```In article <50ka4q\$9ac_002@bzn08.imt.net>, ronman@imt.net (ron newman) wrote:

> Could somebody explain in an intuitive way how to generate the thevenin
> equivalent circuit for something with an input, an output, and a DC offset
> voltage?
>
> What it is is a diode clamp.  Signal goes through a 1K resistor, but the
> output is tied to a diode, which in turn is tied to 5 VDC.  But this 5V
> comes from a voltage divider: the middle of a 2.2K and a 1K resistor, with 15V
> applied at the 2.2K end.
>
> I'm trying to construct the Thevenin equivalent for the ENTIRE circuit.
> I know that you figure Vopencircuit and Ishortcircuit, but at what points?
> At the 5VDC offset?  At the output, which is AC, so constantly changing?
>
> Huh??
> Thanks.
>
> *****************************************************************
> Ron Newman
> "http://www.imt.net/~ronman/animal.htm"
> ronman@imt.net

It sounds like your circuit looks something like this:

15V
1K                              |
Vin ---/\/\/\-------|-------- Vout        /
|                     \ 2.2K
|                     /
----|>|---------------| (4.7V)
diode              /
\ 1K
/
|
gnd

The diode is non-linear, so you can't come up one Thevenin equivalent for
all cases. But there are basically only two cases to consider: when the
diode conducts, and when it doesn't.

When the diode is not conducting, i.e. for Vin < 4.7 + 0.6, then your
Thevenin circuit is trivial:
RT = 1K
VT = Vin ---/\/\/\--------- Vout

For the more interesting case of the diode conducting (Vin > 4.7 + 0.6),
then the diode can be approximated by a DC battery of about 0.6V.
The 2.2K and 1K network can be combined into a Thevenin equivalent of:
1                            1 * 2.2
VT = ----- 15 = 4.7V   and    RT =  --------  = 690 ohms
1+2.2                          1 + 2.2

The circuit now can be redrawn:

1K
Vin ---/\/\/\-------|-------- Vout
|
|  | |         690
---||||-------/\/\/--- 4.7V
+| |-
0.6V

Combining the battery and the 4.7V:
1K
Vin ---/\/\/\-------|-------- Vout
|
|     690
-----/\/\/--- 5.3V

This reduces to the final Thevenin equivalent circuit:

RT
VT  ---/\/\/\------- Vout

where:
1 * .69
RT = --------  = 408 ohms
1 + .69
and
.69
VT = ------- (Vin - 5.3) + 5.3     (remember, only for Vin > 5.3V)
1 + .69

or  VT = .41 Vin + 3.1

For example, if Vin = 6.0V, then VT = .41 * 6.0 + 3.1 = 5.56V :

408
5.56V  ---/\/\/\------- Vout

```

Date: 10 Sep 1996 21:37:55 GMT

Original Subject: Re: Thevenin equiv. circuit question