a high pass filter question

[walid1]

Hi
when we come to design a high pass filter like that formed between stages
Here C is the coupling cap
and R is the i/p impedance of the next stage

the cutoff freq = 1/(2 pi R C)

To make the cutoff freq = say 100 Hz, u can:
take C = 10 uF and R = 160 ohm
OR
take C = 1 uF and R = 1.6 Kohm

You can design it with different values of R and C
MY question is what the differance between them?
thanks

 

[audioguru2]

Hi Walid,
An opamp or transistor can't drive 160 ohms but they can drive 1.6k or 16k.
A 10uF cap costs more and is bigger than a 1uF cap. A 0.1uF cap might cost even less.

If the R is too high and the C is too small then the input of the next stage must be a very high impedance which might pickup interference like mains hum.

 

[walid1]

Thank you guru it is evident and good answer

 

[Kevin Weddle]

I always end up with a bunch of different supply voltages when designing circuits on paper. I think a lot of circuit values are chosen because of constraints. Throw a low value resistance anywhere in the circuit, and it will be seen all the way back to beginning of the circuit.

 

[audioguru2]

Throw a low value resistance anywhere in the circuit, and it will be seen all the way back to beginning of the circuit.

No Kevin. Many circuits have an input that is not affected by the output:
1) An audio amplifier input isn't affected if the output has an 8 ohm speaker connected or nothing connected.
2) An opamp input isn't affected if its output is loaded or even shorted.
3) A common emitter single transistor with its collector grounded has its input not affected by its collector load resistance.

 

[Kevin Weddle]

From what I have seen, it's hard to get gain. Or it's hard to develop the voltage. Or the impedance is reduced, setting up low gain for the previous circuit. Picking up the gain is difficult when the resistance is lowered. Our goal is amplification.

 

[audioguru2]

From what I have seen, it's hard to get gain. Or it's hard to develop the voltage. Or the impedance is reduced, setting up low gain for the previous circuit. Picking up the gain is difficult when the resistance is lowered. Our goal is amplification.

It is easy to get some voltage gain from a transistor circuit and even easier to get lots of voltage gain from an opamp.

A common emitter transistor stage has a voltage gain of RC/(Re + RE).
If the emitter is bypassed to ground with a capacitor to remove RE from the calculation and the circuit has a high impedance load so it isn't loaded down, then its voltage gain is about 173. This is with a 9V supply, operating point with 4.5VDC at the collector and a 10k collector resistor. Negative feedback is needed to reduce the very high distortion at high output levels. The negative feedback reduces the voltage gain and distortion.

Opamps have a typical voltage gain of 200,000 or 1,000,000. They usually have plenty of negative feedback added so that the load resistance doesn't cause their output level to drop. If a high output current is needed then power transistors or Mosfets can be added.

 

[Kevin Weddle]

Take some transistors and drive a 50 ohm load. I want the gain to be about 200. I want 1Vpp on the load. I want the signal source to be an oscillator.

It'll be a very long circuit. Increase the load to 1Kohm, and it will be a lot easier.

So, throw in a low resistance "anywhere", and it will be seen all the way back to the beginning of the circuit.

 

[audioguru2]

Take some transistors and drive a 50 ohm load. I want the gain to be about 200. I want 1Vpp on the load. I want the signal source to be an oscillator.

It'll be a very long circuit. Increase the load to 1Kohm, and it will be a lot easier.

So, throw in a low resistance "anywhere", and it will be seen all the way back to the beginning of the circuit.

What are you talking about, Kevin?
A single transistor can't have a voltage gain of 200 with a load resistance of 50 ohms, but two or three transistors can. When you increase the load resistance to 1k it won't affect the input.
You can short the collector of a transistor with a capacitor to ground and it doesn't affect the input.

An audio amplifier IC can drive an 8 ohm speaker or an open circuit. The negative feedback creates an output impedance of 200 times less than 8 ohms so the load has nearly no effect on the level of the output.

 

[Kevin Weddle]

If you set in place reasonable constraints, it's difficult. Yes, you can say let's stick in a 100VDC supply there, let's use a wide variety of DC supply voltages here and there, let's raise the frequency so we can use some emitter capacitors, let's lower the transistor current so we can up the beta, so on and so forth.

 

[Kevin Weddle]

You will notice that a small radio doesn't produce much volume, and it's not because of the small speaker or the size of the battery.

 

[audioguru2]

A small radio is not very loud because it has a small inefficient speaker, a small amplifier that doesn't need a heatsink and a small battery that can't provide much current for a long time.

A loud radio has a big efficient speaker, an amplifier with a big heatsink and a big battery that can provide a lot of current for a long time. Of course this radio is bigger.

 

[Kevin Weddle]

I belive the component count should be higher too, because it takes more circuitry to go from airwaves to large signal on a low impedance.

 

[audioguru2]

A low power audio power amplifier IC is the same number of parts as a high power audio power amplifier IC. Then the high power IC needs a big heatsink.

If you are making your own power amplifiers then just use darlington transistors on the output of the high power audio amplifier. If you are talking about hundreds of Watts then many power transistors need to be paralleled at the output.