I need some guidance and explanation as to whether I am interpreting the circuit
ID: 2988030 • Letter: I
Question
I need some guidance and explanation as to whether I am interpreting the circuit below correctly. Please read my interpretation below and tell me if I am thinking correctly or not.
The purpose of R4 and R5 is to adjust the gain of the op amp because there is negative feedback coming from the output that is then divided amongst those resistors to determine the output.
The purpose of R1 and R3 is to allow current to flow througth all the diodes and forward bias them. Once they are all forward bias, D2 and D4 will keep the base voltage at the npn darlingtion transitor Q1 two diode drops higher than the input signal. (That is important when the input gets between 0 and 1.7V) On the other hand, D1 and D3 will keep the base voltage at the pnp darlingtion transitor Q2 two diode drops lower than the input signal. (That is important when the input gets between 0 and -1.7V) Those diodes essentialy make sure that at least one transisotor is 'on' at all times and letting current to flow through the load.
If my interpretation is correct, here are some questions I need answered,
1. What is the point of the transistors? This model is suppose to be similar to a class AB amp but if all we want is for there to be some voltage across the load reistor, why do there need to be transistors. Can't the op amp just output a voltage across the load resistor without the transistors?
2. Why is the configuration of the pnp transitor Q2 showing the arrows of the current flow not toward the load resistor? Isn't the current suppose to flow toward the load resistor?
3. What would I have to change on this circuit to allow a 4 ohm load to dump only 18 watts of power?
Explanation / Answer
Your two guesses are correct.
a) Think about the schematics with only the op amp driving the load without transitors. Then R4 should be connected between - In of op amp and Out of op amp. This is the classic arrangement for non-inverting op-amp amplifier (see wiki). The gain is
G = Vout/Vin =(1+R4/R5)
b) R1 and R3 are polarising both bases of Q1 and Q2 (the provide the base current of the transistors AND also open the 4 diodes in series). For BJT to be open they need a finite base current (by contrast with FET). As you said, togheter with the diodes (2 diodes voltage drop =2*0.7 =1.4 V) they just open BOTH transistors when Vout of op amp =0. When Vout of op amp becomes positive Q1 will open more while Q2 will enter the "cut-off" region. When Vout of op amp becomes negative Q2 will open more while Q1 will enter the cut-off region.
Now your questions:
1) The two transistors are connected in the so called "emitter repeater" configuration (collector on the voltage source, emitter on the load). In this configuration transitors DO NOT AMPLIFY VOLTAGES they just AMPLIFY CURRENTS. So that while the voltage on the 4 ohm load R2 will remain the same as Vout of op amp, the current I on the load will be much higher that that provided by the op-amp. The transistrs are necessary to achieve high powers of ouput (P=U*I), since Op amp have only small output currents.
2) Q1 and Q2 are so called PAIR transitors. one is npn (Q1) and the other is pnp (Q2). The direction of the arrow in the emitter shows the direction of positive charges, so while for Q1 positive charges flow from V3 (+) towards R2 to the GND, for Q2 the positive charges flow from GND through R2, towards V1(-).
The direction of current is correct for Q2. From GND (which is positive) towards V1(-) which is negative.
3) To change the power delivered to Rload (R2) you just need to change the voltages V3 and V1.
P = U^2/R so you need to decrease V3 and V1 to decrease power.
Observation: ussualy the peak power is somewhat less than Ppeak =V3^2/R since the transitors have a saturation voltage on CE (collector -emitter). Hence for original V3=V1= 20 V the peak power is less than
20^2/4 =100 W.