Current conveyor as a voltage amplifier

dadod

Member
2006-04-18 3:11 pm
Zagreb
As Satri circuit is a current conveyor used as voltage amplifier why not to simulate it to see what could do that technology.
The voltage gain for a current conveyor is output resistance divided with input resistance. In this schematic it is R1/R12.
Input impedance of current conveyor is very low, so to be used as a line amplifier it needs an input buffer. Here is used a variant of JLH buffer.
This line amplifier does not use global negative feedback and still has very low uniform distortion across full audio band(even father).
A volume regulation should be done varying output resistance, and would need an output buffer too.
I used in this simulation very high power supply voltage as this give lower distortion.
dado

This one is called GainWire and here is the gallery: http://www.diyaudio.com/forums/gallery/showgallery.php?cat=2247
 

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keantoken

Member
2006-08-10 12:32 am
Texas
It needs an output buffer unless you plan on your circuit to have a 4.7k output impedance, more if you need higher gain.

Just use a Wilson mirror, any more than that won't improve linearity because it is Q4/8 B-C leakage that determines linearity once Q5/6 has been cascoded. To get around B-C leakage you need to use Baxandall (or make Rout so low that it is a non-issue). Why the 100V rails? Do you plan on electrocuting curious children? Put some sort of limiter on it at least, to protect proceeding equipment if not innocent children.

Use BC5xxC or BC3x7-40 transistors for the current mirrors, not those japanese types. High-voltage transistors shouldn't be used in current mirrors because of the abysmal quasi-saturation characteristics. This will limit the design in real life, but won't show in the simulation because few models include quasi-saturation parameters. If you still want 100V rails Q4 and Q8 should be japanese types, but try it with 24V rails and BC5xxC transistors instead.

I guess you could tweak the mirror ratios for higher gain, but this would probably worsen distortion and cause offset. It is inverting so you could use an input cap and put the input resistor between input and output to stabilize bias.

Output quiescent is not very predictable with the diode bias. If you are sure the diodes have greater on voltage than the BJT's, you can put a trimmer between the bases of Q7 and Q10 and this will work for bias control. Alternatively, degenerating Q7 and Q10 with 1k each and lowering R12 to 500R will do nothing to hurt linearity and will allow no-trim automatic output quiescent. Might as well go all the way, replacing R12 with a 2k degeneration at each emitter.
 
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dadod

Member
2006-04-18 3:11 pm
Zagreb
It needs an output buffer unless you plan on your circuit to have a 4.7k output impedance, more if you need higher gain.

Just use a Wilson mirror, any more than that won't improve linearity because it is Q4/8 B-C leakage that determines linearity once Q5/6 has been cascoded. To get around B-C leakage you need to use Baxandall (or make Rout so low that it is a non-issue). Why the 100V rails? Do you plan on electrocuting curious children? Put some sort of limiter on it at least, to protect proceeding equipment if not innocent children.

Use BC5xxC or BC3x7-40 transistors for the current mirrors, not those japanese types. High-voltage transistors shouldn't be used in current mirrors because of the abysmal quasi-saturation characteristics. This will limit the design in real life, but won't show in the simulation because few models include quasi-saturation parameters. If you still want 100V rails Q4 and Q8 should be japanese types, but try it with 24V rails and BC5xxC transistors instead.

I guess you could tweak the mirror ratios for higher gain, but this would probably worsen distortion and cause offset. It is inverting so you could use an input cap and put the input resistor between input and output to stabilize bias.

Output quiescent is not very predictable with the diode bias. If you are sure the diodes have greater on voltage than the BJT's, you can put a trimmer between the bases of Q7 and Q10 and this will work for bias control. Alternatively, degenerating Q7 and Q10 with 1k each and lowering R12 to 500R will do nothing to hurt linearity and will allow no-trim automatic output quiescent. Might as well go all the way, replacing R12 with a 2k degeneration at each emitter.

With Wilson mirror only it is very sensitive to R13 R14 resistor adjustment (to get zero V at output) look first screenshot.
I used +-100 V to power current conveyor to get lover distortion, with +-60 V distortion tripled(second creen shot).
I said if used as line amp it needs output buffer too.
I tried different current mirror ration, but as you said, distortion worsen.
Kean I would apreciate if you suggest and show improvements with simulation.
dado
 

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keantoken

Member
2006-08-10 12:32 am
Texas
I think you can get good distortion performance with less heat. An extra transistor to convert Q4/8 to Baxandalls would cost less in the end than a 100V power supply I think.

You need to separate how much distortion comes from the input buffer and how much comes from the "conveyor".

Here's my try. THD is similar, but it has a gain of 22, better than a gain of 4. It also has 2.2k output impedance, slightly better than 4.7k.
 

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keantoken

Member
2006-08-10 12:32 am
Texas
Improved the Baxandall mirrors a bit, and decreased the gain to 10. Perhaps this could be a nice current-drive headphone amp, easy passive EQ. I think that would be its best use. By the time you add an output buffer, the circuit has a lot of transistors, and can't compete with feedback circuits for parts count. With a 2.2k output impedance this circuit's low distortion will be lost on any power amp that doesn't have its own internal buffer.

Due to being virtually passive it is naturally phase-linear, which is one thing that sets it apart from most feedback circuits.

I think calling this circuit is "conveyor" is wrong. As I understand current conveyors conduct current directly from the input to the output, making them inherently non-inverting, and zero input impedance. Most often they are something using a folded cascode. A conveyor could be made very similar to this circuit but without the benefit of output bias self-stabilizing with the input cap.
 

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I think for a buffer your best bet would be to try to convert the tringlotron into a circuit useful for DC. Also pay attention to PSRR.

I am not worried about how to bias Q11 and Q19. This will affect PSRR a lot, but for a NGNFB circuit like this it is better to start with a regulated/filtered supply. Simple C-multipliers should do I think. If this is not enough I recommend MC7xxx regulators from OnSemi, these are much improved versions of the LM devices.

If you replace the diodes and Q7/10 with BC5xxC transistors, I think you will be able to get close to the quiescent you aim for. I've tested these transistors (Fairchild) and found 10 randomly picked ones to be well within 1mV of each other, which is excellent.
 

dadod

Member
2006-04-18 3:11 pm
Zagreb
Output buffer degrade distortion figures, and at higher output more. I tried different output buffer, lineup between others but I am back on simple JLH modified one, similar to the input buffer.
With low inputs as 0.1 V the distortion is very low at all frequencies, degraded a bit by the output buffer, but still excellent and this is no global negative feedback used. A kind of negative feed back is used in the buffers but it is very local.
Where to put volume control? It is possible to use Current conveyor’s output resistor for that, it is of low value and it will need custom volume control.
It is possible to use input volume control of 50k to 100k and those kind are ready made available. The gain between 5 and 10 is a good value for line amplifier, and more the enough to push any output amplifier into clipping.
dado
 

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keantoken

Member
2006-08-10 12:32 am
Texas
If I see correctly you are running Q6 and Q8 at 20mA. These are dissipating 700mW and burning up. These can be C3503/A1381, with the Baxandall they will work equally well. Another option is to use paralleled BC5x0C for Q6 and Q8, with small degeneration resistors so they share equally, and this is fully compatible with the Baxandall as well.

A warning is in order for the current mirrors. At low currents all diode junctions have virtually perfect logarithmic conformance, but at too extreme bias the log conformance will fail. This will cause current mirrors to distort. If so you should try decreasing Iq.

However if the 20mA output Iq works, I think you can reduce distortion by increasing R3 and R4 to 2k and R5 and R6 to 1k. R12 will need readjusted for 3rd harmonic null if desired.
 

dadod

Member
2006-04-18 3:11 pm
Zagreb
This one is closer to the practical implementation. Distortion is as before, used more powerful transistor for Q6 and Q8, output buffer still in class A but with less dissipation.
Kean I triad your suggestion and could not get better result. I think this here is more then good, the sound should be good.
FFT are for 0.1 and 1 input voltage.
dado
 

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keantoken

Member
2006-08-10 12:32 am
Texas
How is PSRR?

You are right it can drive an amp to clipping. It could drive an amp to burning. I hope that is not a "feature". :D

Antiparallel diodes between the output buffer input and output will provide a current limit for the output, but may only work for headphones.

I would limit output voltage to 2V; I see no need for more than that. I would illustrate this with a schematic but I don't have access to a schematic program right now.

Imagine a PNP transistor, emitter at the output. Base at the same point as the base of Q3. Collector to the junction of R3 and R4. Mirror with an NPN for the negative side. Add diodes in series with emitters optionally to prevent excessive Vbe reverse bias. If it is unstable during clipping, try resistors in series with the emitters.
 

keantoken

Member
2006-08-10 12:32 am
Texas
Go to AC simulation. Turn off the AC for the signal source. Right click on a rail source (V1) and make it the AC source. Run the simulation and plot Vout. Plot the equation 1/Vout. Switch rail sources, try again. Finally make both rails AC source and again plot 1/Vout. This will be your PSRR in db. If your rails have a second harmonic at -30db, and your PSRR is 50db, the second harmonic will appear at -80db at the output even if the circuit has very low distortion, because the distortion is conducted from the rails. Rail filtering and/or regulation is mandatory to bring out this circuit's potential. Another reason lower rails could really help. MC7x15 are a good idea I think.
 

keantoken

Member
2006-08-10 12:32 am
Texas
Yep, like that. In chipamp datasheets, PSRR is plotted separately for each rail. It is typically between 60db and 80db.

Is the circuit powered for that plot? It looks like you turned the rail sources into 0V signal sources. You need to enter your rail voltage as the DC offset.

Have you tried the limiter I described in post 14?
 
Nice. Sometimes a diode junction on the rails will generate enough of its own distortion to affect the output through the rail, so we're not out of the woods yet. If this proves an issue I fear a Kmultiplier or chip reg will be necessary. See if the Cmultipliers decrease distortion.

Again, check that V1 and V2 are actually at the correct rail voltage... The PSRR thing only works if the circuit is powered. That doesn't look right to me.