Voltage to Current Converter with Floating Load
Figure 1, shows a voltage to current converter in which load resistor RL is floating (not connected to ground).
Figure 1
The input voltage is applied to the non-inverting input terminal and the feedback voltage across R drives the inverting input terminal. This circuit is also called a current series negative feedback, amplifier because the feedback voltage across R depends on the output current iL and is in series with the input difference voltage Vd.
Writing the voltage equation for the input loop.
Vin = Vd + Vf
But vd » since A is very large,therefore,
Vin = Vf
Vin = R iin
iin = Vin / R
and since input current is zero.
iL = iin = Vin / R
The value of load resistance does not appear in this equation. Therefore, the output current is independent of the value of load resistance. Thus the input voltage is converted into current, the source must be capable of supplying this load current.
Voltage to Current Converter with Grounded Load
If the load has to be grounded, then the above circuit cannot be used. The modified circuit is shown in Figure 2.
Since the collector and emitter currents are equal to a close approximation and the input impedance of OP-AMP is very high, the load current also flows through the feedback resistor R. On account of this, there is still current feedback, which means that the load current is stabilized.
Since Vd = 0
Therefore; V2 = V1 = Vin
Therefore iout = (VCC – Vin) / R
Thus the load current becomes nearly equal to iout. There is a limit to the output current that the circuit can supply. The base current in the transistor equals iout / βdc. Since the op-amp has to supply this base current iout / βdc must be less than Iout (max) of the op-amp, typically 10 to 15 mA.
There is also a limit on the output voltage, as the load resistance increases, the load voltage increases and then the transistor goes into saturation. Since the emitter is at Vin w.r.t. ground, the maximum load voltage is slightly less than Vin.
In this circuit, because of negative feedback VBE is automatically adjusted. For instance, if the load resistance decreases the load current tries to increase. This means that more voltage is feedback to the inverting input, which decreases VBE just enough to almost completely nullify the attempted increase in load current. From the output current expression it is clear that as Vin increases the load current decreases.
Another circuit in which load current increases as Vin increases is shown in Figure 3.
The current through the first transistor is
i = Vin / R
This current produces a collector voltage of VC = VCC – i R = VCC – Vin
Since this voltage drives the non-inverting input of the second op-amp. The inverting voltage is VCC - Vin to a close approximation. This implies that the voltage across the final R is
VCC - (VCC - Vin) = vin
and the output current .
iout = Vin / R
As before, this output current must satisfy the condition, that iout / βdc must be less than the iout(max) of the OP-AMP. Furthermore, the load voltage cannot exceed VCC - Vin because of transistor saturation, therefore iout R must be less than VCC - Vin. This current source produces unidirectional load current. Figure 4, shows a Howland current source, that can produce a bi-directional load current.
The maximum load current is VCC / R. In this circuit Vin in may be positive or negative.
Voltage to Current Converter with Grounded Load
If the load has to be grounded, then the above circuit cannot be used. The modified circuit is shown in Figure 2.
Figure 2
Since the collector and emitter currents are equal to a close approximation and the input impedance of OP-AMP is very high, the load current also flows through the feedback resistor R. On account of this, there is still current feedback, which means that the load current is stabilized.
Since Vd = 0
Therefore; V2 = V1 = Vin
Therefore iout = (VCC – Vin) / R
Thus the load current becomes nearly equal to iout. There is a limit to the output current that the circuit can supply. The base current in the transistor equals iout / βdc. Since the op-amp has to supply this base current iout / βdc must be less than Iout (max) of the op-amp, typically 10 to 15 mA.
There is also a limit on the output voltage, as the load resistance increases, the load voltage increases and then the transistor goes into saturation. Since the emitter is at Vin w.r.t. ground, the maximum load voltage is slightly less than Vin.
In this circuit, because of negative feedback VBE is automatically adjusted. For instance, if the load resistance decreases the load current tries to increase. This means that more voltage is feedback to the inverting input, which decreases VBE just enough to almost completely nullify the attempted increase in load current. From the output current expression it is clear that as Vin increases the load current decreases.
Another circuit in which load current increases as Vin increases is shown in Figure 3.
Figure 3
The current through the first transistor is
i = Vin / R
This current produces a collector voltage of VC = VCC – i R = VCC – Vin
Since this voltage drives the non-inverting input of the second op-amp. The inverting voltage is VCC - Vin to a close approximation. This implies that the voltage across the final R is
VCC - (VCC - Vin) = vin
and the output current .
iout = Vin / R
As before, this output current must satisfy the condition, that iout / βdc must be less than the iout(max) of the OP-AMP. Furthermore, the load voltage cannot exceed VCC - Vin because of transistor saturation, therefore iout R must be less than VCC - Vin. This current source produces unidirectional load current. Figure 4, shows a Howland current source, that can produce a bi-directional load current.
Figure 4
The maximum load current is VCC / R. In this circuit Vin in may be positive or negative.
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