your suggestions sound great, shall try to get wire wound resistor instead of the hair dryer. Am probably going to try get the parts this weekend or something, the wire wound resistors and SSR.
Thanks all for all your assistance!
Shall keep the progress posted here, haha and do hope ya will continue to help should I run into any problems.

Cheers,
Zexclo

 

[KeepItSimpleStupid]

I actually didn't realize you had the supply capable of doing what you need to do, The manual is here. **broken link removed** Take a look at figure 5.4 where it talks about external programming of 0-10V. Also note that the programming source is applied to the positive output and some internal part. What that means is, the negative output cannot be grounded if the programming source is not isolated.

==

The words Series/Parallel are used when you have multiple identical supplies. You don't. The options allow one supply to be the master (control) and one to be the slave. You you can doube the voltage or double the current and use one knob or programming signal from the master.

==

Now, the specs for the controller is another problem, because it doesn't look like it has a 0-10V output. e.g. wrong controller for that type of connection.

Output Options
• Solid-state relay, 0.5A @ 24VÅ (ac) min., 253VÅ (ac) max., opto-isolated, burst fire. With or without contact suppression.
• Open collector: Max. voltage 42VÎ (dc), max. current 1A.
• Switched dc signal: Min. turn-on voltage of 3VÎ (dc) into min. 500Ω load; max.On voltage not greater than 32VÎ (dc) into an infinite load, isolated.
• Electromechanical relay1, Form C, 5A @ 120/240VÅ (ac), 6A @ 28VÎ (dc), 1/8 hp. @ 120VÅ (ac) or 125VA @ 120VÅ (ac). With or without contact suppression. Off-state output impedance with RC suppression is 20kΩ.

So which option do you have?

SSR's can be purchased that turn on with a 3 to 32 V control signal. AC SSR's will turn off at 0V and 0 current without a control signal. AC SSR's used on DC will basically stay stuck ON. SSR's require a heat sink and a thin thermal grease.

These http://www.omega.com/pptst/SSRL240_660.html are AC SSR's.

Here http://www.omega.com/pptst/SSRDC100V.html are some SSR's designed for DC loads.

With an SSR, you would be switching the load on and off.

==

ASIDE: Theoretically, if you had a 0-10V reference), you could switch the supply on and off via the 0-10V control signal. 0 being full off and 10 being full ON.
So, a 0/x (x<10) voltage signal would turn on and off the power supply using the control signals. I can't say one way or the other whether the method would be recommended.

==
I can't comment on the resistor vs light bulb without knowing more. With some light bulbs like the EYC (12 V) and ELH (120 v), the IR (Heat) goes out the BACK of the bulb and the light out the front. I've also use water cooled IR lamps and others for heating.

 

[zexclo]

Earlier, I measured the output signal from the PID controller, it was about 27V.
Using a self-made voltage divider of 2 ohms and 1 ohms, I stepped down to 9V before connecting to the DC power supply.
The DC supply reads the pulses but the light bulb did not light up.

Its rather confusing to me as to the master, slave, series and parallel setting. Plus, will be running this operation continuously for at least a week, hence I gave up on the use of DC supply.

===

Thanks for the recommendations on the SSRs, now that I have decided to remove the DC supply.

Should to confirm, I should be getting the AC SSR? which is https://www.omega.com/pptst/SSRL240_660.html ?
Is there any attention to the SSR which I should pay to regard the SSR for my setup?

I will be using the resistor as its should be safer. With the constant change and pulsing of the voltage, I am worry about the shelf life of the light bulb plus the safety. Am looking at some 50w to 100w wire wound resistors as suggested by Les.

 

[KeepItSimpleStupid]

Probably way over your head but a worthwhile read: https://literature.cdn.keysight.com/litweb/pdf/5989-6288EN.pdf A bible, of sorts.

The DC supply reads the pulses but the light bulb did not light up.

Agreed,it won't based on a lot of stuff.

I need to apologize. I should have been paying more attention I'm used to attacking the problem from the heat end first and that's what I TRIED to do instead of analyzing what you have and what your trying to do.

You linked to the wrong manual, I think: https://www.watlow.com/downloads/en/manuals/series 982 rev.r 01-12-07.pdf is for the WATLOW controller model ( (982C-25CC-KRGG) that you have.

Using PDF page 139, The CC says, two Open Collector isolated outputs. This means a LOT to me and zip to you, the OP (Original Poster).


R is RS232

So, in terms of the controller, we MAY or MAY not be able to salvage something. I know, it's WAY over your head.

Without reading the manual, outputs 1 and 2 are probably "heat/cool" and outputs 3 and 4 are "other stuff such as alarms. The Open Collector output is a nice output for an alarm. The SSR relay sometimes is a nice output to control an AC contactor.

K is K = Solid-state relay, Form A, 0.5A, without contact suppression
This

Output Options
• Solid-state relay, Form A, 0.5A @ 24V~ (ac) min., 253V~ (ac) max.,
opto-isolated, burst fire switching. With or without contact
suppression. Off-state output impedance is 20kΩ with RC suppression, 31MΩ without contact suppression.

tells us, that the SSR is designed for AC.

So, we have a BIG oops on the surface

To control the "power supply", you would have wanted this kind of output for the "HEAT" Output. i.e. Max is hotter.
F = Universal process, 0-5V; (dc), 1-5V; (dc), 0-10 (dc); 0-20mA (dc), 4-20mA(dc), isolated

In other words, you have mess to clean up.

There was/is a lot of stuff you didn't/don't know to be able to make an informed decision and I have to do some reading. (The controller manual)