Common Emitter Active Load Problems

[fobos3]

Hi.

I am trying to design a common emitter amplifier with a current source as an active load. Currently I am simulating it in SPICE using an ideal current source with an NPN transistor. I have the emitter connected to ground, the input signal is applied at the base and the collector as the output.

The problem I am having is that the gain I get is negative. The response is actually a high pass filter with a corner frequency in the GHz range.

I tried introducing feedback but my response is weird again. The gain between the base and the emitter is around 60dB, however, there is a DC offset at the output of the amplifier. The output is given by:
$$V_{o} = V_{be} + A(-V_{in} + V_{be})$$
where A is the absolute value of the voltage gain. The diode drop is around 0.7V so if I set the gain to A=1, I get the expected gain but the signal is centred at 1.4V. As I start increasing the gain the voltage eventually reaches the power rails and I get a DC signal.

I also tried adding a coupling capacitor at the input and having a 1Meg resistor between the collector and the base to provide the DC path for the amplifier. That worked fine but, as expected, the voltage gain starts dropping off as you approach DC.

Any ideas how I can fix that? More precisely, I don't understand what I am doing wrong. I looked up circuits from IC design books and also the 741 op-amp internal circuitry. Both have a common emitter stage where the signal is directly applied to the base.

Regards

 

[Studiot]

I'm not sure I understand the question. The gain of a CE amplifier is negative, in that there is a 180 phase shift. An increasing base voltage results in a decreasing collector voltage.

 

[skeptic2]

The description of your circuit is vague. Can you post a schematic?

 

[fobos3]

circuit

 

[skeptic2]

Thanks for the schematic. Now can you tell us the values of components such as Vin, R1, R2 and I. V probably doesn't do anything (in SPICE) and can be replaced with a wire. It is important to know if you have a DC offset in the Vin parameters and how much it is.

 

[fobos3]

Thanks for the schematic. Now can you tell us the values of components such as Vin, R1, R2 and I. V probably doesn't do anything (in SPICE) and can be replaced with a wire. It is important to know if you have a DC offset in the Vin parameters and how much it is.

Vin is a 1mV sine wave with no offset, V is 10V, and the R1 and R2 resistor are in the kOhm range. I change their ratio to set up the gain.

The problem I am having is that the output is given by:
$$V_{out} = V_{be}-\frac{R_1}{R_2}(V_{in} - V_{be})$$
So, if I set the gain to 1 or 2 my sine wave at the output has an offset of 1.4V and 2.1V respectively. Eventually this offset reaches the positive power rail and I get a DC output. This is for R1 = 10k, 20k and R2 = 10k. The gain between the input at the base of the transistor and the output is 71dB.

 

[skeptic2]

The attachment shows some values that worked for me. Note that the signal must pass through the 10K resistor. That tremendously reduces your gain. The usual way is to couple the signal in through a capacitor.

In the first plot, with a peak input voltage of 1mV the circuit has an unloaded output of 3.3mV or a voltage gain of 3.3.

In the second plot, with the peak input voltage reduced from 1mV to 0.5mV the unloaded peak output voltage is about 1.25V for a voltage gain of 2,500.

Of course you won't be able to realize that gain once you load the output but this gives you and idea of how a simple change to the topography can significantly change the result.

 

[mheslep]

Vin is a 1mV sine wave with no offset, V is 10V, and the R1 and R2 resistor are in the kOhm range. I change their ratio to set up the gain.

The problem I am having is that the output is given by:
$$V_{out} = V_{be}-\frac{R_1}{R_2}(V_{in} - V_{be})$$
So, if I set the gain to 1 or 2 my sine wave at the output has an offset of 1.4V and 2.1V respectively. Eventually this offset reaches the positive power rail and I get a DC output. This is for R1 = 10k, 20k and R2 = 10k. The gain between the input at the base of the transistor and the output is 71dB.

Can you AC couple the output of the amplifier to the load (load impedance unspecified?)?

 

[fobos3]

The attachment shows some values that worked for me. Note that the signal must pass through the 10K resistor. That tremendously reduces your gain. The usual way is to couple the signal in through a capacitor.

In the first plot, with a peak input voltage of 1mV the circuit has an unloaded output of 3.3mV or a voltage gain of 3.3.

In the second plot, with the peak input voltage reduced from 1mV to 0.5mV the unloaded peak output voltage is about 1.25V for a voltage gain of 2,500.

Of course you won't be able to realize that gain once you load the output but this gives you and idea of how a simple change to the topography can significantly change the result.

I see what you've done. Indeed, I have tried that and my problem is the offset at the output that is also present in your simulation. Maybe there is no way to remove it and I'll need to add a differential stage.

Can you AC couple the output of the amplifier to the load (load impedance unspecified?)?

The amplifier needs to be able to work with low frequencies so AC coupling is not an option.

 

[davenn]

.....
The amplifier needs to be able to work with low frequencies so AC coupling is not an option.

thats a contradiction ;)

if the signal has a frequency, then it must be AC and therefore can be capacitor coupled
you really only can have ... DC, pulsed DC or AC
your circuit shows an AC signal source, so what frequency range are you wanting the amplifier to work at ?

Dave

 

[skeptic2]

The only realistic way to accomplish what you want to do is with an opamp and a dual power supply.

R1 & R2 along with C1 & C2 create an artificial ground. R1 & R2 should be equal to each other as well as C1 & C2. The lower the values of R1 & R2 and the higher the values of C1 & C2, the better the ground will be, but of course as you lower the values of R1 & R2, the more power your circuit will draw. The gain of your circuit is determined by the ratio of R4 to R3. This circuit will give you an inverted output, i.e. negative gain. If you need positive gain this circuit can be modified for positive gain.

 

[fobos3]

thats a contradiction ;)

if the signal has a frequency, then it must be AC and therefore can be capacitor coupled
you really only can have ... DC, pulsed DC or AC
your circuit shows an AC signal source, so what frequency range are you wanting the amplifier to work at ?

Dave

I need it to work from 0.05Hz up to about 500MHz

The only realistic way to accomplish what you want to do is with an opamp and a dual power supply.

R1 & R2 along with C1 & C2 create an artificial ground. R1 & R2 should be equal to each other as well as C1 & C2. The lower the values of R1 & R2 and the higher the values of C1 & C2, the better the ground will be, but of course as you lower the values of R1 & R2, the more power your circuit will draw. The gain of your circuit is determined by the ratio of R4 to R3. This circuit will give you an inverted output, i.e. negative gain. If you need positive gain this circuit can be modified for positive gain.

Yeah, I thought so. Thanks for your help.

 

[davenn]

except that you will struggle to find readily available ( if any) Op-Amps that will work over a few 10's of MHz


D

 

[skeptic2]

http://www.ti.com/lit/an/sloa051a/sloa051a.pdf

http://www.analog.com/en/high-speed-op-amps/current-feedback-amplifiers/ad8000/products/product.html