A different voltage divider

  • #1
634
1

Main Question or Discussion Point

I am trying to design a 3 Resistor voltage divider. The center resistor is shorted out using a transistor depending on the input voltage.
If Vin is <1, then center resistor is not shorted.
If Vin is >1 then center resistor is shorted by the BJT, so only the top and bottom resistor will act. (ignoring Vce sat).

But when I simulate, the voltage at the emitter increases when the BJT is turned ON.
I don't understand why.
In the attachment, Vin goes to the +Ve terminal of the buffer.
 

Attachments

Answers and Replies

  • #2
vk6kro
Science Advisor
4,081
40
If the transistor was not conducting, the current through the 100 ohm resistor would be
8 volts / (1K + 2K + 100 ohms ) or 2.58 mA.

If the transistor did conduct (and was perfect), the current through the 100 ohm resistor would be 8 volts / (1 K + 100 Ohms) or 7.3 mA.

So, the voltage across the 100 ohm resistor would increase to 0.73 volts from 0.258 volts.
 
  • #3
dlgoff
Science Advisor
Gold Member
3,827
1,759
Instead of the transistor, perhaps you could use a bilateral switch. It is just a thought and you may not even want to consider one. Any way this one (and there are many others) has the following description:

The NTE6403 is a silicon planer, monolithic integrated circuit having the electrical characteristics of a bilateral thyristor. This device is designed to switch at 8 volts with a 0.02%/°C temperature coefficient and excellently matched characteristics in both directions. A gate lead is provided to eliminate rate effect and to obtain triggering at lower voltages.

The NTE6403 is specifically designed and characterized for applications where stability of switching voltage over a wide temperature range and well matched bilateral characteristics are an asset. It is ideally suited for half wave and full wave triggering in low voltage SCR and TRIAC phase control circuits.
http://docs.google.com/viewer?a=v&q...w63ny&sig=AHIEtbQPDXDZ9Zcku--1UREu2yNJLg7PQQ"
 
Last edited by a moderator:
  • #4
uart
Science Advisor
2,776
9
But when I simulate, the voltage at the emitter increases when the BJT is turned ON.
I don't understand why.
As vk6kro already pointed out, even if the BJT behaved like an ideal switch in this circuit then you'd expect the voltage across R1 to increase (from .26 volts to .73 volts). I don't see how you could be expecting otherwise.

Now what I think you really meant to say was that the voltage across R1 increases by more than what you expected. That's what I would expect to happen with this circuit. The exact value will depend upon how well your software models the opamps current limit but I'd expect the voltage to rise to about 3 or 4 volts (instead of to the expected 0.73). Is that what happened?
 
Last edited:
  • #5
634
1
As vk6kro already pointed out, even if the BJT behaved like an ideal switch in this circuit then you'd expect the voltage across R1 to increase (from .26 volts to .73 volts). I don't see how you could be expecting otherwise.

Now what I think you really meant to say was that the voltage across R1 increases by more than what you expected. That's what I would expect to happen with this circuit. The exact value will depend upon how well your software models the opamps current limit but I'd expect the voltage to rise to about 3 or 4 volts (instead of to the expected 0.73). Is that what happened?
Yup, it increased by 3 to 4v.
Soon after posting, I realized, it was bcoz of the base voltage. The emitter( or R1) was only 0.7v less than the base voltage. Base voltage is nothing but op amp output voltage.

Thanks Vk6kro, dlgoff.
I think a switch is better than BJT.
 

Related Threads on A different voltage divider

  • Last Post
Replies
4
Views
2K
  • Last Post
Replies
15
Views
1K
  • Last Post
Replies
1
Views
2K
  • Last Post
Replies
2
Views
504
  • Last Post
Replies
3
Views
6K
  • Last Post
Replies
6
Views
3K
  • Last Post
Replies
13
Views
7K
  • Last Post
2
Replies
31
Views
13K
  • Last Post
Replies
2
Views
6K
Top