 ### Network # design a limiter circuit

#### Thilzz

Apr 13, 2023
5
The requirements are:

• The input signal of the limiter Vin is a sine wave having a maximum amplitude of 10 V and a frequency of 60 Hz.
• The input signal Vin is fed to the circuit through a 500 Ω resistor.
• If Vin <= -9 V, Vout = 10 V
• If -9 V <= Vin <= 6 V, 0 V <= Vout <= 10 V
• If Vin >= 6 V, Vout = 10 V
These are the needful factors for the limiter.

I should use only the following components

• Diodes with a constant voltage drop of 0.7 V
• Zener diodes operating in the reverse bias region with Vz = 4 V at Iz = 1 mA, and rz = 1 Ω
• Resistor (having any value)
• DC voltage sources with a voltage between -0.7 V and +0.7 V
How to design the circuit using these components? The circuit must be easy to understand and please give me a diagram for the circuit.

#### Harald Kapp

##### Moderator
Moderator
Nov 17, 2011
13,461
We will not do your homework for you - forum policy.
Start by showing us what you have achieved so far and wheree you are stuck. Then we can guide you to find the missing links for the solution.

• davenn

#### Thilzz

Apr 13, 2023
5
We will not do your homework for you - forum policy.
Start by showing us what you have achieved so far and wheree you are stuck. Then we can guide you to find the missing links for the solution.
i design a circuit with a diode bridge for limiting, but i think that was not correct and i can't understand how to design using the requirements given

#### Harald Kapp

##### Moderator
Moderator
Nov 17, 2011
13,461
I think you got the requirements wrong:
• If Vin <= -9 V, Vout = 10 V
• If -9 V <= Vin <= 6 V, 0 V <= Vout <= 10 V
• If Vin >= 6 V, Vout = 10 V

A limiter can comparatively easily limit a positive input voltage to an upper limit and a negative input voltage to a lower limit (marked green).
What it can't do, at least not without active components, is to deliver a positive output voltage for a negative input voltage (marked red).

#### Thilzz

Apr 13, 2023
5
I think you got the requirements wrong:

A limiter can comparatively easily limit a positive input voltage to an upper limit and a negative input voltage to a lower limit (marked green).
What it can't do, at least not without active components, is to deliver a positive output voltage for a negative input voltage (marked red).
but i get the above values. are those wrong?

#### Harald Kapp

##### Moderator
Moderator
Nov 17, 2011
13,461
but i get the above values. are those wrong?
Where do you get these from?
Are they wrong? I can't say, but they are not plausible for a circuit using the components you may use.

#### Harald Kapp

##### Moderator
Moderator
Nov 17, 2011
13,461
Here's the problem in a graphical illustration: The requested Vout is non-monotonous and jumps from 0 V to 10 V for Vin = -9V.
A limiter or clipper (cf. post #4) will output a voltage as shown in the orange curve: Vi = Vout when not limiting, otherwise fixed at a minimum or maximum voltage, depending on Vin.

The specification in post #1 require an output voltage higher than the input voltage for
Vin <= -9 V or Vin >= 6 V. This is impossible with only passive components that you are allowed to use.

Another issue with these requirements is that they are contradictory. Take these two for example:
• If Vin <= -9 V, Vout = 10 V
• If -9 V <= Vin <= 6 V, 0 V <= Vout <= 10 V
What is supposed to happen at Vin = -9 V?
The first statement requires Vout = 10 V in that case (note the "<=" operator).
The second statement requires Vout = 0 V in that case (again, note the "<=" operator).
But Vout cannot be +10 V and 0 V at the same time.

The reasoning above is why I questioned the validity of your requirements.

Feb 19, 2021
673
If you look at limitations :

• If Vin <= -9 V, Vout = 10 V
• If -9 V <= Vin <= 6 V, 0 V <= Vout <= 10 V
• If Vin >= 6 V, Vout = 10 V

I would tackle the 2'ond item first. Than add the clamps, item 1 and 3. Break the design into
parts and solve the parts sequentially so's to speak.

Direct hint, you need to map one range into another on item 2. A way of doing
that is : Hint let R3 = infinity. So you can see I solve this circuit
to have 0 Vout for -9 Vin in and +10 Vout for +6 Vin. But
you have to pick a Vdd value to do the solution. So try it
at 10V. If you get a - value for a R you will have to raise Vdd

What's got me stumped is you are not allowed to use V source in your allowed list, eg. the V2 needed ?

Regards, Dana.

#### Harald Kapp

##### Moderator
Moderator
Nov 17, 2011
13,461
No clamps whatsoever will provide the solution to requirement #1. Note that Vout = +10 V for Vin <= -9 V!
Nor will Vout = 10 V be possible for Vin >= 6 V by any clamping method.

Feb 19, 2021
673
Last edited:
• Harald Kapp

#### Harald Kapp

##### Moderator
Moderator
Nov 17, 2011
13,461
Accepted, good solution for the upper clamp. Still I'm wary about the other side.

#### Thilzz

Apr 13, 2023
5
y
Here's the problem in a graphical illustration:
View attachment 58724
The requested Vout is non-monotonous and jumps from 0 V to 10 V for Vin = -9V.
A limiter or clipper (cf. post #4) will output a voltage as shown in the orange curve: Vi = Vout when not limiting, otherwise fixed at a minimum or maximum voltage, depending on Vin.

The specification in post #1 require an output voltage higher than the input voltage for
Vin <= -9 V or Vin >= 6 V. This is impossible with only passive components that you are allowed to use.

Another issue with these requirements is that they are contradictory. Take these two for example:
• If Vin <= -9 V, Vout = 10 V
• If -9 V <= Vin <= 6 V, 0 V <= Vout <= 10 V
What is supposed to happen at Vin = -9 V?
The first statement requires Vout = 10 V in that case (note the "<=" operator).
The second statement requires Vout = 0 V in that case (again, note the "<=" operator).
But Vout cannot be +10 V and 0 V at the same time.

The reasoning above is why I questioned the validity of your requirements.
yes the values are wrong now i get it

#### Thilzz

Apr 13, 2023
5
No clamps whatsoever will provide the solution to requirement #1. Note that Vout = +10 V for Vin <= -9 V!
Nor will Vout = 10 V be possible for Vin >= 6 V by any clamping method.
The input signal of the limiter Vin is a sinewave having a maximum amplitude of 10V and a frequency of 60Hz. The input signal Vin is fed to the circuit through a resistor 500 ohm. the limiter specifications are when the Vin<=-9v , Vout = 2/6 * Vin + x1 and -9v <= Vin <= 6v , Vout = 2/3 * Vin +x2 and Vin>=6v , Vout = x3. assume x2 = 0v. find the value for x1, x2 and x3

#### Delta Prime

Jul 29, 2020
1,626
Clamp what clamp?
The clipping circuit, also referred to as a limiter,clips off some of the portions of the input signal & uses the clipped signal as the output signal.
The clamping circuit keeps the amplitude of the output signal same as that of the in put signal except that
the D.C.level(offset)has been changed.
Being different from clippers, clamping circuits uses a capacitor
& diode connection.When diode is in its on state,the output voltage equals
the diode drop voltage (ideally zero)plus the voltage source, if any!!

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