Member postings for AJAX

Maybe you are right - it could be a later addition - who knows?

Making a collar would be simple enough but it would be nice to add some form of adjustment or loading (spring?) or at least consider doing so instead of to shim. Some of the decision making may depend on how well I can clean that dial. If I can't make it readable then I'll have to buy or make a new one.

That would make sense as it would allow for some adjustment.

However, the collar is definitely mis-drilled on one side and exits as a single hole on the other.

I would like to come up with a better solution to take up any slack when it is re-assembled.

I was actually surprised to find the needle roller bearings in place as I had planned to shim and/or add such bearings myself.

The next job will probably be a complete strip down of the apron. I know the worm is very badly worn (I had it apart once before) and would like to get this replaced somehow.

Progress made. Two pins found and both tapped out with some encouragement. Neither were willing to press out with the (mostly unsuitable) tools I had available.

The one closest to the end was drilled slightly off centre and the other one, well I can only speculate was drilled by a YTS trainee on his first day at work. Even though it's hidden, I will most likely be remaking that part!

Thanks Old Mart and Pete. I will take a look later this evening. I'm working now and then the kids have to go to bed first...

I have found a working drawing for the top slide leadscrew. I have yet to check, but it looks like it may be the correct drawing. If it is, the leadscrew is drilled and reamed for 5/32" dia taper pins. I may be back later...

Thanks for that, Clive. I agree with your suggestion that it could be a roll pin. I had a light tap previously believing the same but it did not budge. Perhaps it was in a tapered hole as you suggested and I was going the wrong way. Tricky to tell. I will take another look later today.

Regards, Brian

I'm wondering if there is a taper pin there but as I can't actually see it I would have no idea which way the taper ran.

I would like to remove the cross slide handle from the leadscrew which I assumed would be a simple task. I assumed it might be held with a pin or grub screw but on closer inspection I can see neither. However, the light and my eyes aren't great.

I'm hoping there may be another Denford owner who can explain the process.

There is a "through hole" of some sort as demonstrated when I squirted WD40 straight through. The end looks like it may have been tapped with a hammer at some point, so maybe that is what is keeping it in place.

Does anyone have useful ideas for how I might go about improving the appearance and usability of the graduated dial when I can get it off?

In a closely related question, I would be interested in experiences of machining a new plain / slotted cross slide for the purposes of attaching a vertical milling slide. I have access to a Bridgeport-sized milling machine at work but no surface grinder.

I fitted one to my denford lathe for the same reason it has been mentioned it is easier to fit to a boxford. It is easy to fit.

It has worked fine. Not stretched. About average noise.

Due to price I would not fit one unless I had to. Vee belts work just fine for me.

James,

I have only just read your updated question.

Short answer - don't use a transistor. Don't use a MOSFET. Use a Darlington array (ULN2803)

"The ULN2803A device is a 50 V, 500 mA Darlington transistor array. The device consists of eight NPN Darlington pairs that feature high-voltage outputs with common-cathode clamp diodes for switching inductive loads. The collector-current rating of each Darlington pair is 500 mA. The Darlington pairs may be connected in parallel for higher current capability. Applications include relay drivers, hammer drivers, lamp drivers, display drivers (LED and gas discharge), line drivers, and logic buffers. The ULN2803A device has a 2.7-kΩ series base resistor for each Darlington pair for operation directly with TTL or 5-V CMOS devices. "

datasheet

Note that base resistors are in the package so you don't need to add resistors in-line with the microcontroller outputs. You should make use of the internal diodes to clamp any inductive spikes from the loads.

Brian

IRL540N is a logic level MOSFET that may suit.

https://docs.rs-online.com/8d29/0900766b80028e3a.pdf

Edited By AJAX on 29/12/2020 13:06:19

IRF3205

https://www.digchip.com/datasheets/download_datasheet.php?id=436025&part-number=IRF3205

Edited By AJAX on 29/12/2020 13:07:09

Thanks for making me check the datasheet.

https://www.digchip.com/datasheets/parts/datasheet/232/IRF3205-pdf.php

Fig 3. Typical Transfer Characteristics indicates a drain-source current of about 50A with a 5V logic signal from the microcontroller. It may not be a true logic level MOSFET but should be fine for this application.

 VGS(th) Gate Threshold Voltage 2.0 ––– 4.0 V VDS = VGS, ID = 250µA

Check the datasheet for the 2N2222A - https://www.farnell.com/datasheets/296640.pdf

Ic (max) = 800 mA

This is an absolute maximum rating.

The suggested MOSFET is a voltage operated device and for most purposes can be assumed to draw no current on the gate input. As you say, it is capable of switching a large current but it can also be used to switch a very small current too. RDS(on) is very low, only 8 mOhm making it good for power applications.

https://www.digchip.com/datasheets/parts/datasheet/232/IRF3205-pdf.php

Edited By AJAX on 29/12/2020 12:46:12

Port bits ("pins" ) used as outputs must be declared as outputs. Outputs don't have extra "oomph". Inputs configured with internal pullups may be a cause of confusion?

Current is not "absorbed". I realise this is an attempt to explain the situation using layman's terms, but it may be useful for the unaware to read about charge conservation and Kirchoff's current law (KCL).

An ideal MOSFET ("ideal" doesn't really exist) has an infinite gate impedance and will not draw any current. A real world MOSFET such as the IRF3205 has a very high gate impedance and small gate capacitance and so any transitory current (to charge the gate) can be ignored.

In some applications it may be necessary to add a resistor between the microcontroller pin and the MOSFET gate, but in this application it is not required.

Edited By AJAX on 29/12/2020 12:36:05

Another point, most standard 5mm red LEDs have a maximum forward current of about 15 - 20 mA.

Check the data sheet if you have one.

If not, choose a current limiting resistor that limits current to about 10 mA. You can assume Vf is about 2V.

I believe that James wants the LED to act as an indicator for the solenoid (LED on when the coil is energised). The LED therefore needs to be placed in parallel (forward biased) with the solenoid load.

The solenoid is an inductive load and the circuit should be protected with a freewheeling or flyback diode. 1N4001 in reverse bias with the solenoid would be fine.

The use of low-side switching where the load is placed on the collector side of the the transistor "switch" is correct for this application.

Assuming pin 13 outputs a logic high of about 5V and Vbe 0.7V this will draw about 20 mA base current. BC547 is a small signal transistor and not ideally suited to switching solenoids and you'd have to be careful of the expected collector current.

For a switching application I would suggest use of a logic-level N-channel MOSFET. Off the top of my head, an IRF3205 would be fine and would be able to handle substantially more current than you will switch. The MOSFET requires no resistor on the gate. Any charge on the gate due to gate capacitance would discharge (sink) back to pin 13 during logic low.

Do you need a circuit drawn?

Brian

I sold a Rapidor Manchester today that had a modified vice with the fixed jaw at the front and movable at the rear (closest to the driving linkage). That actually made sense especially as the blade was set to cut on the forward stroke.

On the other hand, my Q&S 6H cuts on the pull stroke and has the fixed jaw at the rear. The hydraulic relief gives a very slight lift (or reduction of weight) on the forward stroke.