Circuit Analysis Question - Page 1
Forum section: electronics / beginners

author: kas123
author_rank: Newbie
timestamp: May 23, 2023, 03:22:45 pm
content: I cant solve this question. I would be very happy if you help me.In this circuit VD = 0.7V. R1= 10k R2 = 5k. Find current and V0.

author: wasedadoc
author_rank: Super Contributor
timestamp: May 23, 2023, 06:32:47 pm
content: From VD you can calculate the voltage at D1 anode. From that you know the voltage across R1 and hence the current through R1.From the Voltage at D2 anode (which is the same as the voltage at D1 anode) you can calculate Vo. From Vo you can calculate the current through R2 (which is the same as the current through D2).From the current through R1 and the current in D2 you can calculate the current in D1.

author: wasedadoc
author_rank: Super Contributor
timestamp: May 23, 2023, 07:02:49 pm
content: Quote from: kas123 on May 23, 2023, 06:59:31 pmNo. You asked for help. Are you too lazy to complete it yourself?

author: Andy Watson
author_rank: Super Contributor
timestamp: May 23, 2023, 07:18:02 pm
content: Look at the values of the resistors. Make an approximate guess at Vo.Then realise that the anodes of both diodes will be 0.7V above this value - this should tell you something about the current flowing through D1- Hint: you do not need a calculator to workout \$I_{D1}\$ !

author: sparkydog
author_rank: Regular Contributor
timestamp: May 24, 2023, 04:18:02 pm
content: Simulation is your friendV0= -3.566 V.You didn't clearly specify where you want to measure current, but I ≈ 1.287 mA everywhere except D1, which is experiencing saturation current (that is, current — a verysmallamount of it — is flowingbackwardsacross D1). I can't tell you what that is, because it's a property of the diode that you didn't specify.Note that VDis something of a misnomer. A real diode has a forward voltage, but the actual voltage drop is also current-dependent. I had to fiddle with the parameters to get a drop of (almost) exactly 0.7 V.

author: MrAl
author_rank: Super Contributor
timestamp: May 24, 2023, 04:49:26 pm
content: Quote from: kas123 on May 23, 2023, 03:22:45 pmHello there,I can understand why you might be having a problem with this circuit. That is because it is not the usual analog circuit you probably usually run into. This is actually a switching circuit. To handle that in a general way, we have to think in terms of switching elements like switches which are either 'on' or 'off'. Since the diodes are the switches and they also have a voltage drop, we have to impose some additional voltage sources as well if they are 'on' and when they are 'off' we consider them to be open or have infinite resistance.To start, consider each diode to either have that 0.7v drop when 'on' or be open when 'off'.A general procedure that works with many of these circuits then, given that you can not foresee the answer more simply, is to replace both diodes in turn, first with their 'on' model and then with their 'off' model. When you do all possibly combinations this way, after each combination you analyze the result and see if the current levels match your current configuration.The possible configurations are as follows... D2 off, D1 off. D2 off, D1 on. D2 on, D1 off. D2 on, D1 on.Notice this pattern simply follows the binary count sequence: 00, 01, 10, 11, if 0 is 'off' and 1 is 'on'.Remember when a diode is 'on' it becomes a voltage source of 0.7v (with proper forward voltage polarity), and when a diode is 'off' it is an open circuit.After each configuration, consider the voltages at the two nodes, one to the left of D2 and the other to the right of D2. Then, consider the current though the resistors, and see if the current level matches the configuration such that the now established voltages (or open circuits) would cause the given current to flow (or not flow).This is a very general procedure and assumes that you know nothing else about the circuit by inspection alone. With inspection alone is sometimes hard to visualize which diodes are on and which are off, so we go through all the possible combinations until we find one where the currents and voltages all work out to be correct. This circuit happens to be a bit easier, but you may not notice why if you didn't do many of these kinds of circuits in the past. By using this general procedure, you should be able to solve these circuits. The more diodes the longer it will take, but these little circuits do not take long as they usually have only 1 to 3 diodes or so.If you like i can give you another example and then you can see how this works by example.The main theory behind this is simple, and that is a diode that is 'on' has a forward drop of 0.7v and when 'off' it is an open circuit.We could delve into the exponential model of the diode that would also provide a solution, but that would be overly complicated for this problem. Using a simulator may not teach you might either but you can use that to check your manual results.In many cases you can consider one diode to be on or off, then see if the other diode might be on or off by checking the voltage and current levels. Sometimes you have to also apply a little basic reasoning. For example, if both diodes were 'off' then no current would flow from the +10v source to either ground or -10v, and that would not make any sense because the polarity of the diodes indicates that at least one of them should be on, if not both.Once you do a few of these problems you will probably get the hang of it.

author: Solder_Junkie
author_rank: Frequent Contributor
timestamp: May 24, 2023, 05:29:20 pm
content: While this circuit is simple enough, many of us use Spice circuit modelling software. Such software will show DC current and Voltage at any place in a circuit.LT Spice is free, although I prefer Tina. The lowest cost version of Tina is the student one at €49.https://www.tina.com/tina-student-version/SJ