A resistor with current flowing but no dissipation

[wwoollyyhheeaa]

If an amplifier (audio or RF) has the collector of a transistor for its output electrode and assuming the transistor operates in its active region (not saturated or cut off) the output impedance of the amplifier will be high (several 10s of kilohms). If now a tuned circuit is connected as a load to the amplifier's output and impedance-matched to a real load R (the input impedance of the following amplifier) the apparent load on the collector of the transistor is n sqared x R (n is the turns ratio of the matching transformer). But when the amplifier is working it delivers ALL its output power to R. The apparent load doesn't dissipate any power. Why not? It is resistive (wholly real) and it has current flowing in it.

 

[berkeman]

If an amplifier (audio or RF) has the collector of a transistor for its output electrode and assuming the transistor operates in its active region (not saturated or cut off) the output impedance of the amplifier will be high (several 10s of kilohms).

Why do you say that it has a high output impedance? What if it has negative feedback?

The apparent load doesn't dissipate any power. Why not? It is resistive (wholly real) and it has current flowing in it.

What's an "apparent load"? Do you mean the load resistance reflected back through the turns ratio squared? The real resistor is still dissipating the power...

 

[wwoollyyhheeaa]

hi berkeman. I agree that feedback would alter it but let's suppose no feedback, for now. Yes I agree the load resistor connected to the transistor is the one reflected back through the matching transformer but this reflected one doesn't dissipate any power (forgetting small losses). And I agree the power is dissipated in the resistor on the other side of the transformer. So leaving aside the fact of the load resistor's dissipation being there (because it is not connected direct to the transistor), the reflected resistor does not dissipate. To illustrate what I mean suppose the actual load is some distance away from the transistor and not connected to it by anything that conducts heat, then the heat from the dissipation occurs away from the transistor. But the reflected load resistance, which is connected to the transistor by virtue of the transformer being connected to the transistor does not itself dissipate any heat anywhere near the transistor. It doesn't dissipate anything at all. Does this make sense?. This reflected load is a real quantity with current flowing in it but it dissipated nothing.

 

[berkeman]

Does this make sense?.

Nope, sorry.

 

[wwoollyyhheeaa]

OK, well perhaps the answer to my question is this and hopefully it will explain my problem better. If the resistor is R and its image at the collector is only a virtual resistor, we say it's real because algebraically it is but it's only virtually real. With the distance separating R and the collector as I described the power will be dissipated somewhere away from the transistor whereas the virtual load is across the transformer winding, and this could be right close to the transistor. So a so called "real" resistor isn't real in the everyday sense and it doesn't dissipate.

 

[anorlunda]

I think that the term virtual resistor should be abolished. It is a very fuzzy concept and several people may have several definitions of what it means. For example, your idea of virtual resistor seems to be very different than the one below.

Now, the concept of effective resistor Reff is defined as the (virtual) resistor that makes the real power equal to P=V2/Reff comparing with our reactive circuit we see that Reff=R(1+ω2L2/R), which depends on the frequency, ω.

For other more general reactive circuits, the algebra would be more complicated but the concept would be the same: the effective resistance is the expression giving the (virtual) ohmic passive resistor which would dissipate the same real or active power that the reactive circuit is dissipating. If the load circuit is purely reactive (e.g. a LC series or parallel), then the active power and the effective resistor are both zero.

 

[berkeman]

OK, well perhaps the answer to my question is this and hopefully it will explain my problem better. If the resistor is R and its image at the collector is only a virtual resistor, we say it's real because algebraically it is but it's only virtually real. With the distance separating R and the collector as I described the power will be dissipated somewhere away from the transistor whereas the virtual load is across the transformer winding, and this could be right close to the transistor. So a so called "real" resistor isn't real in the everyday sense and it doesn't dissipate.

You're just over-thinking this. There is no virtual resistor, there is just the transformer and the real resistor. We use math to figure out what the secondary currents and voltages will be, and what the secondary power dissipated will be. We use the same math to figure out what the primary voltages, currents and impedances will be. End of story.

 

[wwoollyyhheeaa]

Hi anorlunda and berkeman. I agree that I must be over thinking this. I understand what both of you say and I was wrong to say otherwise. Thank you for your advice.

 

[berkeman]

No worries. It's good to keep asking questions. As you learn more and more about electronics and EE, you get a better intuition for how things work, and you ask better and better questions. Never stop asking...