differential amplifier with bjt - Page 1
Forum section: electronics / beginners

author: fady232
author_rank: Contributor
timestamp: May 19, 2025, 10:26:09 am
content: Hello, i saw the theory behind the differential amplifier and now im building one.VCC is 12V, VEE = -12VItail = 2mA, so ic on each resistor needs to be 1mA.What i did is first of all calculating the needed resistor in order to have 2mA on R1, so its -VEE-(-VBE)/R1=2*10^(-3) , that does R1=5.65know i know that each ic needs to be 1mA and vc needs to be at half of vcc, in order to have maximum swing, so vc is at 6Vso R2=R3 = 6k and RE is 5.65k, now if u calculate the gain its RC/RE so around 1,06, which is right in the simulationBut when i extract the signal from VD1 and VD2 there is also the D.C component that sum so i will have RC/RE + 6 VDCI tried eliminating it by adding a big cap ( 100uF ) but in the simulation i still see that there is the dc component, can someone help me? thanks

author: fady232
author_rank: Contributor
timestamp: May 19, 2025, 10:48:43 am
content: whats that? i need to add a load?

author: MariuszD
author_rank: Regular Contributor
timestamp: May 19, 2025, 02:24:03 pm
content: Quote from: fady232 on May 19, 2025, 10:26:09 amThis doesn't make any sense.You have 2mA current source. It gives current for both emitters. Resistors cant change that (ideal current source can make any voltage to give 2mA). How much current flows to which transistor depends on Vbe difference.Resistor in series with emitter is used only to set the gain.Quote from: fady232 on May 19, 2025, 10:26:09 amCircuit in simulation is ideal. When capacitor is connected only on one side no current can flow. In such conditions voltage across capacitor is always 0V. So it's the same as without capacitor.

author: Grandchuck
author_rank: Super Contributor
timestamp: May 19, 2025, 03:20:55 pm
content: I took out the negative supply and connected the current source to ground.

author: Grandchuck
author_rank: Super Contributor
timestamp: May 19, 2025, 03:48:11 pm
content: Here is another mod.

author: TimFox
author_rank: Super Contributor
timestamp: May 19, 2025, 04:18:51 pm
content: Quote from: MariuszD on May 19, 2025, 02:24:03 pmA non-ideal 100 uF capacitor would have substantial leakage current, which would force the voltageacrossthe capacitor to zero, so the dc components of the two voltages at the left and right sides of the capacitor would be equal.In simulation with an ideal capacitor, the calculation may set the initial voltage across the part to zero, which would persist.Spice simulation does not allow any node in the circuit that does not have a conductive path to “ground” (reserved node number 0). When simulating RF networks where the dc components were not important, I would fix such nodes by connecting a 1 G\$\Omega\$ kluge resistor to ground. Otherwise, Spice would report an error and not continue. With such a resistor added after your series capacitor, the dc component after the capacitor will go to zero. If you need the actual low-frequency response to be accurate, you need realistic values for C and R.

author: fady232
author_rank: Contributor
timestamp: May 22, 2025, 08:16:58 am
content: Quote from: MariuszD on May 19, 2025, 02:24:03 pmthen what i can do to remove the d.c in real life circuit?

author: MariuszD
author_rank: Regular Contributor
timestamp: May 22, 2025, 09:47:26 am
content: Look at post #2. To remove DC you need RC high pass filter.Resistance can be so high that circuit works like there is no load.

author: TimFox
author_rank: Super Contributor
timestamp: May 22, 2025, 04:00:02 pm
content: Quote from: fady232 on May 22, 2025, 08:16:58 amRemoving DC component:Anycapacitor from the circuit output toanyresistor to ground (0 V) will give a voltage across that resistor whose mean value over time (DC component) is zero.TheRCproduct (time constant) of that network determines the low-frequency response (high-pass filter). You choose those values to achieve your requirements for the low-frequency end of your signal bandwidth.