How to decrease a Square wave ouput

Thread Starter

captoro

Joined Jun 21, 2009
151
Hello,

I have a LTC6903 IC that outputs a square wave signal at 5Vpk-pk.
I need to decrease the amplitude to 1V pk-pk. I tried using a simple voltage divider (with two resistors) put that didnt work. If the values of resistors are too high, in the Kilo ohm range , then there is nothing in the output. If I use low value resistors, then the signal output is really bad.
I need the signal to be that low because I need to modulate it with a AD835, and that IC has an input of 1V pk-pk.
Is there another and better solution to decrease the output voltage to 1V pk-pk and have a clean signal ?
Keeping in mind the LTC6903 is a fast rise signal.

Ken
 

Audioguru again

Joined Oct 21, 2019
4,523
You mentioned the AD835 voltage controlled amplifier that has a max input of +/-1V.
So the squarewave from the LTC6904 must be attenuated to 1V feeding the AD835 as a carrier and the AD835 has a signal for amplitude modulation of the squarewave.

The many harmonics of the squarewave will cause communications interference.
 

LowQCab

Joined Nov 6, 2012
1,913
The LTC6903 is a 5-Volt, single-supply device ........
"" and they operate over a single wide supply range of 2.7V to 5.5V. ""
so you're not talking about 5-Volts-peak-to-peak,
it's actually 2.5-Volts-peak-to-peak if put through a Capacitor, or,
a 0-to-5V Pulse with a 50% Duty-Cycle.

The AD835 requires a Split-Supply of Plus & Minus 5-Volts.

Do You need to be able to input Negative-Numbers ?
.
.
.
 

Thread Starter

captoro

Joined Jun 21, 2009
151
The LTC6903 is a 5-Volt, single-supply device ........
"" and they operate over a single wide supply range of 2.7V to 5.5V. ""
so you're not talking about 5-Volts-peak-to-peak,
it's actually 2.5-Volts-peak-to-peak if put through a Capacitor, or,
a 0-to-5V Pulse with a 50% Duty-Cycle.

The AD835 requires a Split-Supply of Plus & Minus 5-Volts.

Do You need to be able to input Negative-Numbers ?
.
.
.
Hi,
The output of the LTC6903 is 0V to 5V 50% duty cycle. There is no negative input.
 

MisterBill2

Joined Jan 23, 2018
11,556
"Peak to peak" amplitude does not depend on where the zero voltage reference point is relative to the signal. Zero center is usually assumed but not required.
Getting a 1.00 volt amplitude wave with a resistor will require knowing the resistance of the load that the signal is connected to, and then adding a series resistor of four times that resistance. That is presuming that the source resistance is zero, probably not a good guess.
 

Thread Starter

captoro

Joined Jun 21, 2009
151
"Peak to peak" amplitude does not depend on where the zero voltage reference point is relative to the signal. Zero center is usually assumed but not required.
Getting a 1.00 volt amplitude wave with a resistor will require knowing the resistance of the load that the signal is connected to, and then adding a series resistor of four times that resistance. That is presuming that the source resistance is zero, probably not a good guess.
any IC that has a variable voltage output that I can set to 1V ? like a logic gate ?
 

MisterBill2

Joined Jan 23, 2018
11,556
To design a simple voltage divider that will reduce the 5.00 volt square wave to 1.00 volts I need to know the resistance of both the source and the load. To get a voltage divider that will reduce the 5 volt square wave to a 1 volt square wave we can guess that the source resistance is zero and that the load resistance is very high. Then, because precision resistors are not handy, I can build a series string of five 1000 ohm 1% tolerance resistors, and apply the 5 volts to the top of the string and find the one volt across the bottom 1000 ohm resistor.
Realize now that two guesses were made here, and both of them matter quite a bit. If the driving circuit has feedback to keep the voltage at 5 volts independent of load, the the guess of zero source resistance is reasonable, and if the load has a reasonably high input resistance, such as a stereo amplifier, then the high resistance input guess is reasonable. But if it is the 50 ohm impedance input of an oscilloscope, then the input will be much less.
 

dendad

Joined Feb 20, 2016
4,109
You could always put a 1K trim pot on the output, adjust it to 0V out first, then connect the amp and wind it up until you get 1V out.
 

Thread Starter

captoro

Joined Jun 21, 2009
151
To design a simple voltage divider that will reduce the 5.00 volt square wave to 1.00 volts I need to know the resistance of both the source and the load. To get a voltage divider that will reduce the 5 volt square wave to a 1 volt square wave we can guess that the source resistance is zero and that the load resistance is very high. Then, because precision resistors are not handy, I can build a series string of five 1000 ohm 1% tolerance resistors, and apply the 5 volts to the top of the string and find the one volt across the bottom 1000 ohm resistor.
Realize now that two guesses were made here, and both of them matter quite a bit. If the driving circuit has feedback to keep the voltage at 5 volts independent of load, the the guess of zero source resistance is reasonable, and if the load has a reasonably high input resistance, such as a stereo amplifier, then the high resistance input guess is reasonable. But if it is the 50 ohm impedance input of an oscilloscope, then the input will be much less.
In my first post I mentionned the output of the LTC6903 goes directly into the AD835 input. From the datasheet of the AD835 ports are : High input impedances.
 

MisterBill2

Joined Jan 23, 2018
11,556
Although it was not stated, my last post gave a couple of possibilities, depending on the accuracy required. That is, how close to 1.00 volts must the output be? The amount of effort doubles at least with each additional decade of accuracy required, and often the cost as well.
Of course, a simple potentiometer adjusted to the desired output, would be a simple method, adequate for a one-off but not good enough for a production of 10,000 units.
 

ScottWang

Joined Aug 23, 2012
7,178
You can try the circuit below.

Or you can measure the Vout resistor from Ve(Pin e of bjt) and Ground (You should turn off the power, and separate the SVR from the transistor), and then to find two fixed resistors to replace the SVR, if you can find the appropriate values of resistors.

Converting5VSquarewaveTo1V_ScottWang.png
 
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