# Solving Circuits

#### hgmjr

Joined Jan 28, 2005
9,027
Originally posted by celect@Jun 10 2005, 07:05 PM
I'm confused if this zener diode is connected to anode to a neg voltage source means it reversed bais does that mean its open
and i have 0-0.7 = -0.7V and V e
[post=8390]Quoted post[/post]​
Unlike an ordinary signal or power diode, It is the nature of the behaviour of a zener diode that its zener voltage is the voltage at which it begins to conduct when it is reverse biased.

The zener diode is serving to provide a constant voltage that is applied to the base of the transistor. By using the zener in this way the voltage across the emitter resistor is held constant even if there are small variation in the -20V supply voltage. With a constant voltage across the emitter resistor then the emitter current is held constant and if the emitter current is held constant then the collector current is constant. Therefore the transistor acts as a constant current source.

hgmjr

#### hgmjr

Joined Jan 28, 2005
9,027

#### davidand

Joined Jun 2, 2005
43
Originally posted by hgmjr@Jun 10 2005, 08:14 PM
Unlike an ordinary signal or power diode, It is the nature of the behaviour of a zener diode that its zener voltage is the voltage at which it begins to conduct when it is reverse biased.

The zener diode is serving to provide a constant voltage that is applied to the base of the transistor. By using the zener in this way the voltage across the emitter resistor is held constant even if there are small variation in the -20V supply voltage. With a constant voltage across the emitter resistor then the emitter current is held constant and if the emitter current is held constant then the collector current is constant. Therefore the transistor acts as a constant current source.

hgmjr
[post=8392]Quoted post[/post]​
If the zener is providing a constant voltage to the base it states this is a 5v zener connected to the -20 supply I keep thinking i plug -5V - 0.7 to get -4.3V
If I put a voltmeter at the base and to ground i should read 5v

#### hgmjr

Joined Jan 28, 2005
9,027
Originally posted by davidand@Jun 10 2005, 08:48 PM
If the zener is providing a constant voltage to the base it states this is a 5v zener connected to the -20 supply I keep thinking i plug -5V - 0.7 to get -4.3V
If I put a voltmeter at the base and to ground i should read 5v
[post=8394]Quoted post[/post]​
The best way to look the 5V reverse-biased zener is in the same way you treated the three forward-biased diodes in HW#2. As you recall you started from the -10V negative supply and went positive from there to the -7.9V. You could have replaced the three diodes with a 2.1V zener diode and your results would have been the same.

In this problem, you could think of the reverse-biased 5V zener as nearly equivalent to a series of 7 forward biased signal diodes (4.9V).

Does this help?

hgmjr

#### davidand

Joined Jun 2, 2005
43
Originally posted by hgmjr@Jun 10 2005, 09:48 PM
AAC Forum Tutorial link to Zener Diode presentation

The information contained in the AAC Tutorials on the subject of zener diodes is very good. You may want to read the material and see if it helps clear up some of your questions.

hgmjr
[post=8393]Quoted post[/post]​
voltage base = -15 - 0.7 =14.3

#### hgmjr

Joined Jan 28, 2005
9,027
Originally posted by davidand@Jun 10 2005, 09:02 PM
voltage base = -15 - 0.7 =14.3
[post=8396]Quoted post[/post]​
You're close. Check your math. Subtracting a positive number from a minus number does not yield a positive number.

hgmjr

#### davidand

Joined Jun 2, 2005
43
Originally posted by hgmjr@Jun 10 2005, 10:05 PM
You're close. Check your math. Subtracting a positive number from a minus number does not yield a positive number.

hgmjr
[post=8397]Quoted post[/post]​

-15- 0.7 = - 14.3V

#### hgmjr

Joined Jan 28, 2005
9,027
Originally posted by davidand@Jun 10 2005, 09:31 PM
-15- 0.7 = - 14.3V
[post=8398]Quoted post[/post]​
Still not quite right. If you have a calculator see what answer it gives
for -15-0.7 = ?.

#### davidand

Joined Jun 2, 2005
43
Originally posted by hgmjr@Jun 10 2005, 10:40 PM
Still not quite right. If you have a calculator see what answer it gives
for -15-0.7 = ?.
[post=8399]Quoted post[/post]​
-15 - 0.7 = -15.7
I wasn't using the -/+ key

#### hgmjr

Joined Jan 28, 2005
9,027
Originally posted by davidand@Jun 10 2005, 10:05 PM
-15 - 0.7 = -15.7
I wasn't using the -/+ key
[post=8401]Quoted post[/post]​
Very good.

In my experience, nothing will give you more headaches than mishandling signs in electronics.

Now you can apply the technique you used in HW2 to get the emitter current. Once you have the emitter current you have the collector current. Once you have the collector current you have the current that feeds the two transistors connected up as a differential pair.

hgmjr

#### davidand

Joined Jun 2, 2005
43
Originally posted by hgmjr@Jun 10 2005, 11:15 PM
Very good.

In my experience, nothing will give you more headaches than mishandling signs in electronics.

Now you can apply the technique you used in HW2 to get the emitter current. Once you have the emitter current you have the collector current. Once you have the collector current you have the current that feeds the two transistors connected up as a differential pair.

hgmjr
[post=8402]Quoted post[/post]​
Thanks hgmjr for you support, I'm a electrician by trade it's a lot eaiser when I go out and test with a meter. When I started to Study electronics at first was like preparing to go to the dentist, I sure I'll have more headaches soon.

#### hgmjr

Joined Jan 28, 2005
9,027
Originally posted by davidand@Jun 10 2005, 10:28 PM
Thanks hgmjr for you support, I'm a electrician by trade it's a lot eaiser when I go out and test with a meter. When I started to Study electronics at first was like preparing to go to the dentist, I sure I'll have more headaches soon.
[post=8403]Quoted post[/post]​
You hung in there and worked through the three homework problems. With practice, the process will become much easier.

I wish you the very best of luck in your quest to expand your career into the field of electronics.

hgmjr