# Analog switch

Discussion in 'Homework Help' started by Peytonator, Jun 18, 2011.

1. ### Peytonator Thread Starter Active Member

Jun 30, 2008
105
3
Hi,

I can't figure out this problem... Please see the attachment. I'm not even sure whether the "true" statement is in fact true.

How do I go about solving this? I don't know whether the BJTs are in saturation or in the active region... so are they functioning as switches or what?

Surely, if Q1 turns ON, then 12V will appear across the 10k resistor, so Vout = 12V, not Vin?? But obviously the circuit is a kind of buffer, so the output should equal the input.

Thanks

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Last edited: Jun 18, 2011
2. ### beenthere Retired Moderator

Apr 20, 2004
15,815
283
Look at Q2. Think why the Vswitch label is on its input.

Last edited: Jun 18, 2011
3. ### Peytonator Thread Starter Active Member

Jun 30, 2008
105
3
Ok so they're biased in saturation. But how would you know that unless it it said "Vswitch"?

So if it's a switch, when Vswitch is high, Q2 turns on, so the base of Q1 is shorted to ground. So Q1 turns off, and Vout ≠ Vin. Right?

4. ### beenthere Retired Moderator

Apr 20, 2004
15,815
283
More questions - why would Q1 run in saturation is its otput is supposed to be very close to the op amp output? How much current can the op amp push through 3K? Other than switching the base of Q1 to ground or having no effect on the base current, what could Q2 do?

What do you suppose an analog switch function is?

5. ### Peytonator Thread Starter Active Member

Jun 30, 2008
105
3
This question is exactly what is confusing me. Then probably it's NOT running in saturation. Why must the op amp output be close to the output of Q1?

Probably low mA or uA range...

See below.

You want the ac input voltage to appear on the output, when the device is "switched" on, right?

Ok here's attempt two:

Say β = 100. Assume ative region. Then get the thevinin equivalent of Vswitch, 2k and 20k. Thus I get (for Q2)

Ib = 5.67mA

So

Ic = 0.567A.

Ok now I'm stuck. What do I do with this? Do I assume the Vo = Vin and go from there? Isn't that defeating the purpose of the question

6. ### beenthere Retired Moderator

Apr 20, 2004
15,815
283
Look at the circuit practically. If the output on Q1's emitter resistor does not resemble the output of the op amp, what is the purpose of the part of the question that states that to be the case?

We know that Q2 is the switch, thanks to the label. Apply both stated voltages to the Vswitch point and determine how Q2 affects Q1. Are the statements true or false?

7. ### Peytonator Thread Starter Active Member

Jun 30, 2008
105
3
So then if Q2 is a switch, when Vswitch is 0, Q2 is off, and Q1 functions as a normal BJT amplifier. Then the output will indeed follow the input. But if Vswitch = 12V, Q2 is on, and the opposite is true. So the statement is actually false? This is one confusing question

8. ### Ron H AAC Fanatic!

Apr 14, 2005
7,050
657
You got it. The circuit operates just the opposite of what was stated.