Hi all i have just bought a achi ir6500 and im trying to setup the lower pid witch is above mentioned.

Problem is its over shooting the temperature and im really not sure what my values need to be as what comes on the screen is different to the manual.

Any help would be greatly appreciated.

Values atm:

Lck-0000

ARU- 0

AL1- 0002

P- 36

I- 27

D- 0002

R- 0001

SC- 000.0

UO- 0001

P1 020.0

## PID XMTG-8000

### Re: PID XMTG-8000

If you're going to set up a lower PID you need to follow an equation for a closed-loop response. I admit I don't know much about this, but I know that as the phase margin gets smaller your oscillation will get worse. This might help:

In the Sept. 1990 issue of Motion Control, this block diagram of a basic servo and its response formula were published.

In the top diagram, we have the element (A). The action of summing junction is to subtract the feedback signal (F) from the input (C) with the result known as the error signal (E)=C-F.

The Bode diagram (below) shows how open loop gain A in an amplifer/motor combination typically experiences a decrease of amplitude by a factor of 10 for every factor of 10 increase in frequency.

The net effect is that A is also A-90°, since it has a gain factor of A and a phase lag of 90°. This closed loop response [F/C = A/(1+A)]

As A' approaches 1 on the Bode diagram (at 10 rad/sec in the example) the denominator become 1+1 -180°=1-1=0 and F/C becomes infinite.The result of this is severe oscillations. But in order to maintain a stable system, the denominator must not be allowed to approach zero. A commonly accepted design goal is for A' to have -135° of phase shift or less (45° of phase margin) This will result in a 25% overshoot of the closed loop system in response to small step inputs.

In the Sept. 1990 issue of Motion Control, this block diagram of a basic servo and its response formula were published.

In the top diagram, we have the element (A). The action of summing junction is to subtract the feedback signal (F) from the input (C) with the result known as the error signal (E)=C-F.

The Bode diagram (below) shows how open loop gain A in an amplifer/motor combination typically experiences a decrease of amplitude by a factor of 10 for every factor of 10 increase in frequency.

The net effect is that A is also A-90°, since it has a gain factor of A and a phase lag of 90°. This closed loop response [F/C = A/(1+A)]

As A' approaches 1 on the Bode diagram (at 10 rad/sec in the example) the denominator become 1+1 -180°=1-1=0 and F/C becomes infinite.The result of this is severe oscillations. But in order to maintain a stable system, the denominator must not be allowed to approach zero. A commonly accepted design goal is for A' to have -135° of phase shift or less (45° of phase margin) This will result in a 25% overshoot of the closed loop system in response to small step inputs.

### Re: PID XMTG-8000

Sorry. The diagrams didn't post for some reason. If you want to see those, they're on https://www.axcontrol.com/blog/2019/09/ ... -tutorial/(which is where I copied everything after "this might help.") I should've said.

### Re: PID XMTG-8000

Hi guys i wonder what are the electronic components inside the ir6500 ?

What I know so far is

1- PC410 PID temperature controller

2- Rex c-100 pid temperature controller

3- ssr Relay ( but how many relays ? 2 ? )

4- k type thermocouple ( 2 or 1 ? )

5- 180×180mm heating elements

Are there more parts that I missed ?

Thanks

What I know so far is

1- PC410 PID temperature controller

2- Rex c-100 pid temperature controller

3- ssr Relay ( but how many relays ? 2 ? )

4- k type thermocouple ( 2 or 1 ? )

5- 180×180mm heating elements

Are there more parts that I missed ?

Thanks