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'deamplifier'

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a_drunken_dwarf

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hi, I've never taken an official course in electronics, so some of these questions may seem a little obvious. But don't think I'll be insulted if you say stuff that you think is obvious in your replies. It willprobably be helpful. Thanks. So, I need to 'deamplify' a TTL signal so that the output of this 'deamplifying' circuit has a lower voltage (~10 milivolts). As far as I know, the only way to reduce the voltage is to use a transformer. Is this true or is there another way? Are there any transformers that introduce very very little noise? I need a minimum of noise in the deaplification. Is deamplification the right terminology?
 

Roff

Well-Known Member
It's called an attenuator, and you do it with resistors. We could offer more help if you would explain where the ~10mv signal goes, and where the TTL signal comes from.
 

Agent 009

New Member
Yup, I could think of such 'deamplifier' made up of resistors and a npn transistor. The amplifier would be a common emitter, where the output voltage gain would be less than 1.
 

a_drunken_dwarf

New Member
couldn't I just hook up a lot of resistors in series for a high voltage drop? How is an attenuator different from doing this? Does it introduce less noise? Thanks!
 

Klaus

New Member
How about using a single potentiometer instead of the 'lot of resistors'?

Connect the TTL output to one end of the potentiometer. Connect the other potentiometer end to the same common as the TTL signal.
Pick up your 'attenuated' output between the wiper of the pot and the common.
You can then adjust the output to whatever ratio of the input you want.
This works for AC and DC and is linear (use a linear pot).

Pick your pot resistance so that it does not load the TTL signal unduly, in other words, when you have 5V at the output the total current through the pot is less than what the IC output can provide, try a 10K pot to start with.

If your output signal is very small you need to make sure it has shielded wires and the pot case is earthed or you might pick up noise.
 

Optikon

New Member
a_drunken_dwarf said:
couldn't I just hook up a lot of resistors in series for a high voltage drop? How is an attenuator different from doing this? Does it introduce less noise? Thanks!

You could but what does it have to drive? Your "deamplified: load could require an amount of current that your resistor divider would not allow(think V=IR). This is why an transistor makes more sense.

And resistors do not "introduce" noise like active circuits can. They do add some thermal noise just because they contain molecules with electrons in random motion. But I don't get the impression that you are woried about noise at these levels.
 

a_drunken_dwarf

New Member
I'm stimulating neurons in a petri dish. They need about a 50mV stimulation to be activated. The current needed is very low, pAmps.

Even though plain resistors might work (due to the low currents needed), I would still be interested in knowing more about attenuators. So from the 5V to 50mV I guess I would need an attenuation value of 1/100. Does anybody have circuit designs of an attenuator (as far as I've learned, an attenuator is a circuit with resistors and a transistor?), or know of an atenuator IC that has A=1/100? Thanks!
 

Dean Huster

Well-Known Member
A simple attenuator

It's two resistors in series. Let's put them down on paper in schematic form in a vertical line, R1 on top and R2 on bottom. The bottom of R2 is grounded. The top of R1 goes to your voltage source to be dropped or attenuated, in this case, the output of your TTL chip. The junction of the two resistors is the output with respect to ground, i.e., the output is across R2.

Two resistors are used in the design like this rather than just a "bunch of resistors in series" so that you have more control if the load resistance bounces around a bit.

Choose a value out of thin air for R2. You want this resistance to be about 4 to 10 times lower than the load resistance (the petrie dish). We'll pick 100 ohms just for the heck of it. Divide your desired output voltage by the resistance value you've chosen for R2 to get the current through R2 and the rest of the series circuit. So, 10mv divided by 100 ohms is 100µa. Let's assume that the TTL output of the chip is 4v p-p. The voltage across R1 will be the source voltage minus the voltage across R2 or 4 - 10mv which is 3.99 v. The value of R1 will be the voltage across R1 divided by the current through R1 or 3.99 divided by 100µa which is 39,900 ohms. Nearest standard 5% value for this will be 39K ohms. You'll have to use a little trial and error here since we're making a few assumptions, such as the actual output voltage of the chip, etc. But that's the general way do figure the thing. Dropping 5v down to 10mv is a bit of a drastic change, so there's going to be a lot of slop in this circuit.

Dean
 

Optikon

New Member
Re: A simple attenuator

Dean Huster said:
It's two resistors in series. Let's put them down on paper in schematic form in a vertical line, R1 on top and R2 on bottom. The bottom of R2 is grounded. The top of R1 goes to your voltage source to be dropped or attenuated, in this case, the output of your TTL chip. The junction of the two resistors is the output with respect to ground, i.e., the output is across R2.

Two resistors are used in the design like this rather than just a "bunch of resistors in series" so that you have more control if the load resistance bounces around a bit.

Choose a value out of thin air for R2. You want this resistance to be about 4 to 10 times lower than the load resistance (the petrie dish). We'll pick 100 ohms just for the heck of it. Divide your desired output voltage by the resistance value you've chosen for R2 to get the current through R2 and the rest of the series circuit. So, 10mv divided by 100 ohms is 100µa. Let's assume that the TTL output of the chip is 4v p-p. The voltage across R1 will be the source voltage minus the voltage across R2 or 4 - 10mv which is 3.99 v. The value of R1 will be the voltage across R1 divided by the current through R1 or 3.99 divided by 100µa which is 39,900 ohms. Nearest standard 5% value for this will be 39K ohms. You'll have to use a little trial and error here since we're making a few assumptions, such as the actual output voltage of the chip, etc. But that's the general way do figure the thing. Dropping 5v down to 10mv is a bit of a drastic change, so there's going to be a lot of slop in this circuit.

Dean

For VERY low level measurements, you will probably be in trouble noise-wise. Your TTL ouput is going to have LOTS of noise on it. You might want to consider using a quite supply switched through a low charge injection analog switch and just run the switch off of your TTL output (still lots of filtering needed)
 

Agent 009

New Member
I think the fact that resistors have tolerances (+/- 5 % or so) doesn't help much in noise reduction too... Especially with low-level analysis... A percent could excite the neurons or not.
 

Dean Huster

Well-Known Member
Pots as attenuators

Klaus, I forgot to mention that your pot idea for the attenuator, although looking good in theory, would never work. With a 5v input and trying to get a 10mv output, the wiper would be sitting on one of the end tabs and just hopping off onto the resistance track would shoot you 'way over-voltage. You'd have to at least put one fixed resistor in series between the pot and the input signal to get the pot to cover range you'd need.

Dean
 

lordjw

New Member
OPAMP Attenuator

Hi!

If you need very precise attenuation you could experiment with OPAMPS. Operational amplifiers can be used both as amplifiers and attenuators. Attenuation value is set by just two resistors. In inverting configuration 1K and 100K resistors will provide precise attenuation 1/100. Find application note from NationalSemi for basic operational amplifier circuits.

With OPAMPS you will need more electrical skills though. You will need to select amplifier that remains stable and you have to provide amplifier preferably both + and - operating voltages.

If you run into noise/instability problems with resistors it might be a good idea to start experimenting with OPAMPS.

Julius
 
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