Help Coupling 2 cicuits

Thread Starter

DCent

Joined Sep 21, 2012
20
I need a bit of help please.

I have two 9v DC circuits made that alter an audio input.

They're able to work in conjunction with each other only if they have their own battery. The signal goes through each stage and is altered appropriately. When I run them off the same power source neither works. In fact, linking the grounds or the line signal together causes both circuits to stop functioning.

What I want to do is have both circuits powered by one battery (or one 9v wallwart).

I tried using a charge pump that produced 9+/9-/ground but it didn't work. Linking the ground or the hot line shut down the circuit.

I tried using 1:1 transformer in between the circuits but it just introduced a lot of noise (picked up some AM stations as well).

I tried altering the way each circuit gets its power (to make them more compatible) but it took away the sonic character of the 2nd circuit.

Is there a way to make the circuits works together? Daisy-chaining, or power conditioning, isolating, coupling... I'm just throwing words out here as I've no idea. Any help/direction would be greatly appreciated.

The attached image is super simplified but shows the basic issue. One circuit has + to ground, the other has the power independent of the audio signal ground.

 
Not quite sure what your are trying to do here

Could explain the operation of the transistor circuit and op amp circuit?

No resistors on op amp is not normally the case there is normally either positive or negative feedback?

Also are you trying to amplify a signal in the transistor section?
 

Thread Starter

DCent

Joined Sep 21, 2012
20
I'm trying to power both circuits from one source. The circuits drawn are simpler than the the real ones. They illustrate the powering of them. The first is an amplifier circuit, the second a wave shaper.

They work fine together off two batteries but I've exhausted my ideas on powering them from one source (either adapter, or 9v batt.)
 

liquidair

Joined Oct 1, 2009
192
Well, obviously Gnd can't be gnd (as in the first) and -4.5V (as in the second example), so you are going to have to pick one or the other.

Assuming you choose gnd as gnd, then wire the opamp pin 4 to gnd. You then have to create a voltage divider with equal (1M) high value resistors from battery to pin 3 to gnd to create a DC voltage of 4.5V at the opamp pin 3. It goes without saying though that you need an input blocking cap and an output blocking cap.

The other option is to use 4.5V as Gnd. Tie the the same divider as above to + and - of the battery and the output of the divider is now gnd, but the transistor needs to be readjusted to work off of +4.5V. You are going to now have to have a blocking cap on the transistor input and op amp
output.
 
Y sh real po the comp circ

the above is a simplified version of the following sentence

you should really post the complete circuit.

Your Circuit diagrams are absolutely awful, I could probably tell you the answer if i had some idea what the circuits are?

What is that transistor section, is it back to front?

You no doubt will have some dc bias from the transistor messing the other circuit about so i would suggest putting a DC blocking capacitor between output 1 and input 2.

In future i would suggest posting the circuit as what you have posted looks like absolute rubbish
 

liquidair

Joined Oct 1, 2009
192
Ihaveno...tact.


Little harsh maybe? We are all here to help and learn, I don't see why it's necessary to insult someone for posting a question, especially when the problem is clearly shown in the description and diagram provided by DCent. Sure a full schematic would be prefered, but there are various reasons not to include a full schemo, one of which may be to avoid having people on forums not paying attention to what is actually being asked.
 
I do apologize if that was harsh, it certainly was not meant to be.

Im just totally confused by those two circuits, what are they supposed to be doing, if liquidair could please tell me what type of transistor circuit that is and how the operational amplifier works then i would be most grateful.

the emitter is pointing in the wrong direction for a start, it should be pointing towards to ground line if its npn and if its pnp it should be pointing towards the base terminal.

im honestly at a complete loss as to what those circuits are supposed to be doing, thats all.

If i knew what was going on, by way of a schematic, not neccessarily a complete one, then im sure i could provide a more than reasonable answer to the question of how to link two circuits using the one DC supply.

Thanks
 

liquidair

Joined Oct 1, 2009
192
Reading the Dcents first post, the issue is clearly a power issue given the circuits interface to each other perfectly when two batteries are used, doesn't work with one battery, shut down the charge pump and the transformer was noisy. DCent knows that it is a power problem and asked us to look for a solution.

DCent, I too am confused by the diagram, for future reference, any negative supply pin should always point down. I didn't pick up on that in my first post. I assume the transistor is an input buffer for a guitar pedal?

Not only is the problem 1/2 of the battery is shorted through the op-amp, but one circuit is negatively powered the other is positively powered. The last bit isn't a problem as long as you can change at least one of the circuits. Preferably, you rewire the transistor circuit to be conventional (i.e. powered from the + terminal of the battery, - to gnd). Then ground the Pin4 and supply Pin3 with a 4.5VDC reference.

IF you can't alter the transistor circuit, then ground pin 8 and leave pin4 to the battery - terminal. You still need a reference voltage for the op-amp input that is 4.5V (negative in this case with respect to ground).

Just out of curiousity I Have No Education, are you from the U.K? My sisters both married English (both named Simon) and they use "absolute rubbish" a lot!
 

Potato Pudding

Joined Jun 11, 2010
688
The first circuit show a positive ground so the emitter is correctly biased.

I am not comfortable with how it is drawn and I agree that it is confusing.

The first circuit is using a positive ground.
The second circuit is using a 4.5V ground.

I would almost suspect we are being punked, but I have seen worse, and I have no doubt that both circuits work until they are coupled together.

Now to be helpful.

Voltage divider and run the first circuit off of one 4.5V side might work.

A dc blocking capacitor between stages would be my first suggestion.

But first I want to know how you tested the first stage to say that it worked.

How did you detect the signal?

A speaker seems likely since sound was mentioned.

That stage would be a crappy power amp but sound might be detectable - especially with a headphone or a piezo speaker.

Second stage is possibly a follower power amp.

Guessing is not very helpful.

Goes back to needing more information - better circuit diagrams.
 
Hahaha yeah im from the uk!!

Until DCent replies and shows better circuit diagrams im not going to even guess what those circuits are doing!!

Although for the rest of you guys we dont even know if its current or voltage being amplified by the transistor.

If i showed someone a circuit like that to someone in the office they would probably burst out laughing and or fire my ass!!!
 

Thread Starter

DCent

Joined Sep 21, 2012
20
If i showed someone a circuit like that to someone in the office they would probably burst out laughing and or fire my ass!!!
Post:

How do you curl a mustache like this:


Reply: "Wait, how can a mustache just float in space? Where's the face? Where's the nose. I could show that mustache to the guys at the mustache shop and they'd probably laugh which tells you just how well developed their sense of humor is."
 
Last edited:

Thread Starter

DCent

Joined Sep 21, 2012
20
The transistor might be facing the wrong way in the schematic. I could power it with positive.

On the 2nd circuit there I don't know what's going on, but I like what it does, so whatever the powering issue fix is, I'd like to keep the 2nd circuit intact.

Is it 4.5v? or shorting? I'm pretty new to this, so I don't even know.

On the schematic, I kept it simplified because the problem is in the powering.

I wonder, can't I just daisy chain the power (like they do in 9v power cables) and that would work like 2 9volts?
 
look, im going to do you a favour here.

What type of amplifier is it, common base, common emitter, common collector?

what is the desired effect of the operational amplifier?

are you amplifying voltage or current?

post the complete schematic and i will try to walk you through your problem.

Short of that then i honestly think that you will struggle to get the resolution you are looking for as i honestly don't fully understand your circuit as present.

thanks.
 

liquidair

Joined Oct 1, 2009
192
DCent, could you provide what the circuits are supposed to be doing and the application (I know it is audio but you may want to handle the audio differently if it is a guitar signal vs. an active bass vs. a line level signal)? Are these 2 circuits from different schematics you are assembling as building blocks?

It sounds from the response that you are getting confused with the concept of grounding and voltage. The thing to know is that voltage is RELATIVE or the difference in potential between two points. You see ground often labeled as 0V (which caused me a first to see it as this magical hole where all voltages disappear). In fact, it is only 0V compared to the rest of the circuit.

Think of audio and electronics like water in a pipe. Currently, the water is standing still and the top of the pipe is 9V and the bottom of the pipe is gnd. When there is no audio, thus the water is flat, it makes sense that the biggest wave you can get through the pipe is when the water level is at 4.5V (it can go up 4.5V and down 4.5V; 9V total). Any water level above or below the midpoint limits the maximum swing to that level.

So let's look at your circuit 2 for example. Since the battery isn't grounded, it finds it's reference (water level) through the opamp load (which i assume connects to ground) and makes it somewhere between 9V, probably close to the middle. Verify this with a DMM. It also means that the audio is ground referenced and thus somewhere in the middle of the pipe, so we get audio out when we put audio in.

In order for circuit 1 to work since the battery is grounded positively, we have to assume that the water level is somewhere between 0 and -9V standing still (which means you likely have a capacitor at the input of the transistor circuit and its output, not shown of course). Thus you get audio out when you put audio in.

However, when you put the 2 together, watch what happens. Our battery in circuit 1 is grounded at its positive terminal, which means that pin8 is also grounded and so is the op amp's load. The transistor circuit will work now, but not the opamp circuit, do your see what happened? The top of our pipe is 0V and the bottom is -9V for the transistor section like before. But the opamp's pin8 is at 0V along with the water level! It can only pass the negative wave information. The standing water level needs to be -4.5V to work.

So in short (no pun intended <-I wonder how many times that has appeared here), there's many ways to get these 2 circuits to work together, but as shown they are not compatible (which is why I wondered if they are from two separate schematics above). I think why everyone wants the full schematic DCent is that we are concerned about the housekeeping circuitry required, which may present an easy fix as opposed to just telling you to mod the hell out the thing when all your really need is a top of the pipe, a bottom of the pipe, and a water level (preferably close to the middle).
 

Thread Starter

DCent

Joined Sep 21, 2012
20
But first I want to know how you tested the first stage to say that it worked.

How did you detect the signal?
I used a sine wave generator, ran a 1/4" line off it. I attached the pin to "IN" and the shield of the audio cable to ground. Attached another line to the out and put this to an external amplifier.

The first circuit acted as a booster preamp. Brought the signal well above the volume of the bypass signal.

With the second circuit, it added clipping and volume vs the bypass signal.

If you run both circuits off two batteries, you can couple the circuits and get a large boost running into a clipping circuit.
 

Thread Starter

DCent

Joined Sep 21, 2012
20
The first circuit isn't as important. I could change it to match the second part. It's just a gain/volume boost. What's more important to me is keeping the powering method for the OPamp part, which is something I'm not understanding.

If I ground either side of the battery the 2nd circuit stops working.. the sine wave doesn't make it through. With the current wiring (the 1/4" ground being separate from the battery) it works. I don't quite grasp the 4.5v thing or how that's figured out.

It's weird to me that it works without a hitch off 2 power supplies. Can I fake 2 power supplies? Or break up a 9volt adapter signal? Like, when you power multiple guitar pedals off one adapter by daisy chaining... or powering multiple circuits (like cell phones) from one power outlet. How are conflicts solved there? Or how are the circuits separate even though the power is ultimately coming from the same place?

Can whatever allows that be used to make these two circuits works together? Or is there another way for solving it? Can I keep the 2nd circuit powered the way it is?

Thanks for direction. Trying to wrap my head around this, but I'm still very new.
 

Potato Pudding

Joined Jun 11, 2010
688
Ok. I think that helps, quite a bit.

Try attaching circuit circuit one output to a capacitor and the other side of the capacitor to the input of circuit 2.

This is just a test but to minimize attenuation and loss of low end frequency response start with the largest value non polarized capacitor you have on hand.

If that doesn't give you a reasonable propagation then measure the dc levels and polarity across that cap and replace it with a large properly oriented electrolytic.

Chances are the problem that will prevent propagation is a bias flood where the capacitor fills to full voltage. If you measured the dc levels and it had most of 9Volts across it then you need to add a bias current that will bring the voltage across the capacitor to a smaller value, hopefully less than 3 volts.

For bias current - especially in a battery powered circuit, try to keep as small as possible. I would bias the op amp side first. Try and center that side of the capacitor to 4.5 volts. If you have a 1MOhm trimcap that you could place across the battery power leads with the wiper connected to that input that would be a good start.

I don't want to overload you so see how that works for you for now.
 

liquidair

Joined Oct 1, 2009
192
If I ground either side of the battery the 2nd circuit stops working.. the sine wave doesn't make it through. With the current wiring (the 1/4" ground being separate from the battery) it works. I don't quite grasp the 4.5v thing or how that's figured out.
The 2nd circuit's pipe looks like +4.5V and -4.5V and the water level is at 0 so all is good. But when you ground the battery you are telling the circuit that the top of the pipe and the water level to be both 0V, where the bottom becomes -9V. This only negative waves can show up.

Can I fake 2 power supplies?
Yes essentially, and this is what I have been telling you to do and Potato Pudding is alluding is already in the transistor side. Here's what it looks like: http://www.electronics-tutorials.ws/amplifier/amp_5.html

Scroll down to the Class A amplifier schematic and you see something that prolly looks similar to what you are using. R1 and R2 are there to set the transistor's base to some portion of VCC. Essentially this is like a second smaller power supply (but it only needs to supply a small current, so these resistors are typically larger than you'd use in a real supply). Those pictures of the sine wave illustrate my water pipe example, and the output picture shows that the bias or "water" level (about which the sine wave is centered) is the DC voltage at the R1/R2/C1 junction.

Potato Pudding is telling you to put a second capacitor on the output which blocks the DC at the output, but I don't think that cap alone will help your problem because you are wiring the battery to disconnect when the 1/4" jack is removed, no?

Let us know because if so, there's a couple of routes to take to get this working.
 

Thread Starter

DCent

Joined Sep 21, 2012
20
I have input/output caps. Haven't tried the biased power supply. I think I have a lot more reading to do.

I don't need the battery disconnect. I always used wall warts.

As far as those 2 resistors connecting in the Class A amplifier circuit, I have nothing like that in these circuits. The 2nd circuits power supply in literally what you see in the schematic. Nothing coming off of it.

Thanks for the patience.
 
Can you write down in, I don't know, a small paparagraph what you want each circuit to do.

I'm sure I could design something for you or at least show you the circuits and let you work out the values you need!!
 
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