Solving Basic Diode Circuit: Voltage Across Diodes @ 1mA

In summary: Then use the given info to find the product.In summary, the given circuit consists of a voltage source and two diodes in series, with D1 having a voltage of 0.7V and a current of 10mA, and D2 having a voltage of 0.6V and a current of 0.1mA. Using the diode equation, we can find the inverse saturation currents Is for D1 and D2. With the value of VT given, we can then find the individual voltages V1 and V2 at 1mA. Alternatively, we can also find the sum of V1 and V2 by using an expression involving the product of i1 and i2.
  • #1
sisyphus0321
10
0

Homework Statement


Given a basic circuit of a voltage source and two diodes in series:
D1 is .7v @ 10mA
D2 is .6v @ .1mA
If I is 1mA, What is the voltage across the two diodes?

Homework Equations


VT~.026
Vt= 60log (I(t)/Is
I1 produces V1
I2 produces V2
so V2 = V1 + 60(mV)[log (I2/I1)]


The Attempt at a Solution


Just plain lost on how to start. I get that this needs to be turned into a ratio between the two votage/current values across the diodes, but where do I start?
 
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  • #2
sisyphus0321 said:

Homework Statement


Given a basic circuit of a voltage source and two diodes in series:
D1 is .7v @ 10mA
D2 is .6v @ .1mA
If I is 1mA, What is the voltage across the two diodes?

Homework Equations


VT~.026
Vt= 60log (I(t)/Is
I1 produces V1
I2 produces V2
so V2 = V1 + 60(mV)[log (I2/I1)]


The Attempt at a Solution


Just plain lost on how to start. I get that this needs to be turned into a ratio between the two votage/current values across the diodes, but where do I start?

The inverse saturation current Is is different in each diode. Determine Is from the data you have at the currents 10mA and .1 mA and use it to find the voltages at 1mA.
 
  • #3
So solving for V2=V1 + 60mV[log(I2/I1)] I get D1=.64V and D2=.66v than using Kirchoff I can assume that the solution would be 1.3V? I know the whole thing is theoretical anyway because of the lack of resistance but is this the solution?
 
  • #4
sisyphus0321 said:
So solving for V2=V1 + 60mV[log(I2/I1)] I get D1=.64V and D2=.66v than using Kirchoff I can assume that the solution would be 1.3V? I know the whole thing is theoretical anyway because of the lack of resistance but is this the solution?

I would use the expression [tex]I=I_se^{\frac{V}{V_T}}[/tex], using I =10mA and V=.7V, to obtain Is for D1. Then use I = .1mA and V = .6V, to obtain Is for D2.
In reality, there should be a resistance limiting the current, but it is irrelevant, since we don't know the voltage of the source.
 
  • #5
I don't think we can obtain Is1 or Is2 without knowing VT. But I agree, the diode equation is probably the key here.

Note, they seem to be asking for the sum V1+V2. If so, it's not necessary to find V1 and V2 separately, we just need the sum.
 
  • #6
Redbelly98 said:
I don't think we can obtain Is1 or Is2 without knowing VT. But I agree, the diode equation is probably the key here.

Note, they seem to be asking for the sum V1+V2. If so, it's not necessary to find V1 and V2 separately, we just need the sum.
The OP already gave the value of VT: 26mV at 300K.
I don't see how we can find the sum without knowing the indi8vidual values of V1 and V2. Remember that the two diodes have different characteristics.
 
  • #7
CEL said:
The OP already gave the value of VT: 26mV at 300K.
Aha, I missed that. Thanks.

I don't see how we can find the sum without knowing the indi8vidual values of V1 and V2. Remember that the two diodes have different characteristics.
Guess it's a moot point, because knowing VT we can find IS1 and IS2, as you said earlier, and then both V1 and V2 at 1 mA.

So the key is using the 10 mA and 0.1 mA info to find IS1 and IS2, using the diode equation in post #4.

p.s.
Alternative, less straightforward method: one could also write out an expression for V1+V2, and find that it depends on the product i1·i2.
 

FAQ: Solving Basic Diode Circuit: Voltage Across Diodes @ 1mA

How do you calculate the voltage across a diode in a basic circuit?

The voltage across a diode in a basic circuit can be calculated by using Ohm's Law, which states that voltage (V) is equal to current (I) multiplied by resistance (R). In this case, the resistance is the forward resistance of the diode, which is typically around 0.7V for silicon diodes and 0.3V for germanium diodes. Therefore, the voltage across a diode can be calculated as V = I * 0.7V for a silicon diode or V = I * 0.3V for a germanium diode.

What is the significance of a 1mA current in diode circuits?

A current of 1mA (milliampere) is commonly used in diode circuits because it is a safe and practical value for most diodes. It is also a convenient value for calculations and allows for a good balance between accuracy and power consumption.

How does the voltage across a diode change with varying current?

The voltage across a diode remains relatively constant with varying current, as long as the current is within a certain range. This range is known as the forward bias region, and within this region, the voltage across the diode remains approximately constant while the current changes. Outside of this region, the voltage across the diode can vary significantly with changing current.

Can the voltage across a diode ever be negative?

No, the voltage across a diode cannot be negative. This is because a diode only allows current to flow in one direction, from the anode to the cathode. Therefore, the voltage across a diode can only be positive, and any attempt to reverse the voltage will result in very little current flow.

How does temperature affect the voltage across a diode?

Temperature can have a significant impact on the voltage across a diode. As the temperature increases, the forward voltage drop of a diode decreases, meaning that the voltage across the diode will also decrease. This is due to the change in the physical properties of the diode's materials at different temperatures. Therefore, it is important to consider temperature when designing diode circuits to ensure proper voltage regulation.

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