# Homework Help: Circuit design

1. Sep 5, 2006

### EvLer

Hello,
I have this problem: design a circuit using diodes, resistors and voltage sources to obtain the input function.
And the input function is Iin vs Vin plot with following points connected as straight lines: (2V,0mA), (3V,2mA), (4V,6mA).
I guess I need a hint on what kind of topology this might be and also, would I use several voltage sources? Anyway.... i need some help seeing to start this problem.

2. Sep 5, 2006

### chroot

Staff Emeritus
Well, first notice that the current only begins flowing when the input voltage is above 2V. Consider that no current flows through a resistor when both ends are at equal voltage, and you might decide you need a voltage source connected to the input through a resistor.

Next, you might size that resistor for the current you desire.

Next, you need to make sure the circuit doesn't conduct "backwards" for an input voltage less than 2V, which sounds like the perfect task for a series diode.

- Warren

3. Sep 5, 2006

### EvLer

thanks chroot,
so let's say I have Vin in series with R and in series with Volt source and a forward biased perfect diode, then my Volt source has to be a bit greater voltage than Vin to make current 0. So, now if I pick Volt source to be 2.5V and have a voltage drop across diode to be 0.3V so then voltage across R is 0.2V, I can figure out the R to be 100 Ohm, but then what do I need to do to make it work for 4 V as well?
Specifically, if I have a 100 ohm resistor from previous calculations I won't get the current to be 6mA with Vin = 4V and R = 100 Ohm if Vdiode remains 0.3V, does it remain the same?
Some more explanation would be very helpful...

4. Sep 5, 2006

### chroot

Staff Emeritus
A forward-biased diode drops approximately 0.7V. The diode will probably be in series with Vin. If so, just redo your calculations for the correct resistor value with (Vin - 0.7) instead of Vin.

- Warren

5. Sep 5, 2006

### EvLer

don't i need another voltage source besides the Vin? or would voltage drop across the resistor be enough?
the diode has 0.7 V drop even if the Vin changes, right? can i just use 0.7 with no derivation?

6. Sep 5, 2006

### chroot

Staff Emeritus
Yes, you can pretty much use 0.7V as the threshold voltage with no further fuss. If the voltage across the diode is less than 0.7V, it does not conduct. If the voltage is greater than 0.7V, it conducts with negligible resistance.

So if you apply 2V at the input, the diode drops 0.7V, and you're left with 1.3V at the input side of your resistor. Choose a voltage source for the other side of the resistor, and size the resistor to conduct 2 mA per applied volt.

- Warren

7. Sep 5, 2006

### EvLer

that makes sense to me, but the problem is giving points that are not incremental per volt, it's (2V,0mA), (3V,2mA), (4V,6mA), so the currents do not come up to be 2 and 6 if I have a constant value resistor and a voltage source. unless i'm not seeing something....

Last edited: Sep 5, 2006
8. Sep 5, 2006

### chroot

Staff Emeritus
I see what you mean, EvLer -- I was indeed making it too simple. The desired output curve is not linear, but a simple resistor has a linear V/I relationship. What is the circuit supposed to do with an input of 0V? Does it not matter at all what the circuit does at any other points than those that are given? What about the behavior between 2V, 3V, and 4V? Does it have to be linear in the regions between those points?

- Warren

Last edited: Sep 5, 2006
9. Sep 5, 2006

### EvLer

i don't think so, only the points of Vin-Iin (not the diode V-I plot) plot are given and graph stops at the point of (4V,6mA).
yeah, it's linear in between the points, the points are connected by a straight line from one point to another.

it should be not very complicated, this is a fundamentals course :|

Last edited: Sep 5, 2006
10. Sep 5, 2006

### chroot

Staff Emeritus
Okay, well, if the behavior between the points is not relevant, nor is the behavior outside the range 2V to 4V, a simple circuit might simply use two strings of series diodes connected in parallel to the input source.

The first string would contain four diodes, each dropping 0.7V when forward-biased. If the input voltage is below 2.8V, all four diodes will not be turned on, so no current will flow through that branch. If the input voltage is above 2.8V, all four diodes will be turned on, and each will drop 0.7V. Put a resistor in the string and size it to conduct 2mA with the "leftover" voltage.

The second string would contain five diodes and a resistor. That string would not conduct any current until the input voltage is above 3.5V. Size its resistor so the two strings in parallel together carry the desired 6mA of current at 4V input.

There are probably many other topologies that would work, but this is the one that occurs to me at the moment.

- Warren

11. Sep 5, 2006

### chroot

Staff Emeritus
Okay, now I'm confused. You first said that only the points (2V, 0mA), (3V, 2mA) and (4V, 6mA) were given, then you changed your mind and said the points are actually connected by straight lines.

There is a world of difference between having to meet some specific points, and having to make a circuit which is linear between 2V and 3V that seamlessly becomes a different linear circuit between 3V and 4V.

If the circuit must be linear between the points, you can use a voltage sources in each string of diodes to fine-tune the turn-on voltages of each string. You need the first diode string to turn on at 2V, and the second to turn on at 3V. Each string has a resistor sized to give you a slope of 2mA / V.

- Warren

12. Sep 5, 2006

### EvLer

yes, the points are connected by a straight line, so it is a piece-wise linear graph, i guess I should have posted an image of it...
Thanks though, was working on other stuff now back to this :)

ps: would it still be the same slope 2mA per volt?

Last edited: Sep 5, 2006
13. Sep 6, 2006

### Gokul43201

Staff Emeritus
Haven't read this thread in its entirety, but it seems that all that needs making is a circuit with some effective load resistance that is (i) infinite up to some V1, (ii) a constant finite value from V1 to V2 and (iii) a smaller constant value from V2 up. Correct?

How do you make an effective resistance smaller? You throw another resistor in parallel. But you want to "switch on" this parallel resistive path at some fixed voltage, so you need some kind of a "relay" (diode+ voltage source) for that.

Looks to me like there's a pretty simple solution using a pair of diodes and a pair of fixed voltage sources in addition to Vin (but having never seen a problem like this before, I might be totally wrong).

Last edited: Sep 6, 2006
14. Sep 6, 2006

### chroot

Staff Emeritus
Gokul,

That's pretty much the same solution I came up with -- the only difference is that used multiple diodes in series to drop voltage, while you just went the "direct" method and used a voltage source to drop voltage.

- Warren

15. Sep 6, 2006

### EvLer

thanks everyone, I used one diode and resistor + voltage source per string, it worked.