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The goal of this activity is to study the BJT current source or current mirror. Important attributes for current sources include high output resistance with a wide range of voltage compliance and rejection of external variations such as power supply or temperature.
ADALM2000 Active Learning Module
Solder-less breadboard
Jumper wires
2 - 1KΩ Resistors (values matched as close as possible, or measured to 3 digits or better)
2 - small signal NPN transistors (2N3904 or SSM2212)
1 - Dual Op AMP ( such as ADTL082 )
2 - 4.7uF decoupling capacitors
The good way to measure the characteristics of the current mirror is to reuse the same basic configuration that was used in the common emitter BJT curve tracer experiments. The input and output resistors R1 and R2 are now both 1KΩ. Be sure to accurately measure (with the most significant figures possible) the actual values of R1 and R2. This is to insure accurate measurement of the input and output current of the mirror. Iin will be equal to the AWG2 output voltage at W1 divided by the value of R1. Iout will be the voltage measured by scope channel 2 divided by the value of R2. Diode connected Q1 is connected across the base and emitter terminals of Q2.
Figure 1 Current mirror test circuit
Figure 2 Breadboard Connection of Current mirror test circuit
In the current mirror configuration, the op amp serves as a virtual ground at the mirror input (base) node to convert the voltage steps from AWG 2 ( W2 output ) into current steps through the 1KΩ resistor. The collector voltage is swept using a ramp from AWG 1(output W1) set to 3V peak to peak with the offset to 1.5V. VCE of output device Q2 is measured differentially by scope inputs 1+, 1-. The mirror output current is measured by scope inputs 2+. 2- across 1KΩ resistor, R2. If you don't want to use the op-amp configuration the following simplified configuration can be used as well.
Figure 3 Alt, Simple current mirror test circuit
Figure 4 Breadboard Connection of Simple current mirror test circuit
Two identical transistors with the same base to emitter voltage will have the same collector current IC. The second transistor, Q2, in effect mirrors the current in the first, Q1. Remembering the collector current to base emitter voltage relationship for a bipolar transistor:
where IS = the saturation current, and is a constant
VBE is the base emitter voltage
The thermal voltage, KT/q = 25.8 mV at room temperature
Identical transistors by definition have the same IS. In the simple current mirror, both transistors have the same VBE. Thus, both transistors will have the same IC and if base currents are ignored, Iin = Iout. Actually IC1is Iin - (IB1 + IB2).
Plot the two waveforms using the Oscilloscope provided by the Scopy tool.
Figure 5 Current Mirror waveforms, W2 at 10kHz Sample Rate
You are to measure Iin, Rout seen into the collector of Q2, the current mirror gain = Iout/Iin and determine the Norton and Thevenin equivalent circuits for this mirror.
Modify the simple mirror circuit by adding the base current compensation transistor Q3 as shown below in figure 6.
Figure 6 Current Mirror with Base Current Compensation
Repeat the same procedure you followed for the simple mirror circuit.
Figure 8 Current Mirror waveforms, W2 at 10kHz Sample Rate
In addition to the same quantities and graphs, does your data indicate any advantage to this circuit? Any disadvantages?
Add questions more here:
Modify the simple mirror into a Wilson Mirror as shown below in figure 9. Repeat the same procedure you followed for the simple mirror circuit.
Figure 9 Wilson current mirror
Repeat the same procedure you followed for the simple mirror circuit.
Figure 11 Current Mirror waveforms, W2 at 10kHz Sample Rate
In addition to the same quantities and graphs, does your data indicate any advantage to this circuit? Any disadvantages?
Add questions more here:
Modify the simple mirror into a Widlar Mirror as shown below in figure 12. Repeat the same procedure you followed for the simple mirror circuit. In addition to the same quantities and graphs, does your data indicate any advantage to this circuit? Any disadvantages?
Figure 12 Widlar current mirror
Repeat the same procedure you followed for the simple mirror circuit.
Figure 14 Current Mirror waveforms, W2 at 10kHz Sample Rate
1. Use the output impedance of the simple mirror to determine the Early voltage for the NPN transistor.
2. Build a mirror using PNP transistors and use the output impedance of the simple mirror to determine the Early voltage for the PNP transistor.
3. The output impedance of a Widlar current mirror is approximately,
Rout = ro[1 + gmR3]
where:
ro = VAF/IC, VAF is the Early voltage.
gm = IC/VT is the transconductance.
RE is the emitter resistor.
How accurately does this formula predict the output impedance of the Widlar current mirror you constructed? 4. If base currents are not ignored, how is Iout related to Iin in the simple current mirror? 5. If I need a second (or third) copy of Iin how would I make it?
The goal of this activity is to study BJT current source or current mirror with lower input headroom requirements.
2 - 1K Resistors
1 - 150K Resistor (or a 100K? in series with a 47K?)
2 - small signal NPN transistor (2N3904 or SSM2212)
1 - small signal PNP transistor (2N3096)
The diode configuration with nearly zero turn on voltage from activity 2 is used here, in figure 15, to make a current mirror. The current input node at the collector of Q1 (base of PNP Q3) is now much closer to ground compared to the conventional current mirror. What advantages would this have over the conventional mirror?
Figure 15 Low input head room mirror
Repeat the same procedure you followed for the simple mirror circuit.
Figure 17 Current Mirror waveforms, W1 at 10kHz Sample Rate
Ideally the collector of PNP Q3 would be connected to some negative voltage with respect to ground. Try connecting the collector of Q3 to the negative board supply Vn. What happens? Can the input node of the mirror get even closer to ground now?
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