Referring to the attached image, please help to advise are both diodes turned ON. And I appreciate if you can also explain why.

I am not sure as both are having same diodes and resistance.

Thanks.

 

both will conduct, but depending on their resistance will drop different voltages, normally 0.7V each, so the Vo will be 10- 0.7=[9.3] /2 = 4.65V

 

I beg to differ, Vo will be aproximately (10-0.7)*2k/3k=6.2V, and simlation agrees with that.
As long as the current through the diodes is of similar magnitude, you can rearrange the circuit to have a single diode and then the two 2k resistors in parallel, then the 2k load resistor.

 

I am not sure as both are having same diodes and resistance.

Both diodes will be on ONLY if they are exactly the same, otherwise only one would have most of the current through it and the other very little if any current.

 

Kubeek is correct.

The voltage drops 0.7v through each diode, V=-9.3 at each doide-resistor node.

The two parallel 2k resistors is like 1 x 1k resistance. The 2k resistor to ground means V0= 9.3*2k/(1k + 2k) = 6.2v

 

Even if the two diodes are off by 0.1v, you can still calculate the current flow through each route (Assume V0 is what we estimated from the previous = 6.2V, then you have 1.55 mA through one route and 1.50 mA through the other route. Current cannot flow from one 9.3v to the other 9.2v since the diodes block current. Therefore, current will still flow through both routes since the 2k resistor acts as a shock absorber in each bridge. Some curcuit designs prevent current flowing from one leg but not this one.

 

Thanks everyone- understand now that both will turn ON.
I manage to get Vo = 6.2V , Id = 1.5mA.
Thanks so much.

 

Isn't it a practice when parralleling diodes or SCR's that a small resistance in each circuit forces the diodes to share and not allow one diode to hog all the current?

 

Yes, tinkerman. The 2k resistors will make the diodes share. If those 2 resistors weren't there, the diode with the lowest turn on voltage would try to pass all the current, and as it heated up, it would pass very nearly all the current.

 

Yes, tinkerman. The 2k resistors will make the diodes share. If those 2 resistors weren't there, the diode with the lowest turn on voltage would try to pass all the current, and as it heated up, it would pass very nearly all the current.

referring to this circuit, would both diode turn ON?

 

I disagree with Kubeek. Even if the Vf of the two diodes are drastically different, there will be current in both series resistors.

ak

 

I disagree with Kubeek. Even if the Vf of the two diodes are drastically different, there will be current in both series resistors.

...and the two currents will be nearly the same, about 1.55 mA.

 

View attachment 87862

referring to this circuit, would both diode turn ON?

Both diodes will conduct current, but D1 will conduct much more than D2.

 

Here's an LTspice simulation of both circuits.

I added a 0.3V variable voltage in series with D1 to simulate diodes with drastically different forward bias voltages. This only causes a small change in the current division between the two diodes, as expected.

The bottom circuit has a very large difference in the two diode's current, also as would be expected.

 

View attachment 87862
referring to this circuit, would both diode turn ON?

Wow! Pulling my posts up from more than 3 years ago? Impressive!

About Post #10: Assuming a 1N4148 signal diode at room temperature
In the real world, D1 would show a voltage of 0.65 volts at 0.9 ma of current.
If you apply that 0.65 volts to an exactly same diode in series with 680 ohms, some smaller current will flow.
If 100 ua flows through D2, then the voltage across D2 is 0.58 volts and the voltage across 680 ohms is 0.07 volts. Double check that as 0.07 volts across 680 ohms and we have 0.1029 ma. Close enough for government work.

That's the way I see it. D2 is going to carry about a tenth of the total current because diodes have performance curves.

 

.................
That's the way I see it. D2 is going to carry about a tenth of the total current because diodes have performance curves.

Pretty close agreement with my sim where the two diode currents are 125μA and 817μA.

 

Pretty close agreement with my sim where the two diode currents are 125μA and 817μA.

Diodes current are different , but the voltage across both diodes will be the same?

 

In the real world, D1 would show a voltage of 0.65 volts at 0.9 ma of current.
then the voltage across D2 is 0.58 volts

Diodes current are different , but the voltage across both diodes will be the same?

You have to read the answers.

 

Diodes current are different , but the voltage across both diodes will be the same?

No. The voltage across a diode is a logarithmic function of the current.
Here's a typical curve of diode forward voltage versus current.

My simulation gave 490mV and 575mV drop respectively for the two diodes.

 

A similar graph is in the datasheet provided in post #15
Page 2, top left corner of the page.
The diode I chose is a different part number than what crutschow used, but the proportions are about the same.