Hello
(First time poster with rusty familiarity with vaccuum tube audio amplifiers)
I am working with a Siemens 1212C PLC and looking to power a Burkert proportional solenoid via the PLC HSC outputs. Smaller solenoids have been successfully powered this way, but I need a bigger orifice size which requires a bigger solenoid. The maximum rated output of the PLC is 20V @ 500mA (ON) and the Burkert 2875 requires 24V @ 750mA to fully open. (In the smaller valves the 20V maximum is very nearly fully open and an acceptable compromise). Rather than add expensive AD/DA modules, is there a basic current gain MOSFET / transistor amp that I could employ? I'm not that familiar with transistor design, but have added a simple design that may work (?). I have a 24V supply to work with and don't need a big gain ~x2 current, and ideally a small voltage gain (which this design certainly won't provide). It's a 900Hz signal, so I imagine it can be a very simple circuit - added C2 to tune the output to a square wave and C1 to add some local capacitance. No doubt missing a diode.
Can anyone point me in the right direction? I have some IRF540N in the drawer, so if that works for easy prototyping - great!
Siemens 1212C
Burkert 2875
Thanks for any help,
Ben
"Normally" you would use a circuit like this with NCHAN MOSFET -
View attachment 138459
I would advise R1 somewhere in range of 50 - 100 ohms. Using a scope you tune that value to
minimize ringing at MOSFET gate.
D1 is used to protect mosfet from damage due to inductive transients generated by valve
coil.
Your IRF540 OK choice for MOSFET although I might be tempted to use a 200V part
for additional margin under transient stress.
D1 should be a 2A, 200 V rectifier, reasonably fast recovery. In4148 shown simply not adequate.
Regards, Dana.
Similar to what I use in one of our commercial products, which funnily enough uses a (very expensive) Burkert solenoid valve, with the usual 24V requirement.
We use a TK40E06 FET, 1N5395 diode, 10K for R2 - with the gate connected directly to the I/O pin of a PIC.
"Normally" you would use a circuit like this with NCHAN MOSFET -
View attachment 138459
I would advise R1 somewhere in range of 50 - 100 ohms. Using a scope you tune that value to
minimize ringing at MOSFET gate.
D1 is used to protect mosfet from damage due to inductive transients generated by valve
coil.
Your IRF540 OK choice for MOSFET although I might be tempted to use a 200V part
for additional margin under transient stress.
D1 should be a 2A, 200 V rectifier, reasonably fast recovery. In4148 shown simply not adequate.
Regards, Dana.
Do you know the PLC PWM output frequency when running ?
I'd probably use a bipolar transistor rather than MOSFET, as this will give simpler drive and faster switching.
Can you use inverse PWM (0% = max, 100% = off), or does the PLC have the option to invert the output?
The simplest way to retain a positive switched feed to the solenoid (to keep it failsafe) is a PNP power transistor.
Emitter to positive supply, base to PLC output via a resistor, and add a base-emitter resistor to ensure the transistor switches fully off.
A TIP32, MJE15029 or MJE15033 etc. should be suitable.
The base current should be about 75 - 100mA, so eg. 270 Ohm 3W or 5W resistor from the PLC output to base, then a 1K or 470 Ohm between base & emitter.
If the solenoid valve does not have some form of internal snubber, you will need to add one.
A flywheel diode will prevent back EMF from causing damage, but will also retain the circulating current and may make it appear the output is higher than it is.
You can add a resistor in series with the flywheel diode, about equal to the solenoid resistance, to give faster current delay while limiting the negative spike to around 24V.
[Note for others - in any automation system, control outputs should always be positive-switched for safety; in case of wiring damage, the commonest type of short (to ground, the machine metalwork) switches things off and blows fuses, rather than activating valves or relays etc. which could cause malfunction or injury.]
Hi all,
Down under here - so excuse the slow response!
Yup. Them is expensive! Adding an 4~20mA AO module and a Burkert Analogue to PWM module seems way more expensive and counter-intuitive. I was under the impression that FETs need a small resistance at the gate to inhibit resistance? Is that not needed here? The PLC is capable of strong drive obviously.
Thanks very much for this - I have redrawn it for my monkey brain to understand:
View attachment 138471
I have some MUR860 in drawer, and the STP60NF06 is available in store locally. I also added some local capacitance (C1) to help out power supply - needed?
Mine isn't proportional, and only very low frequency, a few hertz at most.I can set this - anything up to 100kHz, this Burkert is specified at 900Hz, and I have driven the smaller units direct at 750Hz.
Thanks again.
The transistor and resistor part is correct, but the solenoid would be in the collector circuit, between the transistor and 0V, with the emitter to +24V.I also drew this as I understand, but assume I have made a mistake(s):
The transistor and resistor part is correct, but the solenoid would be in the collector circuit, between the transistor and 0V, with the emitter to +24V.
The snubber circuit across the solenoid is also correct, just what I was thinking of.
The PLC output would then be low for on and high for off.
If you wanted to keep the PLC output as high = on, you could add the FET stage from the previous diagram, but using a lower rated part, and have that switch the base resistor in the bipolar transistor stage so the overall circuit inverts twice.
An NPN bipolar transistor would also work in that circuit in place of the FET & the resistor values could be somewhat higher if it is use as driver to a second transistor stage, as the currents involved are lower.
The load is supposed to be in the collector, not emitter...Circuit seems to work well although TIP32 not getting saturated,
Circuit seems to work well although TIP32 not getting saturated,
maybe a little excess power getting consumed -
View attachment 138477
Did not know relay L so "picked" 1 mH as a reasonable
value
Regards, Dana.
Keen to understand why the above circuit is not so optimised - is it that the TIP32 is not fully saturated and in the threshold area: therefore dissipating unnecessary heat? Is the above circuit still inverting?The load is supposed to be in the collector, not emitter...
I used the post in post # 10, now saw the additional comments., soThe load is supposed to be in the collector, not emitter...
Remember the 20V was when sourcing 500mA; when sinking the maximum should be higher.Looking at L (relay) current not fully off because PLC cant generate 24 Vdc
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