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  1. R

    Output resistance of equivalent op amp circuit

    If you drive another amp with the voltage at Vo, it will appear to the amp like the Vo has an impedance of Ro || RL || R22 no matter what the value of Vo is. Immittance does not change with output voltage. Ratch
  2. R

    Output resistance of equivalent op amp circuit

    The denominator of the transfer function determines the input and output immittances. The denominator is affected by the circuit component values, not the voltages applied to the circuit. The questions you are asking appear to be related to determining the immittances by the applied source...
  3. R

    Output resistance of equivalent op amp circuit

    If Vin = 0, then no current will be present in the output. Are you saying the book answer is correct, and the calculated answer of ro||R22||RL is wrong? Ratch
  4. R

    Output resistance of equivalent op amp circuit

    Yes, just make sure the transfer function has no polynomial fraction terms in the numerator or denominator. Yes, the factors that make up the denominator are equal to zero, so naturally any term within those factors is going to be the negative of the remaining terms. I like the way this...
  5. R

    Output resistance of equivalent op amp circuit

    I can do that. It is a simple typo on Sedra and Smith's part. There is a dependent voltage source in that provlem that is confusing and bemusing to you. Even so, the applied source method you are using is a clunky way to do it, especially when you have a dependent source. You need to use a...
  6. R

    Question about operational amplifier circuit Dorf exercise 2.5?

    You are right. The problem is wrong, and the given solution is wrong. It is a voltage follower with a gain of 1. If the writers wanted to show a non-inverting amp, then they should have put R1 on the other side of R2. Ratch
  7. R

    Determine the Vcbo of a germanium transistor and the voltage drop along the connected components.

    Doesn't matter. With that much resistance in the base circuit, and no resistance in the emitter circuit, all the 10 ua of Icbo will pass through the b-e junction and raise the Ic to at least 0.5 ma. The b-e junction does not have to be turned fully on. If the 10 ua amps passes through it, it...
  8. R

    Determine the Vcbo of a germanium transistor and the voltage drop along the connected components.

    It will leak 10 ua regardless. But, if you don't have any Rb, and put in some Re, then the collector current will be 10 ua when turned off. On the other hand, if we use your circuit, then all the 10 ua will pass into the b-e junction, and be betatized to to 0.5 ma Ic. I am weary of explaining...
  9. R

    Determine the Vcbo of a germanium transistor and the voltage drop along the connected components.

    That's right. A germanium transistor with those specs will do so. You cannot turn off that internal Icbo generator in the collector of Q1. As long as you reverse bias the collector-base, which you do in the active region, that internal current generator will fire up and send its current into...
  10. R

    Determine the Vcbo of a germanium transistor and the voltage drop along the connected components.

    Sure it will. When the base voltage is zero. 10 ua of current will be present in the collector of Q1. Multiply that by 125 and Q2 will be 1.25 ma. That is sure a lot less than 10 ma it will have when Q1 is turned on. The best course of action is to use a silicon transistor for Q1, so that...
  11. R

    Confused what voltage would be depending on its location?

    Voltage is the energy density of the charge. It is measured in the MKS system as joules/coulomb, which is the definition of a volt. When the energy density of a charge (voltage) at one point is higher than the energy density at another point, a current will be present from the higher energy...
  12. R

    Determine the Vcbo of a germanium transistor and the voltage drop along the connected components.

    Certainly it does. It keeps the Icbo out of the emitter circuit so it does not get betatized. You are correct about that. For the circuit you show above with Rb=100k and Re=0, all the Icbo will be directed through the base-emitter junction and be betatized to a collector current of 4 to 8...
  13. R

    Determine the Vcbo of a germanium transistor and the voltage drop along the connected components.

    We are trying to turn the LED ON and OFF. The circuit is no good if it doesn't do both. This is about the 3rd time I said that Icbo cannot be stopped. It can only be controlled, If he uses the BC557, then the diode current will be down to around 1.75 ma. Probably not enough for the light...
  14. R

    Determine the Vcbo of a germanium transistor and the voltage drop along the connected components.

    There you go again, making wild statements. The datasheet disagrees with you. http://www.nxp.com/documents/data_sheet/BC556_557.pdf Ratch Edit: Ok I see that adding a "C" suffix increases the minimum beta to 420. That means that an it will be even easier to get the transistor to turn on...
  15. R

    Determine the Vcbo of a germanium transistor and the voltage drop along the connected components.

    As I explained before, and which you should know by now. Icbo is an internal current generator within the collector of a BJT whose existence cannot be eliminated or turned off. It is intrinsic to semiconductors. It can only be minimized by the method I have presented. Who said the...
  16. R

    Determine the Vcbo of a germanium transistor and the voltage drop along the connected components.

    Assuming you are saying the Icbo is 14 uamps. No problem. It takes 80 uams to turn on the LED. The 14 uamps is shunted away from the base-emitter of Q1. 14 uamp*125 = 1.75 ma. Not much compared to the ON current of 10 ma, but it is the best you can do with germanium transistors, which are...
  17. R

    Determine the Vcbo of a germanium transistor and the voltage drop along the connected components.

    Why do I get the impression that you are not getting the big picture? Assuming a 1 volt drop across the collector-emitter of Q2 when it is turned on, and 0.7 volts across the LED, that makes 7.3/680 = 10 ma through the LED. The minimum beta of Q2 is 125, so a minimum 10ma/125 = 80 uamps is...
  18. R

    Determine the Vcbo of a germanium transistor and the voltage drop along the connected components.

    Nonsense. Just putting in an emitter resistance does not make an emitter follower. You have to take the signal off the emitter resistor before you can say it is an emitter follower (common collector) ...
  19. R

    Determine the Vcbo of a germanium transistor and the voltage drop along the connected components.

    See post #4 of this thread. Ratch
  20. R

    Determine the Vcbo of a germanium transistor and the voltage drop along the connected components.

    Forget about putting any resistor across Q2. I suggested you put a resistor Re in the emitter circuit of Q1 and take out the 100k base resistor. That means putting a resistor in series with the emitter terminal of Q2, not across the base-emitter terminals like you show in latest drawing. Read...
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