Clippers & Clampers

Clipper Circuits:

Problem: Design a circuit which clips the voltage output at <6V using an ideal silicon diode.

Solution:

Vin+-

R

5.3V

Vout

+

-

Assume the diode is off. In this case, Vout = Vin.

Now, assume the output is large (say, +10V). The diode has more than 0.7V across it, so it turns on.When it turns on, the output is clipped at +6V (0.7V + 5.3V = 6.0V).

Problem: Design a circuit which clips the voltage at the output at >1V.

Solution:

Vin+-

R

1.7V

Vout

+

-

Assume the diode is off. Again, Vout = Vin.

Now, assume Vout is small (say, -10V). Now the diode is forward biased and turns on. The voltage at theoutput is then 1.0V (-0.7V + 1.7V = 1.0V).

Problem: Design a circuit which clips a signal at +1V < Vout < +6V.

Solution: Put the previous two circuits in parallal:

NDSU 9: Clippers & Clampers ECE 321 - JSG

1 January 25, 2011

Vin+-

R

1.7V

Vout

+

-5.3V

When Vout is large (+10V), D1 turns on, D2 turns off, and the output clips at +6V.

When Vout is small (-10V), D2 turns on, D1 turns off, and the output clips at +1V.

When 1V < Vout < Vin, both diodes are off and Vout = Vin.

Zener Diodes

An easier solution is to use a Zener diode. Zener diodes are special diodes which are designed to breakdown at a certain voltage when reverse biased. For example, a 6V Zener diode will have a fixed, 6V dropacross it if you try to reverse bias it by more than 6V. (i.e. a zener diode will dump whatever current isrequired to hold the voltage at 6V. )

If you want to clip an output at <6V, use a 6V Zener diode as follows. If the output tries to exceed 6V,the Zener diode turns on and keeps the output at +6V. Note that a zener diode is still a diode. If theoutput tries to go below -0.7V, the diode turns on like any silicon diode would and clips the output at-0.7V as well.

Vin+-

R

Vout

+

-

6V Zener

If you want to clip an output at > -8V, flip the diode around and use a 8V zener diode.

If you want to clip an output between -8V < Vout < +6V, use two zener diodes:

NDSU 9: Clippers & Clampers ECE 321 - JSG

2 January 25, 2011

Vin+-

R

Vout

+

-

8V Zener

6V Zener

Note that the output will actually clip at -8.7V and +6.7V. If the output tries to become large (say,+10V), the 6V zener turns on and clips at +6.0V. The 8V zener is now a forward biased diode, which hasanother 0.7V drop across it, for a total of 6.7V.

This conficguration is fairly common. If you want to protect an electronic device from abuse, two zenersare often placed across the input pins as shown above. In normal operation, one of the zener diodes is offand the diodes have no affect. If the opeator tries to hit your device with +24V, the zeners turn on. clipthe input at +6.7V, and protect your device.

Clamper Circuits:

Problem: Design a circuit which shifts the DC level of a signal so that the minimum voltage is 0V.

Solution:

Vin+- Vout

+

-0.7V

C

RVd+

-

Vin I1

I2 I3

Vout

If the output is more than 0V, the diode turns off. Assuming C is large (i.e. the resistance of R is muchlarge than the impedance of C, meaning that by voltage division Vout = Vin), the output follows theinput.

If the output is less than 0V, the diode turns on and current flows to the right side of the capacitor. Thiecharges up the capacitor with a positive voltage going left to right.

The output is Vin + Vc = Vout. The output is shifted up until the miniimum output voltage is 0V.

NDSU 9: Clippers & Clampers ECE 321 - JSG

3 January 25, 2011

MultiSim Simulation:

Draw the circuit in Multisim and let it run...

Note that the output (the top trace) is the same as the input, shifted so that the minimum is 0V. If youwant to shift the voltage so that the maximum is -5,

Flip the diode, Change the voltage source. to -5.7V (-5V minus another 0.7V for the diode):

You can also do this in VisSim, but it's a lot harder. You see the mathematics behind the simulation,however.

NDSU 9: Clippers & Clampers ECE 321 - JSG

4 January 25, 2011

VisSim Simulation:

Using voltage node equations, if you treat the diode as a resistor, Rd, then

I1 + I2 + I3 = 0

C ⋅ ddt(Vout − Vin) +

Vout−0.7Rd

+ VoutR = 0

Solving for Vout:ddt(Vout − Vin) = −

VoutCR − Vout

CRd

(Vout − Vin) = ∫ ⎛⎝−VoutCR − Vout

CRd

⎞⎠ dt

Vout = Vin − ∫ ⎛⎝ VoutCR + Vout

CRd

⎞⎠ dt

The resistance of the diode is small when Vd > 0.7 and large for Vd < 0.7. Using the diode equations:

Vd = nVT ⋅ ln ⎛⎝Id

Idss− 1⎞⎠ ≈ nVT ⋅ ln ⎛⎝

IdIdss

⎞⎠

dVddId

≈ Rd =nVTId

Id = Idss ⋅ ⎛⎝exp ⎛⎝VdnVT

⎞⎠ − 1⎞⎠

1Rd≈ dId

dVd= Idss

nVT⋅ exp ⎛⎝

VdnVT

⎞⎠

Rd ≈ 1011 ⋅ exp (−19 ⋅ Vd)

Together this lets you simulate the clamper circuit. A VisSim simulation is show below with R = 1M andC = 100uF. Note that the output (blue) is the input (red) shifted up so that the minimum is approximately0V. To shift it down so the maximum is 0V, flip the diode around.

NDSU 9: Clippers & Clampers ECE 321 - JSG

5 January 25, 2011

NDSU 9: Clippers & Clampers ECE 321 - JSG

6 January 25, 2011