This is a first post from a new member so please bear with me...
That valve seems to be from HWH Corp. They have an online technical school, in lesson 9 i found this..
2-11.2 Hydraulic Solenoid Valves

https://www.hwhcorp.com/ml57000-017.html#_2-6__Transistors

 

If I read the post #5 drawing correctly, the transistor (item 9) is a BDX46, i.e. a PNP Darlington. However, that would be in a TO-126 package, wheres the drawing and pics show a TO-3 package. .

Edit: Looking again at the drawing, I'm now reading the transistor type as BDX66; another PNP Darlington but this time in a TO-3 package. That makes more sense!

I knew some people will used uncommon way to design the circuit, I just thought that with the common method, but the circuit still doesn't complete yet, so do you have the possible wiring for three pins?

 

This is what I think you have :-

 

This is what I think you have :-
View attachment 141559

Since this device only used one power supply and the Green wire is connected to +12V and the White wire is connected to negative of the power supply and then where is the Gnd(c) connected to ... ?

 

The collector of the transistor appears to be connected to the case which will be one side of the power supply, IOW the case is at common potential. which is probably referenced to the -ve side of the supply at some other point..
Max.

 

the White wire is connected to negative of the power supply

I wonder how that was determined? Is it intentionally connected like that, or only under the present fault condition (shorted transistor perhaps)? It would have to be connected to negative (ground), or very close, in order to turn off the transistor if, as I suspect, the white wire is actually the control input. If the transistor is indeed a BDX66, then the top pin in the pic is the base connection.

Edit: 'off' should read 'on'.

 

The TS states that the white wire is negative in post #11. I think we need the TS to tell us if there are realy 3 wires going to the device. Is one leg of the power supply connected to the case and is either the green or the white wire just a control signal. The TS makes this statement in post #15 "Open circuit between - 12V leg and shell, zero Ohms between other leg and shell." Which implies that the green wire is connected to the case and therfore the collector.

Les.

 

Could you measure the resistor and describe the colors, the resistor used 5 colors code then the tolerance could be 1% or less.

The resistor measures 991 Ohms. This is a replacement resistor - it's not possible to look at the originals as they are cased in tough protective sleeve. The original I measured at 992 Ohms.

With reference to post #11 Does one end of the resistor connect to the bottom left hand stud ? (One end of the coil.) When you say "the green wire on the right connects to the bottom leg (Emitter) of the transistor" Is this from the top right coil stud or the bottom left one ? What does the wire from the other stud connect to ? Assuming the transistor pinout is normal for a TO3 transistor then the top pin is the base. The bottom pin the emitter and the case the collector. The fact that the coil only has a resistance of 1.2 ohms suggests that the tramsistor limits the current through the coil as it would take 10 amps (120 watts) if the 12 volts was straight across the coil. I would not expect a solenoid valve to take this much power. If the green power lead (Positive ) is connected to the case (Therfore the collector.) and the white wire power lead (Negative.) is connected to the top pin (Base) of the transitor the circuit does not make sense. How much current does a working valve take ?

Les.

Thanks! Yes, one end of resistor connects to bottom left hand stud. Bottom leg of transistor connects to top right stud. Bottom left hand stud has +12v DC connected to it from DC source. As per photo in post #11, green is +12v DC, white is -12v DC. The studs have phenolic washers on them on both sides of the core (item 22) - the other ends of the studs are the coil ends. I mention this in ref to your "If the green power lead (Positive ) is connected to the case". There's no direct connection between +12v and any casing.

If I read the post #5 drawing correctly, the transistor (item 9) is a BDX46, i.e. a PNP Darlington. However, that would be in a TO-126 package, wheres the drawing and pics show a TO-3 package. .

Edit: Looking again at the drawing, I'm now reading the transistor type as BDX66; another PNP Darlington but this time in a TO-3 package. That makes more sense!

I think you're absolutely right. Attached is the parts list from the drawing post #5. I read item 9 as being BOX 66 and thought it referred to the package type. Sorry to all, if I'd have known more I would have seen it straight off but I hadn't heard of BDX 66.

 

This is what I think you have :-
View attachment 141559

Yes, from reading various replies etc I think that's spot on.

 

I wonder how that was determined? Is it intentionally connected like that, or only under the present fault condition (shorted transistor perhaps)? It would have to be connected to negative (ground), or very close, in order to turn off the transistor if, as I suspect, the white wire is actually the control input. If the transistor is indeed a BDX66, then the top pin in the pic is the base connection.

There's two things that make me think the white wire is not a control input. One is that the gauge on green and white wires is the same. The other is that connecting white to negative terminal on a car battery and green to positive actuates the working valve perfectly. Is that conclusion logical?

 

The TS states that the white wire is negative in post #11. I think we need the TS to tell us if there are realy 3 wires going to the device. Is one leg of the power supply connected to the case and is either the green or the white wire just a control signal. The TS makes this statement in post #15 "Open circuit between - 12V leg and shell, zero Ohms between other leg and shell." Which implies that the green wire is connected to the case and therfore the collector.

Les.

I guess I'm the TS? There is only two wires to the valve, please see photo in post #11. Both green and white wires are of sufficient gauge to make me think they're not control wires. And just a reminder that I think the transistor and diode are blown so multi meter readings can't be taken as indicative of anything normal.

 

connecting white to negative terminal on a car battery and green to positive actuates the working valve perfectly. Is that conclusion logical?

Perfectly logical if the circuit is as in post #23. (See edit to post #26). However, in that circuit the white wire should not be connected directly to ground, because excess (>0.25A) base current could flow and damage the transistor.

 

Does the negative of the 12 volts power supply connect to the machine metalwork ? If it did it would make sense as it then would be connected though the vave casing to the collector of the transistor. This would mean that pulling the white wire to negative would turn the transistor on and most of the current to drive the coil would pass between the collector and emitter. So in effect there would be three conductors to the valve. The green and white wires and the third one being the metalwork of the machine.

Les.

 

That's how I see it , Les; which is why I am doubtful that the white wire is the negative of the supply.

 

I'll put the multi meter across the white wire and metalwork to see. I'm a bit lost with the conversation but not as confused as when the thread started. Thanks!

 

The negative of the power supply does have continuity with the metal manifold. There is no continuity between white wire and manifold body (all power is off). There is continuity (zero Ohms) between green wire and manifold for a few seconds then resistance rises to 4.67K Ohms over a few seconds. I checked another valve port wiring with the same result (except resistance was higher at 6.7K Ohms. I had to unplug the valve to take the reading which I reason in some way accounts for the difference).

I think I might get posts #33 and #34 now. Although I've been connecting the white wire to the negative side of the supply to test, that only means that that needs to happen for the valve to operate. If white carried -12v all the time the valve would be permanently open. Is that right? So my diagram should show a normally open switch on the negative side of the supply?

I'm trying to understand better - if either of the wires was a control wire wouldn't there have to be a relay or something similar to control? I don't really get how control wires come into it as it seems the only thing that physically operates in the valve is the coil armature...?

 

On the assumption that the schematic in post #23 is correct (Which I think it is.) Then The case of the valve would have to be connected to the negative of the power supply and the green wire to to the positive. The valve would then be activated by pulling the white wire towards negative. This would cause the transistor to conduct and apply just under 12 volts across the coil. Assuming your resistance measurement of the coil resistance of 1.2 ohms in post #11 is correct it's current will be 10 amps. (This is higher than I would expect which is why I asked you to measure the current taken by a working valve.) As the a BDX66 transistor is a darington transistor (Which is really 2 transistors in the same package connected in a way that makes the current gain about equal to the product of the current gains of each transistor.) the current through the white wire only needs to be about 10 mA. The way you have been connecting it with negative to the white wire and NO negative to the collector (The casing.) the base emitter junction of the transistor is behaving like a diode (Actually 2 diodes in this case as it is a darlington transistor.) it will be passing the full 10 amps through this junction. This is almost certain to destroy the transistor as the maximum base current rating is only 0.25 amps. (Alec has pointed this out in post #32.) I think you will have problems getting any BDY66 transistors. RS and Farnell do not have them and there are not even any listed on Ebay. You will need to find a more modern equivelent. By the way TS stands for thread starter which is you in this case.

Les.

 

On the assumption that the schematic in post #23 is correct (Which I think it is.) Then The case of the valve would have to be connected to the negative of the power supply and the green wire to to the positive. The valve would then be activated by pulling the white wire towards negative. This would cause the transistor to conduct and apply just under 12 volts across the coil. Assuming your resistance measurement of the coil resistance of 1.2 ohms in post #11 is correct it's current will be 10 amps. (This is higher than I would expect which is why I asked you to measure the current taken by a working valve.) As the a BDX66 transistor is a darington transistor (Which is really 2 transistors in the same package connected in a way that makes the current gain about equal to the product of the current gains of each transistor.) the current through the white wire only needs to be about 10 mA. The way you have been connecting it with negative to the white wire and NO negative to the collector (The casing.) the base emitter junction of the transistor is behaving like a diode (Actually 2 diodes in this case as it is a darlington transistor.) it will be passing the full 10 amps through this junction. This is almost certain to destroy the transistor as the maximum base current rating is only 0.25 amps. (Alec has pointed this out in post #32.) I think you will have problems getting any BDY66 transistors. RS and Farnell do not have them and there are not even any listed on Ebay. You will need to find a more modern equivelent. By the way TS stands for thread starter which is you in this case.

Les.

Great explanation! Thanks LesJones. I think I understand almost all of that. I did miss your request to measure the current from an earlier post. I'll do that today.

 

Hello, I found these on ebay:

KT825G: pnp-darlington BDX66: pnp-darlington
Ic : 20A Ic : 20A
U : 70V U : 60V
P : 125W P : 150W
ß >750 ß > 1000
Ft 4MHz Ft 3MHZ

If the current draw is 10A as thought, does 125W allow enough headroom? I don't know about the beta or Ft values either.....any pointers gratefully received.

 

I think that should be a suitable replacement. While looking at data sheets I did notice that the KT825 (With no "G" at the end comes in a TO220 package so make sure you get the verion with the G which is in a TO3 package. I think the power rating is OK. The minimulm beta value is slightly lower than the BDX66 but I don't think that will be a problem. (The beta value is the gain of the transistor which is the ratio of collector current to base current.) It will just mean the drive current (Via the white wire.) may be slightly higher. What drives the signal to the white wire ? (A PLC, a relay, a transistor or something else.) It may be worth buying more than just the one transistor that you need in case you need some in the future. I can find no other suitable replacements on the Farnell (UK) website. (I assume you are not in the UK or Europe from the times of your posts.)

Les.