# Find $V_{output}$

1 vote
91 views I could not how to find the current in the circuit as the resistance of diode is not given .
But I think the voltage across the diode should always be +5V

asked Mar 27, 2017
edited Mar 27, 2017
Is the answer 2nd ?
Yes the answer is 2nd
5V remains across the diode  alaways to keep it going , so remaining 5 V is output.
Can you please explain it more clearly.

1 vote

The diode only allows current to flow in one direction, as indicated by the triangle/arrow. In this case +ve (conventional) current can only flow to the right through the diode.

When the voltage at the left of the diode is less than 5V, +ve current cannot flow towards the left from the higher voltage, because it cannot flow against the arrow marked in the diode. But when the voltage to the left of the diode is greater than 5V then +ve current can flow to the right through the diode, in the direction of the arrow. The diode then acts like a short circuit wire with no resistance.

This prevents the voltage at the left side of the diode from becoming larger than 5V at any time.

answered Mar 27, 2017 by (28,806 points)
selected Mar 28, 2017 by koolman
You mean it prevents from becoming **less** than 5V at any time.
As when  V(out) is greater than 5 then only the current will flow .
The voltage on the left of the diode is always $\le$ 5V, as the graph shows. The diode prevents the voltage at the left of the diode from becoming greater than 5V, not less than 5V.

The voltage on the right of the diode is only able to affect the voltage on the left when a current can flow through the diode. If +ve current cannot flow left because the voltage there is below 5V, then the voltage on the left cannot be increased by the voltage on the right.
As you said
' But when the voltage to the left of the diode is greater than 5V then +ve current can flow to the right through the diode, in the direction of the arrow.'
Then for current to flow , voltage should be greater .
When the voltage on the left of the diode is infinitesimally greater than 5V, that causes current to flow to the right. But the voltage on the left of the diode remains infinitesimally greater than 5V, it does not increase any further.
Why it does not increase further
It does not increase further because current is flowing through the diode and the resistance of the diode is zero (or infinitesimally small) when current flows to the right. (The resistance is infinite for current flowing to the left.) So the PD across the diode is zero (actually infinitesimally small, negligible), and the voltage on the right is 5V, so the voltage on the left must also be 5V (actually 5V + some very small negligible amount).

The voltage on the left of the 1$k\Omega$ resistor can be higher than 5V, but the voltage on the right of this resistor is clamped at 5V.

When the voltage on the left of the resistor is say 9V then the PD across this resistor is 4V, so the current through this resistor, and through the diode, is 4mA. The PD across the diode is larger than when the current first flows to the right, but it is still negligibly small, because the resistance of the diode is still negligible.
And  in graph we are counting that case also when no current is flowing .
(when potential is less than 5 to the left of diode .)
Yes, the graph shows that the voltage $V_{out}$ does not exceed 5V. Current is flowing through the diode when $V_{out}=$5V.