Vo for the clipper circuit where Vi = 10sin (6280t) and diodes are 0.6 volts.
I think the simple form of the circuit is like this.
if we don't calculate the resistors. Could the output signal be this way?
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2\$\begingroup\$ In your 'simple' circuit Vo = Vi, no matter what is between them. It's not at all like the original. \$\endgroup\$– Bruce AbbottCommented Jun 5, 2020 at 1:58
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\$\begingroup\$ IN your hand drawn schematic Vo = Vi. The diodes and resistors don't do anything. \$\endgroup\$– Kevin WhiteCommented Jun 5, 2020 at 1:58
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\$\begingroup\$ can you help for the truth? \$\endgroup\$– safak26Commented Jun 5, 2020 at 2:03
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\$\begingroup\$ @safak26 More like this, perhaps? \$\endgroup\$– jonkCommented Jun 5, 2020 at 2:23
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1\$\begingroup\$ @safak26 You should be able to just look at it and work out the output. When the signal goes high, it has to drop through one diode (the other is blocking) and overcome 3 V (so 3.6 V total) before it can appear at the output. So that one will be (10-3.6)=6.4 V peak. From 0 V to 3.6 V, it is clipped. On the other half cycle it has to drop through the other diode, 0.6 V, and then experiences a 1/3 resistor divider, so it peaks at 1/3 * 9.4 V or about 3.1V. From 0 V to -0.6V, it is clipped. A double-bump with shoulders, kind of. \$\endgroup\$– jonkCommented Jun 5, 2020 at 2:36
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