can someone help me check this circuit?

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Hi guys! Can you help me to check this circuit to see if is there any problem with it? It is just a simple circuit with a very simple CCS. It is supposingly to work in Class A mode most of the time. Thanks!
 

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hi x-pro, thanks for your reply. I redrew the scematic in LTspice and simulated it. It works well but then it must be biased above 1.4A if not the some parts of the bottom half of the wave will be clipped off. Any idea why?

Also, must the 2N5550 be mounted together with the 2N3055 on the same heatsink? Cause its a TO-92 Package type.

Thanks!
 
Hi,
I must be thick. I find trying to read that schematic with the voltage sources in unusual locatons very difficult to read.

You have posted a DC coupled amplifier.
The source resistance on the inverting input is 22k//100k=18k
the non-inverting sees 22k//0k0=0k0.!!!
That will guarantee an output offset.

The pre-amp or source that you finally connect instead of the idealised voltage source will controll the output offset by substituting a different source resistance. If it is DC blocked the non-inverting source resistance changes to 22k. oops the output offset has gone (unless you had zeroed it with a shorted input).

Do you plan to use a DC coupled pre-amp with a DC servo?

Have you tried moving the lower end of the load from the ground to the drain of M1? What does the simulation say? I may have this mixed up due to not properly understanding this schematic.
 
happyboy said:
hi x-pro, thanks for your reply. I redrew the scematic in LTspice and simulated it. It works well but then it must be biased above 1.4A if not the some parts of the bottom half of the wave will be clipped off. Any idea why?

Also, must the 2N5550 be mounted together with the 2N3055 on the same heatsink? Cause its a TO-92 Package type.

You may change the CCS current by changing the value of R1 on the diagram (i.e. 0.4 Ohm for 1.5A) and if you need substantially more current, reducing R2 value. 2N5550 can be replaced by something like BD139 - however it dissipates less than 100mW and doesn't really need a heatsink.

Cheers

Alex
 
AndrewT said:
Hi,
I must be thick. I find trying to read that schematic with the voltage sources in unusual locatons very difficult to read.

You have posted a DC coupled amplifier.
The source resistance on the inverting input is 22k//100k=18k
the non-inverting sees 22k//0k0=0k0.!!!
That will guarantee an output offset.

The pre-amp or source that you finally connect instead of the idealised voltage source will controll the output offset by substituting a different source resistance. If it is DC blocked the non-inverting source resistance changes to 22k. oops the output offset has gone (unless you had zeroed it with a shorted input).

Do you plan to use a DC coupled pre-amp with a DC servo?

Have you tried moving the lower end of the load from the ground to the drain of M1? What does the simulation say? I may have this mixed up due to not properly understanding this schematic.

Hi Andrew,
I am kind of lost when u mentioned about the output offset. Could you enlighten me further?

The simulation says that it would work without any DC offset so im kind of confused.

I plan to connect this directly to a CDP through a passive preamp.

Thanks!
 

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AndrewT said:
The source resistance on the inverting input is 22k//100k=18k
the non-inverting sees 22k//0k0=0k0.!!!
That will guarantee an output offset.

Hi Andrew,

the schematics is somewhat unusually drawn, however there is nothing wrong with it (IMHO) with the exception of the CCS. The offset due to the resistor imbalance on the inputs depends on the input currents of the opamp. For LT1001 in is only few nA and there is nothing to worry about - maximum offset it could create on the output would be less than 1mV anyway. For some opamps with >0.1uA of the input current that could be a problem, thought.

Cheers

Alex
 
x-pro said:


Hi Andrew,

the schematics is somewhat unusually drawn, however there is nothing wrong with it (IMHO) with the exception of the CCS. The offset due to the resistor imbalance on the inputs depends on the input currents of the opamp. For LT1001 in is only few nA and there is nothing to worry about - maximum offset it could create on the output would be less than 1mV anyway. For some opamps with >0.1uA of the input current that could be a problem, thought.

Cheers

Alex

Hi,
Sorry for the misconception, but the LT1001 is just there for me to do simulations. I intend to use AD826 in the final thing. Would it be a problem with the 826?
 
happyboy said:


Hi,
Sorry for the misconception, but the LT1001 is just there for me to do simulations. I intend to use AD826 in the final thing. Would it be a problem with the 826?

Yes, it would be a problem. AD826 has very high input currents (as do many fast opamps) - upto 10 uA ! In your circuit it would introduce far too much offset. Easy way of getting rid of it without changing opamp or circuit would be to put a coupling capacitor (say 10uF) on the input before R3 and reducing R3 to 18K to balance DC resistance as seen by the inputs.

However I would be careful in using AD826 or similar opamp in a circuit like this. Fast opamps do require much attention to details, including power supply arrangement, layout, feedback resistor values etc. In your circuit 100K and 22K in the FB chain is far too high resistance for AD826.

I would use for example OPA134 to start with in a circuit like this and with little experience of fast opamps.

Cheers

Alex
 
x-pro said:


Yes, it would be a problem. AD826 has very high input currents (as do many fast opamps) - upto 10 uA ! In your circuit it would introduce far too much offset. Easy way of getting rid of it without changing opamp or circuit would be to put a coupling capacitor (say 10uF) on the input before R3 and reducing R3 to 18K to balance DC resistance as seen by the inputs.

However I would be careful in using AD826 or similar opamp in a circuit like this. Fast opamps do require much attention to details, including power supply arrangement, layout, feedback resistor values etc. In your circuit 100K and 22K in the FB chain is far too high resistance for AD826.

I would use for example OPA134 to start with in a circuit like this and with little experience of fast opamps.

Cheers

Alex

Hmm in that case i think i will just use a NE5532 Opamp. Am i right to say that it will not have much DC offset cause the input current is 200nA?

Thanks!
 
happyboy said:


Hmm in that case i think i will just use a NE5532 Opamp. Am i right to say that it will not have much DC offset cause the input current is 200nA?

Thanks!

If you will use a capacitor on the input and change R3 to 18K the offset wouldn't be a problem. To calculate the additional input offset due to input currents you need to multiply input ("bias") current of the opamp to the DC resistance connected to it. In your case 18K*200nA=3.6mV. To reduce the output offset you should make sure that the offsets on both inputs are the same, in this case the output offset (excluding the initial offset of the opamp itself) should be 0. Any difference would be amplified by the gain of the circuit - in your circuit it is about 6. Even if your non-inverting input is grounded the output offset due to the input current would be around 3.6*6=21.6mV which is acceptable for a power amplifier output.

However I would advise you to use the input capacitor as you may have some DC offset coming from your source - this would be amplified accordingly and may become dangerously high for the speaker.

Cheers

Alex
 
x-pro said:

However I would be careful in using AD826 or similar opamp in a circuit like this. Fast opamps do require much attention to details, including power supply arrangement, layout, feedback resistor values etc. In your circuit 100K and 22K in the FB chain is far too high resistance for AD826.

Pardon my lack of knowledge but is there some way to know if the resistors in the feedback chain is suitable?

Thanks!
 
happyboy said:
Pardon my lack of knowledge but is there some way to know if the resistors in the feedback chain is suitable?

As a quick rule of thumb - the parasitic properties of the components preferably should not interfere with the workings of a circuit. These parasitics mostly consist from small capacitances (acros a resistor, on the input of an opamp etc.) Larger the value of the resistors used in the circuit - lower the frequency where such unwanted influence would occur. Generally it is advisable to keep parasitic RC parameters out of the working range for a circuit. For example: AD826 has got the unity gain frequency around 30-50 MHz and 1.5 pF input capacitance. Cut-off frequency for a feedback resistor and the input capacitance would be sensible to keep much (2-5 times) higher than the unity gain frequency. For 1.5 pF you'll need a value of approx. 1K resistor to get a cut-off frequency of 100 MHz. For a feedback resistor value 22K, cut-off frequency would be in the area of 5MHz and that could create a potential stability problem. There are ways of dealing with such difficulties however it is usually not an easy task :)

Cheers

Alex
 
richie00boy said:
It comes down to the input current requirements of the op-amp and how much noise you can tolerate, traded with the dissipation in the resistors.

These are generally secondary effects - I was concerned first about stability of a circuit with a fast opamp. You may add to your list maximum available current from the opamp output as low values of feedback resistors also create a load for the opamp.

Cheers

Alex
 
x-pro said:


As a quick rule of thumb - the parasitic properties of the components preferably should not interfere with the workings of a circuit. These parasitics mostly consist from small capacitances (acros a resistor, on the input of an opamp etc.) Larger the value of the resistors used in the circuit - lower the frequency where such unwanted influence would occur. Generally it is advisable to keep parasitic RC parameters out of the working range for a circuit. For example: AD826 has got the unity gain frequency around 30-50 MHz and 1.5 pF input capacitance. Cut-off frequency for a feedback resistor and the input capacitance would be sensible to keep much (2-5 times) higher than the unity gain frequency. For 1.5 pF you'll need a value of approx. 1K resistor to get a cut-off frequency of 100 MHz. For a feedback resistor value 22K, cut-off frequency would be in the area of 5MHz and that could create a potential stability problem. There are ways of dealing with such difficulties however it is usually not an easy task :)

Cheers

Alex

What i understand from the example is that i must use a feedback resistor that when matched with the input capacitance of the Opamp must be out of the unity gain bandwidth of the opamp? If what i understand is right, then how do i calculate what value of a resistor to use with different input capacitance of different opamp?
 
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