Passive Pre

I'm thinking of building this preamplifier for a first soldering project. Being that there is no electrical current in this project, I should be able to build it without electrocuting myself.....but then again.

I've read the thread looked at photos and am completely confused. I have a black beauty alps 100k pot, which should suffice. I have silver wire, resistors, ect. I just can't figure out how to wire it.

I couldn't read a schematic if my life depended on it. His pictures are good but not detailed enough for a fool like me.

Any help on this one?

link?

Oops....
hehe looks like this one is going to really be a :sigh: please don't laugh at my inexperience though, I'm sure its really easy.

http://www.diyaudio.com/forums/showthread.php?threadid=2608&highlight=

I will try and help out. Hope I do not confuse you. First, let us start out with the schematic of this passive device, which I copied from the diyaudio.com thread. The picture is courtesy of planet-10 (Dave Dlugos). Edit 3/12/2013 - replaced picture with one of my own. Original picture not available any more.



Only one channel is shown, but once you get that done, the other is going to be easy. Also, I am spelling out everything, assuming that you need that level of detail. If not, I apologize in advance.

First, the 100K pot. I think this is a stereo pot, with two rows of solder lugs, three lugs per row. So, one set of three legs will be for the left channel, and the other set of three lugs for the right channel.

The center lug is the wiper, the arrow shown in the figure pointing to the pot. What you have to find out is which of the two outer lugs in each row is the one that is to be connected to ground (the big triangle, pointing down). The other outer lug in each row will be unconnected.

So, hold the pot so that the shaft is pointing towards you, and the solder lugs are pointing up. Turn the shaft fully counter-clockwise. Then connect a digital multimeter (in the resistance measurement setting, at its lowest range) and check the resistance between the center lug (the wiper) and each outer lug in turn. One of the outer lugs will show you a resistance measurement close to zero, the other outer lug will show a measurement of 100Kohms. You need the lug that shows you zero resistance measurement. The other outer lug will be unconnected.

Verify that the outer lug in the same position in the other row shows zero resistance connection to the wiper lug in the row.

Next, we take a look at the RCA jacks for input and output. These are presumably mounted at the back of the box you are going to use for the project. Each RCA jack will have a pin in the center, with a small cup. There will also be a removable metal ferrule that slips over the threaded portion of the jack. The ground connection is made to this ferrule. The signal wire is soldered to the center pin with the cup.

Fix the four RCA jacks to the box. Looking at the box from the back, let us assume that the jacks, in order from left to right are:

Left Input, Left Output, Right Input, Right Output.

Take one 11K resistor (or whatever value you are using). Solder one end of this resistor to the center pin of the Left Input jack. Solder the other end to the center pin of the Left Output jack. Measure and cut a piece of wire to solder between the center lug of one row of lugs on the pot and the center pin of the Left Output jack. Solder this wire to these points. You have now completed the signal wiring on the left channel.

Finally, connect the two metal ferrules on the Left Input and Left Output jacks by soldering a small piece of wire between them. Measure and cut a piece of wire to solder between the outer lug (in the same row of lugs as you connected the wiper) and the connection between the two metal ferrules. Solder this wire. You have now completed the ground wiring on the left channel.

Repeat on the right channel jacks.

Note that I did not connect the ground (outer) lugs at the pot. I kept the left and right channels independent of each other, even their grounds. The amplifier that you will connect to the output of this passive preamp will tie the grounds together.

Hope this helps.

Ashok

Thanks, I put it together on a breadboard and made sure it worked and have wired it together. I used a cheaper Alps pot for this one as I put it in a guitar switch pedal I wasn't using anymore because the switches were shot. Looks and sounds pretty good. I appreciate your help. I now understand a simple schematic, might try a harder passive next.

ashok said:

So, hold the pot so that the shaft is pointing towards you, and the solder lugs are pointing up. Turn the shaft fully counter-clockwise. Then connect a digital multimeter (in the resistance measurement setting, at its lowest range) and check the resistance between the center lug (the wiper) and each outer lug in turn. One of the outer lugs will show you a resistance measurement close to zero, the other outer lug will show a measurement of 100Kohms. You need the lug that shows you zero resistance measurement. The other outer lug will be unconnected.

Click to expand...

Allow me to ask a couple questions here so I can learn as well.

1. Why are we using the 11K resistor? Is that our resistance to the signal?

2. Wouldn't you want to use the other leg in the above quoted paragraph? Wouldn't that give you more resistance to passing signal?

3. Am I looking at this the wrong way? The pot going to ground directly after the input is effectively stopping the signal, hence the necessity of the 0 Ohm you wrote about in the paragraph. If this is the case, again, why do we need the 11K resistor?

Thanks for addressing these questions in advance.

LPD, glad to hear it worked for you.:yes:

This is not just a resistive volume control, it is a voltage divider. With the pot at 0 it will shunt the signal to ground. As the pot increases more of the signal will pass to the outputs.

Ray

50k

A line out should be loaded with 50K Ohms.:yes:

Darius

So Ray, if the pot was 11K it would reduce the volume to half of input level when run wide open correct?

And..if your pot doesn't quite make it to 0 Ohms you're never going to stop all of the sound from getting through. Right?

Oldeurope, are you saying that the fixed resistor should be 50K?

Thanks guys!

1. Not quite. It's a 100k pot that you would be dividing with an 11k input resistor so you'd be looking at losing ~1/9 of the signal, not 1/2. Something like .5dB from that. There is also the output impedance of the source so you'll probably lose a little more but not a ton.

2. Correct. Same thing with a resistive control...if the resistance doesn't go to infinity then you'll get leakage.

As for the other I think the point was to make a generic match to the output of the source, meaning at least 10x whatever might be in front of it. If your CD player has a 300 ohm output impedance then you'll need at least 3k. I think 11k was just a safe value.

Ray

I understand the answer to #1 as was answered. But I don't think I worded my intent correctly. If the POT was 11K, the same value as the resistor, would running the pot WOT split the signal in half?

No, guess it wouldn't would it? It would just choose whichever route it wanted to take. My head hurts now and I've confused myself again.

cabinover said:

So Ray, if the pot was 11K it would reduce the volume to half of input level when run wide open correct?

And..if your pot doesn't quite make it to 0 Ohms you're never going to stop all of the sound from getting through. Right?

Oldeurope, are you saying that the fixed resistor should be 50K?

Thanks guys!

Click to expand...

Take a 50K log pot and make the input of your amp high impedance.

Darius

The passive-pre in the schematic is just another way of implementing a volume control. Here are a couple of pictures that might help.

First is a picture of a conventional potentiometer arrangement:



The source (possibly the CD player) is represented by an ideal voltage source, and an impedance Zs. It is connected across the outer lugs of a pot. The wiper divides the pot into R1 and R2, and they both vary continuously as the pot is turned. The smaller R2 is, the lower the signal that is fed to the amplifier. When R2 = 0, you have shorted the input to the amp. The advantage of this arrangement is that the source always sees a constant load impedance of R1+R2 = the value of the pot. (This assumes that the amplifier's input impedance is much larger than the value of the pot, so that it is essentially an open circuit).

An alternate way of implementing the volume control is:



Now, we have a fixed series resistor R1 (the 11K resistor), and a variable shunt resistor R2. The value of R2 can go from 0 all the way to the full value of the shunt pot. But R1 is always in the circuit, and there will always be some attenuation of the signal. The input impedance that this arrangement presents to the source is not fixed. It will vary from a minimum value of R1 to a maximum value of R1 + R2 (assuming that the amp's input impedance is not a factor).

Hope this helps.

cabinover said:

I understand the answer to #1 as was answered. But I don't think I worded my intent correctly. If the POT was 11K, the same value as the resistor, would running the pot WOT split the signal in half?

No, guess it wouldn't would it? It would just choose whichever route it wanted to take. My head hurts now and I've confused myself again.

Click to expand...

Sorry, my bad for not reading carefully enough. You are correct. If the pot and the fixed resistor are the same then you cut the signal in half.

Stick with Ashok, he's way better at this than me!

Ray

Thanks. I'll never completely understand this stuff but I can follow directions and schematics enough to build one of these passive pres.

Later on tonight I'll try to digest all of the info posted.

Yep my pot has leakage, which I don't like and with my Aleph 30 it has way too much gain. I have a black beauty Alps in my box of goodies but wanted to test pilot a cheapo version to try it out. If I raise the resistance on the shunt will it lower the gain? Will it affect anything else?

LPD said:

Yep my pot has leakage, which I don't like and with my Aleph 30 it has way too much gain. I have a black beauty Alps in my box of goodies but wanted to test pilot a cheapo version to try it out. If I raise the resistance on the shunt will it lower the gain? Will it affect anything else?

Click to expand...

You will get more attenuation of the signal by increasing the fixed series resistance R1, instead of increasing the value of the shunt pot. Here is another look at the volume control.

The voltage at the output of the control is the voltage across the resistor R2. Call this Vout. The voltage at the input of the volume control is the output of the source, and is measured at the junction of Zs and R1 (to ground). Therefore, the ratio of Vout/Vin, which is the gain (or attenuation), is
By increasing R1, Vout will be smaller than before at any given R2.

But there are other things to consider too – namely the input and output impedance of this volume control. Ideally, the input impedance of the volume control should be 10 times (or more) the output impedance of the source feeding it. The output impedance of the volume control should be 1/10th the input impedance of the next device downstream of it.

Since R1 is fixed at 11K, the input impedance of the volume control will have a minimum value of 11Kohms (shunt pot turned all the way down). As the shaft is rotated, the input impedance will increase from its base value of 11K to a maximum value of 11K + (R2 || Zamp). Here Zamp is the input impedance of the amplifier. It will be parallel to R2.

Therefore, if the output impedance of the source is around 1Kohm or less, we would have met the minimum input impedance criterion. Modern CD players should easily meet this criterion.

The output impedance of the volume control is a little bit more complicated. The input impedance of the amp (the Aleph30 in this case) is 47 Kohms. So, you would like the output impedance of the volume control to be around 5K or less.

The output impedance of the volume control is determined by looking into the volume control from the terminals of the amplifier. So, you will see the resistance R2 to ground. In parallel with R2, you will see the series combination of ZS and R1. The voltage source, being ideal, is treated as a short-circuit. So, the output impedance is R2 || (ZS + R1).

If R2 = 0 (pot turned all the way down), Zout = 0. With the pot turned all the up (wide-open) and ignoring ZS, Zout = 100*11/111 = 9.91K which is about 1/5th the input impedance of the Aleph.

But, the maximum Zout = 9.91K is at the wide-open position. At normal listening levels, R2 will be a lot smaller, and the effective Zout is probably going to be OK for satisfying the 1:10 rule. For example, a 50% mechanical rotation on a 100K log pot will give a value of R2 = 10% of the full value. Therefore R2=10K. Therefore, Zout = 10*11/21 = 5.23K. Close to the ideal 1:10 ratio.

But you are not getting a very usable range with the 100K pot. If you increase R1 from its present value of 11K to say 20K, Zout at 50% rotation = 10*20/30 = 6.67K. Not as good as 5.23K (using R1 = 11K) but not a whole lot worse either – a ratio of 1:7, instead of the ideal 1:10. But you will get a more usable range out of the pot.

Hope this helps. Passive preamps can be difficult to get right.


Ashok

Edit: Formula for Vout/Vin assumes no amp connected, or input impedance of the amp is much higher than R2.

Thanks,
I may have to read your post a few times to absorb, but I think I understand the fundamentals.

I wired one of these together today, using components excactly as spec'ed. I used a 100K Noble stereo volume pot, and a couple 11K Shinkoh tantalum resistors ('cause they sound so exotic!).

It's got a lot of gain, at least in my setup, but it seems to sound pretty nice.

One thing that worked out real nicely is the box I put it in. I gutted an old data switching box, and didn't have to drill a single hole!

BTW, I now know what this thing will sound like if you accidentally run the signal through backwards... I've got to label the jacks!