phono load resistors
#1 phono load resistors
Just a quick question.
If trying primary loading for my cart, do i need to replace the 47k in the phono with 1M?
Why do i need the 1M and why isnt this 1M resistor acting as a large loading resistor?
Not sure whether to just take out the 47k and leave it like that or fit the 1M's ive read about elswhere.
(WD phono III clone by the way and 1:10 tx's)
Thanks.
If trying primary loading for my cart, do i need to replace the 47k in the phono with 1M?
Why do i need the 1M and why isnt this 1M resistor acting as a large loading resistor?
Not sure whether to just take out the 47k and leave it like that or fit the 1M's ive read about elswhere.
(WD phono III clone by the way and 1:10 tx's)
Thanks.
-
- Old Hand
- Posts: 780
- Joined: Thu Jun 07, 2007 2:32 pm
- Location: sheffield
#2
loading resistors are connected from signal to earth
so the SMALLER the value of the resistor, the BIGGER the effect it has
by increasing the value of the resistor on the input of your amp i.e. on the secondary; you can place more of the loading on the primary
at least that how i understand it i.e not gospel !
so the SMALLER the value of the resistor, the BIGGER the effect it has
by increasing the value of the resistor on the input of your amp i.e. on the secondary; you can place more of the loading on the primary
at least that how i understand it i.e not gospel !
#3
Still dont get it.
At the moment i have 10k on the secondary (in the phono)
Means i have 100r seen by the cart.
I want to try loading the primary so ill take out the 10k and put 100r on the primary.
But, if i put 1m on the secondary (to replace the 10k) and 100r on the primary why isnt that loading the cart at 10,100r?
Thats what i dont get.
At the moment i have 10k on the secondary (in the phono)
Means i have 100r seen by the cart.
I want to try loading the primary so ill take out the 10k and put 100r on the primary.
But, if i put 1m on the secondary (to replace the 10k) and 100r on the primary why isnt that loading the cart at 10,100r?
Thats what i dont get.
-
- Old Hand
- Posts: 780
- Joined: Thu Jun 07, 2007 2:32 pm
- Location: sheffield
#4
isn't the 10k is in parallel with the 47K on the input to your phono stage ? so the effective load across the secondary is less than 10K
and you would replace the 47k with 1M; not the 10K
and you would replace the 47k with 1M; not the 10K
#5
No, i replaced the 47k with the 10k.
There is only the 10k resistor loading the cart, in the phono stage, loading the secondary.
I'd forgotten about resistors in parallell.
My maths is wrong, but the theory stays.
So, putting 1m in the phono doesnt actually load the cart right?
Is it too high a value to have any effect?
The cart will only see the smaller resistor?
There is only the 10k resistor loading the cart, in the phono stage, loading the secondary.
I'd forgotten about resistors in parallell.
My maths is wrong, but the theory stays.
So, putting 1m in the phono doesnt actually load the cart right?
Is it too high a value to have any effect?
The cart will only see the smaller resistor?
-
- Old Hand
- Posts: 780
- Joined: Thu Jun 07, 2007 2:32 pm
- Location: sheffield
#6
relatively it will have a negligible effect
but you need to leave something across the input to your phono stage
but don't ask me why !
but you need to leave something across the input to your phono stage
but don't ask me why !
#8
This is a quote from mike, this is whats been nagging me, i just couldnt find the reply earlier.
'I might suggest, take out the 47k's, have the secondaries act as grid bias for the first stages, and put your load resistor at the cartridge end. So it's whatever it should be for that cartridge. Seems the simplest option to me unless anyone else knows different '
Mike, so the input load resistors (47k as standard) are also doing grid bias?
Is it ok to run no resistors here at all and rely on the secondaries as you suggest?
Therefore not needing the 1m resistor.
What are otehr people doing when loading the primary, are you all using these 1M resistors?
'I might suggest, take out the 47k's, have the secondaries act as grid bias for the first stages, and put your load resistor at the cartridge end. So it's whatever it should be for that cartridge. Seems the simplest option to me unless anyone else knows different '
Mike, so the input load resistors (47k as standard) are also doing grid bias?
Is it ok to run no resistors here at all and rely on the secondaries as you suggest?
Therefore not needing the 1m resistor.
What are otehr people doing when loading the primary, are you all using these 1M resistors?
#9
Graeme
Yes I'm using 1Meg in place of the 47K and loading on the primary.
The reason as I understand it is the 1meg ties the grid to the ground, in effect it negates the secondary loading.
There was a lot on this on the old forum HERE.
Andrew was the first to really play with it and demonstrate it a Witham.
Yes I'm using 1Meg in place of the 47K and loading on the primary.
The reason as I understand it is the 1meg ties the grid to the ground, in effect it negates the secondary loading.
There was a lot on this on the old forum HERE.
Andrew was the first to really play with it and demonstrate it a Witham.
Regards
Gerry
Gerry
- Mike H
- Amstrad Tower of Power
- Posts: 20189
- Joined: Sat Oct 04, 2008 5:38 pm
- Location: The Fens
- Contact:
#10
Guilty as charged, I did write that.Graeme wrote:This is a quote from mike, this is whats been nagging me, i just couldnt find the reply earlier.
'I might suggest, take out the 47k's, have the secondaries act as grid bias for the first stages, and put your load resistor at the cartridge end. So it's whatever it should be for that cartridge. Seems the simplest option to me unless anyone else knows different '
No, the secondary winding is.Mike, so the input load resistors (47k as standard) are also doing grid bias?
I would have thought so. Only reason I can see why not is so it presents some sort of load for the secondary current to go down. Otherwise there's a possibility the freq. response could be uneven.Is it ok to run no resistors here at all and rely on the secondaries as you suggest?
Therefore not needing the 1m resistor.
What are otehr people doing when loading the primary, are you all using these 1M resistors?
What is the turns ratio (or Voltage step-up ratio) of the transformer?
"No matter how fast light travels it finds that the darkness has always got there first, and is waiting for it."
- Mike H
- Amstrad Tower of Power
- Posts: 20189
- Joined: Sat Oct 04, 2008 5:38 pm
- Location: The Fens
- Contact:
#12
1:10
In that case if there's 1M on the secondary that will make the primary 10k
Turns ratio ^2 = impedance ratio
Could put 4.7M on the secondary and then the primary will be 47k
Is the cart MC?
In that case if there's 1M on the secondary that will make the primary 10k
Turns ratio ^2 = impedance ratio
Could put 4.7M on the secondary and then the primary will be 47k
Is the cart MC?
"No matter how fast light travels it finds that the darkness has always got there first, and is waiting for it."
#13
yes. I have a couple and will hopefully be adding another mc cart to the stable too.
I am generaly aiming for 100r load but want to try down to 50r too.
I can leave the 47k in the phono and fit 100r on the secondary too to give around 45r total if i remember right.
I am generaly aiming for 100r load but want to try down to 50r too.
I can leave the 47k in the phono and fit 100r on the secondary too to give around 45r total if i remember right.
- Mike H
- Amstrad Tower of Power
- Posts: 20189
- Joined: Sat Oct 04, 2008 5:38 pm
- Location: The Fens
- Contact:
#14
OK that was what I was trying to get at, what impedance does the cart need
In that case, 10k on the secondary will get 100R on the primary
So 47k will get 470R on the primary
So will need to add 127R on the primary to get it down to 100
In that case, 10k on the secondary will get 100R on the primary
So 47k will get 470R on the primary
So will need to add 127R on the primary to get it down to 100
"No matter how fast light travels it finds that the darkness has always got there first, and is waiting for it."
#15
Yeah, i found a couple of handy calculators on-line that tell you totals for resistors in paralell/series and the same with caps.
Saves me having to work it out
For now it will probably stay with the 10k secondary load as i think my time will be put towards this attenuator first.
Saves me having to work it out
For now it will probably stay with the 10k secondary load as i think my time will be put towards this attenuator first.