Modifying Maplin preamp kit

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I should start by saying that I have a very basic knowledge of electronics, and that this is the first electronics project I have worked on; I also have a very limited budget.

My brother has a pair of active monitors (Alesis M1Active MKII), and multiple input sources. He would like to be able to switch between sources, and control their volume, without unplugging and and fiddling around with the monitors.

I initially thought that it would be okay to simply wire inputs through a double pole rotary switch, and then through a dual potentiometer, however on further reading it looked like this would cause problems with impedance matching. The monitors have an input impedance of only 20 kΩ, and runs of cable of up to 5 meters might be used between the passive pre'amp' and the monitors. In order to rectify this it looks like I should use a unity gain buffer amplifier, so that the monitors see a low output impedance from the preamp. What I hope to be able to do is modify this preamp kit from Maplin so that the gain is lowered to unity,
Low-Noise Stereo Preamp Kit : TR Kits : Maplin Electronics

and schematic,
http://www.maplin.co.uk/media/pdfs/N47FL.pdf

I would like to use the kit as it is cheap, readily available to me, comes with a PCB, and according to the specs should run off of a 5V power supply which I can get very cheaply from a wall USB charger, however with my limited knowledge this might not be an appropriate solution.

So, what I want:
  • Modify the Maplin preamp so that it works as a buffer 'amplifier'

What I don't understand:
  • How does the Maplin preamp use ground as the negative voltage rail to the opamp, do audio signals always remain positive above ground, or is this some black magic I don't understand?
  • What difference does it make to put the opamp before or after the voltage pot, how does this effect the input and output impendance of the preamp?
  • From my limitied knowledge of opamps, would taking values of R1 = R2 (and R5 = R6) from the schematic, create an opamp with unity gain? What are the consequences on the rest of the componentes in the circuit? Is it okay to short R1 and R2 to create neglibable resistance?

If my solution sounds unfeasible, unnecessarily difficult, or just plain stupid, can you suggest a better method?

Thank you for taking the time to read this, I would greatly appreciate if someone could help me out a bit.
 
A virtual ground is created by R3/R4 (R7/R8), creating an offset voltage on the outputs of the IC equal to the ratio of these resistors
Placing the gain control after the amp reduces the signal and the noise of the amp
If R1=R2 gain would be 1
The problem with this circuit is that it inverts the signal (not desirable), the resistors, to reduce noise, should be metal film, not carbon film. The supply voltage should be much higher than 5V to give you some headrooom
E
 
Mockymoose, thanks for your information. I can get metal film resistors for 26p each (0.42 Canadian dollars) to replace those which come with the kit, and I found a 12V power supply which is currently unused. If the only implication of the inverting amplifier is that the output signal will be 180 degrees out of phase than that is acceptable to me.

Would it be better to short R1 & R2, or choose some value of resistor to use for them? Does this choice affect the input impedance and power requirements of the circuit?
 
26p for a resistor seems a bit steep unless they are 1%. Go for the 5% MF types.
R2:R1 give you the gain of the amp, here 100 (40dB). To allow for a useable range for the adjustment of the pot you should have some gain. Maybee 2k2 Ohm for a linear pot and 10k for a log pot to give you 0dB at midpoint pot setting
Good luck. E
 
I would like to use the kit as it is cheap, readily available to me, comes with a PCB, and according to the specs should run off of a 5V power supply which I can get very cheaply from a wall USB charger, however with my limited knowledge this might not be an appropriate solution.
Sorry, that looks totally unsuitable.
Maybe it's supposed to be a microphone preamp or something.
  • Input impedance = 1K.
  • Voltage gain of 100, with the volume control pots after the gain stage.
  • The volume control / gain control consists of 2 separate trimmers mounted on the PCB.
If you want to use it, you'd have to change the resistors to get unity gain and a sensible input impedance.

The volume control potentiometers are in the wrong place electrically and mechanically, so it would be best to leave them out and just use this as a unity gain buffer after a passive preamp.

It won't work with a 5V supply. They claim it works with a supply of 3V to 25V, but it doesn't. It uses a 5532 chip which needs at least 10V, preferably more. (+-15V is normal).

There must be better preamp kits out there.
 
Thank you godfrey,

There must be better preamp kits out there.

Unfortunately I have already bought this kit, I will give it my best to try and modify it for my requirements, however if I am unable to do this a recommendation of a better kit would be greatly appreciated.


If you want to use it, you'd have to change the resistors to get unity gain and a sensible input impedance.

Would using a value for R1 = R2 = 1k be okay for this, I don't get where the input impedance enters the circuit, I expected to see a resistor from live to ground before the opamp. I'm pretty sure I don't understand this properly.


The volume control potentiometers are in the wrong place electrically and mechanically, so it would be best to leave them out and just use this as a unity gain buffer after a passive preamp.

I didn't expect to use the given potentiometers at all, and instead get a dual case mount one and connect it to the PCB with wires in place of the other two, however from what your saying this doesn't sound like a good idea. Would it be best to put the input signal through the potentiometer, than into the LIN on the PCB, and to short the contacts between C3 and LOUT?
 
Unfortunately I have already bought this kit, I will give it my best to try and modify it for my requirements.....
Cool. :)

I didn't expect to use the given potentiometers at all, and instead get a dual case mount one and connect it to the PCB with wires in place of the other two, however from what your saying this doesn't sound like a good idea.
No, you had the right idea. As you said here:
I initially thought that it would be okay to simply wire inputs through a double pole rotary switch, and then through a dual potentiometer......
....and then into the Maplin preamp.

Would it be best to put the input signal through the potentiometer, than into the LIN on the PCB, and to short the contacts between C3 and LOUT?
Yes, but there's a couple of choices with the output:
  1. Leave the pot there, but take the output from the junction between C3 and the pot, instead of from the middle of the pot.
  2. Replace the pot with a resistor, and take the output from the junction between C3 and the resistor.
  3. Leave the pots there and use that part of the circuit as is, just turn the pots all the way up.
  4. Like (3) above, but short the contacts as you suggested.

Would using a value for R1 = R2 = 1k be okay for this, I don't get where the input impedance enters the circuit, I expected to see a resistor from live to ground before the opamp.

The gain = R2 / R1, so you probably want them equal.

R1 sets the input impedance, so I would recommend at least 10K, probably more. Bear in mind the low frequency rolloff is determined by R1 and C1. With 10K and 1uF, the bass will roll off below about 16Hz (ignoring the volume control pot). Doubling R1 to 22K will get you down to 8Hz.

I'd probably use a 10K volume control pot and set R1 = R2 = 22K. Maybe others will chip in with their opinions.

btw, I'd recommend using 1% metal film resistors - they're cheap enough. You'll probably have to buy a packet of 10 (another good reason to use all the same value), so you'll have some leftovers. It might be a good idea to use a couple of those to replace the pots that came with the Kit.

Here's a pic so we can see what we're talking about. I've filled in the original values that came with the kit.

[OT]
What was anybody thinking when they designed this? As supplied, the frequency response rolls off below about 160Hz. :no:
[/OT]
 

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Re: supply voltages for a 5532, I'd have to disagree. I have a little Btech BT928 headphone amp using one (the Philips variety, I think), and that seems quite happy amplifying at 4.5 volts, though obviously maximum output amplitude is quite modest. 3 volts gets kinda noisy and distorted, so that's too little.

10k pot and R1 = R2 = 22k (MF) seems about right. With the given values of R3/R4, C4 should be more like 220n (film). When reducing them to 47k/47k for better DC balance, increase to 470n.
 
Dear all, I have put together the circuit using R1 = R2 = 22K, and have made sure that the volume pots which came with the kit have been turned up full. This allows the signal to pass through when unpowered, however when powered the opamp seems to get rid of the signal, muting the output. I am currently using a battery to power the circuit, however I previously used a wall power supply (12V , 300mA) to power the circuit; when I did this the opamp started smoking!

Do you think the opamp is dead? Will the wall power supply try and 'push' 300mA through the circuit, frying the opamp, or would it just provide a voltage and allow whatever current might be drawn from it by the resistance across it?
 
I have got a brand new opamp, connected it to a 9V battery, carefully checking the polarity; this results in the output signal being muted. When I use batteries with lower voltages the signal is not 'muted' as much, and the signal becomes very distroted.

I have checked for dry joints, and the values of resistance in the circuit with a basic mutlimeter. I don't have access to a function generator, oscillascopeor more advanced multimeter to check the capcitors.

Any idea why this is happening!?
 
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Possibly ,

R2 and R6 need to be 1K..if the gain is to high it will distort.

What are you feeding it with..ie what is the input connected to?

You must use a smooth DC supply ie like your battery or it will hum badly.
Polarity is very important..wrong polarity will burn out the IC..

If the input level is two high then it will distort no matter what you do!

You also say you are driving powered speakers with this?
Are you sure that you need a preamp with gain?
Are you sure that just via a volume pot (no preamp) to the input of the speakers will not work?
Normaly you would have the volume control on the input to the preamp board and the output out to the power amp.
This means that the IC is driving the cable to the main amp and not being wasted in the volume pot.

A wall wart power plug is not smooth enough for a supply of a preamp...the output of a plug in power supply is rubbish<<realy bad..if you can find a laptop power supply with correct polarity and Voltage it will work OK. (get it working with a battery first)..Always put a fuse in the supply to any project you build.<<that includes connection to low voltage laptop or battery supplies. At least it will protect something like a laptop PSU if you short it out. To protect polarity you can fit a Diode in line with the + supply like this.. battery + -I>I- to project. Hope it all goes OK.. :)


Regards
M. Gregg
 
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Dear Gregg,

Both R1 and R2 are set to 22k, therefore the gain should be unity, or 0 dB, so I don't think this is the reason for the distortion. The potentiometer to control volume is attached before the inputs shown on the schematic for the kit.

The only reason I wanted to use active electronics is so that the output impednace could be lowered to drive the signal along quite long cables (~5m) to the powered speakers.

The input is a Behringer UCA202, which has 400 ohm output at 2 dBV max.
 
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Joined 2010
OK how have you wired the input volume pot,

Is it wiper to the preamp or wiper to the input?
What value is it?

Are you sure that when the IC burnt out it did not drain the battery<<<?
IE battery output is now low? or is it a new battery?

If you connect the pot on the input direct to the output with circuit board disconnected ie to powered speakers is it distorted or is it clean with volume control working (no pre amp in the line)

Regards
M. Gregg
 
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Joined 2010
So as above; wiper of pot on the input connected to output no circuit connected does the volume work with a clean signal to the powered speakers?

Please hit your refresh page button every now and then to show further posts after you have posted... :)

Regards
M. Gregg
 
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So as above; wiper of pot on the input connected to output no circuit connected does the volume work with a clean signal to the powered speakers?

Yup, the volume pot works as expected, with a clean undistorted signal, when the opamp is unpowered, however the signal is a lot quieter than when the input is directly connected to the powered speakers.
 
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