AKG C12 A / B Version Build

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Swap that 3k3 for a 6k8 (oddly enough, what my earlier guesstimate was), or stick 2 x 3k3 in series there.

I don't know what the lorlin looks like that you are using. Look at the schematic and look at the pin diagram of the lorlin. Reconcile the two.

You should have 56k resistors between the consecutive "throw" connections, the bottom one also grounded, the top one also to B+. The "pole" connection of the lorlin goes to the 1Meg.

If your switch is something like this - http://www.lorlin.co.uk/PDF/CK.pdf - then you should have 56k resistors 1 to 2, 2 to 3, 3 to 4, 4 to 5, etc. up to 8 to 9.  "A" to the 1 Meg. 1 to ground and 9 to B+ (or vice versa).
 
Hi Matt,

I've wired it up as you've said and now nothing happens?
Have swapped the connections round on pin1 and 9 and I get nothing happening when going through the 9 selections?
whats going on? ive tested at each step and the voltages look good, just nothing happens when i turn the lorlin?
voltage is still about 127v.

Regards

Spence.
 
B+ should remain unchanged.

The voltage at the "A" connection on the lorlin should change (the voltage at the 1Meg) as you move the switch position.
 
Thank for that! I understand now! Lol oops!
I have even voltages on the Lorlin!!! Going from 15v up to 124.4v!
Just waiting for the 100R pots to arrive then I can start to test H+,
tomorrow I'll start putting together the body and braising the headbasket to it.
Need to make the internal metal plate to hold the pcb and attach the 7 pin plug to it.
Then need to make a mould for the rubber capsule mount and then mount the capsule.

Spence.
 
Yes, sorry Matt, 0v is first position then 15v, 30v etc up to 124v. B+ will probably drop a little once tube

connected so hopefully fine.

Thanks so much for you help with this! Nearly there now!

Regards

Spence.
 
One more slight problem!
I'm wanting to run the B+ to two separate feeds for two mics.
I'm slightly confused of how when running two feeds, with one Lorlin turned down the other Lorlin doesn't work?

Any ideas?

Regards

Spence.
 
Spencerleehorton said:
One more slight problem!
I'm wanting to run the B+ to two separate feeds for two mics.
I'm slightly confused of how when running two feeds, with one Lorlin turned down the other Lorlin doesn't work?

Any ideas?

Regards

Spence.
Look at your wiring. Something must be wrong.

I'm not sure how - given that from the pictures you have built 2 separate PCBs, one for each power supply.

You have PSU-A generating a B+ which goes to Mic A and also drives the 330k bleeder resistor A and the top position of Lorlin A. You then have PSU-B generating a B+ which goes to Mic B and also drives the 330k bleeder resistor B and the top position of Lorlin B.

The switch positions should not affect B+, nor should they affect each other.

Or have you only got 2 separate PSUs for H+ and a shared one for B+ ?  If so, you'll need to adjust your RC filter resistors to compensate for twice the current draw on B+. Then you'd have both Lorlins strapped in parallel from B+ to ground, but their pole connections should be separate, one going to each mic.

I don't mean to be rude, but if you're having troubles with stuff like this, I worry for your safety wiring anything up to mains transformers. Perhaps some revision of Ohm's law and Kirchoff's laws is in order?
 
Yes I worry all the time with me wiring transformers!!! Lol.



I have 1 x pcb for B+ and 2 x pcb for H+.
Sorry to be such a pain, but which ones are the RC filter resistors? 15k, 15k and 6.8k are you referring to?
So I need to allow for 4.8mA right? 2.4mA each feed?

Regards

Spence.
 
2.4mA for each Mic
0.27mA for each Lorlin
0.36mA for the 330k bleeder resistor.

5.7mA total.

Those resistors (the RC filter resistors - yes, the ones by the reservoir caps, the 15k, etc.) are going to have to come down in value quite a bit I think. Then, if you want the same power supply ripple on b+ (you do) you'll have to up the size of the capacitors to compensate, so that each R*C comes to about the same (or larger is OK) value as before.

Yes, you need the 1Meg and 1uF on both Lorlin wipers.
 
right, ok so i have to up the 47uf values to ? i have put the first two 47uf at 400v, the second two are 250v. will i need to take these up to around 100uf? is there a calculation for this or do we just assume doubling the current we should double the UF?

my calculation for 5.7mA - 120v / 0.0057 = 21,000ohm

i'm at 36,800ohm with the (15k,15k,6.8k)

so would it be best to have 10k, 10k, 1k?

am i guessing about right?

regards

Spence.
 
No. You're mixed up again. Or at least I think you are.

This isn't particularly exact, because we don't have pure DC all the way through, but it will work as a rough starting point. Exact values may have to be tweaked in circuit.

Assuming you were getting approx 120V with a single mic dummy load, the lorlin and the bleeder in place then you had 2.4 + 0.36 + 0.27 mA = 3mA.

Those 3mA were running through your 15k + 15k + 6k8 = 36k8. So, they were dropping 3mA x 36k8 = 110V across them. They had 120V at the mic end, so must have had 120V + 110V = 230V at the rectifier end under load.

Now you want to draw 5.7mA and still have 120V at the mic end, so still 110V dropped across the new resistors. This is pretty close to the 120V in your calculation, so my apologies if this is actually what you were thinking.

110V / 5.7mA = 19.3k

Two 6k2 and one 6k8 is close. You may need a little less, you may need a little more. 100uFs will work just fine.
 
ah ok great stuff,
i have re wired the lorlins and now have separate switches working fine, under load with 47k and was getting 120v exactly!!
will change other resistors and caps tomorrow as a bit shattered after beavering away at these mic bodies.
just got to design an inner section that the capsule can mount on now and nickel plate them.

regards

Spence.
 

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