Building a replacement MTX PSU

About original Memotech hardware.
gunrock
Posts: 19
Joined: 28 Oct 2020 21:17

Re: Building a replacement MTX PSU

Post by gunrock »

Actually Mark, I retested this morning the diode reverse polarity protection and the reading was 0.03 to 0.04V. I used the LM2577 dialled to 9.8V and repeated, the reading was the same. Must have been my eyes late last night :roll:

Using the higher voltage meant that my load test was much better, that 6V led really shone! 8-)

Now I have to start wiring up. This is the bit I'm most nervous about, because as great as all the photos you have produced are, there's a lot of shrink tubing, which obscures things a bit (obviously *not* a criticism). So I need to refer to the schematic more and I'm an almost complete noob with electronics, thus I'm concerned I'll misread it.

I also know I'm going to forget to put shrink tube on something and have to unsolder one end to fit it on! :lol:

Steve
gunrock
Posts: 19
Joined: 28 Oct 2020 21:17

Re: Building a replacement MTX PSU

Post by gunrock »

Ok, so I have a question :?: It seems I have foolishly bought the wrong type of LED; instead of a Kingsbright L-53gd I bought a Kingsbright L-7113id-5v, which is a 5v LED with a built-in resistor.

I can't work out how on the one you specified, there's need for a 1.2K resistor and that's because I can't understand the scematic with my dunce level of electronics knowledge. How much current feeds the LED? 16V from the LM2577, to the center tap (series joined dual secondary) which is rated what? The datsheet says 2.0 forward voltage and 10mA forward current, but TBH I'm lost :( :arrow: :oops:

I'm trying to work out if I can use my LED with built-in resistor and whether I will need extra resistance still. I've learnt a lot the last few days (center tapping dual secondary windings, for one!) but this is twisting my melon, man!

Please help. :oops:
User avatar
1024MAK
Posts: 644
Joined: 24 Dec 2012 03:01
Location: Looking forward to summer, in Somerset, UK

Re: Building a replacement MTX PSU

Post by 1024MAK »

LEDs are light emitting diodes. Like a normal diode, once the forward voltage gets to the ‘turn on’ value, they start to conduct current. With a silicon diode, this is about 0.6V. With a LED, it’s between about 1.8V to 3.6V depending on the type/colour.

By forward voltage, I mean a positive voltage on the anode (A) compared to the cathode (K) terminals/leads.

Once a diode is conducting, a very small increase in voltage results in a VERY BIG increase in current. With a normal diode this is not normally an issue, as it feeds some other load.

But with a LED, normally we only want the LED to light up, and there is no other load. So we have to find another way to limit the current. The easiest and cheapest way is to include a resistor in series with it. Either wired to the anode terminal, or to the cathode terminal. Hence this is what I did.

So what happens if no resistor is used? Well, once the voltage across the LED gets to it’s forward voltage (in this case let’s say 2V), it will try to maintain about 2V across itself. But here the supply voltage is 16V. So the current will increase and increase and increase until something goes pop! In practice the LED will glow brightly for a very brief instant and then it will never work ever again.

To calculate the value of the resistor, you use this formula:

(Vs - Vf) / If = R

Where Vs is the DC supply voltage, Vf is the forward LED voltage, If is the forward LED current that you want, and R is the value of the resistor that you need.

Alternatively you can use this formula to work out the current with a particular resistor:

(Vs - Vf) / R = If

With a standard brightness green LED, the forward voltage is about 2V, so Vf = 2V. The recommended forward current should be 20mA or less, so this gives the maximum If.

And yes, the DC supply voltage in the MTX PSU circuit for the LED is 16V DC.

So the formula becomes:
(16V - 2V) / 1200Ω = 11.6mA

A 5V LED often just contains a built in resistor. Now, it’s unlikely that the manufacturer will tell us the value of resistor that they used. But they do give a typical current. In this case, I believe it’s 13mA.

So I work out that they have used a resistor of about 220Ω.

If you wire the 5V LED in place of an ordinary LED in this circuit along with the 1.2kΩ resistor, the forward current will be about 9.9mA. So the LED will still work, and it should still be a reasonable brightness.

So no need to panic!

The configuration of the transformer secondary windings that I am using is to use the two separate secondary windings to form a centre tapped secondary. I have done this because it’s the arrangement that the power supply circuitry in the MTX requires. The middle / centre tap is effectively the MTX 0V/GND.

I hope this helps.

Mark
gunrock
Posts: 19
Joined: 28 Oct 2020 21:17

Re: Building a replacement MTX PSU

Post by gunrock »

Hi Mark,

That is very helpful, yes, thank you. Normally, in the case where I hit a knowledge gap that I can't seem to fill by some research, I try to make an educated guess and carry on (or obtain the right component!) but this being a PSU, I'm being way more cautious.

I don't want to create a dangerous device nor kill my new Memotech, so I'm being very careful.

I'm back at work now, so progress might be (even) slower, but will keep you posted.

Thanks
Steve
Post Reply